141

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 _____°

142

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

« on: May 08, 2019, 09:50:30 PM »

Sorry by piggybacking over your original post, but as far as I read, everything on FE accelerates upwards.  This includes "everything" in reference to the ground, atmosphere, the Sun, Moon, stars, no matter the altitude to sea level.  Considering everything goes upwards at the same acceleration rate, there is no difference on the nominal acceleration based on altitude to sea level.  Imagine a high altitude balloon, sit on some atmospheric gases at 50km of altitude.  The balloon is sit quietly there because it is being pushed upwards at the same acceleration imposed to the trash can at my backyard.   

Using well known, measured, tested and used for decades formula, the ER gravity at 50km above sea level would use the formula g1/g2 = (R2/R1)², where R is the radius of the planet where the gravity g1 or g2 is measured.  So, consider g1 = gravity at sea level and Radius of Earth on equator to be 6400km, then 9.8/g2 = (6450/6400)² = 9.8/g2 = 1.0078125² = g2 = 9.8/1.015686035, result g2 = 9.648m/s² (at 50km of altitude over equator).  That means a reduction of gravity in the order of 1.5%, or, 0.153m/s².   Using a second formula that calculates Height (of fall) for a specific time and gravity, h=t²*g/2, we can find a mass can fall through 50.067km in just 809 seconds (not considering air friction), final velocity of 445.6km/h.  What all this means?  That balloon at 50km of altitude, with acceleration of only 9.648m/s², less than the earth itself, would comes down and eventually hit the ground in 809 seconds.   Of course that for several reasons the balloon will not splat on ground, but with different acceleration reates the balloon will not stay at 50km of altitude, it would be at much lower altitude trying to compensate the differences of acceleration and pressures differences between the atmosphere and balloon. Eventually it would touch ground, since speed differences don't forgive.

So, the only way to consider this possibility on FE, is if upwards acceleration is constant 9.8m/s² at any altitude, what was already proven otherwise.  A 10kg solid metal mass at sea level will weight only 9.988km (12g less) at La Paz (Bolivia) 3640m above sea level, using the same precise scale, this proves without questions the gravity changes by altitude. 

This test can be repeated, exercised, duplicated inside airplanes, top of mountains, etc.  As a reference, such 10kg solid metal mass at sea level would weight only 9.968km, 32g less, inside an airplane flying steady at 10km of altitude over equator. FErs can do this test anytime.

I am here considering a very linear gravity acceleration over the whole planet, what is not.  Several precise measurements were made using special satellites for that (NASA Grace (twin), and European GOCE), and found changes all over the globe, due different mass densities.

But wait... there is something here, I was wrong.  It is not everything that accelerates at 9.8m/s² on FE world.  We don't. My chair doesn't. My trash can doesn't.  Only the earth does that, only the flat planet does that and pushes everything along.  Everything else is just sit on top of it.  No matter how dip I dig a hole and remove dirt and rocks, they don't accelerate, they still sit over the land.  I can go into very dip caves and find rocks on the floor of the caves, what means what accelerate upward on FE world is the very very deep base.  Even the volcanos's lava doesn't float in space, they sit and runs down the mountain's surface, it means, not even molten lava is being accelerated upward by this strange force.  So, someone care for explain?

143

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.

144

Flat Earth Theory / Re: Universal Acceleration - how does the ISS maintain an orbit?

« on: May 08, 2019, 05:32:28 PM »

since we do not have access to space to study its properties, or the technologies used, it is a pointless question for us.

Hmmm...

Who from "us" went there to test, survey and confirm, so the 9 (to be a short list) statements below are surely unquestionable solid truth?

1.) Moon's own light
2.) Invisible object (antimoon) hiding Moon's light for the phases
3.) Antarctica being militarized for no access at all
4.) The fantastic dome existence
5.) Land distances on FE map South hemisphere
6.) Despite FE perspective and vanish point, the sun keeps the same diameter from sunrise to sunset due atmosphere lens effect
7.) Satellites and ISS do not exist
8.) GPS and GeoStat comm satellites are just radios and antennas on balloons
9.) RE Airplanes would fly straight to space if pilot doesn't pitch down constantly

145

Flat Earth Theory / Re: The length of the day on a flat earth

« on: May 07, 2019, 10:00:36 PM »

Today in Central Florida we have 13 hours and 23 minutes of visible Sun.

Sunrise in Orlando, FL, USA: 6:40 AM
Sunset in Orlando, FL, USA: 8:04 PM
Wednesday, May 8, 2019 (EDT)

Yesterday I observed the magnificent Sunset right at my backdoor at 280° exactly at 8:04pm, see the record below.



On June 21st, it will be 13 hours, 57 minutes and 40 seconds... from 6:28am to 8:26pm, it is almost 14 hours of visible Sun.

On the FE map, it will represent almost 210° of the circle with latitude 28.5°N.

It means, an FE observer at latitude 28.5°N would be able to see the sunrise at 36° and the sunset at 324°, but it is not where I will see the Sun at the same latitude 28.5°N, it will rise at 63°ENE and set at 297°WNW, the angle difference is brutal, due the FE flat circular equator line.

So, or my compass and Timeanddate.Com are wrong, or something else is wrong.

146

Flat Earth Theory / Re: Universal Acceleration - how does the ISS maintain an orbit?

« on: May 07, 2019, 08:31:21 PM »

A couple of years ago, the ISS would cross over Central Florida at a specific time few minutes after Sunset, when the Sun even being under the horizon for the Floridians, would be illuminating the ISS 400km up in orbit.  That was a not so rare, but difficult to repeat on such exact variables, so it was announced vastly with days in advance on newspaper, local TV and internet.  I was out there sit waiting... and with a clockwork precision, there she comes from Southwest right over my house, disappearing 3/4 of the way to Northeast (when it loses solar reflection).  It was a bright beautiful thing, not so far, same distance from Central Florida to Miami, but small enough to allow us to think "there are people inside there".   I saw it, naked eye.  I had my good binoculars with me, but didn't use it, it was moving fast.   

If you are interested to actual "see" the ISS flying over Central Florida, use the following web-link to know in anticipation.
https://spotthestation.nasa.gov/sightings/view.cfm?country=United_States&region=Florida&city=Orlando#.XNHpnOhKiUk

The following web-link is to locate yourself in the world and then pinpoint when ISS will be visible in your evening skies:
https://spotthestation.nasa.gov/sightings/

For the ones that didn't understand yet how this work:
It is very difficult to see ISS during the day, or night, its external lights become so dim for us to see it.
But there is a special time window (minutes) right after sunset, when the sky becomes dark enough but the angle of the sun right below the horizon still illuminating the ISS 400km above, so it becomes like a mirror, reflecting sunlight down to us, in contrast with the dark sky.  It becomes bright enough to see it very well, impossible to think it is an airplane or another satellite.  It is really bright, more than Venus.   You just need to go outside, sit down, use a compass to pinpoint exactly where in the horizon it will appear and where it will go disappear, so you can keep looking on that sky stretch, can not miss it.

147

Flat Earth Theory / Re: Theory/Model Request

« on: May 07, 2019, 07:59:30 PM »

174*10^15 watt / 3*10^8 m/s = 58 * 10^7 N spread out over the entire surface of Earth.
Let's assume that the sunlight hits an area of pi*6000*6000 km² = 113,000,000 km²
That means, sunlight is pushing us down with roughly 58*10^7 / 113*10^6 = 5.1 N/km²
Your theory has been disproven.

Oh darn!  I thought I would be famous.  But even with the Sun much closer and focusing all its energy upon a spot area?  May be you can recalculate for millions more Newtons/km² ?

148

Flat Earth Theory / Re: Polaris & Alpha Crucis visibility

« on: May 07, 2019, 05:03:31 PM »

Stack, congratulations by your post #1000.

Anyway, in all my long life I always found only two things impossible to be expressed by correct mathematical form;

1) The animal emotions
2) A cult

When you crash your car, you may explain it by science, using kinetic energy, vectors, speed, mass, space, time, impact, energy released, resultant vectors, mass deformation, rotation, heat, etc, and even providing math calculations for that (car's manufacturers use lots of those calculations to understand and construct a better and safer vehicle), OR, you can say it was the result of evil forces in action. 

In certain ways, even animal emotions can be anticipated, programmed, calculated, promoted, manipulated. 

A cult can only be followed.

Below some of thousands of pictures at Internet showing that even under heavy fog, lights get dim and smaller at distance, never enlarged and stronger. When the air/atmosphere becomes denser by any reason (smoke/fog) it refracts and disperses strongly the photons, in some cases even blocking them, this is more pronounced in high energy / high frequency photons - this is why car's fog-light is usually yellow.  There is not a single scientific proof that an illuminated image can become enlarged or brighter based on "light intensity" or how far it is.  I also would love to see any proof of that in math or optics form, or evidence reproduced in laboratory.

149

Flat Earth Theory / Re: Theory/Model Request

« on: May 07, 2019, 04:11:39 PM »

Brain Storming is good, it allows both FEs and REs to participate and try to help each other in the issue.

I would go for the idea of the Sun projecting "Repulsive Radiance" (RR™) towards Earth down below.

You can think of "Repulsive Radiance" (RR™) as the same of atomic nuclear force, it attracts electrons that has opposite polarity, but does not expunge the protons tied together, even having the same polarity.  The repulsion itself is proved by the fact that no electron plunges and it is consumed by the nucleus - even being strongly attracted, it is also strongly repulsed.  The scientists also have plenty of calculations and formulas to explain the Sun's flares and winds that can blow matter far away, the case of particles and gases from comets, forming the long trails.

So, "Repulsive Radiance" (RR™) is caused by the extreme energy from the Sun projected towards Earth, it repulses Earth and everything on it downwards, while the Sun itself becomes repealed by the masses below, so it stays up at a distance, like floating.

I could, in time, come up with a physics formula, trying to show Repulsive Radiance (RR™), may be something like that:

RrV = (Rra x m ÷ Ea) x eV x sin(a) ÷ d²

RrV = Radiance Repulsion Vector
Rra = Radiated Radial Area in km²
m = sum of all masses
Ea = Earth Area in km²
eV = electron-volts, energy generated by the Sun
a = angle of radiation vs angle of Earth
d = distance between the Sun and the Earth

Note that I did not include "steradians" in the calculation, due considerations of a non spherical Sun or Earth in this case study.
This (initial) formula could explain why the Sun stay afloat above Flat Earth, as well, why everything is pushed down against the solid solid Earth underneath, considering not only the edge is solid ice, but the bottom of the flat earth is also a very tick ice base.

Of course I don't believe in a single word I post above, but a brain-storm process relies exactly on that, go out of your mind and find possible alternative explanations.  If Einstein was under effect of some acid when he came up with the idea of space and time could deform, become plastic, elastic and dilute itself under presence of mass, I don't know, but there is a chance. If this was the case, hours later Einstein probably thought "wow buddy, that is a great possibility". Most musicians and artists sometimes need a "stimulae" in order to reach nirvana and get out of the box.  For decades, "brain-storm" means exactly that, you need to turn off all your locks and even the logical, psychological and cultural bias and let the brain to process junk, like a dream.  It may looks strange, but I already found path for a complicated Sin(x) calculation 24 bits routine I was doing in assembly language, while dreaming, woke up in a jump, rush to the computer.   It is not rare for me to find answers while showering, try to explain that.

So, may be the above formula could be part of a larger explanation about day & night on FE.  At least now we could explain the "push" of gravity from above and how the Sun floats.  Oh, I forgot.  The Repulsive Radiance is also reflected by the Earth's surface back towards the Sun, and that is what makes it move forward in a circular motion towards the West at angular speed of 15° per hour.  So that also explains the Sun's movement above.  That also, could be the basis for ocean tides.

Now no one could say I am a polarized insensitive RE, not at all, see, no FE even thought about RR™ up to now, and here I am, giving it for free.

150

Flat Earth Theory / Re: Contradictions in Flat Earth Theory: Constellations and Universal Accelleration

« on: May 07, 2019, 02:34:36 PM »

I think we need to let aside personal euforia and go a little bit more academic and scientific on this:

Ending in "-tion" is used to form nouns meaning "the action of (a verb)" or "the result of (a verb)". Words ending in this suffix are almost always derived from a similar Latin word.   So, "gravitation", is the action of "gravity" upon any mass.

https://en.oxforddictionaries.com/definition/-tion

Merrian-Webster explains that "gravitate" is to move under the influence of "gravitation".
https://www.merriam-webster.com/dictionary/gravitate

There is no definition anywhere about gravitation being "apparent attraction by gravity", also nothing defined anywhere that it is related to electromagnetic causes.

Even, there is a lot to study and learn to do about gravity, it is an effect of a mass sliding to a less dense space, means "space deformation" and CAN be caused by any mass, as far as we know, doesn't mean the space deformation cannot be caused otherwise.

Also, Merrian-Webster definition of "gravitation"
https://www.merriam-webster.com/dictionary/gravitation

Oxford-dictionary, gravitation:
https://en.oxforddictionaries.com/definition/gravitation

151

Flat Earth Theory / Re: Planetary retrograde motion

« on: May 06, 2019, 10:12:22 PM »

Based on the size of the planet and our own minuscule size over it, the tinny part we can see will always looks flat, no matter if Earth is flat or oblate spherical, also, the bottom of the clouds are flat since they are literally SIT over a cushion of warm atmosphere, it is like foam floating over water, and that, would not differ from happen on Flat Earth or oblate spherical one.   The FE statement about this is the same as saying:  "Oh, this car runs on gasoline, using an Otto engine, over four rubber tires, so for unquestionable truth, it is a Ford".   I would like to read how a FET believe the bottom of clouds should be on a RE.... round?

152

Flat Earth Theory / Re: Polaris & Alpha Crucis visibility

« on: May 06, 2019, 10:00:06 PM »

You can assume RE if you wish. But if you assume FE, the OP argument doesn't hold, it doesn't lead to a contradiction. See this https://wiki.tfes.org/Distance_to_the_Sun

Quoting from a second link from the link above:
""IT is well known that when a light of any kind shines through a dense medium it appears larger, or magnified, at a given distance than when it is seen through a lighter medium. This is more remarkable when the medium holds aqueous particles or vapour in solution, as in a damp or foggy atmosphere. Anyone may be satisfied of this by standing within a few yards of an ordinary street lamp, and noticing the size of the flame; on going away to many times the distance, the light upon the atmosphere will appear considerably larger. This phenomenon may be noticed, to a greater or less degree, at all times; but when the air is moist and vapoury it is more intense. It is evident that at sunrise, and at sunset, the sun's light must shine through a greater length of atmospheric air than at mid-day; besides which, the air near the earth is both more dense, and holds more watery particles in solution, than the higher strata through which the sun shines at noonday; and hence the light must be dilated or magnified, as well as modified in colour.""

I really wonder where it is written, tested and confirmed  "well known that the image of an object (photons) through a denser medium, becomes larger".
Based on optics (that also I know very well), the only way for that to happen is if the denser medium to have a special shape, the case of any mid school optics teaching, with enlarging lenses, any optical ocular lens to enlarge small objects, any basic idea of microscope and even telescopes objective + ocular lens to refocus exactly over the retina of the observer. Careful adjustments for focus are necessary.   

Six important points,

1)  I wonder how with several different temperature and moisture in the atmosphere all year round, the Sun's size at sunset is exactly and always the same as at noon time, no matter what, this can be measured with the most sensitive angular optical instrument.

2) Based on this "atmosphere enlarge the Sun", by itself it breaks down completely the FE "vanishing and perspective".  If the Sun becomes larger as it goes further through the lens effect of the atmosphere, it must even be super-large visible at midnight. 

3) Also, because not only the Sun, but everything around it would be enlarger at sunset, the angular speed of the Sun at sunset would be much faster than on noon time.  Any small solar image projector, even a cardboard with a pinhole projecting the Sun on a wall at sunset can show its angular speed to be perfectly the same as on noon-time projecting its image on the floor.

4) Using any optical spectrometer and measuring the low frequency selection (>650nm), the solar intensity at sunrise, noon or sunset will be exactly the same in minute detail of Lux or Candelas.  The point here is, if the solar disc was enlarged by some magical effect of the atmosphere at sunset, it will have light intensity reduced in the same factor of image enlargement (that is pure optical science), so, if the Sun is further, thus reduced in size and brightness and then enlarged, its brightness will not increase, its low frequency radiation will continue to be smaller than at noon.  By filtering only low frequencies (>650nm wavelength),  you get only a little bit of yellow, all RED and some InfraRED.  Those frequencies doesn't change intensity from sunrise to sunset, you can measure them steady all day long.  The upper part of the spectrum change intensity, for example, Blue and UV (<520nm wavelength) are very much increased from 10am to 3pm due atmospheric refraction.  The "no change" in intensity for >650nm proves the solar distance and size never changed substantially as it would happen in FE.

5) Pictures and videos showing the Sun at sunset view by high altitude balloons and airplanes, didn't show the Sun smaller, by being watched by less dense atmosphere and then not under the "effect of being enlarged".  The atmosphere pressure at sea level is 101 kPa, at 10km of altitude is around 26.5 kPa, a person could not breath at such altitude, the atmosphere is very cold and dry, no water droplets in suspension.

6) What you see in the street pole light under heavy fog is the refraction of the light inside of the tinny floating drops of water, in most cases creating a kind of halo, that does not (I repeat) DOES NOT enlarge the lamp (showing the edge of the glass lamp enlarged), the halo and refraction does NOT create a perfect image of the lamp enlarged.  We can photograph a perfect edge of the Sun, including the sunspots, not like a halo. And more, the afternoon atmosphere is not fumed by a dense and heavy fog, and even if it happens, not everyday.

Here is street lamps with refraction halo under heavy fog, you can not see any edge of the lamps, the refraction on the suspended water is so heavy;



Here is the Sun at sunset, even trough cloudy sky closer to horizon, the edge is solid, no halo, no refraction, precise focus, even in the band with denser clouds, it didn't show any difference in size.



So, the quote above from the weblink is just trying to satisfy and justify by a non scientific observation what really happen in the RE converted to FE, and obscuring a failure on the FE sunrise and sunset.   The "change in color" to redish at sunrise and sunset is exactly what happens with low frequency photons being able to go through denser atmosphere, the same when coincidental position of the Earth and the Moon receiving more of those photons frequency than direct sunlight for few minutes, creating the effect as known as "blood Moon", nothing else than simple optics in the RE world.  Interesting is that even Blood Moon can be predicted exactly by numbers, using science and physics of the RE.

Then, I would really love to see where it is tested and documented about the expression I mark in bold and underline in the quote text from the link above.  "It is well known...".   It is NOT well known.  See, if that would be true then everything you see outside through your car's windshield will be enlarged.  Also, everything you see through your home window's glass will be enlarged.   According to optics (I repeat, that I know very well), to enlarge an image you need to have a correct shape lens, at a certain distance from the object, and at another certain distance from your eye.  It is not just a denser medium in the middle that will cause the enlargement.   The atmosphere shape at sunset is totally incorrect to cause any image enlargement, the focus distance from our eyes is also totally wrong.

For anyone wanting to learn more about optics, rays of light, mirrors and lenses, visit any of the below weblinks, there you will be able to play with lenses and find out what you need to enlarge any image into your retina.  You will find out the sunset atmosphere can not produce such effect, far from it.
https://ricktu288.github.io/ray-optics/
It is really a nice experience.  There is also an extension for Chrome:
https://chrome.google.com/webstore/detail/ray-optics-simulation/egamlemiidmmmcccadndbjjihkcfiobh?hl=en

Several other simulators:
https://www.comsol.com/models/ray-optics-module
https://www.edumedia-sciences.com/en/node/69-lenses-and-mirrors
https://arachnoid.com/OpticalRayTracer/
https://ophysics.com/l.html

153

Flat Earth Theory / Re: Polaris & Alpha Crucis visibility

« on: May 05, 2019, 01:57:17 AM »

the bright absolute -4.14 magnitude star on the Crux Constellation

Just to be clear, the absolute magnitude of a star is the brightness it would have as viewed from a standard distance of 10pc (32.6 lightyears) and not from Earth. At that same distance the Sun would shine at only +4.8 magnitude making it quite difficult to see with the naked eye, especially from urban areas.

The observed magnitude of Alpha Cruxis (Acrux) is 0.77 which is still a respectably bright star only slightly fainter than Vega.  Magntiude -4.14 is closer to the apparent brightness of Venus.

You are quite correct.  My mistake on that, it is indeed magnitude around 0.77.

154

Flat Earth Theory / Re: Polaris & Alpha Crucis visibility

« on: May 05, 2019, 01:33:43 AM »

From the posted link:
"If the Southern Cross is a circumpolar cluster of stars, it is a matter of absolute certainty that it could never be in-visible to navigators upon or south of the equator. It would always be seen far above the horizon, just as the "Great Bear" is at all times visible upon and north of the equator. "

One should notice that the Crux Constellation is not positioned vertical from the South Pole as "almost" Polaris is on the North Celestial Pole.
As "out of luck", the south has not a easily visible polar star or constellation, the Crux is the easier recognized and closer we have to the South Pole, it is out of alignment, so the entire constellation rotates around a common empty center (at naked eye) on the South Celestial Pole.

From wikipedia:
"At magnitude +5.42, Sigma Octantis is barely visible to the naked eye, making it unusable for navigation, especially by comparison with the much brighter and more easily visible Polaris. Because of this, the constellation Crux is often preferred for determining the position of the South Celestial Pole."

So, Sir James Clarke Ross could indeed not see Sigma Octantis better than myself at naked eye.  I was only able to see it using a 6" Schmidt–Cassegrain telescope in a very clear and dark night.  Great Sir James, an early Antarctica explorer in 1840.  On my first cruise trip over Atlantic waters I took a good Pentax binoculars, not telescope since the vacation ship rocks enough to make you crazy with a telescope.  I initially thought I woud be able to see Atlantic's sky better than from dark home, silly me, the ocean mist and high humidity atmosphere (almost a constant thin fog) make it horrible and dirt to even get close to see the Orion's Nebula, that I can easily see (faint) form my backyard. Nothing better than deep Arizona. I am not surprised that Sir James would have difficult to spot easily the very small and dim Sigma Octantis aboard the rocking HMS Erebus on high seas.  He was a fantastic Royal Navy explorer and deserves our greatest respect.

I was born in the very south of Brazil, live there until my 37 years old, loved astronomy and gazed the southern skies whenever was clear.  I was never surprised by NOT SEEING the Crux Constellation at clear night sky.  Returning home any time at night it was always there, as a friend to make my companion, I could almost say the right time by its rotation in the sky. 

Below a picture showing how to locate the South Pole based on the Crux Constellation.  You need to extend the long line of the cross four and a half times and then vertical down will be the South Pole.   This demonstrate the constellation is not exactly over the South Pole and it "rotates" with 24 hours period.

No strong navigation beacon would be so much used for centuries, if not visible all the time in the southern hemisphere night sky.  That is a fact.

But lets not deviate from the other questions I posted, including how come a FE observer at the Equator CAN see the Sun (or any other star) after raising on the East or almost setting on the West horizon 141000 km away, but an observer in Rio de Janeiro CAN NOT see Polaris at North at only 12910 km away.



From: https://www.constellation-guide.com/constellation-list/crux-constellation/
"Crux is not visible north of +20° in the northern hemisphere, and it is circumpolar south of 34°S, which means that it never sets below the horizon. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia."

From: https://earthsky.org/favorite-star-patterns/the-southern-cross-signpost-of-southern-skies
"When to see the Southern Cross from the N. Hemisphere At 35 degrees south latitude and all latitudes farther south, you can see the constellation Crux – otherwise known as the Southern Cross – at any hour of the night all year around. In that part of the Southern Hemisphere, the Southern Cross is circumpolar – always above the horizon."

155

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.

   

156

Flat Earth Media / Re: Heliocentric Speed Change Problem

« on: May 04, 2019, 04:34:41 PM »

Thank you, QED.   I wish we could have only good days in life.

157

Flat Earth Media / Re: Heliocentric Speed Change Problem

« on: May 03, 2019, 09:12:38 PM »

Years ago there was a student that came up with a intriguing question. He said: 

A car is running at 60km/h in a perfect straight concrete road, all four tires are touching the ground.
The bottom of the tires that touches the ground are unquestionable stopped in reference to the ground.
The upper part of the tires are speeding at twice the speed of the car.
The front of each tire, speeds as the car, but reducing speed, they need to stop when touching the ground.
The back of each tire, speeds as the car, but increasing speed, they need to be double speed on top.
The question was:  how the tire stands this constant acceleration and breaking once per revolution, without damaging? and why we don't feel such "bumping" speed change in the tires?

It took a while to make the students to "see the magic". 
Our brain is unable to by itself catch the idea of certain physical forms or movements.
The only way to start to grasp such light, is thinking; after the car/airplace accelerated and reach some steady speed, it flew few inches in the air.
During this minute, even that the tires are continuing rotating and "almost touching" the ground, and coincidentally they seems to be touching the ground and stopped while doing that, they are not.
In reference to the car and shaft, the tires are in a perfect circular motion, no breaking, no accelerating whatsoever. The centrifugal force on the rubber is perfectly balanced.  What is breaking is the reference to the ground, and as a reference it does not, and can not change anything in the rubber movement.  The surface and angular speed of any part of the tire rubber is exactly the same in reference to the shaft and car.



The same, the illusion of Earth's different speed on the illusion of forward and backward motion due its spinning.  You can find a ton of formulas and calculate several variables and came up to nice conclusions, but in real, it doesn't happen, you can't even measure it, because it doesn't exist.  It doesn't change a bit if the Earths is moving forward or backward in its path around the Sun, the spinning around its own axis cancels any angular vector.  The centrifugal force is equal based on the frame of reference, and doesn't change if you have the pivoting point moving in a constant speed.  What you have is a constant centrifugal force, that in case of Earth's is minuscule compared to the space deformation sliding vector.  Of course, you can have differences IF the pivoting point speed changes, that is not the case of Earth.

You can make a nice and easy experiment.  Remote the front bicycle wheel from the bicycle, cut 20 pieces of a light string 50 cm each, find 20 objects like large metallic nuts or solid 50+g objects (same size rocks or even little bags of sand), tie the objects to the end of the strings, and tie the other side of each string to the bicycle wheel equally spaced by 18 degrees.   Now, hold the wheel horizontally and rotate it until all the strings and objects spread aside in an angle around 45° by the centrifugal force, so you will be able to see any difference of speed, if they speed up the string will climb to a higher angle, if they slow down, the string and the object will go down.  Now, while it rotates happily, and holding the wheel in a way you could walk, start to walk forward holding the rotating wheel, keep an eye on the angle of each string.  Adjust your walking speed in order for the "backward motion" coincide with the ground, so they appear as stopped, and tell me if any string performing the "backward" movement goes down due low speeding or the forward movement would climb in angle for having a greater speed.  They would not.  If you run much fast, or attach the wheel to the hood of your car and run, the air movement will interfere with the results. 

158

Flat Earth Theory / Re: Celestial Gravitation

« on: May 03, 2019, 08:12:59 PM »

The Twin Grace Satellites mapped the planet's gravity.  They orbit the planet at the same altitude (286 miles), one just 137 miles behind another following the exact same path.  When the leading satellite finds a place with pronounced space deformity (strong gravity, even if little), it slides a bit into such less dense space and accelerates a bit. The opposite is true, when finding a place where gravity is weaker (less space deformity), the leading satellite slows down its speed.



Both satellites keep track of their distance precisely, and based on minute changes in such distance they log abnormalities in the space deformation, caused by uneven mass down on the planet.   Several different reasons the planet may present differences in mass, material composition, mountains, valleys, water, etc.   The images on this NASA link are exaggerated to demonstrate the survey results.'

https://nasaviz.gsfc.nasa.gov/11234

The European satellite GOCE also mapped the planet changes in space deformation caused by differences in mass below.  This satellite helped to track movement of lava underground, and melting glaciers. It helped to produce the first high resolution map of the boundary between the Earth's crust and mantle.  To increase resolution, the satellite was lowered to an altitude from 158 to 146 miles.

https://www.space.com/18575-gravity-satellite-lowers-orbit.html

159

Flat Earth Theory / Re: Gravity

« on: May 03, 2019, 04:21:47 AM »

Of course I could answer your questions, with pleasure.
Problem is, I am not here to waste my precious time (and I am old, no much time to waste) teaching physics, mostly just to show my knowledge, this is not a competition, sorry, the time of my life to show a bigger stick is gone long ago.  My intention here is to clarify issues about flat earth.  It is clear that this discussion will go on and on, and it is not the scope of this forum, so research yourself about space deformity by mass and sliding vectors. 
Gravity is not a force - Albert Einstein said that, who am I to go against it?
I wonder if someone could demonstrate force without energy being applied.

For the ones wanting to learn about sliding vectors, I present you Einstein's Field Equation formula:

Sorry moderators, I stop here.
Cheers.

160

Flat Earth Theory / Re: Gravity

« on: May 02, 2019, 09:32:47 PM »

Your weight is not 80kg, that is your mass. Your corresponding weight is about 800N. That is how you calculate it. It is the magnitude of the local gravitational force on you. The gravitational acceleration is found by taking your weight and divisions by your mass.

Again, gravity is not a force, it is a sliding vector (you will be reading more and more about it, pay attention).
By the way, in Brazil we started to learn physics right at the first semester of middle school, not high school.
Also, if you think gravity IS a force, you are first going against Albert Einstein (Nobel prize in physics), Carl Sagan, and several other astrophysicists, good luck on that.

Gravity is most accurately described by the general theory of relativity (proposed by Albert Einstein in 1915) which describes gravity not as a force, but as a consequence of the curvature of spacetime caused by the uneven distribution of mass. The most extreme example of this curvature of spacetime is a black hole, from which nothing—not even light—can escape once past the black hole's event horizon.[3] However, for most applications, gravity is well approximated by Newton's law of universal gravitation, which describes gravity as a force which causes any two bodies to be attracted to each other, with the force proportional to the product of their masses and inversely proportional to the square of the distance between them.

But we now understand that gravity as a force is only part of a more complex phenomenon described the theory of general relativity.  While general relativity is an elegant theory, it’s a radical departure from the idea of gravity as a force.  As Carl Sagan once said, “Extraordinary claims require extraordinary evidence,” and Einstein’s theory is a very extraordinary claim.  But it turns out there are several extraordinary experiments that confirm the curvature of space and time.

===

A body undergoing geodesic motion feels no forces acting upon itself. It is just following what it feels to be a "downward slope through spacetime" (this is how the bending affects the motion of an object). The particular geodesic an object wants to follow is dependent upon its velocity, but perhaps surprisingly, not its mass (unless it is massless, in which case its velocity is exactly the speed of light). There are no forces acting upon that body; we say this body is in freefall. Gravity is not acting as a force.

Dr Jolyon Bloomfield


===

Einstein’s discovery was based on a series of thought experiments. Consider an astronaut floating in space, away from any source of gravity, and that same astronaut free falling in a planet’s gravity. The similarity of both experiences is uncanny. The astronaut must glide or sit still until affected by an external force. If an astronaut falls or floats without any knowledge of his location, say, in an enclosed lift, he cannot distinguish whether the lift is floating in deep space or through a building on Earth. In both cases, he is essentially weightless. However, if he does not experience any force, why does a free-falling astronaut accelerate? In Newtonian mechanics, this is paradoxical, as it contradicts Newton’s second law of motion – the magnitude of acceleration is proportional to the applied external force.

Einstein suggested that objects aren’t pulled by massive objects, but rather pushed down by the space above them. According to General Relativity, matter warps the fabric of not only space but time as well, collectively known as the continuum of space-time. The fabric is like a grid of tightly strung rubber bands; when a massive object pushes and stretches them downward, the deformed rubber bands push objects under them. The theory implied that smaller objects weren’t pulled towards massive objects but were traveling on a downward slope, as the space in the latter’s vicinity was warped by its large mass. A free-falling body, therefore, follows the straightest possible path in space-time.

Einstein developed this theory on the assumption that the laws of physics must appear the same to every observer. This is also true for planets revolving around the Sun. Orbiting planets follow the shortest path around the Sun to minimize energy. This path is an ellipse, the most efficient path in the gravity well of the Sun… but what about the astronaut’s acceleration?

Einstein’s geodesic equations signify that acceleration is a product of curved space-time. His equation explains how curvature accelerates a falling object. In the absence of curvature, the body would move in a straight line with a constant velocity, unless this motion would be disrupted by an otherwise external force. However, the most interesting aspect of the equation is the absence of mass in its expression. The magnitude of acceleration is independent of the falling body, just as the equivalence principle would demand (if you drop a hammer and a feather on the surface of the moon, they would drop at the same time).

Akash Peshin
 
===

Albert Einstein, who won the Nobel Prize in Physics in 1921, contributed an alternate theory of gravity in the early 1900s. It was part of his famous General Theory of Relativity, and it offered a very different explanation from Newton's Law of Universal Gravitation. Einstein didn't believe gravity was a force at all; he said it was a distortion in the shape of space-time, otherwise known as "the fourth dimension" (see How Special Relativity Works to learn about space-time).

Basic physics states that if there are no external forces at work, an object will always travel in the straightest possible line. Accordingly, without an external force, two objects travelling along parallel paths will always remain parallel. They will never meet.