How far is Sun in FE model? Definitely more than 5000km.

Quote from: gazza711 on December 11, 2015, 10:20:00 AM

lets just make this easy-how do you know the sun and moon are the distances we are told?simple question?

For FE model, I do not know. Therefore I am using what I know to find values.

RE model is not the topic of this discussion.

Quote from: gazza711 on December 11, 2015, 10:20:00 AM

you say the sun moves 9000km at noon horizontally?really-ive seen an aeroplane move about 100km at 8-11km up @885km/hour.so if you sized down the apparent size of the sun by 3000,you would still get the same answer.you are told the sun is 149600000km away which happens to be 3000 times further than a the FE theory.well,how do you get sunrays at sharp angles from a light source so far away.and its size is 267000 times the diameter of the FE 52km.its all waffle.i mean the sun and moon just happen to be the same size.

How is that related to the topic of this discussion?
ur asking how fe'rs come up with a different size and distance.well no matter what the real distance and diameter is,at sunrise,you should be able to see the light hours before sunrise.so it cant be the distance and diameter re'rs believe /state.so i will answer ue question soon.thx

Quote from: rabinoz on December 11, 2015, 07:35:02 PM

I am going right back to what I see as a completely false premise in the OP.
You say "According to wikipedia, the angular size of the sun is 31'6'' to 32'7''."  That seems to be true enough,
but then you go one to say, "This is enough to calculate both diameter and the distance to the Sun."  That is completely incorrect, because it seems to say that the sun's apparent size varies throughout the day.
Apart from some possible distortion due to diffraction when near the horizon the sun's apparent size does NOT change measurably throughout the day.  The change in the sun's apparent size is over the course of a year due to the ellipticity of the earth's orbit.
One would think that the with the FE model of the Sun's motion that it should change, but it does not.

I assumed the apparent size changes as this is natural for things that are close to you but move closer/further. Changes that we observe are based on daily motion from the noon to the sunset. Since the FE model hand-waves about seasons and day/night cycles (especially for the case of 24h day in Antarctica), I have no other point of referrence.

Keeping the size of the Sun would "destroy" the entire argument, but I wanted to give it a chance and hence my assumption.

Quote from: rabinoz on December 11, 2015, 07:35:02 PM

The distance to the FE sun has been "calculated" by Voliva, Rowbotham and I guess a few others. Jroa has referred to the Voliva calculation, and in other posts I have mention others.

Rowbotham in his book does not give any specific value. He just shuffles few numbers and claims the Sun is no further than 4000 miles. The remaining "calculations" are flawed - see previous posts.

I believe we can measure the size of the Sun directly.

Assuming the Sun is 32' 00" of arc in the sky:

• at 3000 mi up, it is 27.92547 mi in diameter
• at 5000 mi up, it is 46.54245 mi in diameter
• at  93M mi up, it is   865,690 mi in diameter

On the equinox, on the equator, we just need to place vertical poles along the equator, those distances apart and measure the shadows when the Sun is directly between them (3rd pole in the middle). If they are say the 27.9 mi apart and have no shadows, the Sun is that size and 3000 mi up!

If the Sun is 93,000,000 mi away, then all the poles shadows (except the middle one) should point inward.

Am I missing something?

Also another thing bothers me.

In the FE model the sun and moon are the same size, and same distance away. That would mean during a Lunar eclipse the umbra would be huge, and the penumbra would be small.

But we don't see that. We see a small umbra and large penumbra. That shows the sun is much further away than the moon, and much larger.

Also another thing bothers me.

In the FE model the sun and moon are the same size, and same distance away. That would mean during a Lunar eclipse the umbra would be huge, and the penumbra would be small.

But we don't see that. We see a small umbra and large penumbra. That shows the sun is much further away than the moon, and much larger.

• The lunar eclipse happens during a FULL Moon. The Moon has to be on the opposite side of the Sun in the sky.
• The Sun also has a "spotlight" hitting the Earth and does not shine on the Moon (therefore self-illuminating), therefore has nothing to do with the penumbra and umbra in the FE model. It is a 100% Moon ONLY thing that needs to be explained.
• RET/heliocentric models explain the umbra as the light from the Sun going through the Earth's atmosphere  - therefore fuzzy and turning the light reddish (like a sunset) and is about 4x the diameter of the Moon.
  
  Consider this image of a lunar eclipse. You can see the light from the larger Sun hitting the smaller Earth. It creates penumbra and umbra shadows on the Moon:
  
  
  
  It is hard to photograph because the FULL Moon is VERY bright and the total eclipse is VERY dim. The last image demonstrates the brightness changing and the Moon going from white/gray to red.
  
  The very bright Moon at the top is the FULL Moon. As the Moon enters the Penumbra, it gets dimmer. When it enters the Umbra, it gets dimmer and redder.
  

Quote from: yobbo on January 01, 2016, 02:09:08 PM

Also another thing bothers me.

In the FE model the sun and moon are the same size, and same distance away. That would mean during a Lunar eclipse the umbra would be huge, and the penumbra would be small.

But we don't see that. We see a small umbra and large penumbra. That shows the sun is much further away than the moon, and much larger.

Actually, that would not be correct. Keep in mind several things:

• The lunar eclipse happens during a FULL Moon. The Moon has to be on the opposite side of the Sun in the sky.
• The Sun also has a "spotlight" hitting the Earth and does not shine on the Moon (therefore self-illuminating), therefore has nothing to do with the penumbra and umbra in the FE model. It is a 100% Moon ONLY thing that needs to be explained.
• RET/heliocentric models explain the umbra as the light from the Sun going through the Earth's atmosphere  - therefore fuzzy and turning the light reddish (like a sunset) and is about 4x the diameter of the Moon.
  
  Consider this image of a lunar eclipse. You can see the light from the larger Sun hitting the smaller Earth. It creates penumbra and umbra shadows on the Moon:
  
  
  
  It is hard to photograph because the FULL Moon is VERY bright and the total eclipse is VERY dim. The last image demonstrates the brightness changing and the Moon going from white/gray to red.
  
  The very bright Moon at the top is the FULL Moon. As the Moon enters the Penumbra, it gets dimmer. When it enters the Umbra, it gets dimmer and redder.
  

• The Sun and Moon need to be the approximately the same size.
• The Moon must be lower than the Sun so it can make a solar eclipse.
• One HUGE GINORMOUS problem is if the Sun and Moon were close together VERTICALLY in the sky, the Moon (34′6″ max) would TOTALLY block the Sun (31′31″ – 32′33″) creating a solar eclipse all around the world at once. This does not happen. Per the picture below, although the Sun and Moon have the same ANGULAR size in the sky (~0.5°) they are clearly not the same size or close together in the sky so they cut such a narrow path across the Earth. (http://www.greatamericaneclipse.com/best-places-to-view/)
  
  
• The Moon MUST be flat or people around the Earth would not see the same image.
• THE problem with a flat Moon is Lunar Libration. Although we see 50% of the Moon at any given time, we can see upwards of 9% more based on the location of the observer Norway/Sydney or moonrise/moonset, and the Moon has a 5.1° orbital inclination and a 1.5° axial tilt. Also apogee/perigee affects what can be seen. This can not be described by a flat disk unless the "skin" of the Moon moves across it's surface.(https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Libration)
• As the Moon approaches the Sun (whether or not it eclipses it), depending on the albedo (reflective surface brightness) of the Earth, the dark side of the the phase shadow (NOT the back side of the Moon that we NEVER see), can be lit up by Earthshine. It is not clear why/when this happens on the FE model since the Moon is always some 3000 mi up (look up "Earthshine Images" - https://en.wikipedia.org/wiki/Planetshine).
  
• The Moon must be self-illuminating or the Sun would illuminate it all the time and the phases would not work.
• Also what is not immediately clear is what the solar eclipse will look like with the Sun's "spotlight" in effect, especially in Antarctica (assuming the N.Pole is in the center of the disk as opposed to the S.Pole):
  
• Other than these, in the simplest sense (as discussing this with a child), yes, one disk blocking the light of the other disk would make sense.

Also what is not immediately clear is what the solar eclipse will look like with the Sun's "spotlight" in effect, especially in Antarctica (assuming the N.Pole is in the center of the disk as opposed to the S.Pole):

The Moon must be self-illuminating or the Sun would illuminate it all the time and the phases would not work.

Again, read how the angular size of Mars is completely wrong...

http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

brouwer, let us go back to your first message.

There is a difference in applying angular diameter calculations re: spherical bodies, as opposed to applying them to objects in the shape of a disk.

Here is the formal definition: The angle subtended at the observer by a diameter of a distant spherical body which is perpendicular to the line between the observer and the center of the body.

For a disk-shaped body we have this formula, using the graphic http://wpcontent.answers.com/wikipedia/commons/thumb/4/49/Angular_dia_formula.JPG/400px-Angular_dia_formula.JPG :

@ = 2 arctan (1/2 x d/D), if D is much larger than d, then we can approximate by @ = d/D
@ = angular diameter

Let me remind you that indeed the sun has the shape of a disk.

Impossibility of a round Sun shape:

The atmospheric pressure of the sun, instead of being 27.47 times greater than the atmospheric pressure of the earth (as expected because of the gravitational pull of the large solar mass), is much smaller: the pressure there varies according to the layers of the atmosphere from one-tenth to one-thousandth of the barometric pressure on the earth; at the base of the reversing layer the pressure is 0.005 of the atmospheric pressure at sea level on the earth; in the sunspots, the pressure drops to one ten-thousandth of the pressure on the earth.

The pressure of light is sometimes referred to as to explain the low atmospheric pressure on the sun. At the surface of the sun, the pressure of light must be 2.75 milligrams per square centimeter; a cubic centimeter of one gram weight at the surface of the earth would weigh 27.47 grams at the surface of the sun. Thus the attraction by the solar mass is 10,000 times greater than the repulsion of the solar light. Recourse is taken to the supposition that if the pull and the pressure are calculated for very small masses, the pressure exceeds the pull, one acting in proportion to the surface, the other in proportion to the volume. But if this is so, why is the lowest pressure of the solar atmosphere observed over the sunspots where the light pressure is least?

Because of its swift rotation, the gaseous sun should have the latitudinal axis greater than the longitudinal, but it does not have it. The sun is one million times larger than the earth, and its day is but twenty-six times longer than the terrestrial day; the swiftness of its rotation at its equator is over 125 km. per minute; at the poles, the velocity approaches zero. Yet the solar disk is not oval but round: the majority of observers even find a small excess in the longitudinal axis of the sun. The planets act in the same manner as the rotation of the sun, imposing a latitudinal pull on the luminary.

Gravitation that acts in all directions equally leaves unexplained the spherical shape of the sun. As we saw in the preceding section, the gases of the solar atmosphere are not under a strong pressure, but under a very weak one. Therefore, the computation, according to which the ellipsoidity of the sun, that is lacking, should be slight, is not correct either. Since the gases are under a very low gravitational pressure, the centrifugal force of rotation must have formed quite a flat sun.

Near the polar regions of the sun, streamers of the corona are observed, which prolong still more the axial length of the sun.

If planets and satellites were once molten masses, as cosmological theories assume, they would not have been able to obtain a spherical form, especially those which do not rotate, as Mercury or the moon (with respect to its primary).



Solar Atmosph. Pressure as a Function of Depth (official science information)

Depth (km) % Light from this Depth Temperature (K) Pressure (bars)

0 99.5 4465 6.8 x 10-3
100 97 4780 1.7 x 10-2
200 89 5180 3.9 x 10-2
250 80 5455 5.8 x 10-2
300 64 5840 8.3 x 10-2
350 37 6420 1.2 x 10-1
375 18 6910 1.4 x 10-1
400 4 7610 1.6 x 10-1

This table indicates that the solar atmosphere changes from being almost completely transparent to being almost opaque over a distance of about 400 km. Notice also that in this region the temperature drops rapidly as we near the surface, and that the pressure (measured in bars, where one bar is the average atmospheric pressure at the surface of the Earth) is very low - generally 1% or less of Earth surface atmospheric pressure.


Now, let us go back to the very subject of your thread.

http://evildrganymede.net/rpg/world/angular_diameters.pdf

To calculate the angular diameter, you need to know the diameter of the object in question, and how far away it is from the observer.


Here is where each and every scientist (from Picard, official chronology of history, to today) makes the mistake: they will use the following data, diameter 1392000 km, distance 149600000 km (for the sun-earth system).


You made the same mistake.

According to wikipedia, the angular size of the sun is 31'6'' to 32'7'', but they used the same wrong data taken from the textbooks on heliocentricity.

Moreover, the assumptions made by the official figures offerred by textbooks on astronomy (including the work attributed to J. Picard), rely on the very wrong ideas about stellar parallax:

http://web.archive.org/web/20150321094726/http://www.realityreviewed.com/Negative%20parallax.htm


Here is a classic work on the angular size of Mars, how the assumptions made by the figures offered by official astronomy, are absolutely wrong:

On the angular size of Mars:

http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4

Quote from: jroa on November 06, 2015, 12:51:10 AM

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

Mmmmmm thats a bit rich from you, I do remember you presenting your 'calculations' on the diameter of the Sun that I think constituted as drivel...I suppose it takes one to know one....Come on jroa...you used google to get some of your facts, but where did you get the Sun Earth distance from? where was your acceptable calculation for that one?

Quote from: Nightsky on August 10, 2016, 12:15:09 AM

Quote from: jroa on November 06, 2015, 12:51:10 AM

I don't think that anybody even understands what your rant means.  I was tempted to move it to the AR section when I first saw it, but left it just in case someone else can make any sense out of your drivel.  If you want people to actually respond, it would help to make an intelligible post to start with.

Mmmmmm thats a bit rich from you, I do remember you presenting your 'calculations' on the diameter of the Sun that I think constituted as drivel...I suppose it takes one to know one....Come on jroa...you used google to get some of your facts, but where did you get the Sun Earth distance from? where was your acceptable calculation for that one?

I told you how to calculate the distance to the sun in the exact same thread that you are referencing.  Did you even read the thread, or did you just see that I posted in it so you decided to harass me in every thread?

The sun could be any distance. The fact that you and your cronies harass people is deplorable.

The sun could be any distance. The fact that you and your cronies harass people is deplorable.