The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: Brouwer on November 02, 2015, 11:23:51 PM
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Hi FES!
I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.
The topic I'd like to discuss is how far the Sun is.
(you can skip this post and go to http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321 (http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321) for more details).
According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.
Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball). Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.
Notion:
d - the distance to the Sun
R - radius of the Sun
Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).
In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).
Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.
So the result is far from 5000km. Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.
How far the Sun really is?
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Any comments?
This simple math basically proves the Sun can't be as low as FET claims. Unless you prove I am wrong...
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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.
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If you're going to try to use your math, explain your process rather than just claiming results. You might be right or wrong, probably wrong, but until you actually explain what you're done...
"we have the relation d=104.131456R," makes no sense until you can actually show your calculations, rather than claiming results.
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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.
Why is it a rant? Just needs more explanation.
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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.
Why is it a rant? Just needs more explanation.
He could actually explain his point, instead of just hitting random letters and numbers.
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I don't think that anybody even understands what your rant means.
If they ever bothered to make a proper picture, what I wrote would be easy to explain.
Let me explain triangles and the math a little more precise.
We have a right triangle with side (d^2+2dR)^(1/2) and legs: R and d+R. If A is the angular diameter, then A/2 is the angle between (d^2+2dR)^(1/2) and d+R. R is the radius of the Sun. Then we have
sin A/2= R/(d+R)
so
d/R=(1-sin A/2)/(sin A/2).
Now take A=31.6' to obtain
d/R=107.7909868
and A=32.7' to obtain
d/R=104.131456
This gives d between 104.131456R and 107.7909868R.
The rest is simple. We have yet another right triangle, with sides 104.131456R, 107.7909868R and 9003.1632km (107.7909868R is the hypotenuse). This triangle is formed by the observator point, the Sun at noon and at set/rise. From Pythagoras we get R, and then d.
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We have a right triangle with side (d^2+2dR)^(1/2)
2dR.
Just what.
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Hi FES!
I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.
The topic I'd like to discuss is how far the Sun is.
According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.
Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball). Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.
Notion:
d - the distance to the Sun
R - radius of the Sun
Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).
In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).
Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.
So the result is far from 5000km. Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.
How far the Sun really is?
If they liked/understood math and geometry, they wouldn't believe the earth is flat in the first place
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2dR.
Just what.
Can you be more specific? I do not have mindreading skills.
If they liked/understood math and geometry, they wouldn't believe the earth is flat in the first place
You may assume that people know math (trig) in suffucient level, but it turns out that most of them tend to forget. I can explain things in more details, but I expect readers not to show ignorance.
It is not only understanding, but also proper application showing that there is something wrong.
Also, posting things like that:
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.
just proves one thing: some people here are willing to discuss FE stuff, but they do not even put an effort in reading and checking numbers. It is a shame that a moderator (someone who should present a generic, proper behaviour style on the forum) simpy wanted to throw my posts calling them "drivel". If he spent few minutes, he wouldn't call them "random letters and numbers".
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2dR.
Just what.
Can you be more specific? I do not have mindreading skills.
Where the hell did you get that? You're either multiplying the distance to the Sun by its diameter, which is pointless, or you've inexplicably tried to use twice the distance to the Sun.
Try explaining a post rather than handwaving through a series of assertions.
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Try explaining a post rather than handwaving through a series of assertions.
We have a right triangle.
Hypotenuse is R+d.
One of legs is R.
Pythagoras theorem says that 2nd leg is sqrt((R+d)^2-R^2).
Did you even bother to draw a triangle? This is completely elementary "math" (actually, it is just calculations).
This lenght actually does not matter, It is here just for completion and can be skipped without ruining the rest of the argument.
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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 (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 (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 (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 (http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4)
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Let me remind you that indeed the sun has the shape of a disk.
That does not make any difference. 5000km remains too small. I repeated calculations for the disk and obtained ~320km diameter and ~33600km distance. Basically the same result as for ball model.
Unfortunately, disk shape suspended 5000km above the surface would make no sense, as it would be seen as an ellipse near the set or rise time, unless it somehow faces towards us. But a disk shape cannot face toward every observer on the Earth at the same time in such a way each observer sees a disk. This is possible for ball shape.
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.
Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'. Using a better equipment you can increase precision. And find a proper interval. Very small intervals work against 5000km.
In case of the Sun with fixed size, suspended 5000km above the surface, the noticable change of the angular diameter would be significant. At noon it would be 2x larger than at set/rise. Anyone would notice that huge difference.
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Let me remind you that indeed the sun has the shape of a disk.
That does not make any difference. 5000km remains too small. I repeated calculations for the disk and obtained ~320km diameter and ~33600km distance. Basically the same result as for ball model.
Unfortunately, disk shape suspended 5000km above the surface would make no sense, as it would be seen as an ellipse near the set or rise time, unless it somehow faces towards us. But a disk shape cannot face toward every observer on the Earth at the same time in such a way each observer sees a disk. This is possible for ball shape.
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.
Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'. Using a better equipment you can increase precision. And find a proper interval. Very small intervals work against 5000km.
In case of the Sun with fixed size, suspended 5000km above the surface, the noticable change of the angular diameter would be significant. At noon it would be 2x larger than at set/rise. Anyone would notice that huge difference.
The distance to the sun is 149Mkm.
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Hypotenuse is R+d.
Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.
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Unfortunately, disk shape suspended 5000km above the surface would make no sense...
Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.
Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'.
Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax). And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.
I can dismiss the 149million km Sun figure immediately using the faint young sun paradox, not to mention the double forces of attractive gravitation paradox...
Again, read how the angular size of Mars is completely wrong...
http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4 (http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4)
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Unfortunately, disk shape suspended 5000km above the surface would make no sense...
Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.
Or you can measure this directly on the sky. No diameter and distance required. Measuring gives approximately 32'.
Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax). And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.
I can dismiss the 149million km Sun figure immediately using the faint young sun paradox, not to mention the double forces of attractive gravitation paradox...
Again, read how the angular size of Mars is completely wrong...
http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4 (http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4)
I think your link, although it probably has some applications, probably does not apply to astronomy. If so, please answer the following question.
How can I see the light from a small flashlight (1" in diameter) a mile away?
Given:
- 1 mi = 5280' = 63,360"
- 24 arcsec = 0.006667°
Max height = tan(0.006667°) x 63,360" = 7.37"
You should not see the light from the flashlight. Max distance I should see it is 8594" = 716'
Please explain...
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Please explain...
I think there should somewhere be permanent post which encourages users to not start debating with sandokhan. It really lives in its own world and never really listens anyone but goes on and on and on with its half baked stories. Just my 2 cents.
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Agreed. The sun cannot possibly orbit at an orbit of 3000 miles (4800 km) above the Earth.
So how high is it?
The point of this thread is to answer this question, possibly giving a final and correct answer. It is clear that 5000km is wrong which is what, based on personal observation, most FET give and believe.
Do not forget that in each such measurement certain assumptions are made (just as J. Picard, official chronology of history, had to make re: stellar parallax).
I know that. I assumed that the change in size is due to the Sun moving away during the day, nothing more.
And if no diameter and distance are required, then you might get a similar reading for a DIFFERENT set of figures: a smaller Sun, and a much lower orbit.
You do not. Actually, all you need is any round object of known size that you can suspend in air and a little bit of math to figure, how far you need to be from this object to see it with proper angular diameter.
The main question remains open. How far is the Sun in FE model? I hope you can present more than just a number, because explaining the value is the most important part.
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brouwer, you were asked several questions by thebigone to which you provided no answers, so far. Basic questions about your calculations.
Actually, all you need is any round object of known size that you can suspend in air and a little bit of math to figure, how far you need to be from this object to see it with proper angular diameter.
You should do some homework before you come here to post your messages.
Let us go to the textbook on angular size/diameter actual calculations.
The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.
How large an object appears depends not only on its size, but also on its distance.
Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.
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You are correct. You can not determine size or distance from just the angular size of the sun. You need its actual diameter or its actual distance to calculate the other.
You will probably need an "Eratosthenes experiment" but with at least 3 and preferably 5-10 locations. This will determine the shape of the Earth (Flat or Round) based on where the measurements converge, then based on that you can find the distance to the Sun, then the diameter of the Sun - probably with a good deal of accuracy (like <1%). This can probably also be applied to the Moon as well using a Full Moon.
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If they ever bothered to make a proper picture, what I wrote would be easy to explain.
Let me explain triangles and the math a little more precise.
We have a right triangle with side (d^2+2dR)^(1/2) and legs: R and d+R. If A is the angular diameter, then A/2 is the angle between (d^2+2dR)^(1/2) and d+R. R is the radius of the Sun. Then we have
mistaken assumptions.
Your initial premises are wrong.
The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole.
The sun moves up and down as well as in and out.
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Hypotenuse is R+d.
Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.
Sorry that I missed your post. I measure d to be the distance to the surface to the Sun, not to the center of the Sun. The triangle I am talking about has legs R and (d^2+2dR)^(1/2) and hypotenuse is R+d. I hope that should clarify the stuff.
The way I measure d (to the core or to the surface) will not change the final result at all (the value of R will remain the same).
The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.
Which is all what I need - how much of the sky is covered.
Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.
It is crutial, but we are do not need the distance and size to start with. They are found using simple trig.
mistaken assumptions.
Your initial premises are wrong.
The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole.
The sun moves up and down as well as in and out.
To keep the angular size in measurable range, objects moving up and down/in and out should change their size to perfectly match obesrvable angular size.
Can you present your numbers/math then? Mine is based on 6h time shift (from noon to set) and assumptions that seemed natural. I would like too see a different math that would help me understanding, how it is possible that the fixed size object can keep its angular diameter from such low distance.
For instance, assuming the sun is really 5000 km above, it moves 9000km away and say 1000km up/down, it should look significantly smaller (about twice as smaller angular size). But it does not.
I cannot seem to understand the distance, but I am the only one who presented any numbers and calculations. Handwaving and saying stuff without supporting it with citations/math/other sources is not what I call a reliable argument.
There are more paradoxes with the distance, but I figured I will stick to the observable angular size. If you want, I can present the other one.
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Hypotenuse is R+d.
Just what. Did you draw the triangle? The only way R+d is going to give a relevant straight line, let alone a hypotenuse, is if you measure to the core of the Sun, which is nonsenical.
I'm not going to trust your calculations if you can't get a basic step correct.
Sorry that I missed your post. I measure d to be the distance to the surface to the Sun, not to the center of the Sun. The triangle I am talking about has legs R and (d^2+2dR)^(1/2) and hypotenuse is R+d. I hope that should clarify the stuff.
The way I measure d (to the core or to the surface) will not change the final result at all (the value of R will remain the same).
The angular sizes of objects show how much of the sky an object appears to cover. Angular size does not, however, say anything about the actual size of an object.
Which is all what I need - how much of the sky is covered.
Therefore, contrary to your armchair opinion, DISTANCE IS CRUCIAL TO CALCULATE THE ACTUAL SIZE OF AN OBJECT.
It is crutial, but we are do not need the distance and size to start with. They are found using simple trig.
mistaken assumptions.
Your initial premises are wrong.
The sun does not rotate exactly parallel to the surface of the earth nor does it stay exactly equidistant from the North Pole.
The sun moves up and down as well as in and out.
To keep the angular size in measurable range, objects moving up and down/in and out should change their size to perfectly match obesrvable angular size.
Can you present your numbers/math then? Mine is based on 6h time shift (from noon to set) and assumptions that seemed natural. I would like too see a different math that would help me understanding, how it is possible that the fixed size object can keep its angular diameter from such low distance.
For instance, assuming the sun is really 5000 km above, it moves 9000km away and say 1000km up/down, it should look significantly smaller (about twice as smaller angular size). But it does not.
I cannot seem to understand the distance, but I am the only one who presented any numbers and calculations. Handwaving and saying stuff without supporting it with citations/math/other sources is not what I call a reliable argument.
There are more paradoxes with the distance, but I figured I will stick to the observable angular size. If you want, I can present the other one.
The distance has been proven, why are you trying to disagree when you are guessing at numbers.
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and hypotenuse is R+d.
Here's a hint. When I directly question where on earth you get a ridiculous measure of the hypotenuse from, you need to do a bit more than just repeat it. If you are not measuring to the center of the Sun then R+d as you've defined them mean nothing.
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The distance has been proven, why are you trying to disagree when you are guessing at numbers.
I am not guessing any numbers. I am making assumptions and use simple math to derive the outcome. It does not matter whether the Sun changs its distance daily by a small amount, it does not matter if it does that when chaning seasons, it does not matter if the orbit is ellipctic. You repeat the math and obtain the same thing - 5000km is definitely not enough.
Here's a hint. When I directly question where on earth you get a ridiculous measure of the hypotenuse from, you need to do a bit more than just repeat it. If you are not measuring to the center of the Sun then R+d as you've defined them mean nothing.
There is no difference in measuring to the center and to the surface. It is just setting a convention for this calculation only...
(http://i.imgur.com/tHeEfYm.jpg)
tangent^2+radius^2=(d+radius)^2
There the mysterious d+R comes, and so (d^2+2dR)^(1/2). Got it?
If I take d to be measured to the point B, I will have sin A/2 =R/d and then I can repeat the entire argument as I did in 1st post to obtain the same value of R.
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To show how wrong the argument from wiki is:
Thomas Winship, author of Zetetic Cosmogony, provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface if one assumes that the earth is flat --
On March 21-22 the sun is directly overhead at the equator and appears
45 degrees above the horizon at 45 degrees north and south latitude. As
the angle of sun above the earth at the equator is 90 degrees while it is
45 degrees at 45 degrees north or south latitude, it follows that the angle
at the sun between the vertical from the horizon and the line from the
observers at 45 degrees north and south must also be 45 degrees. The result
is two right angled triangles with legs of equal length. The distance between
the equator and the points at 45 degrees north or south is approximately 3,000
miles. Ergo, the sun would be an equal distance above the equator.
We can repeat the same experiment (same date). The Sun remains overhead at the equator. But this time instead of 45 degrees we take 5 degrees (this can be seen close to the North Pole). The math says that if one of legs is 3000 miles and the angle between hypotenuse and the other leg is 5 degrees, then the other leg has almost 34290 miles. This is how far the observer would have to be from the equator to see the Sun at 5 degree angle. Which is nonsense as the distance from the equator to the NP is significantly smaller.
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I see no more replies.
Why?
The distance is one of FET's foundation and noone is willing to discuss it? Noone is willing to provide anything that could eventually solve the issue I have noticed?
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I see no more replies.
Why?
The distance is one of FET's foundation and noone is willing to discuss it? Noone is willing to provide anything that could eventually solve the issue I have noticed?
If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
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Hi Everyone.Im brand new here-so gonna make an impact.i thought the sun and moon were 32miles in diameter and 3000miles from us.Its either that or 93million miles away.
Simple as that. I do however have the best proof the world is flat-who interested.i will be searching to see if this has been mentioned before and I think I can explain the star trails too!
This is the best forum yet and ive totally annoyed the naked scientists-theres only 6 of them I think!
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This entire topic seems pointless to me. It's like creating a post to ask "How far is L.A. from N.Y." - the distance is already know with an acceptable precision.
Yes, the distance between the Sun and Earth is well known for some time now. There's variation, but never http://www.universetoday.com/14437/how-far-is-earth-from-the-sun/ (http://www.universetoday.com/14437/how-far-is-earth-from-the-sun/)
There's no use in debating over facts. They won't change.
"The difference in distance between Earth's nearest point to the Sun in January and farthest point from the Sun in July is about 5 million kilometers (3.1 million miles). Earth is about 147.1 million kilometers (91.4 million miles) from the Sun at perihelion in early January, in contrast to about 152.1 million kilometers (94.5 million miles) at aphelion in early July. "
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what are the chances the sun and moon are the same size.the fact that you cant distinguish a 2 cars from 10 km away,but you can see a satellite in the sky which is 300km in the sky apparently with the naked eye.something like that.
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what are the chances the sun and moon are the same size.the fact that you cant distinguish a 2 cars from 10 km away,but you can see a satellite in the sky which is 300km in the sky apparently with the naked eye.something like that.
This can help. http://www.space.com/6870-spot-satellites.html (http://www.space.com/6870-spot-satellites.html)
You're welcome!
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what are the chances the sun and moon are the same size.the fact that you cant distinguish a 2 cars from 10 km away,but you can see a satellite in the sky which is 300km in the sky apparently with the naked eye.something like that.
You are mistaken. Someone tried to prove this type of argument, that some highly regarded research showed that the human eye can only resolve to 24 arcseconds. This may be true for some things but not LIGHT. I pointed out that we can see a 1" flash light a mile away. It should not be visible past 716'. Laser beams - many miles away.
Stars are many light-years away and we can see them. Galaxies are millions of light-years away and we can see their light. The Hubble is even resolving stars in them (https://i.ytimg.com/vi/loXDVGi_lK0/maxresdefault.jpg).
So, your argument is flawed. Get a flash light and check.
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
Let me repeat myself for sake of completness. This is just a composition of my posts from the beginning:
Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball). Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.
The angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.
Notion:
d - the distance to the Sun (to the surface)
R - radius of the Sun
The math:
We have a right triangle with side (d^2+2dR)^(1/2) and legs: R and d+R. If A is the angular diameter, then A/2 is the angle between (d^2+2dR)^(1/2) and d+R. R is the radius of the Sun.
Picture: http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1727720#msg1727720 (http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1727720#msg1727720)
Then we have
sin A/2= R/(d+R)
so
d/R=(1-sin A/2)/(sin A/2).
Now take A=31.6' to obtain
d/R=107.7909868
and A=32.7' to obtain
d/R=104.131456
This gives d between 104.131456R and 107.7909868R.
In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).
Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.
For the disk shape of the Sun:
I repeated calculations for the disk and obtained ~320km diameter and ~33600km distance. Basically the same result as for ball model.
...
To show how wrong the argument from wiki is:
Thomas Winship, author of Zetetic Cosmogony, provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface if one assumes that the earth is flat --
On March 21-22 the sun is directly overhead at the equator and appears
45 degrees above the horizon at 45 degrees north and south latitude. As
the angle of sun above the earth at the equator is 90 degrees while it is
45 degrees at 45 degrees north or south latitude, it follows that the angle
at the sun between the vertical from the horizon and the line from the
observers at 45 degrees north and south must also be 45 degrees. The result
is two right angled triangles with legs of equal length. The distance between
the equator and the points at 45 degrees north or south is approximately 3,000
miles. Ergo, the sun would be an equal distance above the equator.
We can repeat the same experiment (same date). The Sun remains overhead at the equator. But this time instead of 45 degrees we take 5 degrees (this can be seen close to the North Pole). The math says that if one of legs is 3000 miles and the angle between hypotenuse and the other leg is 5 degrees, then the other leg has almost 34290 miles. This is how far the observer would have to be from the equator to see the Sun at 5 degree angle. Which is nonsense as the distance from the equator to the NP is significantly smaller.
This entire topic seems pointless to me. It's like creating a post to ask "How far is L.A. from N.Y." - the distance is already know with an acceptable precision.
Yes, the distance between the Sun and Earth is well known for some time now. There's variation, but never http://www.universetoday.com/14437/how-far-is-earth-from-the-sun/ (http://www.universetoday.com/14437/how-far-is-earth-from-the-sun/)
There's no use in debating over facts. They won't change.
"The difference in distance between Earth's nearest point to the Sun in January and farthest point from the Sun in July is about 5 million kilometers (3.1 million miles). Earth is about 147.1 million kilometers (91.4 million miles) from the Sun at perihelion in early January, in contrast to about 152.1 million kilometers (94.5 million miles) at aphelion in early July. "
Can you refrain from posting unrelated things? This thread is devoted to FE model and the distance in this model only.
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what are the chances the sun and moon are the same size.the fact that you cant distinguish a 2 cars from 10 km away,but you can see a satellite in the sky which is 300km in the sky apparently with the naked eye.something like that.
This can help. http://www.space.com/6870-spot-satellites.html (http://www.space.com/6870-spot-satellites.html)
You're welcome!
r u a FE'r or not?
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
3000 miles, and here is how it was calculated.
http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal (http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal)
(http://blog.modernmechanix.com/mags/qf/c/ModernMechanix/10-1931/globe/xlg_globe_3.jpg)
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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.
Why is it a rant? Just needs more explanation.
He could actually explain his point, instead of just hitting random letters and numbers.
LOL
"Math I don't understand? Must just be his ranting and not my own ignorance"
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The earth's position can be tracked and seen in the Martian sky, and the Sun in nowhere close to it. Therefore, that flat earth model is (again, and again...) refuted.
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
3000 miles, and here is how it was calculated.
...
I have seen this. I even made a comment in previous post. I will repost this with more details.
Take three observers. For 1st the Sun in overhead at the equator. For 2nd the Sun is at 45 degrees angle (so we have the same setup as the picture you posted). For 3rd the Sun is at 5 degrees (actually, 30 is enough (ask people from Norway/Sweden/Finland), but 5 shows a bit more extreme error; 5 degrees is close to the North Pole). 1st and 2nd are 3000 miles away, so it follows that the Sun is 3000 miles far. Now, with that knowledge, 3rd can calculate his distance from 1st. The math says that if one of legs is 3000 miles and the angle between hypotenuse and the other leg is 5 degrees, then the other leg has almost 34290 miles. So, the North Pole is, based on this experiment, at least 34290 miles away from the equator. But it is not.
The problem with the way the distance was calculated in the source you posted is that it gives different values for different latitudes (or, equivalently, different angles objects cast on the surface).
If we take just two observers and repeat the experiment with 1st situatet at the equator and second moving from the equator to the NP, you will receive values that start from >billions close to the equator and going to just few miles close to the NP.
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Where did you get the 5 degrees from and how did you test the angle of the sun?
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Viewed from Mars, the sun is much farther away than the moon is from Earth. Since FE claim a sun as near as the moon, it is false. The sun that illuminates earth and mars is the same. Right, flat earthers? Anyone disagree?
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Viewed from Mars, the sun is much farther away than the moon is from Earth. Since FE claim a sun as near as the moon, it is false. The sun that illuminates earth and mars is the same. Right, flat earthers? Anyone disagree?
When did you perform this experiment from Mars? Oh, are you relying on what NASA says? You do realize that they get caught lying quit often?
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I did some calculations for the suns highest position on my latitude (57.7°N) during equinox. According to round earth, it should be 32° (90-58 (rounding), no accounting for refraction, which should be minimal when the sun is at it's highest position in the sky) and according to flat earth it would be almost 38° (Distance to equator = 4000 miles (rounded). Height of sun in FE model over equator = 3100 miles (after correcting the measurement in the article, and rounded from a slightly higher number). arctan = 3100/4000 = 37.8°). So the differences between a flat earth and a round earth are measurable here. Even more measurable further north.
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Viewed from Mars, the sun is much farther away than the moon is from Earth. Since FE claim a sun as near as the moon, it is false. The sun that illuminates earth and mars is the same. Right, flat earthers? Anyone disagree?
When did you perform this experiment from Mars? Oh, are you relying on what NASA says? You do realize that they get caught lying quit often?
Any legit examples of them getting caught intentionally lying, or are you just going off of what the bullshit conspiracy sites say they're lying about?
You do know that there has been multiple people here on Earth who has caught a video of the ISS and other large Satelites Transiting the moon, right?
There's also multiple sites and apps that allow you to see when visible satellites will flyby over head, and if you look carefully enough in the sky you can see them(provided it's dark out and light pollution doesn't block it out). What's your excuse for that one?
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You can't use Google to look up examples of NASA lying?
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I did some calculations for the suns highest position on my latitude (57.7°N) during equinox. According to round earth, it should be 32° (90-58 (rounding), no accounting for refraction, which should be minimal when the sun is at it's highest position in the sky) and according to flat earth it would be almost 38° (Distance to equator = 4000 miles (rounded). Height of sun in FE model over equator = 3100 miles (after correcting the measurement in the article, and rounded from a slightly higher number). arctan = 3100/4000 = 37.8°). So the differences between a flat earth and a round earth are measurable here. Even more measurable further north.
Thanks for providing that example. I was referring to ~60 degree N in my post, and you have just provided an even greater example of it. The refraction factor can be skipped, as based on wiki, it just a fraction of a degree (less than 1 degree anyway). The result according to ~3100 miles value is as predicted - the Sun is higher than it can be observed during the equinox.
Where did you get the 5 degrees from and how did you test the angle of the sun?
The place does not matter. All you need is relatively low angle (5 was just an example, which can be observed near the North Pole, as my post stated). Testing the angle is just measuring it. With the 5 degree example, an error of 1 degree will not produce any number close to ~3000 miles.
It is clear that the ~3000 miles explaination is flawed. The height of the Sun (angles) exposes it.
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Also, can you guys keep your discussion about the Mars and other bodies, including ISS, fakery or random things and conspirancy theories unrelated to the topic off this thread? Start your own thread and talk about anything you wish to, and even go off-topic if you feel that is the best thing to do.
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Come one guys. Any other ideas?
The question is simple, yet noone has provided any verifiable answer yet. Only jroa decided to provide anything valuable to the topic, but what he quoted was quickly rejected due to an easy observation.
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Viewed from Mars, the sun is much farther away than the moon is from Earth. Since FE claim a sun as near as the moon, it is false. The sun that illuminates earth and mars is the same. Right, flat earthers? Anyone disagree?
When did you perform this experiment from Mars? Oh, are you relying on what NASA says? You do realize that they get caught lying quit often?
I am assuming NASA information - in this case, images and a video taken from Mars showing us the Sun, as well as the Mars missions - as true. This is not a statement they made about something.
Since NASA is a government-agency I believe this material is audited. Government is paying for all of it, right? Better be real material.
And I also object to the "fact" that they lie a lot. That's a red herring. The fact that they lie does not contradict anything: The material is audited anyway.
Also, NASA has too much failures in its past. If there was a conspiracy, why would they "stage" these? Serves no purpose.
Unless you can falsify those images and the video, they stay as truth. The American government wouldn't accept those missions for so many time without a way of checking its veracity.
Adding insult to injury (for FE): NASA is not the only agency/company/organization that sent stuff to the red planet.
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Also, can you guys keep your discussion about the Mars and other bodies, including ISS, fakery or random things and conspirancy theories unrelated to the topic off this thread? Start your own thread and talk about anything you wish to, and even go off-topic if you feel that is the best thing to do.
Mars is a linchpin here. Rovers on surface providing images and the fact that the same sun illuminates the Earth and Mars - and the Moon, too - is enough evidence to, again, refute the flat earth "theory": It proves the Sun's distance to Earth is much more than the Moon's. This refutes a thousand-kilometer distance between Earth and Sun.
As I said, this is entirely relevant to this topic. Since mars receive less light from the Sun - It even looks smaller on Mars POV, implying greater distance - means the Heliocentric model is correct - Mars is farther away from the sun than the Earth - and their orbits too. And that automatically disproves (again and again) the flat earth model.
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Viewed from Mars, the sun is much farther away than the moon is from Earth. Since FE claim a sun as near as the moon, it is false. The sun that illuminates earth and mars is the same. Right, flat earthers? Anyone disagree?
When did you perform this experiment from Mars? Oh, are you relying on what NASA says? You do realize that they get caught lying quit often?
I asked you - as well as any flat earther - a simple question. You could had it answered, but you didn't. On purpose, perhaps.
Again: Is the Sun that illuminates the Earth the SAME Sun that illuminates our Moon, and that SAME Sun illuminates Mars? Yes, or no?
If not, what illuminates the Moon and/or Mars? Since they receive "light" from space, what could be the Source of their light, if not the same Sun that illuminates Earth?
Dodging the question (as you just did) is like saying you don't have a clue about what's being discussed here. And as so, any argument you defended automatically loses credibility...
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As I said, this is entirely relevant to this topic. Since mars receive less light from the Sun - It even looks smaller on Mars POV, implying greater distance - means the Heliocentric model is correct - Mars is farther away from the sun than the Earth - and their orbits too. And that automatically disproves (again and again) the flat earth model.
Not really. In heliocentric model Mars seen from Earth changes is angular size significantly. Taking average distances (149m and 227m from the Sun), when aligned, they are 78m km apart. When Mars is seen right above the horizon at midnight, it is 171m km away. This gives a bit more than double change in angular size. Something that would work pretty well for FE model.
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Also, can you guys keep your discussion about the Mars and other bodies, including ISS, fakery or random things and conspirancy theories unrelated to the topic off this thread? Start your own thread and talk about anything you wish to, and even go off-topic if you feel that is the best thing to do.
Mars is a linchpin here. Rovers on surface providing images and the fact that the same sun illuminates the Earth and Mars - and the Moon, too - is enough evidence to, again, refute the flat earth "theory": It proves the Sun's distance to Earth is much more than the Moon's. This refutes a thousand-kilometer distance between Earth and Sun.
As I said, this is entirely relevant to this topic. Since mars receive less light from the Sun - It even looks smaller on Mars POV, implying greater distance - means the Heliocentric model is correct - Mars is farther away from the sun than the Earth - and their orbits too. And that automatically disproves (again and again) the flat earth model.
Oh, yeah, all of the evidence. Like, this picture of the curiosity rover on Mars. Pretty cool picture, except for one thing; who took the picture? Perhaps NASA forgot that little detail before they released it.
(http://s1.ibtimes.com/sites/www.ibtimes.com/files/styles/pulse_embed/public/2015/01/28/curiosity-selfie.jpg)
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Perhaps NASA forgot that little detail before they released it.
(http://s1.ibtimes.com/sites/www.ibtimes.com/files/styles/pulse_embed/public/2015/01/28/curiosity-selfie.jpg)
Perhaps you, as the moderator, should stop making posts unrelated to the topic. Or you can explain to the average Reader how does the picture of Mars can help determining the distance from the surface of the Earth to the Sun?
The question asked in this topic remains unanswered for 2 weeks. Yet it is quite simple. I do now know how to measure the distance. Does anyone know how to do it?
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This is a composite picture ... and before you scream "fake" ... NASA, of course, provides the raw pictures too ... about 50 pictures are put together, taking the parts without robot arm in them from each.
(http://www.nasa.gov/images/content/713348main_pia16457-673.jpg)
The robot arm has a specific "selfie" routine for all the necessary positions, that was developed even before it was send to Mars.
(http://www.nasa.gov/images/content/713478main_pia16458b-43_946-710.jpg)
Can we go back to the original topic now ...
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This is a composite picture ... and before you scream "fake" ... NASA, of course, provides the raw pictures too ... about 50 pictures are put together, taking the parts without robot arm in them from each.
(http://www.nasa.gov/images/content/713348main_pia16457-673.jpg)
The robot arm has a specific "selfie" routine for all the necessary positions, that was developed even before it was send to Mars.
(http://www.nasa.gov/images/content/713478main_pia16458b-43_946-710.jpg)
Can we go back to the original topic now ...
Convenient how the NASA people put together that excuse when they were asked to explain the picture. ::)
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Perhaps NASA forgot that little detail before they released it.
(http://s1.ibtimes.com/sites/www.ibtimes.com/files/styles/pulse_embed/public/2015/01/28/curiosity-selfie.jpg)
Perhaps you, as the moderator, should stop making posts unrelated to the topic. Or you can explain to the average Reader how does the picture of Mars can help determining the distance from the surface of the Earth to the Sun?
The question asked in this topic remains unanswered for 2 weeks. Yet it is quite simple. I do now know how to measure the distance. Does anyone know how to do it?
I am not the one who changed the subject. You and Kirk Johnson started talking about Mars missions. I pointed out that NASA may be lying about Mars as well. Why do you pretend that FE'ers are the ones changing the subject? This sounds like a deceitful debate tactic, but it is one that can easily be defeated, much like the rest of your RET.
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Convenient how the NASA people put together that excuse when they were asked to explain the picture. ::)
Yeah. Very convenient that they acted as normal persons providing evidence when they were asked for it - unlike some people around here (You and other "FE belivahs")
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Again. If the sun that illuminates Mars is the same that illuminates Earth, FE and DE fantasies are automatically disproved.
The distance from Mars and Earth is much greater. If the sun can be seen from Mars, then that "Sunset occurs because the sun in far away" is automatically disprove. The FE Sun-Earth distance also is disproved.
So there's no choice for FE believers: They must discredit every mission, always.
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I would respond to Kirk Johnson, but the roundies would accuse me of derailing, once again, even thoughh he is the one going on and on about Mars in a thread that has nothing to do with Mars.
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I am not the one who changed the subject. You and Kirk Johnson started talking about Mars missions. I pointed out that NASA may be lying about Mars as well. Why do you pretend that FE'ers are the ones changing the subject?
Do you realize you lied in this post? Multiple times.
This sounds like a deceitful debate tactic, but it is one that can easily be defeated, much like the rest of your RET.
The debate tactic you seem to follow.
I would respond to Kirk Johnson, but the roundies would accuse me of derailing, once again, even thoughh he is the one going on and on about Mars in a thread that has nothing to do with Mars.
To sum up this and the previous comment, you have shown yourself as a hypocrite and a liar. Congratulations, you have just been promoted to my top worst moderator I have ever seen. You could have stopped the off-topic part a long ago, but you did absolutely nothing.
And we have 4th page with the unanswered question: How far is the Sun in FE model? 3000 miles // 5000 km does not work.
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I am not the one who changed the subject. You and Kirk Johnson started talking about Mars missions. I pointed out that NASA may be lying about Mars as well. Why do you pretend that FE'ers are the ones changing the subject?
Do you realize you lied in this post? Multiple times.
Really? How did I lie when you two were having this conversation that I responded to with a Mars picture? OK, I will admit that it was mostly Kirk who derailed the thread into a Mars discussion, but you can't blame me when I respond to him on topic.
Also, can you guys keep your discussion about the Mars and other bodies, including ISS, fakery or random things and conspirancy theories unrelated to the topic off this thread? Start your own thread and talk about anything you wish to, and even go off-topic if you feel that is the best thing to do.
Mars is a linchpin here. Rovers on surface providing images and the fact that the same sun illuminates the Earth and Mars - and the Moon, too - is enough evidence to, again, refute the flat earth "theory": It proves the Sun's distance to Earth is much more than the Moon's. This refutes a thousand-kilometer distance between Earth and Sun.
As I said, this is entirely relevant to this topic. Since mars receive less light from the Sun - It even looks smaller on Mars POV, implying greater distance - means the Heliocentric model is correct - Mars is farther away from the sun than the Earth - and their orbits too. And that automatically disproves (again and again) the flat earth model.
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Really? How did I lie when you two were having this conversation that I responded to with a Mars picture? OK, I will admit that it was mostly Kirk who derailed the thread into a Mars discussion, but you can't blame me when I respond to him on topic.
This is really a pathetic excuse.
I appreciate your contribution to the thread (posting the 3000 miles image with explaination), but nothing more. Please do not waste more of my time in this thread unless you have something to say that covers the topic.
Same thing to Kirk Johnson, who started this off-topic trash. Same to everyone else...
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
Since apparently the sun is 3100 miles up (according to FET) and the south pole is 6200 miles south of the equator, the sun would be approx 26.6° up.
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
Since apparently the sun is 3100 miles up (according to FET) and the south pole is 6200 miles south of the equator, the sun would be approx 26.6° up.
- I was using rough numbers, but your numbers work fine.
- The sun does not look like it is 26.6° up anyways.
- Changing the numbers does not disprove the concept.
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I would respond to Kirk Johnson, but the roundies would accuse me of derailing, once again, even thoughh he is the one going on and on about Mars in a thread that has nothing to do with Mars.
Mars is not derailment. Observations from Mars are proof that the distance Sun-Earth is much higher than some thousand miles - The same discussion a topic by the title "How far is Sun in FE model? Definitely more than 5000km." has.
Discussion about NASA faking missions, that's derailment.
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
I would use the North Pole image insead, as based on FET, the South Pole (geographic) does not exists.
For the Sun being approximately 1 diameter of itself above the horizon, which is ~0.50 angle, for the object 5000 km above the surface, the distance (horizontal) to it should be around 573,000 km.
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
I would use the North Pole image instead, as based on FET, the South Pole (geographic) does not exists.
For the Sun being approximately 1 diameter of itself above the horizon, which is ~0.50 angle, for the object 5000 km above the surface, the distance (horizontal) to it should be around 573,000 km.
Exactly!!! In all single disk FE models, this picture, taken from Antarctica, is IMPOSSIBLE. Pictures of the S. Celestial Pole (center of the south star trails) are IMPOSSIBLE because the SCP can not be seen from above the disk... and so on...
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
I would use the North Pole image instead, as based on FET, the South Pole (geographic) does not exists.
For the Sun being approximately 1 diameter of itself above the horizon, which is ~0.50 angle, for the object 5000 km above the surface, the distance (horizontal) to it should be around 573,000 km.
Exactly!!! In all single disk FE models, this picture, taken from Antarctica, is IMPOSSIBLE. Pictures of the S. Celestial Pole (center of the south star trails) are IMPOSSIBLE because the SCP can not be seen from above the disk... and so on...
sorry-who took these pics.must have been someone special to get permission.they were taken from Antarctica facing south-odd.makes sense if they were facing north.
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
3000 miles, and here is how it was calculated.
http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal (http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal)
...
Brouwer and Master_Evar have basically said what I was going to say. Basically, like the "Eratosthenes experiment", two points, conveniently equidistant (i.e. +/- 45°) are insufficient to draw conclusions. Different latitudes would make different angles. Look at my pictures from the S.Pole above, ~0.5° would either push the sun out 344,000 mi or at the equator, 6225 mi from the poles, the Sun would be 54.3 mi up. In other words, the 3000 mi height value in your example means exactly nothing. This type of experiment can not determine the height of the Sun directly. IT CAN, with several points, determine the SHAPE of the Earth (it is the reverse of the "Eratosthenes experiment"). Once the shape of the Earth is known, using the values from the experiment, you CAN determine the height of the Sun.
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
I would use the North Pole image instead, as based on FET, the South Pole (geographic) does not exists.
For the Sun being approximately 1 diameter of itself above the horizon, which is ~0.50 angle, for the object 5000 km above the surface, the distance (horizontal) to it should be around 573,000 km.
Exactly!!! In all single disk FE models, this picture, taken from Antarctica, is IMPOSSIBLE. Pictures of the S. Celestial Pole (center of the south star trails) are IMPOSSIBLE because the SCP can not be seen from above the disk... and so on...
sorry-who took these pics.must have been someone special to get permission.they were taken from Antarctica facing south-odd.makes sense if they were facing north.
These were taken from Antarctica facing north obviously.
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Here is a picture of the Sun during an equinox taken in 30 minute intervals from near the S. Pole:
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/sunseriesagain.jpg)
(http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/ (http://astro.uchicago.edu/cara/vtour/pole/dome/life/sun/))
If the Sun is above the equator at equinox, and Antarctica is some 4000 mi south of the equator, and the Sun is some 4000 mi up, on a Flat Earth model, wouldn't the Sun be approximately 45° up?
Since apparently the sun is 3100 miles up (according to FET) and the south pole is 6200 miles south of the equator, the sun would be approx 26.6° up.
- I was using rough numbers, but your numbers work fine.
- The sun does not look like it is 26.6° up anyways.
- Changing the numbers does not disprove the concept.
I know. Just so that the FE's doesn't nitpick on the numbers.
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Is this how starting a serious debate on FES forum ends?
Is there literally no Flat Earther able to discuss the topic?
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Is this how starting a serious debate on FES forum ends?
Is there literally no Flat Earther able to discuss the topic?
the problem is that re's wont admit there are simple flaws in the re model. that's why we are on here in the first place.its a shame re's don't look past what the world has drilled into them.you must remember that superior people aren't cleverer-they just make you look dumber.the face of the moon looks the same is every country.so that cant be right in the re model.
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the problem is that re's wont admit there are simple flaws in the re model. that's why we are on here in the first place.its a shame re's don't look past what the world has drilled into them.you must remember that superior people aren't cleverer-they just make you look dumber.the face of the moon looks the same is every country.so that cant be right in the re model.
Thank you for posting irrelevant to the topic information.
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the problem is that re's wont admit there are simple flaws in the re model. that's why we are on here in the first place.its a shame re's don't look past what the world has drilled into them.you must remember that superior people aren't cleverer-they just make you look dumber.the face of the moon looks the same is every country.so that cant be right in the re model.
The little problem here is that I fail to see any FE supporters posting any of these "simple flaws in the re model"
Where is there any RE problem with "the face of the moon looks the same is every country.so that cant be right in the re model."? That is no problem at all for the RE model.
1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
2) The moon appears to rock slightly as seen from earth (lunar libration) show over time about 59% of surface.
So many so-called problems seem be like this, simply a failure of a FE supporter to understand some aspect of the rotating globe model. I have seen it far worse on YouTube FE videos. I think that most posters on YouTube have a far poorer understanding of both models than most FE supporters on this site.
Mind you at last count there were at least three quite different FE models!
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
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So many so-called problems seem be like this, simply a failure of a FE supporter to understand some aspect of the rotating globe model.
And here we have a good example of a misunderstanding;
1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
The angular size of the moon is indeed about 0.5 degrees but two observers looking from opposite sides of the globe (one seeing the moon newly risen while the other seeing it about to set) would be observing the moon from 2 degrees of separation.
This 2 degrees is the aproximate angular size of the earth viewed from the moon.
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
Be reasonable, I said "angle" hoping it would be clear I meant the angle of the face the moon shows to us.
I didn't think anyone would for a moment think the size changed by a factor of five.
Most people would know that the face we see does change a little due to both parallax and libration.
(My spell checker has kittens over libration - wants to call it liberation!)
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
You misunderstand astronomy.
The size of the Moon as viewed from the Earth, 0.5°, doesn't change much (depends on how far the Moon is from the Earth). This like the sun being 0.5° also, hence Solar Eclipses (it also changes very little based on the Earth's elliptical orbit).
We see 50% of the surface all the time. But we can see more surface, up to the 9° because viewing the Moon from say Alaska/Norway and Argentina/Australia introduces a viewing angle, east (moonrise) to west (moonset) changes the angle slightly too, the Moon's orbit is inclined ~5°, its axis is tilted ~1.5°, etc. See the pictures and explanation here (especially the 2nd paragraph) (http://www.stargazing.net/david/moon/moonlibration.html (http://www.stargazing.net/david/moon/moonlibration.html)).
BTW, Lunar Libration, per the photos above, can not be explained by a flat disk.
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
You misunderstand astronomy.
The size of the Moon as viewed from the Earth, 0.5°, doesn't change much (depends on how far the Moon is from the Earth). This like the sun being 0.5° also, hence Solar Eclipses (it also changes very little based on the Earth's elliptical orbit).
We see 50% of the surface all the time. But we can see more surface, up to the 9° because viewing the Moon from say Alaska/Norway and Argentina/Australia introduces a viewing angle, east (moonrise) to west (moonset) changes the angle slightly too, the Moon's orbit is inclined ~5°, its axis is tilted ~1.5°, etc. See the pictures and explanation here (especially the 2nd paragraph) (http://www.stargazing.net/david/moon/moonlibration.html (http://www.stargazing.net/david/moon/moonlibration.html)).
BTW, Lunar Libration, per the photos above, can not be explained by a flat disk.
Oh, so you have nothing and you will spout gibberish when you are cornered. Good to know.
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
You misunderstand astronomy.
The size of the Moon as viewed from the Earth, 0.5°, doesn't change much (depends on how far the Moon is from the Earth). This like the sun being 0.5° also, hence Solar Eclipses (it also changes very little based on the Earth's elliptical orbit).
We see 50% of the surface all the time. But we can see more surface, up to the 9° because viewing the Moon from say Alaska/Norway and Argentina/Australia introduces a viewing angle, east (moonrise) to west (moonset) changes the angle slightly too, the Moon's orbit is inclined ~5°, its axis is tilted ~1.5°, etc. See the pictures and explanation here (especially the 2nd paragraph) (http://www.stargazing.net/david/moon/moonlibration.html (http://www.stargazing.net/david/moon/moonlibration.html)).
BTW, Lunar Libration, per the photos above, can not be explained by a flat disk.
Oh, so you have nothing and you will spout gibberish when you are cornered. Good to know.
Where is the definitive answer for the distance to the sun?
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1) The moon looks almost the same from all countries. The most the angle can change due to location on earth is around 2 deg.
Considering that the average angular size of the moon is around .5 degree, 2 degrees is a lot of change.
You misunderstand astronomy.
The size of the Moon as viewed from the Earth, 0.5°, doesn't change much (depends on how far the Moon is from the Earth). This like the sun being 0.5° also, hence Solar Eclipses (it also changes very little based on the Earth's elliptical orbit).
We see 50% of the surface all the time. But we can see more surface, up to the 9° because viewing the Moon from say Alaska/Norway and Argentina/Australia introduces a viewing angle, east (moonrise) to west (moonset) changes the angle slightly too, the Moon's orbit is inclined ~5°, its axis is tilted ~1.5°, etc. See the pictures and explanation here (especially the 2nd paragraph) (http://www.stargazing.net/david/moon/moonlibration.html (http://www.stargazing.net/david/moon/moonlibration.html)).
BTW, Lunar Libration, per the photos above, can not be explained by a flat disk.
Oh, so you have nothing and you will spout gibberish when you are cornered. Good to know.
Like JRoweSkeptic (I believe you are him also), you don't understand astronomy. Don't embarrass yourself like he did in my telescope alignment thread. This is not gibberish as anyone who has a telescope or camera with a zoom lens would tell you. The pictures of the Moon are real. All FE models are fantasies that can not explain the pictures (what we see and photograph). Those pictures can be taken by anyone with a camera with a zoom lens so you argue from ignorance or stupidity.
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the problem is that re's wont admit there are simple flaws in the re model. that's why we are on here in the first place.its a shame re's don't look past what the world has drilled into them.you must remember that superior people aren't cleverer-they just make you look dumber.the face of the moon looks the same is every country.so that cant be right in the re model.
Thank you for posting irrelevant to the topic information.
anytime!
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FET is falling apart.
It seems that noone is able to present any valid number with the reference to the proof.
You do not know the distance, so how then can you discuss things like day/night, seasons, sunsets/sunrises and other optical effects?
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Using the distane of 9000 km I derived earlier, and assuming 5000 km distance, I did some math to determine the refractive index of whatever could bend the light from the Sun. The bending is such that the Sun 9000 km off would seem to be setting/rising. Using Snellius law (index of vacuum = 1) I found the index of the other side to be around 1.144. If you take into account that air index is ~1.0003, we have a serious deviation. Even liquid helium has only 1.025.
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FET is falling apart.
It seems that noone is able to present any valid number with the reference to the proof.
You do not know the distance, so how then can you discuss things like day/night, seasons, sunsets/sunrises and other optical effects?
They can't. That was my point in telescope alignment. All it is is hand-waving and making bold vague untrue statements.
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As you can see I have even tried applying some light bending through atmosphere (or atmoplane) and I have obtained quite ridiculous refractive index. The distance is set to match sunset at proper distance/position.
O would really like to see more than hand-waving.
In comparision, even if from FET poinf of wiev, "numerous assumption" made to round model lead to results, numbers obtained throught this method matches observations and make perfect sense under RE model. But FE model assumptions and calculations make, at least up to this point, no sense.
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FET is falling apart.
It seems that noone is able to present any valid number with the reference to the proof.
You do not know the distance, so how then can you discuss things like day/night, seasons, sunsets/sunrises and other optical effects?
But, I did give you a valid number as well as the repeatable way it was calculated. Oh, I forget, you roundies ignore anything that goes against your beliefs and simply stomp around the forum screaming, "no body gives me what I demand!" It must suck to be a roundy.
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Brouwer - These guys are idiots, first reply is an attack, followed by more attacks. They don't believe in maths, because God forbid it proves the Earth is round.
Here is a Question, if the reason we don't see very far is because it is the limit of our eyes and not the roundness of the earth (due to the hump blocking our line of sight) Then why the hell do we see something "5000km" away. Surely if you go on a mountain you should be able to see VERRYY far. Not to mention, the farther the sun goes from us, somehow, it goes below the horizon and at the same size due to "atmoplanic pressure".... never heard pressure effecting light before - when asked for how FE get their answers, there is no source. When RE use maths and gets answers, they say we are calming the results. gg
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But, I did give you a valid number as well as the repeatable way it was calculated.
That is true. But that does not mean the experiment itself provides a correct number. You should be able to produce the same distance to the Sun not only from 45o, but using any angle (and trigonometry). But you will not as it was presented multiple times in this thread.
Oh, I forget, you roundies ignore anything that goes against your beliefs and simply stomp around the forum screaming, "no body gives me what I demand!" It must suck to be a roundy.
You must love wasting your time. Most of your post has nothing to do with the actual topic.
Also, with the above you show how close-minded you are. If you cared enough you would know that I work in FET setup in this thread and I try to find a valid way to calculate the distance. This thread has nothing to do with believing in RE model and numbers. You are just blind for this.
Here is a Question, if the reason we don't see very far is because it is the limit of our eyes and not the roundness of the earth (due to the hump blocking our line of sight) Then why the hell do we see something "5000km" away. Surely if you go on a mountain you should be able to see VERRYY far. Not to mention, the farther the sun goes from us, somehow, it goes below the horizon and at the same size due to "atmoplanic pressure".... never heard pressure effecting light before - when asked for how FE get their answers, there is no source. When RE use maths and gets answers, they say we are calming the results. gg
Good point actually. The Sun is visible thanks to the strength of the light. It keeps it size because, somehow (noone can explain that), it can magnify objects. Going below the horizon can be due to light bending, but I tried using this to determine refractive index. I received basically impossible value.
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Brouwer - These guys are idiots, first reply is an attack, followed by more attacks. They don't believe in maths, because God forbid it proves the Earth is round.
Here is a Question, if the reason we don't see very far is because it is the limit of our eyes and not the roundness of the earth (due to the hump blocking our line of sight) Then why the hell do we see something "5000km" away. Surely if you go on a mountain you should be able to see VERRYY far. Not to mention, the farther the sun goes from us, somehow, it goes below the horizon and at the same size due to "atmoplanic pressure".... never heard pressure effecting light before - when asked for how FE get their answers, there is no source. When RE use maths and gets answers, they say we are calming the results. gg
There is not very much dense air above your head. Straight up, it probably averages less than 10 miles worth of equivalent air at sea level. Of course you can see for thousands of miles if there is relatively little air in between you and the object you are looking at. Why do I have to explain this to you like you are a second grader? I thought we were all adults here having an adult discussion? I guess I thought wrong, though. ::)
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Brouwer - These guys are idiots, first reply is an attack, followed by more attacks. They don't believe in maths, because God forbid it proves the Earth is round.
Here is a Question, if the reason we don't see very far is because it is the limit of our eyes and not the roundness of the earth (due to the hump blocking our line of sight) Then why the hell do we see something "5000km" away. Surely if you go on a mountain you should be able to see VERRYY far. Not to mention, the farther the sun goes from us, somehow, it goes below the horizon and at the same size due to "atmoplanic pressure".... never heard pressure effecting light before - when asked for how FE get their answers, there is no source. When RE use maths and gets answers, they say we are calming the results. gg
There is not very much dense air above your head. Straight up, it probably averages less than 10 miles worth of equivalent air at sea level. Of course you can see for thousands of miles if there is relatively little air in between you and the object you are looking at. Why do I have to explain this to you like you are a second grader? I thought we were all adults here having an adult discussion? I guess I thought wrong, though. ::)
Okay, well, let us say above us is the sun (Mid day), and there is pure clean air - we will see the beautiful sun shining down upon us... Now, when the sun is along the horizon we still see the sun. Oh wait, we are looking out towards the horizon with THOUSANDS of miles of air between the Sun and our eyes - shit, that means we can see through thousands of miles of air after all.. contradiction? Yes, At it's best. Thanks for proving my point.
Moreover, during sun set, the sun illuminates the underside of clouds, how is that explained in flat earth, how does the light go under clouds and if you look a those clouds from above, they are dark. Hmmm
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The sun is always above the clouds so it is impossible to light the bottoms of clouds directly. Doing so goes against all physical proof or any sane logic. In the ball Earth theory all clouds are curved around the ball by gravity and the light rays approach the Earth in straight parallel lines from millions and millions of miles away. There are no clouds directly lighted by the sun from the bottom, period.
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The sun is always above the clouds so it is impossible to light the bottoms of clouds directly. Doing so goes against all physical proof or any sane logic. In the ball Earth theory all clouds are curved around the ball by gravity and the light rays approach the Earth in straight parallel lines from millions and millions of miles away. There are no clouds directly lighted by the sun from the bottom, period.
If there is not then explain the phenomenon occurring at sunsets and sunrises where at least it seems that bottoms of the clouds are directly lighted by the sun. If it is not as it seems then what it is?
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Please do not change the topic. There is separate thread for clouds illuminated from below.
If you claim the Sun is always above, that is ok. But the thread is about the distance.
Please stay on topic.
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I have read how to find a distance to the Sun from "Earth Not a Globe" (section 3, starting from page 79). It fails to give any precise number, it just states that it is "less than 4000 miles", "proved to be under 4000 miles", and then "less than 4000 miles" again.
So, if "the distance to the Sun [...] may readily be measured using plane trigonometry", why would the author skip it? Maybe because it would give him few different numbers instead of one, the same value?
The idea used there is flawed for the same reason as jroa's.
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
3000 miles, and here is how it was calculated.
http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal (http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal)
(http://blog.modernmechanix.com/mags/qf/c/ModernMechanix/10-1931/globe/xlg_globe_3.jpg)
Right back on Nov 15 Jroa was kind enough to answer all you questions as to how the Flat Earth distance to the sun was calculated!
I am a little surprised that something a little more recent has not come to light. I have noted, however, that if the same observations were made from other baselines, the answers would have been rather different. In the following one observer is at the equator, with the sun directly overhead, the other on the same line of longitude, but various distances north. As before, for 45° with a separation of 3000 miles we get the same answer as before, but for other base lines the answers are rathere different! Why was 45° chosen and not some other?
Latitude Sun Elev Base Line Sun Ht
22.5° 63.4° 1500 mi 3621 mi
45.0° 45.0° 3000 mi 3000 mi
67.5° 33.7° 4500 mi 1864 mi
90.0° 26.6° 6000 mi 0 mi
So, will the real height please stand up.
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Oh, yeah, all of the evidence. Like, this picture of the curiosity rover on Mars. Pretty cool picture, except for one thing; who took the picture? Perhaps NASA forgot that little detail before they released it.
(http://s1.ibtimes.com/sites/www.ibtimes.com/files/styles/pulse_embed/public/2015/01/28/curiosity-selfie.jpg)
Excuse this repeated off topic post, but Jroa's accusation does need a reply!
When you say "NASA forgot that little detail before they released it.". Well, maybe NOT!
You are exactly the same as so many Youtube™ FE posters. As soon as you find something you don't understand your first reaction is just to assume it's a fake! Now most reasonable people would, I think, do at least a little Google search to find how this "mysterious" thing was done. But, not you!
It is easy "just use a selfie stick!" - oops, but why can't we see the stick.
Well just look up HOW it was down. Not too difficult in principle - take two pictures from different angles and "Photoshop" the stick out.
On Curiosity Rover it was nothing like that easy, for a start the camera did not have anywhere near a wide enough angle of view to see all of the Rover.
Best just read how they did it in: http://www.planetary.org/blogs/emily-lakdawalla/2015/08191059-curiosity-self-portrait-history-belly-pan.html (http://www.planetary.org/blogs/emily-lakdawalla/2015/08191059-curiosity-self-portrait-history-belly-pan.html) Yes, it was "Photoshopped" (maybe not exactly that program) but you know what I mean!
This reference includes a video animation of the gyrations the Rover's arm went through to get all the pictures that had to be merged into a "panorama".
Next time Jroa, just look around a bit before you accuse anyone of a big mistake. Really, I think NASA are MUCH smarter at these things than you are, as proved by this photo. They did it - you thought it could not be done!
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It is been over 1 month since I started this discussion.
So far the problem is not yet solved. Therefore I see no point in discussing the following things in any thread on this forum:
1. Day/time cycles.
2. Seasons.
3. Sun sets/rises.
4. Clouds illuminated from below.
5. The Moon (its size and distance).
etc
I came here for discussion and possible solution to FE model. I wanted to learn something. So far you gave me nothing that could possibly work...
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It is been over 1 month since I started this discussion.
So far the problem is not yet solved. Therefore I see no point in discussing the following things in any thread on this forum:
1. Day/time cycles.
2. Seasons.
3. Sun sets/rises.
4. Clouds illuminated from below.
5. The Moon (its size and distance).
etc
I came here for discussion and possible solution to FE model. I wanted to learn something. So far you gave me nothing that could possibly work...
This is not really surprising, is it? FE believers never provided those answers since the FE society began. You think one month is too much? Try their entire existence ;D ;D ;D ;D ;D
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It is been over 1 month since I started this discussion.
So far the problem is not yet solved. Therefore I see no point in discussing the following things in any thread on this forum:
1. Day/time cycles.
2. Seasons.
3. Sun sets/rises.
4. Clouds illuminated from below.
5. The Moon (its size and distance).
etc
I came here for discussion and possible solution to FE model. I wanted to learn something. So far you gave me nothing that could possibly work...
hey bud.bet re'rs cant explain waxing and waning moons and the cycle and that would disprove fe;rs anyday.i suggest the infidels study the moon for 1 month.it will shock you!!!
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With the ball Earth theory you can not even see the Sun. You can only see the light that impacts the Earth. Take a look, you do not see the light on the back side of the Sun. You do not see the light of the Sun that is not directed at the Earth. From millions of miles away as the rays/wave go out evenly only a small area the diameter of the Earth hits the earth, not the rest. The ball Earthers think that tiny spot is the sun, it is not. The ball Earthers by their theory have never seen the sun in it's entirety only a small disk of it. The ball Earther's would have to use art work to show a complete Sun. When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
You can see the Sun on the flat Earth, but not on the theoretical ball Earth.
(http://i65.tinypic.com/1dy5nc.gif)
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With the ball Earth theory you can not even see the Sun. You can only see the light that impacts the Earth. Take a look, you do not see the light on the back side of the Sun. You do not see the light of the Sun that is not directed at the Earth. From millions of miles away as the rays/wave go out evenly only a small area the diameter of the Earth hits the earth, not the rest. The ball Earthers think that tiny spot is the sun, it is not. The ball Earthers by their theory have never seen the sun in it's entirety only a small disk of it. The ball Earther's would have to use art work to show a complete Sun. When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
You can see the Sun on the flat Earth, but not on the theoretical ball Earth.
For the life of me, I simply cannot fathom what you are talking about. Maybe someone with far more intelligence can translate this!
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Newton's law of universal gravitation follows an inverse-square law, as do the effects of electric, magnetic, light, sound, and radiation phenomena.
https://en.wikipedia.org/wiki/Inverse-square_law (https://en.wikipedia.org/wiki/Inverse-square_law)
Just do a search for Luminosity and Brightness and how much of the sun's energy is at the Earth. A lot of the sites have the equations wrong , so you have to sort it out for yourself. Short answer is that it is impossible for the Sun to be 83 million mies away from the Earth. By their own ball Earth theory it does not work out.
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ronxyz, please stop spamming the thread with irrelevant to the topic information.
THIS THREAD IS ABOUT DETERMINING THE DISTANCE TO THE SUN IN FE MODEL USING FE ASSUMPTIONS/BASE, NOT COMPARING RE WITH FE TO DETERMINE WHICH ONE IS CORRECT. IS THAT TOO DIFFICULT TO UNDERSTAND?
When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
Please explain how do you determine it is 3000 miles away.
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Newton's law of universal gravitation follows an inverse-square law, as do the effects of electric, magnetic, light, sound, and radiation phenomena.
https://en.wikipedia.org/wiki/Inverse-square_law (https://en.wikipedia.org/wiki/Inverse-square_law)
Just do a search for Luminosity and Brightness and how much of the sun's energy is at the Earth. A lot of the sites have the equations wrong , so you have to sort it out for yourself. Short answer is that it is impossible for the Sun to be 83 million mies away from the Earth. By their own ball Earth theory it does not work out.
No, I think the short answer is that you don't know what you are talking about!
When I try it, based on http://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-outside-earths-atmosphere (http://www.pveducation.org/pvcdrom/properties-of-sunlight/solar-radiation-outside-earths-atmosphere) I get:
Solar Radiation Outside the Earth's Atmosphere
Hsun at the sun's surface 5.96E+07 W/m^2
Radius Sun 695,500 km
Distance to Sun 149,600,000 km
Hsun in space at earth 1288W/m^2
Which agrees pretty well with the figure of arount 1kW/m^2 use in solar power calculations.
BTW The sun is about 93 million mies away from the Earth, 83 million mies.
Now, maybe YOU could get back to how the 3,000 mile figure is determined for the FE model.
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ronxyz, please stop spamming the thread with irrelevant to the topic information.
THIS THREAD IS ABOUT DETERMINING THE DISTANCE TO THE SUN IN FE MODEL USING FE ASSUMPTIONS/BASE, NOT COMPARING RE WITH FE TO DETERMINE WHICH ONE IS CORRECT. IS THAT TOO DIFFICULT TO UNDERSTAND?
When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
Please explain how do you determine it is 3000 miles away.
Spam-ming, surly you jest. I gave real science answers to you query. Take some time to study it and you will see the truth and logic of it of it. Just because you do not like the answer it does not make it spam. The sun is not millions and millions of miles away as science proves it is not.
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With the ball Earth theory you can not even see the Sun. You can only see the light that impacts the Earth. Take a look, you do not see the light on the back side of the Sun. You do not see the light of the Sun that is not directed at the Earth. From millions of miles away as the rays/wave go out evenly only a small area the diameter of the Earth hits the earth, not the rest. The ball Earthers think that tiny spot is the sun, it is not. The ball Earthers by their theory have never seen the sun in it's entirety only a small disk of it. The ball Earther's would have to use art work to show a complete Sun. When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
You can see the Sun on the flat Earth, but not on the theoretical ball Earth.
(http://i65.tinypic.com/1dy5nc.gif)
Your image and understanding are not correct.
- Light rays do not only come out of the center of the Sun but its entire surface up to the edge, as we view from Earth.
- The Sun is MUCH bigger than the Earth. Its light rays converge on the Earth. You can test this with a light bulb (Sun) and a ping pong ball (Earth - smaller than the light bulb) and see what shadow is cast on a wall. With the right distances between the light bulb, ball, and wall, you should see the penumbra and umbra (see below)
- 50% of the sphere sun is shining light on the Earth.
- The Sun is rotating, so we eventually see all of it - 26.24 days as viewed from the Earth.
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:
(https://cdn4.dogonews.com/images/bd4375c5-a3a3-4b98-bcca-530eda20a23d/21_lunar_eclipse.jpg)
(http://eclipse.gsfc.nasa.gov/OH/image/TLE2014Apr15-GMTw.gif)
(http://www.astrosurf.com/comolli/ecl08i.jpg)
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.
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If you would please post your point in a single, concise question, I will be happy to provide you with an answer.
Sure. Let me ask the question once again.
How far is the Sun in FE model? Can you provide valid proof/link/calculations/source for that?
3000 miles, and here is how it was calculated.
http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal (http://blog.modernmechanix.com/5000-for-proving-the-earth-is-a-globe/3/#mmGal)
Back in this post of November 15, 2015 Jroa gives a clear answer to where the 3,000 miles comes from - some calculations by Voliva.
The a similar "measurement" is presented in "Zetetic Astronomy, by 'Parallax' (pseud. Samuel Birley Rowbotham), [1881], at sacred-texts.com".
In his CHAPTER V. "THE TRUE DISTANCE OF THE SUN" he gives essentially the same method, but with measurements done at London Bridge and Brighton, England. He gives a sun height of about 700 miles. The trouble was that his locations are only 50 miles apart, and his equipment was primitive, so his angles were not accurate.
His figures were:
| Sun Alt | Base line | Sun Height |
| London | 61.0° | | |
| Brighton | 64.0° | 50 miles | 751 miles |
The calculations are mine, but the result is near enough to Rowbotham's. I tried the same thing, for the same date, but in 2015.
The sun altitude angles can be found from "SunEarthTools" the results look more consistent with current FE figures.
| Sun Alt | Base line | Sun Height |
| London | 60.30° | | |
| Brighton | 60.98° | 50 miles | 3200 miles |
Showing that the Rowbotham's problem was simply too primitive measurements.
It does, however put the cat among the pigeons if the baseline is extended. I took Brighton and Papa Stour, a little town on the Shetland Is, just for convenience - they have almost the same longitude.
| Sun Alt | Base line | Sun Height |
| Shetlands | 50.44° | | |
| Brighton | 60.98° | 660 miles | 2432 miles |
Now, these figures are from http://www.sunearthtools.com/dp/tools/pos_sun.php (http://www.sunearthtools.com/dp/tools/pos_sun.php), so are based on Globe Earth data (which nobody, except Charles Bloomington, seems to doubt). To be fair some actual measurements should be done by some keen Flat Earth supporters to verify them. Ideally two or more locations with the same longitude as far apart as possible should be used.
Disclaimer: Everyone knows I am not a supporter of the FE, but a debate where the FE supporters don't have a firm basis is no point.
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Spam-ming, surly you jest. I gave real science answers to you query. Take some time to study it and you will see the truth and logic of it of it.
I studied the answer and found it to be wrong/flawed. The only thing you are doing here is blindly repeating the same statement without any actual proof/examination/experiment. Learn the meaning of science/proof before acting like a parrot that blindly repeats what he wants to believe. From us two, you are the one spamming threads and posting nonsense everywhere like few other trolls did.
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Hi FES!
I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.
The topic I'd like to discuss is how far the Sun is.
(you can skip this post and go to http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321 (http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321) for more details).
According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.
Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball). Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.
Notion:
d - the distance to the Sun
R - radius of the Sun
Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).
In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).
Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.
So the result is far from 5000km. Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.
How far the Sun really is?
lets just make this easy-how do you know the sun and moon are the distances we are told?simple question?
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.
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With the ball Earth theory you can not even see the Sun. You can only see the light that impacts the Earth. Take a look, you do not see the light on the back side of the Sun. You do not see the light of the Sun that is not directed at the Earth. From millions of miles away as the rays/wave go out evenly only a small area the diameter of the Earth hits the earth, not the rest. The ball Earthers think that tiny spot is the sun, it is not. The ball Earthers by their theory have never seen the sun in it's entirety only a small disk of it. The ball Earther's would have to use art work to show a complete Sun. When you see the complete Sun it is because it is only 3000 miles away and 32 miles across.
You can see the Sun on the flat Earth, but not on the theoretical ball Earth.
(http://i65.tinypic.com/1dy5nc.gif)
Your image and understanding are not correct.
- Light rays do not only come out of the center of the Sun but its entire surface up to the edge, as we view from Earth.
- The Sun is MUCH bigger than the Earth. Its light rays converge on the Earth. You can test this with a light bulb (Sun) and a ping pong ball (Earth - smaller than the light bulb) and see what shadow is cast on a wall. With the right distances between the light bulb, ball, and wall, you should see the penumbra and umbra (see below)
- 50% of the sphere sun is shining light on the Earth.
- The Sun is rotating, so we eventually see all of it - 26.24 days as viewed from the Earth.
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:
(https://cdn4.dogonews.com/images/bd4375c5-a3a3-4b98-bcca-530eda20a23d/21_lunar_eclipse.jpg)
(http://eclipse.gsfc.nasa.gov/OH/image/TLE2014Apr15-GMTw.gif)
(http://www.astrosurf.com/comolli/ecl08i.jpg)
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.
Isaac Newton is right you are wrong. No where did I say the suns energy only comes from the center. It radiates out in all directions like a wave, a ripple n a pool. The illustration you are using is a simplistic explanation and not how light actually comes from the sun. My example is the same exact thing explained and demonstrated by Isaac Newton. You fail, period.
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Hi FES!
I got some interest in this theory recently, I read lots of stuff but it looks like there are few things that are plain wrong based on assumptions.
The topic I'd like to discuss is how far the Sun is.
(you can skip this post and go to http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321 (http://www.theflatearthsociety.org/forum/index.php?topic=64777.msg1729321#msg1729321) for more details).
According to wikipedia, the angular size of the sun is 31'6'' to 32'7''. This is enough to calculate both diameter and the distance to the Sun.
Assumptions:
1. Earth is flat.
2. Sun has fixed physical size (a ball). Apparent change in size of the Sun is caused by the Sun moving away from us. Assuming disc shape the results are similar to the ones below.
3. Calculations are done for the observer on the equator, during 12h long day.
4. Whenever I write "=", I approximate. I tried minimizng rounding errors.
Notion:
d - the distance to the Sun
R - radius of the Sun
Using simple trigonometry we can find that, during the noon, where the Sun is above us, hence it is the closest and has the largest angular size, we have the relation d=104.131456R.
Same calculations lead us to d=107.7909868R for the case where the Sun is the furthest away, setting (or rising).
In 6h, the Sun will move 9003.1632km away horizontally (1/4th of the circumference, adjusted to straight line - a diagonal of a square).
Using Pythagoras theorem we find that R=323.2912km, and so d=33653.7747km.
So the result is far from 5000km. Such far distance automatically causes massive problems with day/night times, but I leave that off this thread.
How far the Sun really is?
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.
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.
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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.
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?
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.
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.
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Keeping the size of the Sun would "destroy" the entire argument, but I wanted to give it a chance and hence my assumption.
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.
Rowbotham in his book "The Earth is not a Globe" in "CHAPTER V.THE TRUE DISTANCE OF THE SUN." P 102-104 says "Then measure in the same way the vertical line D, S, and it will be found to be 700 miles. Hence it is demonstrable that the distance of the sun over that part of the earth to which it is vertical is only 700 statute miles.".
But, yes I completely agree, his angle measurements were high inaccurate and the calculations were rough. Actually had he accurate instruments he would have got close to the 3,000 miles often quoted.
But of course the whole premise is wrong. Voliva's or "Rowbotham's" method gives different answers for each latitude spacing we choose, as I have tried to demonstrate earlier.
Of course no measurement like this can give consistent answers for the obvious reason!
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Rowbotham in his book "The Earth is not a Globe" in "CHAPTER V.THE TRUE DISTANCE OF THE SUN." P 102-104 says "Then measure in the same way the vertical line D, S, and it will be found to be 700 miles. Hence it is demonstrable that the distance of the sun over that part of the earth to which it is vertical is only 700 statute miles.".
Either I missed that or I forgot. Thanks for the quote.
Of course no measurement like this can give consistent answers for the obvious reason!
Hence the discussion. If anyone has any other approach/idea for the FE model, post them!
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Perhaps this image will help in determining the height of the Sun:
(http://i40.tinypic.com/1zp48dh.jpg)
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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.
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
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.
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.
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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?
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FE model fails to explain one of its basics.
You can "find" holes in globe model, but you are unable to notice major flaws, errors and inconsistencies in RE model. Irony.
Globe model can explain the entire math behind the distance to the Sun, Moon and other bodies. FE cannot provide anything but hand-waving. Good to know...
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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.
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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.
In an animate drawing of the FE Model the moon and the sun are shown as the same size and the same distance from the earth, in the same orbital path and always 180 degrees apart. Another FE statement is that the moon and the sun travel at slightly different speeds. If this was so at some time there would be a collision of the moon and the sun. And the moon's light is caused by self[-illumination and not from the light of the sun. The phases of the moon caused by variations in this self-illumination. The moon and the sun are both the same size-32 miles in diameter- and the same distance from the earth - 3000 miles. Some FE
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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:
(https://cdn4.dogonews.com/images/bd4375c5-a3a3-4b98-bcca-530eda20a23d/21_lunar_eclipse.jpg)
(http://eclipse.gsfc.nasa.gov/OH/image/TLE2014Apr15-GMTw.gif)
(http://www.astrosurf.com/comolli/ecl08i.jpg)
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.
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The last picture looks like the sun shooting a beam on the building.
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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:
(https://cdn4.dogonews.com/images/bd4375c5-a3a3-4b98-bcca-530eda20a23d/21_lunar_eclipse.jpg)
(http://eclipse.gsfc.nasa.gov/OH/image/TLE2014Apr15-GMTw.gif)
(http://www.astrosurf.com/comolli/ecl08i.jpg)
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.
Ok sorry had a brain fart this morning. I meant a solar eclipse.
Now put your reasoning on a solar eclipse. With the sun and moon being the same size and same distance.
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For a solar eclipse on the traditional FE model:
- 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/ (http://www.greatamericaneclipse.com/best-places-to-view/))
(http://static1.squarespace.com/static/53c358b6e4b01b8adb4d5870/t/55eb4972e4b01fb4a6ed6285/1441483156395/TSE2017_US.jpg?format=500w)
(http://www.mreclipse.com/Special/image/SEDiagram1c.JPG) - 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 (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 (https://en.wikipedia.org/wiki/Planetshine)).
(http://en.es-static.us/upl/2013/04/PANSTARRS-moon-earthshine-Matt-Burt-spring-2013.jpg) - 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):
(http://i40.tinypic.com/1zp48dh.jpg) - 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.
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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):(http://i40.tinypic.com/1zp48dh.jpg)
It is not clear how far the Sun is hence it is not clear how the spotlight Sun would create any of the above pictures. Especially if the Sun is much further than alleged 3000 miles.
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So it is certain then, the sun is much larger than the moon. And much further away than the moon.
How does FE stand now?
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The Moon must be self-illuminating or the Sun would illuminate it all the time and the phases would not work.
There is such nice thing as spectroscopy. Everyone can analyze light produced by some body and determine is it hot or not and what components it is consisted of. You can do it with moonlight and if Moon is supposed to be self-illuminated then we can easily determine that it is hot and quite similar to the Sun.
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Again, read how the angular size of Mars is completely wrong...
http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4 (http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4)
Following said link I found claims that Mars is 72 million kilometers away from Earth.
In reality actual distance between Earth and Mars depends on orbital positions.
Closest is 225 million and farthes 401 million kilometers.
Using Mars' angular diameter in formulas that utilize 72 million kilometers will always give wrong result.
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Your turn. See the calculation of the Sun's altitude.
Angular diameter of the Sun can be simply measured. You don't need distance and radius.
But you might need welder's mask.
Set point of view, set ruler at measurable distance from that point of view, read the size of the Sun at the ruler.
If D is eye to ruller distance, and d is measured Sun's appearance on the ruler,
then angular diameter is 2 * ARCTAN (d / 2D).
Since the angle is small, it can be simplified to ARCTAN (d / D).
Having welder's mask, if you want more precise measurement, you can use some stick with
known length to hold calliper at the known distance from eye (from dark glass of the mask).
When you measure it, read this, especially the second part:
https://www.theflatearthsociety.org/forum/index.php?topic=58309.90
Reply #110, where you can find a way to calculate the distance of Sun from Earth's surface.
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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?
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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 (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 (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 (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 (http://www.freelists.org/post/geocentrism/The-resolution-of-Mars,4)
Sorry to say what utter tripe.... where did you get this from? lets have some references or did you make a recent visit?
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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?
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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?
Just so that I can't be accused of debating under the Q&A, I will just give a reference to: Re: Lets talk about the Elephant. « Reply #93 on: Today at 07:54:31 PM », (https://www.theflatearthsociety.org/forum/index.php?topic=67631.msg1808893#msg1808893)
but just asking, if you are going to rely on Voliva's method how do you know that the sun is not at 0 miles, 3,128 miles or 3,891 miles?
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The sun could be any distance. The fact that you and your cronies harass people is deplorable.
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The sun could be any distance. The fact that you and your cronies harass people is deplorable.
So no 5000km? If you are unsure of such basic thing, how can you discuss further implications regarding Sun's position?
If any distance, why not 150m km?
If any distance, why do you claim 5000km derived from flawed math?
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The sun could be any distance. The fact that you and your cronies harass people is deplorable.
Yes, it is quite deplorable that you simply cannot face the fact that there is no foundation for the flat earth's claim that the sun is 5,000 km high!
At will grant that you at least reply to the posts, no other flat earther seems willing to say anything about the topic at all!