Sun and Moon

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COLGeek

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So, I just read through this entire thread, again.

Can a line be drawn directly between the Sun and Moon at any given time? Yes.

Will the relative direction vary in relation to the Earth? Yes.

Picture the 3 objects as the points on a triangle. As we are talking about moving/tilting objects, there will be variations in perspective.

With all this being true, what is the purpose of this discussion? What point are you trying to make, Pies?

Maybe with that information, we'll better understand how to respond.
 
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rod

Oct 22, 2019
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So, I just read through this entire thread, again.

Can a line be drawn directly between the Sun and Moon at any given time? Yes.

Will the relative direction vary in relation to the Earth? Yes.

Picture the 3 objects as the points on a triangle. As we are talking about moving/tilting objects, there will be variations in perspective.

With all this being true, what is the purpose of this discussion? What point are you trying to make, Pies?

Maybe with that information, we'll better understand how to respond.
My opinion COLGeek. Pies is using flat earth arguments here to claim the Sun is much closer to Earth and much smaller than heliocentric solar system astronomy documents.
 

rod

Oct 22, 2019
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FYI, there are plenty of sources on the geometry of solar eclipses including the August 2017 event, e.g. https://www.discovermagazine.com/the-sciences/solar-eclipse-geometry

"Two Types of Shadows
If the sun were a point source, like one of the nighttime stars, the moon would cast only one kind of shadow. Instead, the sun stretches 0.5° across, so even during total solar eclipses, some of its light passes either above or below the moon, creating a less-dense shadow called the penumbra. Only where the moon blocks all the light from the sun — in its dark inner shadow called the umbra — can people on Earth see a total solar eclipse. Anywhere in the penumbra, the eclipse will be partial, but the percentage of the sun covered will increase as you get near the umbra. Unfortunately, the umbra is small, no more than a hundred miles in diameter. On the other hand, the penumbra measures more than 4,000 miles across."

This is an example how astronomy can calculate based upon the size of the umbra and penumbra, where folks will see a total solar eclipse event or partial. Annular solar eclipses take place too but the Moon is farther away from Earth then so does not cover the nearly 0.5 degree angular size for the Sun.

From another source I use that calculated the geometry for the August 2017 total solar eclipse we have:

"Figure 2: The flat-earth ‘model’ cannot successfully explain solar eclipses. First, the sun and moon appear to be the same size to an observer on the earth at the place where the dotted lines come together. The moon must be somewhere between those two lines. But, the size of the shadow will be a mere 114 km at its maximum extent during the Aug 2017 eclipse. In order to generate a shadow that size, and to appear the same size as the sun, the moon has to be very close to the earth and very small: 12.5 km away and 0.13 km in diameter. This simply cannot be true, so the flat-earth ‘model’ must be wrong."

"According to most flat-earth adherents today, the sun is only 5,000 km (3,000 miles) away. To explain both the apparent size of the moon and the observable size of its shadow, there is only one place it could be in the flat-earth model: the “dual-constraint lock point” in Figure 2. It turns out that the moon must be only 12.5 km (7.5 miles) above the earth! And its diameter must be only 0.13 km (about 427 feet)!"
 
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Jan 19, 2021
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Changing the scale to actual will only change the size of the umbra and penumbra.
[/QUOTE]
Yes the penumbra would disappear due to the massive distance.
 
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rod

Oct 22, 2019
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Post #55 looks like confusion over distance vs. angular size of the Sun and Moon in their sky positions during a solar eclipse even with different distances from Earth. The information in post #54 is correct. The Moon's umbra will be limited in extent just like the penumbra because of the Moon's angular size, physical diameter and distance from Earth. This is seen at eclipse events at Jupiter too, e.g. Io or Ganymede shadows crossing over Jupiter are solar eclipse events at Jupiter. The geometry can be calculated accurately and predicted accurately for telescope observations.

Post #55 ignores all the Mercury and Venus transit observations using telescope since the 1700s to measure the astronomical unit, thus distance between Earth and Sun. This is the solar parallax, another well established measurement in heliocentric solar system astronomy.
 
Jan 19, 2021
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In answer to COLGeek #51
You agree that a line can be drawn directly between the Sun and Moon ( a line that is perpendicular to the moon's illumination).
Referring to your triangle analogy, adding a third point as a viewing position, will not change the perpendicular relationship between the moon and the sun.

The point I am trying to make is that the perpendicular line between the moon and the sun is not observed from Earth (the third point of your triangle).
 
Jan 19, 2021
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In answer to Rod #52
I'm not attempting to calculate the sun's distance only that it can't be 93million miles if there is a penumbra. I will have to leave the maths up to those who can work out the new distances and maybe even the diameters if they are affected.
 
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rod

Oct 22, 2019
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Pies, you should use better sources I feel in your post here. https://astronomy.stackexchange.com/questions/35635/during-an-eclipse-how-big-is-the-shadow-of-the-moon-on-the-earth

Specifics for angular sizes, distances, and diameters are all used in real astronomy :)

"...where Rm,Re,Rs are the radii of the Moon, Earth and Sun respectively, dm is the distance Moon-Earth, de is the distance Earth-Sun.

We can investigate several cases by varying the distances dm and de, in accordance with the eccentricities of the orbits...Now the maximum radius of the penumbra is obtained when..."
 
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Jan 19, 2021
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Post #55 looks like confusion over distance vs. angular size of the Sun and Moon in their sky positions during a solar eclipse even with different distances from Earth. The information in post #54 is correct. The Moon's umbra will be limited in extent just like the penumbra because of the Moon's angular size, physical diameter and distance from Earth. This is seen at eclipse events at Jupiter too, e.g. Io or Ganymede shadows crossing over Jupiter are solar eclipse events at Jupiter. The geometry can be calculated accurately and predicted accurately for telescope observations.

Post #55 ignores all the Mercury and Venus transit observations using telescope since the 1700s to measure the astronomical unit, thus distance between Earth and Sun. This is the solar parallax, another well established measurement in heliocentric solar system astronomy.
No there is no confusion on my part. The angular distance between the sun and the moon is almost zero due to the 93million miles. I accept that the moon doess not always eclipse the sun exactly and that there will be minor variations in the shadow and penumbra.
 

rod

Oct 22, 2019
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FYI. The math at the link in post 59 shows how to calculate both umbra size and penumbra size based upon different distances for the Earth-Moon as well as distance for the Sun. Your sources are promoting *error*. The distance to the Sun can be measured and is known today accurately, e.g. https://www.space.com/parker-solar-probe-approach-sun-cycle-january-2021

Reviewing telescope observations from the 1700s for Venus and Mercury transits show the solar parallax, that demonstrates the distance to the Sun. Multiple, independent studies can show a good distance measurement to the Sun today, just like multiple distance measurements showing the distance to the Moon that are independent, e.g. lunar parallax, lunar laser ranging measurements, ham radio operators showing the distance to the Moon, etc.
 
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COLGeek

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In answer to Rod #52
I'm not attempting to calculate the sun's distance only that it can't be 93million miles if there is a penumbra. I will have to leave the maths up to those who can work out the new distances and maybe even the diameters if they are affected.
Huh? Is your concern the Sun's average distance from Earth? Are you saying this isn't true?
 

COLGeek

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In answer to COLGeek #51
You agree that a line can be drawn directly between the Sun and Moon ( a line that is perpendicular to the moon's illumination).
Referring to your triangle analogy, adding a third point as a viewing position, will not change the perpendicular relationship between the moon and the sun.

The point I am trying to make is that the perpendicular line between the moon and the sun is not observed from Earth (the third point of your triangle).
Sure, an imaginary line between two objects in three-dimensional space is always possible. Not seeing this as a terribly hard concept to grasp (or agree on). Why is the point particularly relevant?
 
Jun 1, 2020
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No there is no confusion on my part. The angular distance between the sun and the moon is almost zero due to the 93million miles.
That angular size of the Sun is something anyone can measure. It isn't an illusion caused by our atmosphere or a psychological hiccup. Objective measurements show it is a little more than 30 arcminutes. It happens to vary with our distance from it as we swing in our elliptical orbit.

Any object that appears as a disk (ie extended object) will have a portion of the emissions arrive to the observer at varying angles, with a maximum angle between any two rays of about 30 arcminutes. Thus, when the Moon only partial blocks the Sun, we then see those rays from the limb, and we don't see any rays from the center of the disk. This necessarily makes the Sun less bright than normal. Does this not make sense that a penumbra is a must for any extended object? The Sun is not a laser emitter with all rays parallel. If it were, then space probes would see it turn dark as they traveled outside this laser beam.

I accept that the moon doess not always eclipse the sun exactly and that there will be minor variations in the shadow and penumbra.
When it doesn't fully eclipse, then the umbral zone is too short to reach Earth's surface. The amount of shadow is easy to see just by looking at how much Sun is actually eclipsed. The more the disk is covered by the Moon, the darker is our sky. The Sun, however, is so incredibly bright that even a small amount of disk exposed brightens the sky above. Also, atmospheric scattering helps illuminate the sky.
 
Jan 19, 2021
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In reply to Rod #59

I will make a retraction at this point and concede that a penumbra will be produced and agree that if the light source were a point light then there would be no penumbra.

Thank you for the link Rod - it was one I didn't see when I searched.

I have read the paper and see a problem although I am no mathematician!
The distance from the earth to the sun has been ignored except in calculating the the variation in the moon's orbit.

I have created a drawing using the paper's calculated (not actually observed or measured) diameters for the umbra and penumbra.

The drawing shows that Sun would have to be either 1.239million miles in diameter or only 62.6million miles away.

 
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rod

Oct 22, 2019
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Pies seems to show misconceptions in post #67. "I have read the paper and see a problem although I am no mathematician! The distance from the earth to the sun has been ignored except in calculating the the variation in the moon's orbit."

I am no mathematician either Pies but do see the variance for Earth at aphelion and perihelion distances from the Sun in the algorithm. The penumbra width is calculated using Earth's perihelion and aphelion distances from the Sun in the heliocentric solar system model. You have an interesting drawing in the .jpg image.

Where did the 0.798 degree conical angle of the umbra to the Sun come from? Other sources indicate 0.53 degree value here so this will change the diameter of the Sun in your .jpg drawing.

But for sake of discussion, let me *assume* the drawing is accurate (with no sources shown for the math formulas used in the .jpg image). What angular size measurement should we see for the Sun in the sky today using arcminutes? This is a simple test of your argument here. The test here using trigonometry will allow the true diameter of the Sun to be measured based upon its distance from Earth. I checked the math here. Using your Drawing10.jpg size for the Sun at 1.239 million miles in diameter, the angular size for the Sun as seen from Earth today ~ 48 arcminute size at 1 AU distance from Earth. So Pies, this is a simple test of your drawing and argument.
 
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Jan 19, 2021
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Where did the 0.798 degree conical angle of the umbra to the Sun come from? Other sources indicate 0.53 degree value here so this will change the diameter of the Sun in your .jpg drawing.

But for sake of discussion, let me *assume* the drawing is accurate (with no sources shown for the math formulas used in the .jpg image). What angular size measurement should we see for the Sun in the sky today using arcminutes? This is a simple test of your argument here. The test here using trigonometry will allow the true diameter of the Sun to be measured based upon its distance from Earth. I checked the math here. Using your Drawing10.jpg size for the Sun at 1.239 million miles in diameter, the angular size for the Sun as seen from Earth today ~ 48 arcminute size at 1 AU distance from Earth. So Pies, this is a simple test of your drawing and argument.
In reply to Rod #68
I have gone through my calculations again and produced a new drawing with a slightly different result due to a typo with the original penumbra diameter - sorry about that!

The diameters of the umbra and penumbra were taken from the paper on this URL https://astronomy.stackexchange.com/questions/35635/during-an-eclipse-how-big-is-the-shadow-of-the-moon-on-the-earth
From my perspective there are many variations that can change the results but I suspect that they won't make that much difference to my calculated Sun diameter which is 1.68488million miles. This is 1.93 times larger.

Since you are unlikely to accept that result I have produced yet another drawing working backwards from the accepted dimensions for the Sun dia.
I used a triangle calculator at this URL https://www.calculator.net/triangle-calculator.html?vc=&vx=&vy=238854&va=&vz=1117&vb=90&angleunits=d&x=74&y=18

Working backwards using the accepted diameter of the sun but keeping the same penumbra (twice the diameter of the moon) we get an umbra of 1117miles which is 7.5 times larger.

Something is not adding up!
https://www.dropbox.com/s/c5rz8ysxd2951ii/Drawing 10.jpg?dl=0

 
Jan 19, 2021
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Where did the 0.798 degree conical angle of the umbra to the Sun come from? Other sources indicate 0.53 degree value here so this will change the diameter of the Sun in your .jpg drawing.

But for sake of discussion, let me *assume* the drawing is accurate (with no sources shown for the math formulas used in the .jpg image). What angular size measurement should we see for the Sun in the sky today using arcminutes? This is a simple test of your argument here. The test here using trigonometry will allow the true diameter of the Sun to be measured based upon its distance from Earth. I checked the math here. Using your Drawing10.jpg size for the Sun at 1.239 million miles in diameter, the angular size for the Sun as seen from Earth today ~ 48 arcminute size at 1 AU distance from Earth. So Pies, this is a simple test of your drawing and argument.
In reply to Rod #68
I have gone through my calculations again and produced a new drawing with a slightly different result due to a typo with the original penumbra diameter - sorry about that!

The diameters of the umbra and penumbra were taken from the paper on this URL https://astronomy.stackexchange.com/questions/35635/during-an-eclipse-how-big-is-the-shadow-of-the-moon-on-the-earth
From my perspective there are many variations that can change the results but I suspect that they won't make that much difference to my calculated Sun diameter which is 1.68488million miles. This is 1.93 times larger.

Since you are unlikely to accept that result I have produced yet another drawing working backwards from the accepted dimensions for the Sun dia.
I used a triangle calculator at this URL https://www.calculator.net/triangle-calculator.html?vc=&vx=&vy=238854&va=&vz=1117&vb=90&angleunits=d&x=74&y=18

https://www.dropbox.com/s/60y9aw3kjafkodo/Drawing%2011.jpg?dl=0
Working backwards using the accepted diameter of the sun but keeping the same penumbra (twice the diameter of the moon) we get an umbra of 1117miles which is 7.5 times larger.

Something is not adding up!
 

rod

Oct 22, 2019
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Pies, thanks for the links and drawings in post #70. I do agree, *something is not adding up!*. Today I went out and observed the Sun and sunspots AR2797 and AR2798 reported at spaceweather.com using my 90-mm refractor telescope at 40x with glass, white light solar filter. It is interesting that as you rework your calculations, now we have a Sun with diameter 1.68488 million miles in diameter. That means today when I viewed the Sun using my telescope, the angular size should be > 62 arcminutes in size. The telescope true field of view ~ 78 arcminute across and the Sun was no where near the size you calculate using your *formulas*. The sunspot sizes are <=18 arcsecond size too.

Can you provide the arcminute size of the Sun I should have viewed in my telescope today and the distance the Sun is from Earth? Simply changing the Sun's size from 1.293 million miles across to the new value of 1.68488 million miles across does not add up. It is too easy to check and I did today. Here are sites I use to monitor changes on the Sun for telescope observations:

 
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COLGeek

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Apr 3, 2020
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In reply to Rod #68
I have gone through my calculations again and produced a new drawing with a slightly different result due to a typo with the original penumbra diameter - sorry about that!

The diameters of the umbra and penumbra were taken from the paper on this URL https://astronomy.stackexchange.com/questions/35635/during-an-eclipse-how-big-is-the-shadow-of-the-moon-on-the-earth
From my perspective there are many variations that can change the results but I suspect that they won't make that much difference to my calculated Sun diameter which is 1.68488million miles. This is 1.93 times larger.

Since you are unlikely to accept that result I have produced yet another drawing working backwards from the accepted dimensions for the Sun dia.
I used a triangle calculator at this URL https://www.calculator.net/triangle-calculator.html?vc=&vx=&vy=238854&va=&vz=1117&vb=90&angleunits=d&x=74&y=18

https://www.dropbox.com/s/60y9aw3kjafkodo/Drawing%2011.jpg?dl=0
Working backwards using the accepted diameter of the sun but keeping the same penumbra (twice the diameter of the moon) we get an umbra of 1117miles which is 7.5 times larger.

Something is not adding up!
I have asked a couple times now. What point are you trying to prove? Please, share.
 
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Jan 19, 2021
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Don you have me at an advantage here! When I started on this forum I pointed out that I was not an astronomer but someone who just happened to notice a strange moon - earth anomaly. Although there have many exchanges over this nobody has come up with an answer That means I have to conclude that the moon is illuminated by some means other than the sun.

Regarding my second thread re solar eclipses this again is just my simple no-expert basic question. I do not know what you mean about 'arcseconds'. is this just another way to express an angle?

I get the impression that although we have a great deal of data about the sun, moon, earth, eclipses etc I see that that there is no connection between how eclipses are created and the positions of the sun, moon and earth - they seem to be two separate issues and nobody has ever connected the dots between them. This is what I have attempted do by extrapolating backwards from what is observed and directly measured eclipse umbras and penumbras to the size and position of the sun.

If my reckoning has shown up some discrepancy regarding the sun then the same can also be said about the moon's physical properties. When you couple this with my original question about how the moon is illuminated it seems to have opened a whole new can of worms!

I think what has happened is that ideas and theories have been expounded over hundreds of years about astronomy that it has become very top heavy and looking unstable. That's just my opinion as an outsider looking in.

There is a saying that is incredibly true 'It is very easy to get people to believe a lie but incredibly difficult to get them to believe they have been lied to'. I'm not necessarily saying that astronomy is full of lies but it could be full of misunderstandings and assumptions that have become so deep seated that to challenge them is viewed as heresy through cognitive dissonance.
 
Jan 19, 2021
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I have asked a couple times now. What point are you trying to prove? Please, share.
I assume you mean my original question about how the moon is illuminated.
My point therefore is obvious! Since nobody has yet offered an explanation as to what I observed I have to conclude that the moon cannot be illuminated by the sun so it must be illuminated by some other means. Don't ask me what the 'other means' is as that is something for others to figure out.
This is a case of me saying 'The Emperor appears to have no clothes on!'
 

COLGeek

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I assume you mean my original question about how the moon is illuminated.
My point therefore is obvious! Since nobody has yet offered an explanation as to what I observed I have to conclude that the moon cannot be illuminated by the sun so it must be illuminated by some other means. Don't ask me what the 'other means' is as that is something for others to figure out.
This is a case of me saying 'The Emperor appears to have no clothes on!'
No. Nope. Nyet. Nein. Nothing but the Sun and some reflected light from Earth.

Anything else is imaginary.
 
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