Total Solar Eclipse

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Jan 19, 2021
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I cannot access the diagram but I assume it is the same old diagram where the Sun is shown bottom right and the Sun's rays are shown at an angle to "prove" an incorrect lunar phase. Pies could not defend this and disappeared.

Can we please be protected from what imho are repeated nuisance posters?

Cat :)
I sent you a very clear photo showing the moon and sun. I also suggested that you enlarge the top left area where the moon was to reveal that the angle of the phase did not match the position of the sun. This you do not appear to have done therefore I'm attaching it again.
 

rod

Oct 22, 2019
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Pies new drawing in post #24 shows the Moon diameter = 7926 miles, error upon error keeps piling up here. Radar measurements of the Moon show nothing like this, e.g. the Army 1946 radar efforts and recent like https://skyandtelescope.org/astronomy-news/green-bank-tests-new-planetary-radar/

No matter what, a lunar umbra 0.53 deg is close to 32 arcminute size on Earth. Nothing like this in astronomy and not supported by radar measurements or lunar laser ranging measurements for the size of the Moon shown by Pies. There is a long history in astronomy here including the lunar parallax from telescopes dating back to the 1700 and 1800s.

I must confess here that Pies new diameter for the Moon of 7926 miles across is better than the FES wiki showing 32 miles across :) Going back to my post #22 and the 1963 book on the Moon, it says "The moon is 2,160 miles in diameter..." on page 9.
 
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The result of my calculations show that the umbra would have to be very slightly smaller than the diameter of the Moon at the Earth's equator due to the very small conical angle.
. The umbral size depends mainly on the distance the moon is during the eclipse. The 2024 umbra will be much larger than the prior in the U.S. [It may look smaller while crossing Texas. ;]

It is not difficult to understand why text books show the Sun's rays as being virtually parallel with this tiny 0.53deg conical angle.

Yes, it took a closer look for me to see that your drawing 2 had an angle. A 1/4 deg angle gives a ratio of ~ 230 (distance/half disk radius of sun).
 

rod

Oct 22, 2019
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Keep in mind Helio, Sun rays 3.jpg shows the Moon diameter is 7926 miles across :) The drawing shows the umbra size is 0.53 degrees too on Earth. This is all wrong. Umbra sizes fall within the range ~ 1 to 9 arcminute angular size on Earth during a total solar eclipse using the lunar distance 384401 km as the mean.
 
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If the astronomical information available on the Internet is fraught with inaccuracies then can someone then please provide me with the latest all singing and all dancing figures for the Sun dia and Moon dia and the distance between.
I would imagine that any natural fluctuations in the Moon's orbit is not likely to make a significant difference to the calculations. I'm happy to recalculate on whatever the two extremes of distance are likely to be in order to see what difference they could make to the umbra size.
I'm struggling to believe why this is so complicated!
 

rod

Oct 22, 2019
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Pies, in your post #30, your math you use will tell you the physical size of the Sun, the distance to the Sun from Earth, the Moon's physical size, and distance from Earth, but this may not be the correct solution though for the correct answers as modern astronomy uses. That is the whole point of your discussion as I see it. You are trying to show that the modern, heliocentric solar system astronomy does not know the true distances and sizes here for the Sun and Moon, same as the Flat Earth Society Moon and Sun wiki do. The umbra size is calculated and known before the total solar eclipse event takes place, e.g. the 2017 total solar eclipse as published by many, including the path covered over the Earth and time predictions when the umbra will cover a specific location on Earth so folks living in that area can see a total solar eclipse event. The umbra size will always be a smaller area on Earth than the diameter of the Moon or the surface area of the Earth. This involves much spherical trigonometry, differential calculus for the shape of the Earth, Moon's orbit, Earth's rotation velocity, Moon orbital velocity, etc. No matter though, the umbra size will be much smaller area and as measured from the mean distance of the Moon, 384401 km, the umbra size projected on Earth falls within 1 to 9 arcminute range or so. Your math suggest an umbra size on Earth closer to 32 arcminute angular size is what astronomy should observe, calculate and predict. This is something that is not documented or observed in astronomy. Just to demonstrate, the Earth as measured from the Moon using 384401 km mean distance, the Earth's angular size is about 108 arcminute across or close to 1.8-degrees. In astronomy, there are no giant Moon umbra(s) moving over the Earth during a total solar eclipse as your math shows.

This report may help, https://www.space.com/17638-how-big-is-earth.html so the Earth diameter is 12756 km and observed from the Moon at 384401 km, angular size ~ 114 arcminute. The Moon's umbra on Earth during a total solar eclipse will be a much smaller angular size than 108 to 114 arcminute, but not 32 arcminute size. Smaller still using the correct solution.
 
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Pies, in your post #30, your math you use will tell you the physical size of the Sun, the distance to the Sun from Earth, the Moon's physical size, and distance from Earth, but this may not be the correct solution though for the correct answers as modern astronomy uses. That is the whole point of your discussion as I see it. You are trying to show that the modern, heliocentric solar system astronomy does not know the true distances and sizes here for the Sun and Moon, same as the Flat Earth Society Moon and Sun wiki do. The umbra size is calculated and known before the total solar eclipse event takes place, e.g. the 2017 total solar eclipse as published by many, including the path covered over the Earth and time predictions when the umbra will cover a specific location on Earth so folks living in that area can see a total solar eclipse event. The umbra size will always be a smaller area on Earth than the diameter of the Moon or the surface area of the Earth. This involves much spherical trigonometry, differential calculus for the shape of the Earth, Moon's orbit, Earth's rotation velocity, Moon orbital velocity, etc. No matter though, the umbra size will be much smaller area and as measured from the mean distance of the Moon, 384401 km, the umbra size projected on Earth falls within 1 to 9 arcminute range or so. Your math suggest an umbra size on Earth closer to 32 arcminute angular size is what astronomy should observe, calculate and predict. This is something that is not documented or observed in astronomy. Just to demonstrate, the Earth as measured from the Moon using 384401 km mean distance, the Earth's angular size is about 108 arcminute across or close to 1.8-degrees. In astronomy, there are no giant Moon umbra(s) moving over the Earth during a total solar eclipse as your math shows.

This report may help, https://www.space.com/17638-how-big-is-earth.html so the Earth diameter is 12756 km and observed from the Moon at 384401 km, angular size ~ 114 arcminute. The Moon's umbra on Earth during a total solar eclipse will be a much smaller angular size than 108 to 114 arcminute, but not 32 arcminute size. Smaller still using the correct solution.

Rod many thanks for your astronomical approach to calculating the Moon's umbra during a total solar eclipse.
With respect much of what you said in your reply was muddying the waters in trying to do a simple calculation.
Regarding the relative velocities of the Moon and the earth, I cannot see how they can affect the model at a given instant in time.
I'm talking about at the given instant when the total solar eclipse is considered total then the positions of the Sun, Moon and Earth are fixed for that given instant.

We have 3 spherical objects of a known size and known distances between them. It is then a simple piece of maths to calculate a particular angle.

This is the position I was working from. I attempted to calculate the conical angle produced between the Sun and Moon using figures that I got from the Internet albeit slightly inaccurate. But the huge distance between the Sun and the Moon and the conical angle cannot change that much from adjusting to suit modern figures.
Therefore at that given instant of totality, and using my figures, whether accurate or not, the geometry must be correct. Its just a simple piece of maths.
 
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We have 3 spherical objects of a known size and known distances between them. It is then a simple piece of maths to calculate a particular angle.
Yes. It should be easy. IMO, only the Earh’s crvature makes it trickier, but this causes a slightly larger umbra and penumbra.

The Moon increases its orbital distance from perigee by more than 10%, which is the key to those umbral sizes Rod is giving, and no doubt accurate. The large umbra coming in 2024 will be due to a closer distance with the Moon than most other eclipses.

It may help to note that when the Moon is near apogee, only an annular eclipse is possible.

It’s amazing that the Moon so closely matches the Sun’s disk size as the odds are very much against it.



This is the position I was working from. I attempted to calculate the conical angle produced between the Sun and Moon using figures that I got from the Internet albeit slightly inaccurate. But the huge distance between the Sun and the Moon and the conical angle cannot change that much from adjusting to suit modern figures.
. You may be making this too complicated.

Using the well established radius of the Moon of about 1738 km, multiplying this by the tangent factor for 1/4 degree of 229, yields a base distance of ~ 398,300 km. This distance would be the point where the umbra size becomes zero. So when the Earth is farther than this (e.g. apogee) then there is no umbra (annular eclipse) but if closer, then an umbra will be found if given a true sysygy.

iPad
 
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I sent you a very clear photo showing the moon and sun. I also suggested that you enlarge the top left area where the moon was to reveal that the angle of the phase did not match the position of the sun. This you do not appear to have done therefore I'm attaching it again.
The arrows indicating light did not point to the Sun.
The image of the Moon was twisted.

I have no time to discuss utter nonsense,

Cat /////
 
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In using fixed values, I calculate that the maximum umbral diameter (@ perigee) will be ~ 300 km (190 miles). This uses 0.25 degrees. The Earth’s curvature is not that great over that small diameter, but it would make the umbra slightly larger.

But the actual solar angle will vary with Earth’s orbital position; when at aphelion a larger umbra will be seen.
 
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rod

Oct 22, 2019
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Helio, very good in your post #33 and use of trig :). Yes, Pies solution results in an umbra somewhat larger than 0.5 degrees on the surface of the Earth in the diagrams. Just how long will a total solar eclipse last for a given location on Earth if the umbra is more than 0.5-degrees across sitting on that spot? If most total solar eclipse lunar umbra are some 1 to 9 arcminute in angular size on the surface of the Earth vs. 0.5 degrees or more that Pies gives us, the time of totality for a given location will be very different. We will have very long total solar eclipse events for any given location that the giant umbra passes over :)
 
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At least all seem to accept the ~1/2 degree angle, so only the distance the Earth has within the shadow’s cone needs to be known in order to calculate the resulting shadow size.
 
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rod

Oct 22, 2019
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Helio, your 300 km diameter umbra size in post #35 on Earth as observed from the Moon at 384401 km, is about 2.67 arcminute in angular size. The larger the area of the umbra, the longer the time of totality can become, that is a critical measurement in total solar eclipses. Pies is using an umbra size on Earth as large as the diameter of the Moon to argue against astronomy. The time of totality will be very long if this were true and folks will immediately see this like the August 2017 total solar eclipse across the USA where nothing like that was observed and documented. Here is what MS BING says about this total solar eclipse.

"The above animation shows a correctly scaled view of the Earth-Moon system during the August 21, 2017, eclipse. By the time the umbra reaches Earth it has been reduced in size from ~2,200 miles in diameter to only about 100 miles in diameter. The penumbra, however, has widened to roughly twice its original size."

No one observed an umbra the size of the Moon moving across the USA during this total solar eclipse. As viewed from the Moon, the umbra size moving across the USA would be close to 1.5 arcminute angular size when the Moon is at its mean distance from Earth.
 
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I’m unclear why the angular size of the umbra as seen by a lunar observer is important? Given the shadow‘s angle, only where the Earth sits in the shadow will determine the umbral diameter.
 

rod

Oct 22, 2019
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I’m unclear why the angular size of the umbra as seen by a lunar observer is important? Given the shadow‘s angle, only where the Earth sits in the shadow will determine the umbral diameter.
I showed this to illustrate the angular size differences. Pies umbra size is the diameter of the Moon on Earth so nearly 2200 miles across on the Earth, thus similar to 0.5 degree size used. The larger the umbra size on Earth, the longer totality can become. There is min and max issues that need to be defined, including min duration for a total solar eclipse and max duration as observed on any area on Earth. An umbra near 2200 miles across on Earth, how long do you think totality will last?
 

rod

Oct 22, 2019
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https://www.quora.com/How-do-I-calculate-the-diameter-of-the-shadow-umbra-cast-by-the-Moon-to-the-Earths-surface-during-a-solar-eclipse, this site has some math. You can see this comment, "The illustration you have used is massively out of scale. For a better idea of the scale try this web site When light from the sun arrived at Earth's orbit it's rays are parallel to a very good approximation so the shadow cast by the Moon on the Earth at total eclipse is the same size as the Moon itself. If you want to see what that looks like DSCOVR has taken some good pictures"

Pies is presenting the same argument here in this discussion. So if the umbra during a total solar eclipse is the same size as the Moon on Earth, we will enjoy very long total solar eclipse events :) My post #31 calls this correctly in what is taking place here on the forums. Astronomy does not know the true size or distance to the Moon or Sun according to Pies methodology. My opinion. Pies should acknowledge this, what the real intent here is all about.
 
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I showed this to illustrate the angular size differences. Pies umbra size is the diameter of the Moon on Earth so nearly 2200 miles across on the Earth, thus similar to 0.5 degree size used. The larger the umbra size on Earth, the longer totality can become. There is min and max issues that need to be defined, including min duration for a total solar eclipse and max duration as observed on any area on Earth. An umbra near 2200 miles across on Earth, how long do you think totality will last?
I see. But I think pies has accepted that parallel rays aren’t correct. I suspect he may think that the tiny angle is close enough to produce a large umbra, but the trig is easy enough to demonstrate the contrary. Am I right, pies?
 
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Further, using trig, for a max. umbral projection, the Earth’s curvature would only increase the umbra by about mile along the curved surface.
 
Jan 19, 2021
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Pies, in your post #30, your math you use will tell you the physical size of the Sun, the distance to the Sun from Earth, the Moon's physical size, and distance from Earth, but this may not be the correct solution though for the correct answers as modern astronomy uses. That is the whole point of your discussion as I see it. You are trying to show that the modern, heliocentric solar system astronomy does not know the true distances and sizes here for the Sun and Moon, same as the Flat Earth Society Moon and Sun wiki do. The umbra size is calculated and known before the total solar eclipse event takes place, e.g. the 2017 total solar eclipse as published by many, including the path covered over the Earth and time predictions when the umbra will cover a specific location on Earth so folks living in that area can see a total solar eclipse event. The umbra size will always be a smaller area on Earth than the diameter of the Moon or the surface area of the Earth. This involves much spherical trigonometry, differential calculus for the shape of the Earth, Moon's orbit, Earth's rotation velocity, Moon orbital velocity, etc. No matter though, the umbra size will be much smaller area and as measured from the mean distance of the Moon, 384401 km, the umbra size projected on Earth falls within 1 to 9 arcminute range or so. Your math suggest an umbra size on Earth closer to 32 arcminute angular size is what astronomy should observe, calculate and predict. This is something that is not documented or observed in astronomy. Just to demonstrate, the Earth as measured from the Moon using 384401 km mean distance, the Earth's angular size is about 108 arcminute across or close to 1.8-degrees. In astronomy, there are no giant Moon umbra(s) moving over the Earth during a total solar eclipse as your math shows.

This report may help, https://www.space.com/17638-how-big-is-earth.html so the Earth diameter is 12756 km and observed from the Moon at 384401 km, angular size ~ 114 arcminute. The Moon's umbra on Earth during a total solar eclipse will be a much smaller angular size than 108 to 114 arcminute, but not 32 arcminute size. Smaller still using the correct solution.
Thanks Helio for this response. I now feel that I am getting somewhere with this thread.
I'm not quite sure what you are describing here so I have produce another drawing hopefully understanding what you say and allowing for the various orbital differences. I will have to consult the internet to provide me with these figures. If I happen to choose the wrong ones I would assume that any differences would be minimal. I have used your accepted diameter of the Moon in these calculations.

I will still try and keep this as simple as possible and not get overawed by to much technical information - I am a simple man at heart!
I do realise that the when it comes to the 'exact' diameter of the umbra that various other factors will come into play e.g. the curvature of the Earth, it position on the Earth, the moon's orbital differences and the Earth's orbital differences.

I especially appreciate your comment about the extraordinary coincidence of the Moon's diameter closely matching the Sun's during a solar eclipse. However an explanation for this is likely to be found in the realms of philosophy!

With resect to all the other comments I am not avoiding them as I know from past experience that trying to answer all the different aspects in a discussion only leads to confusion and going off at a tangent. I want to be able to put this simple 3 sphere question to rest by knowing that the geometry is accurate. Once there is agreement on this then I will be happy to discuss the broader implications.

As you can see from my drawing that conical angle of the light ray that produces the umbra change very little for the extreme positions of the Moon relative to the Sun. The resulting umbra, in these scenarios is going to be just slightly smaller than the Moon's diameter at the equator. I understand that anywhere else the umbra will be a different size but, IMHO not hugely significant.
https://www.dropbox.com/s/bju7skmo6knn9lp/Helio data.jpg?dl=0
 

rod

Oct 22, 2019
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There is nothing in the new drawings in post #46 that shows the angular size of the umbra on Earth during a total solar eclipse, just the Sun light cone shape hitting the New Moon phase from the Sun, thus the angular size of the Moon is presented. The drawings fail to extend the umbra shadow of the Moon, at the point of origin (the Moon, thus somewhat larger than 0.5 degrees at the Moon), all the way to the surface of the Earth and show how that umbra size shrinks at it approaches the Earth. This has been the thrust of Pies argument, the umbra size on Earth during a total solar eclipse cannot be valid, thus astronomy does not know the true size of the Moon or distance to the Moon from Earth. That is why I showed the shrinking angular size of the umbra on Earth in other posts, i.e. 1 to 9 arcminute angular size on Earth as viewed from the Moon vs. point of origin size, about 0.5 degrees or slightly larger.

Whether Pies acknowledges or not, the methodology Pies uses in this discussion is all flat earth teaching.

Reference, 'Solar Eclipse Calculation Basics: Umbral Size', https://flatearthinsanity.blogspot.com/2017/06/solar-eclipse-calculation-basics-umbral.html
 
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rod

Oct 22, 2019
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In using fixed values, I calculate that the maximum umbral diameter (@ perigee) will be ~ 300 km (190 miles). This uses 0.25 degrees. The Earth’s curvature is not that great over that small diameter, but it would make the umbra slightly larger.

But the actual solar angle will vary with Earth’s orbital position; when at aphelion a larger umbra will be seen.
Helio, to go along with this post. The Sun’s angular diameter (arcminute and arcsecond size) changes throughout the year during perihelion and aphelion, observable and measurable using telescopes. "As the distance varies, the apparent size of the sun changes, the visible diameter ranging from 32'.6 in January to 31'.5 in July.", Navigation and Nautical Astronomy, Dutton, 1926, updated 1951 US Naval Academy, p. 237.

The same is true for the Moon's arcminute angular size as it orbits from perigee to apogee around the Earth. All of this plays into calculating solar eclipses too. The Pies methodology is inadequate to address these changes in solar eclipses or lunar eclipses. In plain language, when the Moon's arcminute size is smaller than the Sun's during a solar eclipse event alignment, we on Earth will not see a total solar eclipse but could see an annular solar eclipse if in the correction location on Earth. Using the 3 body discussion. the distance of the Moon and Sun are dynamic, changing as Earth revolves around the Sun and the Moon revolves around the Earth as both move around the Sun. The arcminute sizes change for the Sun and Moon in this orbital dance in the heavens. As the orbital dance continues, solar eclipses will have dynamics too like total or annular events. The Moon's umbra at point of origin (the Moon) is going to be near 0.5 degrees or so as viewed from the angular size of the Moon in our sky but the umbra shrinks down as it approaches Earth to a much smaller size and area on Earth during a solar eclipse event. Astronomy does know the true distance to the Moon, the Moon's true diameter size, the Sun too. This is the witness of solar eclipse events and lunar eclipse events documented in astronomy today. My opinion. The flat earth community today seeks desperately to deny this witness in astronomy and that is what we have here in the forums in this entire discussion.
 
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COLGeek

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However an explanation for this is likely to be found in the realms of philosophy!

Once there is agreement on this then I will be happy to discuss the broader implications.
These are the underlying issues that all can clearly see as the true argument here and why the physical reality of the problem being ignored is frustrating to members.

This thread is quickly running its course...
 
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rod

Oct 22, 2019
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We are told in text books that for all intents and purposes that the sun's rays can be considered parallel. This being the case the during a total solar eclipse, logically, the moon's umbra would have to be the same diameter as the moon. Is there a flaw in this suggestion?

See my post #48. *logically, the moon's umbra would have to be the same diameter as the moon..." Yes, the umbra shadow is the same size as the Moon, at the point of origin, the Moon where the shadow of the Moon begins. On its way to Earth, things change dramatically during a solar eclipse event so the umbra is smaller when hitting the Earth. :) For those following this discussion, here is a note I received from a source :)

"Flat-earthers get this argument wrong two contradictory ways. First, flat-earthers argue that shadows are always larger than the bodies casting the shadows. For objects that are larger than the light source, this is the case. Most artificial light sources, such as light bulbs, are relatively small, so this usually is the case. However, for light sources that are larger than the objects casting shadows, this is not the case. If these people looked at shadows of birds or planes, they'd realize that those shadows get smaller as the altitudes of the birds or planes increase, and eventually their shadows disappear.

The second argument is that if the sun's rays are parallel (which is the assumption of the Eratosthenes experiment), then the moon's shadow should be the same regardless of the distance. This confuses the angular diameter of the sun with the average direction of solar rays..."
 
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