Total Solar Eclipse

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Jan 19, 2021
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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?
 

COLGeek

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Can't you provide a source for this assertion? I would suggest that we not recycle the previous thread's discussion. That conversation more than ran its course.

 
Jan 19, 2021
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The reason I started this thread again was because it was never concluded satisfactorily last time.
Regarding a source for this assertion, what assertion are you referring to? Do you mean the sun's rays being parallel or that the umbra would be the same diameter as the moon? Another member of the forum offered a link, which I cannot locate, to a diagram showing phases of the moon and the sun's rays as being parallel.

I have a signed copy of Patrick Moore's book 'TV Astronomer' and on page 20 you can see a typical illustration showing parallel sun rays.

Nearly all the the diagrams in text books show parallel sun rays. However when a diagram is shown of a total solar eclipse the sun is always shown to be very much larger and closer to the moon/earth. These sun rays are shown to be angular due to the false scale of the diagram thus creating an umbra smaller than the moon. I'm saying that parallel sun rays is what you would expect if the total solar eclipse diagram was drawn to scale and therefore the umbra would be the same size as the moon.
 

COLGeek

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Part of the issue, previously discussed, deals with spherical objects, moving through space in their various orbits/distances apart.

In that context, what does rays moving parallel mean? "Parallel" implies a 2D comparison/relationship, vice the 3D arrangement of these bodies that are in constant motion.
 
Jun 1, 2020
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Not all the rays are parallel. As in the other thread, if an object appears as a disk then how could they all be parallel? How could a glowing orb emit only parallel rays? The eye sees the Sun, and Moon, as a disk. The max. angle from center is ~ 1/4 deg. Any rays at an angle from the rays emitted from the center are no longer parallel to those rays, right?

It may help to draw the eclipse to scale on paper. The angles will still be the same. The rays from either side of the Sun’s limb will cross over into the parallel-ray only umbra, thus producing a penumbra.
 
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Interestingly, since the Sun is a ball of gas, we see deeper into its photosphere when looking at the central portion of the solar disk (~ 200 km). But at the limb, that same depth is at an angle so we only see the upper part of the photosphere.

Looking at the center of the disk, we are seeing light from a deeper and hotter region. The central zone is 6390K and the limb is only 5000K. This is called the CLV (Center to Limb Variation), which involves more than observed temperatures.

So.... since radiation is a 4th power law, then the central zone emits ~ 2.5x more radiation than the limb. That means, finally, that more sunlight is close to parallel than one might expect.
 
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I am not an astronomy enthusiast to level that the members of this forum are. I am simply asking what I consider to be a reasonable and logical question. Therefore in response to the above replies I have produced a drawing to explain my question.

In drawing 1 I have shown that the Sun's rays radiate out from its centre in all directions. I have calculated the angle of the ray from the centre of the Sun to the diameter of the moon = 0.533deg. This angle is small and that is why I believe the textbook illustrations refer to the Sun's rays being considered parallel when reaching the Earth.

In drawing 2 I have drawn the angle to scale. If you then continue to project this angle from the Moon to the Earth then it would produce a slightly larger umbra than the diameter of the moon. To me this is just a mathematical exercise. If there is something wrong with the calculations and resulting the extrapolation then I will be very happy to stand corrected.

https://www.dropbox.com/s/i6jan96mkt4z985/Sun rays.jpg?dl=0
 

rod

Oct 22, 2019
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Interesting diagram in post #7. Starting with the Sun ray emerging from the center and traveling to the Moon, the ray expands to 0.533 degrees. What is the width of the solar ray in cm at the center of the Sun when the ray departs, what is the rate of width expansion per cm as the ray travels away from the Sun, and how many cm traveled to the Moon?

I am using c.g.s. units.
 
Jan 19, 2021
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I don't understand what you are asking for!
The drawing simply shows that a Sun ray being emitted from the centre of the Sun traveling 93million miles to eventually cover the moon's diameter (3,475 miles) would subtend an angle of 0.533deg.
Therefore the Moon's umbra projected at this angle from the Moon onto the Earth's surface would have to be slightly larger that the diameter of the Moon.
 

rod

Oct 22, 2019
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In post #9, nothing discloses the original diameter of the Sun ray when it departs from the center of the Sun. The Sun ray will expand in diameter as distance traveled from the Sun. So I use c.g.s. units to define here. You would need to know the initial diameter of the Sun ray in cm and apply a dispersion metric to that ray as it travels a specific number of cm to the Moon. Nothing in your post discloses this. Consider that a Sun ray with 0.533 degree width has a specific cm size for a given distance. The starting value in sun ray size is much smaller so we need a cm to cm conversion metric here and this will disclose the distance traveled from the Sun too. Here is an example from my stargazing experience. I use mini maglites when observing. I can shine that light onto the ground by my telescope and see a small circle covered but if I shine that same light across my fields to the horse barn, that light diameter is now much large in area and size.
 

rod

Oct 22, 2019
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Here is something more to consider. A sun ray that travels 1 au and is 0.533 degree across, that diameter is about 1919 arcsecond size. 0.533 deg at 1 au converts to a diameter close to 1.4E+6 km in size, not the size of the Moon.
 
Jan 19, 2021
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I do not understand why you are unable to grasp a simple piece of trigonometry?
Will it make any difference to you if the ray of light started from the surface of the Sun?
I cannot see why the light ray's starting diameter has anything to do with it.

What is so difficult about understanding a simple diagram showing what kind of umbra the Sun would generate on the Earth from the moon? This would be the same for any light source on any object. It's shadow will depend on the size of the light source, the size of object, the distance between the light, the object and the shadow surface. I can't believe that I have to explain it to you like this!

If I didn't know better I would say that you were winding me up!
 
Jan 19, 2021
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I do not understand why you are unable to grasp a simple piece of trigonometry?
Will it make any difference to you if the ray of light started from the surface of the Sun?
I cannot see why the light ray's starting diameter has anything to do with it.

What is so difficult about understanding a simple diagram showing what kind of umbra the Sun would generate on the Earth from the moon? This would be the same for any light source on any object. It's shadow will depend on the size of the light source, the size of object, the distance between the light, the object and the shadow surface. I can't believe that I have to explain it to you like this!

If I didn't know better I would say that you were winding me up!
Please ignore this reply. I will resubmit a new drawing.
 
Jan 19, 2021
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My apologies for the incorrect drawing previously submitted and the confusion it caused.. I hope that this new drawing will explain better.
The size of the umbra in a total solar eclipse would be just slightly less than the diameter of the moon.
 

rod

Oct 22, 2019
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Okay, we have a new diagram here in post #14 :) Consider this. https://www.timeanddate.com/eclipse/umbra-shadow.html, "How Large Is the Moon's Umbra? The size of the area on the Earth's surface covered by the Moon's umbra during a total solar eclipse depends, amongst other things, on the Moon's current distance from Earth. The smaller the distance, the larger the umbra. If the Moon is at its closest to Earth (its perigee) during the eclipse, the Moon appears larger in the sky. In that case, the umbra's path across the Earth's surface typically has a width of roughly 150 km (90 mi) at the Earth's equator. At higher latitudes, the Sun's rays hit the Earth's surface at a shallower angle, so the umbra's size grows accordingly. During some total solar eclipses, the umbra's path width reaches over 1000 km (600 mi) at the poles. If the eclipse occurs when the Moon's distance is greater, the tip of the Moon's cone-shaped umbra (see illustration) may only just reach the Earth's surface during parts of the eclipse, meaning that its diameter is close to zero..."

This means the umbra size during a total solar eclipse has a min and max size. Easy to convert into arcminutes here. Using 384401 km for mean lunar distance, then we have an umbra varying from about 1.3 arcminute in size on Earth to close to 9 arcminute size on Earth. The diagram showing umbra size 0.527 degrees is clearly wrong. The diagram shows an umbra > 31 arcminute size. In order for the umbra to be > 31 arcminute angular size on Earth, the Moon must also be very, very close to Earth too :) In astronomy, we do not observe umbra angular sizes like this on Earth during total solar eclipses.
 
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COLGeek

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For the sake of us all, re-hashing the previous thread is not beneficial to the community. I ask that something new be added to the conversation or this will be closed.

Previously, no amount of explanation sufficed to address Pies' questions and this new thread seems to be taking a similar course already.
 
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rod

Oct 22, 2019
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https://wiki.tfes.org/Moon, if folks look at this Moon wiki from Flat Earth Society, the Moon is about 32 miles in diameter and about 3,000 miles away or 5,000 km. Using those metrics you will get Pies values for the umbra size on Earth for a total solar eclipse :) Pies in post #14 said "The size of the umbra in a total solar eclipse would be just slightly less than the diameter of the moon."

This works using the FES Moon wiki dimensions and distances :)
 

COLGeek

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https://wiki.tfes.org/Moon, if folks look at this Moon wiki from Flat Earth Society, the Moon is about 32 miles in diameter and about 3,000 miles away or 5,000 km. Using those metrics you will get Pies values for the umbra size on Earth for a total solar eclipse :)
Interesting information. Pure nonsense and nothing of scientific value, but interesting in how anyone with a rational/logical bone in their body could believe such delusional hogwash.
 
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rod

Oct 22, 2019
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Ref. COLGeek post in #18. We can resize the entire solar system using FES wiki for Moon and Sun and distances. All the planets including Jupiter become much smaller than 32 miles across, much closer too and the Earth becomes the largest body, perhaps in the entire universe too, think 2D flat disk giant size :) Changing distances, shrinking physical size in diameters, and working out angular resolution sizes can be much fun :)
 

COLGeek

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Apr 3, 2020
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Ref. COLGeek post in #18. We can resize the entire solar system using FES wiki for Moon and Sun and distances. All the planets including Jupiter become much smaller than 32 miles across, much closer too and the Earth becomes the largest body, perhaps in the entire universe too, think 2D flat disk giant size :) Changing distances, shrinking physical size in diameters, and working out angular resolution sizes can be much fun :)
Understood. A fun imaginary drill based on a flawed premise. :p
 
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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 :)
 

rod

Oct 22, 2019
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FYI, an observation from me. I have two old books. The How and Why Wonder Book of the Moon dated 1963 and Navigation and Nautical Astronomy by Dutton, 1926 and used thru 1951 at the Annapolis Naval Academy. Star navigation, stellar parallax, eclipses, converting lat and long into correct distances on a spheroid Earth, etc. It is amazing to see so much science changed by Internet groups today, somehow these older books got the distance to the Moon from Earth and physical size all wrong but the Internet and social media is here to save the day :)
 
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rod

Oct 22, 2019
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FYI. Here is a quote from the 1926-1951 Naval Academy book I cited in post #22. I think COLGeek will enjoy :) "864. Physcial characteristics of the moon. - When the Army, early in 1946, announced that it had made radar contact with the moon, interest in the possibility of some day making a trip to the moon increased...", page 275-276. Go Army! radar measurements tell us much about the distance between Earth and the Moon :)
 
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Jan 19, 2021
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I have never mentioned or need to refer to the FES. It is quite disengenuous to suggest such as an argument against me. You speak of being rational - that is all I'm trying to be.

Thanks for the link to the solar eclipse illustration which, by its NTS nature, gives a very exaggerated illustration of what happens during a total solar eclipse.

My attached new drawing shows the figures I used for my calculations of the Moon's umbra. I don't have a problem with a penumbra.

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.

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.

 
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