Calling Astro...Please come in and explain how this works???..It's freaking me out.....

That's a fucking lie. I already demonstrated the actual velocities that will be achieved by the shadow due to the moon's orbital velocity. I already calculated the exact path of the shadow and showed the speed it should have.

Forget all of your software crap,

and handing over your brains to a machine,

isn't it obvious that the moon's shadow on the Earth due solely to the moon's motion will travel at the rate of the moon's angular speed as that angular speed appears on Earth? This is the crux I think of where the debate is.

if you think it does not travel at this speed then present a your proof that it does not, and what speed do you think it travel at?

In simple terms, the moon's orbital velocity is greater than the earth's rotational velocity. On average, the moon's orbital velocity is about 3679 km/hr, whereas the earth's rotational velocity is only about 1675 km/hr. That's a difference of about 2000 km/hr. Due to foreshortening effects, the apparent velocity of the moon's shadow across the earth's surface can be even higher than that in terms of the ground-relative velocity on the surface of the earth.

I developed a spreadsheet for calculating the position of the moon relative to earth some time ago, using a book published decades ago:

I later expanded this to include calculations for the position of the sun as well, using equations that date back to the famous astronomer Simon Newcomb in the 19th century. Using this spreadsheet I've now worked out the math for calculating the approximate coordinates of the umbral shadow of the moon on the surface of the earth:
[link to dropcanvas.com]
The approximate longitude and latitude of the shadow are given in cells G20 and G21. For calculating the shadow's position, earth was treated as a perfect sphere with a radius of 6371 km, but it's good enough for our purposes here (the topocentric coordinates of the sun and moon as seen by the observer position given in the input section does take into account the oblate shape of the earth).

I calculated the approximate shadow coordinates at half hour intervals during the August 21st eclipse and loaded the resulting latitude and longitude into google earth:
[link to drive.google.com (secure)]
From 17:00 UT to 19:00 UT at half hour intervals, here is approximately how far the shadow traveled, and as you can see it IS in a west to east direction:
17:00-17:30 2765 km
17:30-18:00 1585 km
18:00-18:30 1349 km
18:30-19:00 1378 km
That's an average 2698 km/hr velocity at the shortest distance traveled from 18:00-18:30 UT. This is due to foreshortening as well as the fact that the shadow is not traveling along the equator, so the earth's rotational velocity is a bit lower across the region the shadow is traveling. You can check the spreadsheet for yourself, it took weeks to create the original moon calculations spreadsheet but it's all there. No cheating or conspiracy involved. And you can see that the moon's own latitude and longitude position over the earth is moving westward as the eclipse progresses, just as expected.

"But what is this," you say? How can the moon's position over the earth be moving from east to west yet its shadow be moving from west to east? Well you have to realize that the distance the moon has to travel in one orbit around the earth is a lot greater than the distance any point on the earth travels in one rotation per day. In that sense, the day is of course much shorter than one orbit of the moon. But that's not the question one is answering when one asks what the path of the moon's shadow will be during the eclipse; the shadow is traveling along the surface of the earth and its ground-relative velocity is generally faster than the rotational velocity of the earth.

This is why when looking at total solar eclipse paths, whether the prediction was generated by NASA, myself, or some long dead astronomer from over a hundred years ago, you'll see they progress in a west to east direction:
[link to eclipse-maps.com]
Note that you can see this month's eclipse predicted on the above map, published in 1887. The open triangle denotes the start of the eclipse, the filled triangle denotes the end.

Are you blind, are you not reading my posts, or are you just too fucking stupid to understand it? Which is it?


How many truly gifted people do you know that use this kind of language? Or need to?

Especially if they are not agenda driven, and only come here to help clarify things for the lesser minded in his supposed field of expertise....

No, it's not obvious. It will NOT "travel at the rate of the moon's angular speed as that angular speed appears on Earth?" That's just it, for that statement to be true, the line would have to extend from the center of the earth to the center of the moon as in the diagram on the right. The shadow would have to magically always point towards earth. What that line actually represents is the sub-moon point, the point on earth that the moon is directly above. That point IS moving from east to west constantly, which is why the moon can be seen moving through our sky from east to west. At the same time, the moon's shadow extends directly away from the sun, which in the diagram means it points from right to left constantly, regardless of where the earth is. The moon's shadow is therefore moving at the moon's orbital velocity and that is faster than the earth's tangential speed of its rotation.

So here is the issue; and stop posting that stupid video and cluttering everything up every time you respond bwahahahaha.

Sure, I understand what you are saying, and my claim that the speed of the shadow due solely to the moon's motion will be the angular speed of the moon is an approximation;

at one point in the eclipse, when the centers of the earth moon sun do line up, this will be exactly the case.

And this implies that in some portion (and maybe the entire portion) of the eclipse's duration, the shadow should travel from the East to the West because the angular speed of the Earth's rotation will be greater than the angular speed of the moon's orbit around the Earth.

Do you agree or disagree with that and why?

bro, you are posting a bunch of junk again. It appears you are trying to obfuscate.

You don't understand what I am asking, so here it is again.

In the scenario, such as the model I use for my proof, where the Earth Sun Moon are all on the same plane and the moon's shadow is traveling completely along the equator (can you use your mind to envision this scenario), at some point in the eclipse, the Earth Sun Moon centers will be perfectly lined up,

And this implies that in some portion (and maybe the entire portion) of the eclipse's duration, the shadow should travel from the East to the West because the angular speed of the Earth's rotation will be greater than the angular speed of the moon's orbit around the Earth.

Do you agree or disagree with that and why?

You keep wanting to force me into the video on the right, that is not where I am.

During that instance of the eclipse (in my specific scenario) when the three centers are aligned in the earth moon sun system, the component of the shadow's motion due solely to the moon's motion, will be moving at the rate of the moon's angular speed in the West to East direction.

Do you agree with that or not, and if not, why not?

Ok, confirmed, Astromut Fuk is a fraud.

Wrong. The sun is sufficiently far that the light rays arrive at the earth-moon system parallel.

Wrong. That is an effect of perspective, but the rays coming through the clouds leading back to the sun are in fact as parallel as the train tracks that seem to converge into the distance.

Ok, confirmed, Astromut Fuk is a fraud.

Nothing really matters
Anyone can see
Nothing really matters to me

So you have to be a troll, right?