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

The thread was a violation of the rules from the start, you're lucky I haven't already deleted it for being a callout thread. You have yet to address my math.

This is all obfuscation bullshit bro, go and debunk my concentric circles proof, or stfu!

This is all obfuscation bullshit bro, go and debunk my concentric circles proof, or stfu!

Hey genius, that video is irrelevant to my proof because the video does not take into account the rotational angular speed of the Earth and the angular speed on the moon.

You claimed that my video does not account for the different angular motions of the earth's rotation vs the moon's orbit. It absolutely does. From the source code:
"moon_theta -=math.radians(0.04)
observer_theta -=math.radians(1.2)"
The moon is orbiting at an angular rate 30 times slower than the earth's rotation in my diagram, rounded up from the difference between the synodic period of the moon's orbit vs the rotation of the earth. Where is your retraction?

This is all obfuscation bullshit bro, go and debunk my concentric circles proof, or stfu!

Already did. It's not obfuscation, it's math that's over your stupid fucking head. Concentric circles? Bahahahahahahahaha! On the left is a diagram of how the shadow actually travels faster than a point on the rotating surface of earth (red square) even though it takes longer for the moon (green square) to complete one orbit of earth than it takes for the earth to rotate one time. On the right is how a dipshit like you thinks this geometry works, which would require the moon's shadow to always point towards the earth as it orbits. That's not how it works.

Here's another one I can't seem to wrap my mind around- The accepted diameter of the Earth and moon are 7,917 and 2,158 miles respectively; if the shadow cast by the moon during a total solar eclipse is between 60-70 miles across on the surface of the Earth, WHY does the umbra of the Earth encompass the moon in its entirety during a total lunar eclipse?

Still waiting on that retraction, or are you a dishonest piece of shit?

Using the same math for the umbral shadow of the earth at the distance of the moon as I used for the umbral shadow of the moon at the distance of the earth I find that the moon's shadow is about 9200 km diameter at the distance to the moon during the 2017 eclipse. That's greater than the 3474 km diameter of the moon. So yes, total lunar eclipses are possible. You can see it all laid out in my spreadsheet.
[link to drive.google.com (secure)]

I think both of these illustrations/animations are incorrect- in the first or 'proper' one the light source seems to be emanating from a general direction, there is no 'vanishing point' (the sun).

If this model were correct, it doesn't seem that we would ever see 'rays' coming through the clouds, all leading to the Sun-

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.

if your program accounts for it then something is wrong with your program bwahahahahahahaha.

Since you are such an expert on this, and have programmed a simulation, explain what is flawed in my proof. Here is my proof again:

Draw two concentric circles. The inner circle represents the earth's surface circle; think of it as the equator circle. And the outer circle is the moons path. And think of it as looking down at it from the north pole direction. Now draw a dot outside both of the circles to represent the sun. Now draw a line that connects the center of the two concentric circles with the dot that represents the sun. That line will intersect the Earth circle where the shadow begins and the outer circle where the moon is initially located. Now from that initial line draw two more lines that both go threw the center of the concentric circles with one line 0.6 degrees counter clockwise from the initial line and the other that is 15 degrees counter clockwise away from the initial line. At the beginning of the eclipse, the shadow of the moon is where the initial line that you drew intersects the Earth's circle as I have previously said. In one hour, that point on the Earth's surface has traveled to where the 15 degree line intersects the Earth's circle, while the moon is located where the 0.6 degree line intersects the moon's orbit circle. Now draw a line that goes through where the 0.6 degree line intersects the moon's orbit circle and the sun, and continue that line down to the Earth's surface. Notice that that line intersects the Earth's circle to the West of the place on the Earth's circle where the shadow started. Therefore the shadow has traveled from East to West.

You made a false claim about my program, I provided the source code and proved you wrong. You need to retract your claim. If something is wrong with my program, then prove it. The source code is right here:
[link to drive.google.com (secure)]
[link to drive.google.com (secure)]


In your head you're picturing it like the diagram on the right where the shadow is west of the red dot, but it's actually the diagram on the left which is correct. The shadow is traveling east of the red dot faster than the red dot rotates. The velocity of the moon is faster than the velocity of earth's rotation, this falls out of the simulation even though the sim is based purely on the ANGULAR rates of the two because the orbit of the moon is to-scale with the size of the earth in my sim.

Well, to be clear, it is ALL perspective as we are observing the phenomena via our eyeballs.

Astro can't handle the truth.

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.