Eclipse 21.08.17 will debunk the Globe

It's already been answered multiple times by myself and others on this thread. It's also clearly shown in the video linked in my first reply to this thread (watch the video for a real scale answer to your question if you don't like the carousel analogy).  That you and others like RIF cannot "phantom" it is of no consequence. The reality remains, the model makes sense to those that can visualise the scale and the difference between angular velocity/orbital speed, and my simple analogy of the carousel holds. Again, you may as well jump up and down challenging me to disprove 2+2=5. If you don't believe 2+2=4, nothing I can say will change your mind.

Ah, as expected !!
Your globe has made a 15 degree turn over the course of only ONE hour.
Do you understand the position of North America after ONE hour ?
It has moved 15 degrees of a full 360 degrees rotation.
The moon only has completed 0.55 degrees of it's orbit.


Your spinning globe becomes your enemy, because even a 15 degree rotation is way to much.....
Now either mock me with your 2+2=5 comments, or show me why my thoughts are wrong.

Quote from: Zammo on July 28, 2017, 01:42:32 AM

It's already been answered multiple times by myself and others on this thread. It's also clearly shown in the video linked in my first reply to this thread (watch the video for a real scale answer to your question if you don't like the carousel analogy).  That you and others like RIF cannot "phantom" it is of no consequence. The reality remains, the model makes sense to those that can visualise the scale and the difference between angular velocity/orbital speed, and my simple analogy of the carousel holds. Again, you may as well jump up and down challenging me to disprove 2+2=5. If you don't believe 2+2=4, nothing I can say will change your mind.

Ah, as expected !!
Your globe has made a 15 degree turn over the course of only ONE hour.
Do you understand the position of North America after ONE hour ?
It has moved 15 degrees of a full 360 degrees rotation.
The moon only has completed 0.55 degrees of it's orbit.


Your spinning globe becomes your enemy, because even a 15 degree rotation is way to much.....
Now either mock me with your 2+2=5 comments, or show me why my thoughts are wrong.

Umm...  I see you are serious.    That's  sad.   Why didn't  you multiply the angular velocity by the radius of the moons orbit?     ( hint: use radians not degrees)   

The speed that the shadow moves across the earth  is related to the orbital velocity,  not the angular velocity,  although the two are related  remember  l = r*theta.  so v=r*theta/sec   


EDIT: Rab,  explained it quite clearly as well as others have already done in this thread.

Quote from: Resistance.is.Futile

https://en.m.wikipedia.org/wiki/Orbit_of_the_Moon

So your claim that the Moon's velocity is greater than the Earth's rotation is irrelevant regarding the Solar Eclipse.

No, who said that? But you cannot compare velocity (a linear measure if unqualified) and rotation (an angular measure).

And this is the crux of the whole matter. At the latitude of the eclipse in August, the surface velocity of the earth is about 766 mph.
But the moon's radius will be about 231,700 miles on Aug 21 so, while it is rotating at a lower angular velocity, it's linear velocity, of about 2054 mph, is quite a lot faster than the earth's surface velocity.

So, the velocity of the moon's shadow across USA will be roughly 2054 - 766 mph (1288 mph) west to east.

dutchy plz: Look up angular velocity on the interwebs and try to wrap your head around why and how much the orbital velocity is affected by it for different diameters. That's grammar school stuff, really... 

Quote from: Zammo on July 28, 2017, 01:42:32 AM

It's already been answered multiple times by myself and others on this thread. It's also clearly shown in the video linked in my first reply to this thread (watch the video for a real scale answer to your question if you don't like the carousel analogy).  That you and others like RIF cannot "phantom" it is of no consequence. The reality remains, the model makes sense to those that can visualise the scale and the difference between angular velocity/orbital speed, and my simple analogy of the carousel holds. Again, you may as well jump up and down challenging me to disprove 2+2=5. If you don't believe 2+2=4, nothing I can say will change your mind.

Ah, as expected !!
Your globe has made a 15 degree turn over the course of only ONE hour.
Do you understand the position of North America after ONE hour ?
It has moved 15 degrees of a full 360 degrees rotation.

What distance does 15 degrees of the earth's rotation cause the ground to move? Try around 1280 km.

The moon only has completed 0.55 degrees of it's orbit.

What distance does 0.55 degrees of the moon's orbit cause the shadow to move? Try around 3700 km.

As we have been saying all along, the moon's shadow moves faster than the surface velocity of the earth.

Your spinning globe becomes your enemy, because even a 15 degree rotation is way to much.....
Now either mock me with your 2+2=5 comments, or show me why my thoughts are wrong.

Nope, our spinning globe does not becomes my enemy!

So, no problem at all - it happens just as NASA and all the references say.

Quote from: Zammo on July 28, 2017, 01:42:32 AM

It's already been answered multiple times by myself and others on this thread. It's also clearly shown in the video linked in my first reply to this thread (watch the video for a real scale answer to your question if you don't like the carousel analogy).  That you and others like RIF cannot "phantom" it is of no consequence. The reality remains, the model makes sense to those that can visualise the scale and the difference between angular velocity/orbital speed, and my simple analogy of the carousel holds. Again, you may as well jump up and down challenging me to disprove 2+2=5. If you don't believe 2+2=4, nothing I can say will change your mind.

Ah, as expected !!
Your globe has made a 15 degree turn over the course of only ONE hour.
Do you understand the position of North America after ONE hour ?
It has moved 15 degrees of a full 360 degrees rotation.
The moon only has completed 0.55 degrees of it's orbit.


Your spinning globe becomes your enemy, because even a 15 degree rotation is way to much.....
Now either mock me with your 2+2=5 comments, or show me why my thoughts are wrong.

As I've pointed out, it can be somewhat counter-intuitive to understand how comparing the angular velocities and surface velocities of 2 objects can provide very different results.

This is so funny.

These "physics experts" are throwing around angular velocity this surface velocity that and orbital velocity errors along with "your degrees and radians are mixed up."

What bologna!


These "experts" cant even point out the error in dutchys garden model.

Closest theyve come is this

Nope, our spinning globe does not becomes my enemy!

So, no problem at all - it happens just as NASA and all the references say.

Wow! Sounds like DENIAL to me. Not a qualified "physics expert"

They also continually badger people for not knowing the difference between angular and surface velocity, and instead of kindly explaining the differences, they simply call people "simpletons and flattards"

Here are some excerpts from the internet, explaining a bit of physics definitions.

In physics, the angular velocity of a body is the rate of change of its angular displacement with respect to time, and in three-dimensional space is a pseudovector quantity that specifies the rotational speed of an object and the orientation of the rotating.

A number of quantities in physics behave as pseudovectors rather than polar vectors, including magnetic field and angular velocity.

In physics and mathematics, a pseudovector (or axial vector) is a quantity that transforms like a vector under a proper rotation, but in three dimensions gains an additional sign flip under an improper rotation such as a reflection. Geometrically it is the opposite, of equal magnitude but in the opposite direction, of its mirror image.

https://en.m.wikipedia.org/wiki/Angular_velocity

https://en.m.wikipedia.org/wiki/Pseudovector

Look up angular velocity

That's grammar school stuff, really...

Wow! Sentinel your grammer school must have been really advanced.

Why dont you solve the N-Body problem for us all?

Y'know the one that calculates the movement of 3 celestial bodies!

Because its still unsolved, and youde do mankind a big favor in being able to calculate lightspeed without zooming into a planet.

Oh, if only space travel wasnt just a FAIRYTALE FROM THE MAGIC KINGDOM OF LIES AND DECEIPT.

Whats a space ship like anyway?

Heres a quote from my very favorite, clinically insane, hired by NASA anyway astronaut, Pete Conrad after flying his first "space" mission.

"It was like eight days in a garbage can." Conrad [The astronaut] remembers.

http://articles.latimes.com/1996-02-11/news/ls-35175_1_lunar-surface/2

Yes, space ships, akin to garbage cans...

This is the same guy who when asked for a stool sample, shat in a box and gave it to the doctor with a red ribbon on top.

I KID YOU NOT.

So keep flinging your poo, "physics experts" of this website. Youre obviously trying to get NASAs attention arent you, being as nutty and rude as you are.

They also continually badger people for not knowing the difference between angular and surface velocity, and instead of kindly explaining the differences, they simply call people "simpletons and flattards"

Simply put, angular velocity is how many degrees (or radians, if you prefer) a reference point rotates per unit time.  Surface (or linear) velocity is the distance a reference point point moves per unit time while rotating.

Do we really need to go through the math comparing the relative rotational speeds and distances traveled between the earth and moon yet again?

I will confess that this question tripped me up for a while.  It actually is pretty counterintuitive, and the explanatory videos are generally lacking.

It did all finally click for me (maybe it shouldn't have taken so long...) but even still I struggle with how to explain it clearly.  I can envision the visual aid that would make this make sense, perhaps I will try to draw it out later.

The critical thing to remember is that it's not a simple matter of the moon's shadow falling directly beneath it.  Relative to the observer, the shadow will be coming in at some angle, and this angle has to do with the position of the sun due to the earth's rotation (i.e. the time of day) as well as the position of the moon (which most seem to agree is moving Eastward).

It doesn't matter that the moon rises East and sets West, because the sun is ALSO setting in the West and therefore when they line up (1:30 pm for me), the shadow formed is "pointing" to me from somewhat to the West.  Neither will actually be over my head at that time.

I may try to work on a visual aid--I can see it in my head but none of the videos I've seen so far really show what I'm seeing.

Quote from: Sentinel on July 28, 2017, 05:37:22 AM

Look up angular velocity

That's grammar school stuff, really...

Wow! Sentinel your grammer school must have been really advanced.

Why dont you solve the N-Body problem for us all?

Y'know the one that calculates the movement of 3 celestial bodies!

Because its still unsolved, and youde do mankind a big favor in being able to calculate lightspeed without zooming into a planet.

I'm well aware of the N-body problem, but why do come up with that for the task at hand as it doesn't bear any significance for an solar eclipse when the three bodies involved are well within a defined, measured and predictable set of movement relative to each other?
Do you even know what that problem is about for real? 

Quote from: Zammo on July 28, 2017, 01:42:32 AM

It's already been answered multiple times by myself and others on this thread. It's also clearly shown in the video linked in my first reply to this thread (watch the video for a real scale answer to your question if you don't like the carousel analogy).  That you and others like RIF cannot "phantom" it is of no consequence. The reality remains, the model makes sense to those that can visualise the scale and the difference between angular velocity/orbital speed, and my simple analogy of the carousel holds. Again, you may as well jump up and down challenging me to disprove 2+2=5. If you don't believe 2+2=4, nothing I can say will change your mind.

Ah, as expected !!
Your globe has made a 15 degree turn over the course of only ONE hour.
Do you understand the position of North America after ONE hour ?
It has moved 15 degrees of a full 360 degrees rotation.
The moon only has completed 0.55 degrees of it's orbit.


Your spinning globe becomes your enemy, because even a 15 degree rotation is way to much.....
Now either mock me with your 2+2=5 cyomments, or show me why my thoughts are wrong.

Ah, as expected. You still don't get it. But that's ok. The moon rises in the East and sets in the West, and its shadow during a solar eclipse moves West to East as expected. Again, your inability to comprehend this does not change the fact, nor alter that it makes complete sense to those that can appreciate the difference between angular velocity and orbital/rotational linear velocities.

Thanks for the replies, but i am still waiting for my answer.
Now that could be totally on me and my invalid ways of expressing me.
From the start of the topic i can understand the differences be angular velocity and linear velocity, that was and is not the point.
I want to know how i can test it in a scale model as i have shown in a previous post.
What i don't understand is how this could work in your model.

the moon's umbra moves from west to east.
After 6 hours (moon's umbra generous duration due to it's slightly oval path  ?)
6 x 0.55° = 3.3° moon's completed part of it's orbit
6 x 15° = 90° earth's completed part of it's orbit

So when i start in my model globe-moon-sun  set up.....(earth 30cm diameter , moon 9,5m away, and 60m orbit around the miniature globe, sun 3.5km away)
The moon's umbra is pointed towards the pacific.
After 6 hours the earth has made a 90° turn.
The moon has moved 3.3° from it's orbit.

In my garden set up i would have to turn the globe and the pacific ocean is now placed 90° counterclockwise. (23,5 cm)
The moon would have been 55 cm further on it's 60m orbit (2,33 times faster)

what do i have to do to make sure the umbra is ahead of earth's spin after 6 hours (or insert the corrected lenght of the duration of the eclips when needed)
Because when i see the scale model and the sun 3.5 km away and the movent at the beginning and the end of the eclips in my garden model there is no way at all that the moon's umbra can travel west to east.
Please an answer in relation to my garden model  !!!!!!!!!!!!!!!!!!!!!!!!!

And please not a ''highschool related answer please''

How does my garden set up look like at the beginning of the eclipse
How does my set up look like after 6 hours.

If you fail to explain my scale model, don't bother to reply in scientific jargon.
I am specialised in music and related things and can explain everything to a non insider without using specific jargon that makes me look smarter than i am.
Explaining so that everyone understands it is the real task.

So please explain my scale set up, at the beginning and the end of the eclipse !!

Thanks for the replies, but i am still waiting for my answer.
Now that could be totally on me and my invalid ways of expressing me.
From the start of the topic i can understand the differences be angular velocity and linear velocity, that was and is not the point.
I want to know how i can test it in a scale model as i have shown in a previous post.
What i don't understand is how this could work in your model.

the moon's umbra moves from west to east.
After 6 hours (moon's umbra generous duration due to it's slightly oval path  ?)
6 x 0.55° = 3.3° moon's completed part of it's orbit
6 x 15° = 90° earth's completed part of it's orbit

I could not care less about your model - stick to real life.

Relative to the sun the moon takes roughly to 29 days to orbit the earth, so, ignoring the ellipticity, the moon orbits ³⁶⁰/_{(24 x 29)} = 0.517 degrees per hour.

The average distance to the moon is 385,000 km, so the moon's orbital velocital is ^{(385,000 x 0.517 x π})/₁₈₀ = 3476 km/hr.

The sun is so much further from the earth that the moon's shadow will move at almost the same velocity and in the same direction as the moon.

Now at latitude 40°N (near enough to the eclipse's path) the surface velocity of the earth is close enough to ^{(40000 x cos(40°))/24} = 1277 km/hr.

So the moon's shadow, moving across the earth at 3476 km/hr greatly outpaces the surface velocity of the earth of 1277 km/hr.

And this is what all the references, except that Flat Earth video, have been saying.

PS The moon velocity is not exactly the value calculated because the orbital eccentricity has been ignored.

Thanks for the replies, but i am still waiting for my answer.
Now that could be totally on me and my invalid ways of expressing me.
From the start of the topic i can understand the differences be angular velocity and linear velocity, that was and is not the point.
I want to know how i can test it in a scale model as i have shown in a previous post.
What i don't understand is how this could work in your model.

the moon's umbra moves from west to east.
After 6 hours (moon's umbra generous duration due to it's slightly oval path  ?)
6 x 0.55° = 3.3° moon's completed part of it's orbit
6 x 15° = 90° earth's completed part of it's orbit

So when i start in my model globe-moon-sun  set up.....(earth 30cm diameter , moon 9,5m away, and 60m orbit around the miniature globe, sun 3.5km away)
The moon's umbra is pointed towards the pacific.
After 6 hours the earth has made a 90° turn.
The moon has moved 3.3° from it's orbit.

In my garden set up i would have to turn the globe and the pacific ocean is now placed 90° counterclockwise. (23,5 cm)
The moon would have been 55 cm further on it's 60m orbit (2,33 times faster)

what do i have to do to make sure the umbra is ahead of earth's spin after 6 hours (or insert the corrected lenght of the duration of the eclips when needed)

6 hours is reasonably close to the duration of the eclipse. First penumbral contact is 15:46:51 UT and last is 21:04:23 UT, so it's 5h 17m 32s long.
 

Because when i see the scale model and the sun 3.5 km away and the movent at the beginning and the end of the eclipse in my garden model there is no way at all that the moon's umbra can travel west to east.
Please an answer in relation to my garden model  !!!!!!!!!!!!!!!!!!!!!!!!!

And please not a ''highschool related answer please''

How does my garden set up look like at the beginning of the eclipse

We'll center a 6-hour simulation near mid-eclipse, which is 18:25:32 UT (call it 18:30 UT).

If we look at a globe from south of it, then west on the globe corresponds to west in cardinal direction in the real world, so, in an effort to reduce confusion, let's presume your model is set up like this, based on your numbers, which look reasonable:

Sun is 3.5 km due south of a 30-cm diameter earth globe. Moon starts 9.5m south of the center of the globe and 27.5 cm west of the line between the centers of the globe and sun. Presuming the globe has the 23.5° axis tilt, and north up, orient the axis of he globe so that it is leaning mostly west and a some south. To be completely accurate, the model moon should also be a few cm north of the globe-sun line, but let's not worry about that.

Start time is 15:30 UT. If we presume the sun is on the meridian at 0° longitude at exactly 12:00 UT (close enough), then 3.5 hours later, at the start of our simulation, the sun is on the meridian at 52.5° W longitude (15°/hr * 3.5 hr = 52.5°). So rotate the globe so that French Guyana's meridian (on the north coast of South America) is directly toward the model sun, due south.

Since the sun is so far away, the axis of the moon's shadow will be, for all practical purposes, parallel to the north-south line through the centers of the globe and model sun. As such, the center of the shadow falls 27.5 cm west of the center of our 30-cm-diameter globe, passing nearest the Pacific Ocean.

Good so far?

Three hours later, at the time of greatest eclipse, move the moon 27.5 cm east so that it's due south of the earth and directly in line with the sun. Also rotate the globe 45° eastward so that 97.5° W longitude (off the west coast of westernmost Guatemala) is directly in line with the sun. In the time the earth was rotating 45° from west to east (French  Guyana, close to the equator, has rotated about 12 cm - half of the 23.5 cm you gave above for 6 hours - from west to east), the center of the shadow has moved 27.5 cm from west to east. Clearly, the shadow is moving to the east faster than locations on the surface of the earth are rotating in that direction, so the shadow moves across the surface from west to east.

Still good?

How does my set up look like after 6 hours.

At 21:30 UT, the moon has moved another 27.5 cm toward the east and the globe has rotated another 45° toward the east, so that longitude 142.5° W (somewhere out in the eastern Pacific) is facing the sun, and the moon's shadow is 27.5 cm east of the globe, passing nearest the central Atlantic Ocean, after having moved from west to east across the face of the globe.

Explaining so that everyone understands it is the real task.

So please explain my scale set up, at the beginning and the end of the eclipse !!

You're right that explaining sometimes counterintuitive stuff like this in an understandable way can be a challenge.

In recap, at your model's scale, the center of the shadow started west of the earth, traveled 55 cm from west to east across the globe, and ended east of the earth in the simulated 6 hours. In that time, the earth rotated 23.5 cm, also from west to east. Since the shadow was moving eastward at about twice the speed as points on the surface of the globe, the shadow moved across the face of the globe from west to east.

I'd like to try and break this down to simple terms the way I understand it. Hopefully it's not too high school.

Let's discuss what we will see from our perspective. We will see the moon rise in the east, like normal. Then the sun will rise in the east, also normal. Then, fairly early in the afternoon, the sun will "overtake" the moon. Since the moon will be directly between some of us and the sun, we get to witness an eclipse. Because of the relative size and distances of the sun and moon, the moon will "block out" the sun.  The sun will continue on, and set in the west, then the moon will follow.

The shadow moving from west to east seems to be the point of discussion, but is explained very easily.

On the RE model, this makes perfect sense. From our perspective, we won't be moving. The sun and moon will. Since the moon revolves in the same direction the earth rotates, it will appear to be traveling slower. They will both be traveling east to west, but with the sun passing (from east to west) behind the moon, the shadow will travel from west to east. It's that simple.  All of the math has already been explained, but even without the math, we can see this happen.

On the FE, the sun and moon revolve around the North Pole. The moon takes roughly an hour longer to make a complete circle, so it is traveling slower.  We won't be moving. The sun and moon will. They will both be traveling east to west, but with the sun passing (from east to west) behind the moon, the shadow will travel from west to east.

Same results from our perspective. This event doesn't prove either model. 

As further thought, it will be interesting to see some photographs from the Southern Hemisphere. I suspect the the angle of perspective will give credence to both theories as well. My guess is; the event will look like two similar sized objects passing close together, not too far from the earth's flat surface. I'm also sure that math will again be able to "prove" that they are vastly different sized objects, with vast distances between them and us.

Thanks for the replies, but i am still waiting for my answer.
Now that could be totally on me and my invalid ways of expressing me.
From the start of the topic i can understand the differences be angular velocity and linear velocity, that was and is not the point.
I want to know how i can test it in a scale model as i have shown in a previous post.

Well, you could try building your scale model and see what happens when you test it.

What i don't understand is how this could work in your model.

the moon's umbra moves from west to east.
After 6 hours (moon's umbra generous duration due to it's slightly oval path  ?)
6 x 0.55° = 3.3° moon's completed part of it's orbit
6 x 15° = 90° earth's completed part of it's orbit

In my garden set up i would have to turn the globe and the pacific ocean is now placed 90° counterclockwise. (23,5 cm)
The moon would have been 55 cm further on it's 60m orbit (2,33 times faster)

By Jove, I think that you might be getting it.

what do i have to do to make sure the umbra is ahead of earth's spin after 6 hours (or insert the corrected lenght of the duration of the eclips when needed)
Because when i see the scale model and the sun 3.5 km away and the movent at the beginning and the end of the eclips in my garden model there is no way at all that the moon's umbra can travel west to east.
Please an answer in relation to my garden model  !!!!!!!!!!!!!!!!!!!!!!!!!

And please not a ''highschool related answer please''

How does my garden set up look like at the beginning of the eclipse
How does my set up look like after 6 hours.

Draw a line from your earth to your sun.

At the moment of eclipse totality, the earth, moon and sun should all be on that same line and the umbra of the moon's shadow should be centered on the earth.

Three hours before that alignment, the moon should be about 1.65 degrees (27.5 cm) west of that center line moving east.  Since the sun is very far away and 1.65 degrees is a fairly small angle, the moon's umbra can considered to be pretty much parallel to that center line for the next 6 hours.

I hope that you can figure out the rest from there.

I figured it out. Wasn't very hard but I need a shape to explain it. When I get back home hopefully I'm gonna make one.

How will this Eclipse be possible on your Heliocentric model?

The producer in this video gives a detailed description of how and why the upcoming solar Eclipse is impossibe on the Heliocentric Globe Model.

Let's try again.
You ask "How will this Eclipse be possible on your Heliocentric model?"
Maybe you can only understand pictures, so try this.

Quote from: Resistance.is.Futile on July 26, 2017, 02:51:59 PM

How will this Eclipse be possible on your Heliocentric model?

The producer in this video gives a detailed description of how and why the upcoming solar Eclipse is impossibe on the Heliocentric Globe Model.

Let's try again.
You ask "How will this Eclipse be possible on your Heliocentric model?"
Maybe you can only understand pictures, so try this.

Flat Earth: Eclipse Special, Sly Sparkane

Gotta love that video. Clear and concise. Money shot for West to East shadow starting from 4:40. Chile sees eclipse twice on flat earth model. 😂😂😂

If they don't get it now one must assume they are intentionally dishonest and lying.
Fingers crossed though at least one of them comes to reason and admit she/he was wrong. Be brave.

Quote from: Arealhumanbeing on July 28, 2017, 10:41:00 AM

Quote from: Sentinel on July 28, 2017, 05:37:22 AM

Look up angular velocity

That's grammar school stuff, really...

Wow! Sentinel your grammer school must have been really advanced.

Why dont you solve the N-Body problem for us all?

Y'know the one that calculates the movement of 3 celestial bodies!

Because its still unsolved, and youde do mankind a big favor in being able to calculate lightspeed without zooming into a planet.

I'm well aware of the N-body problem, but why do come up with that for the task at hand as it doesn't bear any significance for an solar eclipse when the three bodies involved are well within a defined, measured and predictable set of movement relative to each other?
Do you even know what that problem is about for real? 

Still no answer from Arealhumanmoron. Interesting.

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

While you're pondering, perhaps you should look into how FET would explain a solar eclipse with as much detail as you've been given about RET solar eclipses.

How will this Eclipse be possible on your Heliocentric model ?

The producer in this video gives a detailed description of how and why the upcoming solar Eclipse is impossibe on the Heliocentric Globe Model.

Your Strange Heliocentric Religion is False.

That's right... you believe something by some unknown person whose credibility is non-existent.  You do that.

Tell me, who is the real sheep?

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

So you obviously looked up angular velocity on the interwebs as suggested by me?
I'm so proud of you, keep it up. 

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

Good on you Dutchy. It was a slog, but we got there. Lots of things "work" in the current heliocentric model. Like, all of them.

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

The time from first to last penumbral contact is 5 hours 18 minutes (beginning and end of the partial phase from anywhere on the surface of the earth). The time from first to last umbral contact is 3 hours  14 minutes (beginning and end of totality from anywhere on the surface).

Times from this summary: <large GIF image>. [Linking instead of embedding because it takes a lot of space on the page.]

Six hours was a convenient, while not unreasonable, time span to consider the model. Where did four hours come from?

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

I learned a lot in the process, so thanks for asking!

Even Jarenism gets it.

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

The Eclipse of 21.08.17 is expected to have a shadow width 68 miles as in this quote. Quote But even when the moon is at the closest point to Earth in its elliptical orbit, the umbra is but a dark "dot" measuring no more than 170 miles (274 km) across. And many times, the shadow width is considerably smaller than that. During the Aug. 21 total solar eclipse, for example, the shadow will average only about 68 miles (109 km) wide. From: How Long Will the 2017 Solar Eclipse Last? Now, I wonder if you could explain how a moon about 32 miles in diameter could cast a 68 mile diameter shadow from a 32 mile diameter sun. From what I can see the geometry would be something like the diagram on the right. It would seem to that there is no possibly way for the umbra for exceed the the size of the moon. And it's not just the 68 mile width of shadow for this eclipse, but eclipses can have the umbra up to 170 miles (274 km) across.

So, please put as much thought into the weaknesses in your flat earth model as you put into trying to refute the Globe.

Quote from: dutchy on July 31, 2017, 12:14:20 PM

Thanks to those taking the time to explain the eclips from a heliocentric perspective.
Am i convinced ? Not really, but i understand now that the eclips ''works'' in the current heliocentric model, althaugh it has still many questions that i am getting into. The excessive 6 hours timespan instead of four, the shape and size of the shadow etc.

But since i did not consider eclipses important, i really did not look into the phenomena before this topic started.

Again thanks for putting in time and effort to explain it to me !!!!

The Eclipse of 21.08.17 is expected to have a shadow width 68 miles as in this quote. Quote But even when the moon is at the closest point to Earth in its elliptical orbit, the umbra is but a dark "dot" measuring no more than 170 miles (274 km) across. And many times, the shadow width is considerably smaller than that. During the Aug. 21 total solar eclipse, for example, the shadow will average only about 68 miles (109 km) wide. From: How Long Will the 2017 Solar Eclipse Last? Now, I wonder if you could explain how a moon about 32 miles in diameter could cast a 68 mile diameter shadow from a 32 mile diameter sun. From what I can see the geometry would be something like the diagram on the right. It would seem to that there is no possibly way for the umbra for exceed the the size of the moon. And it's not just the 68 mile width of shadow for this eclipse, but eclipses can have the umbra up to 170 miles (274 km) across.

So, please put as much thought into the weaknesses in your flat earth model as you put into trying to refute the Globe.

Another question is how an observer within the umbra could witness the sun's corona when the sun and moon are the same size and relatively close together?  It looks like annular eclipses are pretty much out of the question.

Even Jarenism gets it.

He seems to be one of the brighter bulbs out there, no doubt.
The abuse he receives in the comments by the hardliners though...