ZIGZAG of the Full Moon - Irrefutable argument against the rotation of the earth

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cikljamas

  • 2061
  • Ex nihilo nihil fit
At the edge of the Arctic circle there are many occasions in December (and even in June) when we can observe the moon (and measure the speed of it's motion) throughout substantial part of the day. There are even occasions when the moon at the edge of the Arctic circle doesn't set at all (throughout the day).

For example in a few days (9th June) the moon is going to rise above Fairbanks-Alaska (64,5 degr. N) at 23 h 38 min., and the moon is going to be visible above Fairbanks continually for more than 17 hours (the moon sets at 5 h 24 min. PM next day : 10th June)...



All we have to do is to measure the speed of the FULL moon above Fairbanks in these two periods of the day :

-- 5 minutes around midnight
-- 5 minutes around Noon the next day

If there is no substantial difference in the speed of the FULL moon in these periods the earth is at rest (there is no rotation of the earth)!!!

The thing is very simple :

HC maniacs claim that the moon travels in earth's orbit in direction WEST-EAST (which is an opposite direction of it's "apparent" motion in the sky !?!?!?).

The speed of moon's motion (according to HC maniacs) is about 3600 km/h.

Now, our measuring of the speed of the FULL moon around midnight will show SUBSTANTIAL difference in comparison with the speed of the FULL moon around Noon the next day IF THE EARTH ROTATES on it's axis.

-- 5 minutes around midnight (when the earth allegedly rotates in the same direction with respect to the moon's alleged orbital WEST-EAST motion) the "apparent" speed of the moon will be :

3600 km/h (speed of the moon) - 800 km/h (rotational speed of the earth at 64,5 degr. N) = 2800 km/h

- 5 minutes around Noon the next day (when the earth allegedly rotates in the counter direction of the moon's alleged orbital WEST-EAST motion) the "apparent" speed of the moon will be :

3600 km/h (speed of the moon) + 800 km/h (rotational speed of the earth at 64,5 degr. N) = 4400 km/h

THE FINAL RESULT WOULD BE :

Around the midnight the moon would apparently move MUCH FASTER towards WEST since the apparent speed of it's "real" EASTWARD motion would be SUBSTANTIALLY slower (in relation to the result of measurement which would be carried out around Noon).

Around Noon the moon would apparently move MUCH SLOWER towards WEST since the apparent speed of it's "real" EASTWARD motion would be SUBSTANTIALLY faster (in relation to the result of measurement which would be carried out around midnight previous day).

Let me see any kind of a sane attempt of refutation of this argument...

IN ADDITION (just a reminder) :

“The Moon presented a special math problem for the construction of the heliocentricity model. The only way to make the Moon fit in with the other assumptions was to reverse its direction from that of what everyone who has ever lived has seen it go. The math model couldn’t just stop the Moon like it did the Sun, that wouldn’t work. And it couldn’t let it continue to go East to West as we see it go, either at the same speed or at a different speed. The only option was to reverse its observed East to West direction and change its speed from about 64,000 miles an hour to about 2,200 miles an hour. This reversal along with the change in speed were unavoidable assumptions that needed to be adopted if the model was to have a chance of mimicking reality." -Bernard Brauer
« Last Edit: June 06, 2017, 04:21:22 AM by cikljamas »
"I can't breathe" George Floyd RIP

Just like all your other crappy arguments, this one fails just as hard.

STOP USING LINEAR VELOCITIES!!!
That is not what is observed. Instead, use angular velocities.

Due to the distance from Earth to the moon, it will be pretty much the same.
Earth is rotating at approximately 15 degrees an hour. Thus you can use that as a starting point.
Just accepting your value of the speed of the moon, at 3600 km/hr, at a distance of roughly 400 000 km.
Assuming the path is circular, you can relate the angular velocity to the tangential speed by:
v=omega*r, where typically omega is in radians per second, but as velocity is in km/hr, it would be radians per km.

That means we have the relation:
3600 km/hr=omega*400000 km
Thus omega=(36/4000) radians per hour=0.009 radians per hour=0.5 degrees an hour (note: this means it takes roughly 30 days for an orbit, so that math checks out).

And that means the moon should appear to move at a rate of roughly 14.5 degrees an hour.

But of course, this doesn't let us figure out the difference, and it gets more complicated when dealing with the barycentre and the actual path, so I will simplify to make the math easier.
The easiest way to do this is consider the moon rotating about the point close to it and the point furthest away.

So that means instead of r=400 000 km, we have r=400 000 + 6371 km and r=400 000 - 6371 km, and we then find the difference between them.

If you do the math on that, the difference works out to be 0.016 degrees per hour, so for your 5 minute observation, the difference is 0.0014 degrees out of 1.25 degrees, or roughly 0.1%.

So no, no massive difference is expected.

But yet again, what about on a flat Earth?

well in this case the moon is MUCH closer.

When it is directly overhead (I know, it doesn't get this extreme) it is a mere 5000 km away moving at 1600 km/hr. When it is on the other side of the world, it is over 20 000 km away, yet moving at the same speed.

So once again, this is actually a good argument AGAINST the flat Earth.

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rabinoz

  • 26280
  • Real Earth Believer

It's really no mystery or problem!
Even old Hipparchus, back around 150 BC understood it well enough to make a pretty good calculation of the earth to moon distance!
These old Greeks were really pretty smart, see Hipparchus of Nicaea, Motion of the Moon

The earth rotates once in 24 hours (with respect to the sun) in an anti-clockwise direction when looking down on the North Pole.
And the Moon orbits  the earth in about 29.5 days (again with respect to the sun)  in an anti-clockwise direction when looking down on the North Pole.

So, to us, it appears that the moon goes around us in a clockwise direction when looking down on the North Pole, but on each revolution, the earth has to catch up to it a little more, so the apparent period is about 24 hours plus 40 minutes.

One or other of these "simulators" might help follow it. The distances earth-sun and earth-moon are, of course, not to scale.
          University of Nebraska, Lunar Phase Simulator
and
          University of Nebraska, Three Views Simulator.
In each case you can make a few selections, the start amimation.

Hope it helps.


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cikljamas

  • 2061
  • Ex nihilo nihil fit
4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!
"I can't breathe" George Floyd RIP

4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!

go to a school and learn how to calculate speeds of objects in circular motion.

the funniest think about your calculation here was:
you have 2 numbers with a certain ratio
than you divide both numbers with a constant value.
and you are have to point out that the ratio is the same.

that shows me what your level of math knowledge is.

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cikljamas

  • 2061
  • Ex nihilo nihil fit
4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!

go to a school and learn how to calculate speeds of objects in circular motion.

the funniest think about your calculation here was:
you have 2 numbers with a certain ratio
than you divide both numbers with a constant value.
and you are have to point out that the ratio is the same.

that shows me what your level of math knowledge is.

I don't have to be a great mathematician (like you) in order to prove that the earth is at rest.
But you have to be a great magician (like JackBlack) in order to prove that the earth is in motion.
You noticed well that it was kind of funny (from mathematical point of view, especially from magician's point of view) pointing out something so trivial, however, mathematical procedures (especially magician's procedures) demand clarity, especially if you are chronically drunk (as most of you (HC maniacs) are)...
« Last Edit: June 06, 2017, 09:00:44 AM by cikljamas »
"I can't breathe" George Floyd RIP

At the edge of the Arctic circle there are many occasions in December (and even in June) when we can observe the moon (and measure the speed of it's motion) throughout substantial part of the day. There are even occasions when the moon at the edge of the Arctic circle doesn't set at all (throughout the day).

For example in a few days (9th June) the moon is going to rise above Fairbanks-Alaska (64,5 degr. N) at 23 h 38 min., and the moon is going to be visible above Fairbanks continually for more than 17 hours (the moon sets at 5 h 24 min. PM next day : 10th June)...



That's not what that table says. It indeed has moonrise at 23:38 on June 9th, but moonset will be at 05:24 on the 10th. That's 5:24 AM the next day; 5:24 PM would be 17:24. The moon will be up continuously for only 5h 46m the night of June 9/10 from Fairbanks.

Oops...

If you want to see a full moon up for a long from Fairbanks, you need to find the one closest to the December solstice. Do you know why?

Quote
<flawed argument refuted by JackBlack>

IN ADDITION (just a reminder) :

“The Moon presented a special math problem for the construction of the heliocentricity model. The only way to make the Moon fit in with the other assumptions was to reverse its direction from that of what everyone who has ever lived has seen it go. The math model couldn’t just stop the Moon like it did the Sun, that wouldn’t work. And it couldn’t let it continue to go East to West as we see it go, either at the same speed or at a different speed. The only option was to reverse its observed East to West direction and change its speed from about 64,000 miles an hour to about 2,200 miles an hour. This reversal along with the change in speed were unavoidable assumptions that needed to be adopted if the model was to have a chance of mimicking reality." -Bernard Brauer

Who is Bernard Brauer? He seems as confused about this as you are. You need to find more reliable sources for information.

Yet another "irrefutable" argument easily refuted.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!

go to a school and learn how to calculate speeds of objects in circular motion.

the funniest think about your calculation here was:
you have 2 numbers with a certain ratio
than you divide both numbers with a constant value.
and you are have to point out that the ratio is the same.

that shows me what your level of math knowledge is.

I don't have to be a great mathematician (like you) in order to prove that the earth is at rest.
But you have to be a great magician (like JackBlack) in order to prove that the earth is in motion.
You noticed well that it was kind of funny (from mathematical point of view, especially from magician's point of view) pointing out something so trivial, however, mathematical procedures (especially magician's procedures) demand clarity, especially if you are chronically drunk (as most of you (HC maniacs) are)...

you are simply wrong.
first you did the math wrong
and second, the earth is spinning. it is proven with lots of experiments and even proven visually by astronauts.

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cikljamas

  • 2061
  • Ex nihilo nihil fit
That's not what that table says. It indeed has moonrise at 23:38 on June 9th, but moonset will be at 05:24 on the 10th. That's 5:24 AM the next day; 5:24 PM would be 17:24. The moon will be up continuously for only 5h 46m the night of June 9/10 from Fairbanks.

Oops...

If you want to see a full moon up for a long from Fairbanks, you need to find the one closest to the December solstice. Do you know why?


O.K. (it happens, but don't worry, my argument still holds), then we are simpoly going to wait 6 months (until December - i shouldn't have pointed out such a trivial thing, again ...lol), my argument will still be verifiable and valid :




IN ADDITION (just a reminder) :

“The Moon presented a special math problem for the construction of the heliocentricity model. The only way to make the Moon fit in with the other assumptions was to reverse its direction from that of what everyone who has ever lived has seen it go. The math model couldn’t just stop the Moon like it did the Sun, that wouldn’t work. And it couldn’t let it continue to go East to West as we see it go, either at the same speed or at a different speed. The only option was to reverse its observed East to West direction and change its speed from about 64,000 miles an hour to about 2,200 miles an hour. This reversal along with the change in speed were unavoidable assumptions that needed to be adopted if the model was to have a chance of mimicking reality." -Bernard Brauer

Who is Bernard Brauer? He seems as confused about this as you are. You need to find more reliable sources for information.

Yet another "irrefutable" argument easily refuted.
This argument is not a refutable argument, since this is an IRREFUTABLE argument (i shouldn't have pointed out such a trivial thing, again...lol)
« Last Edit: June 06, 2017, 09:38:50 AM by cikljamas »
"I can't breathe" George Floyd RIP

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sokarul

  • 17483
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What FE model can explain your "argument"?
Sokarul

ANNIHILATOR OF  SHIFTER

Run Sandokhan run

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cikljamas

  • 2061
  • Ex nihilo nihil fit
What FE model can explain your "argument"?

This argument proves that the earth is at rest (no matter what shape of the earth turns out to be), this is going to be the best time for measuring non-existent speed's differences of moon's "apparent" westward motion within one polar arctic day (from Fairbanks) :



 
« Last Edit: June 06, 2017, 09:57:54 AM by cikljamas »
"I can't breathe" George Floyd RIP

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sokarul

  • 17483
  • Discount Chemist
So no?
Do you understand the moon can be up all day in the RET?

Were you going to edit the OP as to hide your inability to tell time?
Sokarul

ANNIHILATOR OF  SHIFTER

Run Sandokhan run

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cikljamas

  • 2061
  • Ex nihilo nihil fit
So no?
Do you understand the moon can be up all day in the RET?

Were you going to edit the OP as to hide your inability to tell time?

Is that all you have to say against validity of my irrefutable argument?

No, i won't edit my OP, since it was a minor error (which means : an error of no big importance - oops, i shouldn't have pointed out such trivial thing, again...lol). Why it was a minor error? Because i was right when i said that there are many occasions when we can observe the moon from the edge of the Arctic circle quite substantial amount of time (most of the day), and sometimes even ALL DAY LONG...and i corroborated that truth with additional moonrise/moonset tables for Fairbanks.
"I can't breathe" George Floyd RIP

So no?
Do you understand the moon can be up all day in the RET?

Were you going to edit the OP as to hide your inability to tell time?

Is that all you have to say against validity of my irrefutable argument?

No, i won't edit my OP, since it was a minor error (which means : an error of no big importance - oops, i shouldn't have pointed out such trivial thing, again...lol). Why it was a minor error? Because i was right when i said that there are many occasions when we can observe the moon from the edge of the Arctic circle quite substantial amount of time (most of the day), and sometimes even ALL DAY LONG...and i corroborated that truth with additional moonrise/moonset tables for Fairbanks.
I'm not understanding how seeing the moon all day is an argument against a round earth

What FE model can explain your "argument"?

This argument proves that the earth is at rest (no matter what shape of the earth turns out to be), this is going to be the best time for measuring non-existent speed's differences of moon's "apparent" westward motion within one polar arctic day (from Fairbanks) :

Table showing full moon up all day in December, 2021 from Fairbanks

I think we all agree that there won't be significant change in the moon's rate of motion across the sky as it crosses the meridian and as it passes the opposite side of the sky. You seem to think this is evidence that the earth is stationary, but it isn't. Your calculations are wrong.

If the huge effect (57%) you think exists near the arctic circle were real, it would be even more pronounced at mid to low latitudes. We would notice the moon's apparent motion vary significantly as the moon rises, crosses the meridian, and sets. Instead, the moon moves at an almost constant rate across the sky the whole time it's up, varying by a small amount due to parallax, accumulating up to about 2° over a period of about 12.5 hours, "speeding up" the moon the most, by a maximum of roughly 0.25°/hr, as it crosses the meridian viewed from the equator, and approaching zero as it rises and sets. Meanwhile, the rotation of the earth produces an apparent motion of 15°/hr, and the moon's orbital motion reduces that by about 1/2°/hr, so the maximum effect of parallax is almost 60 times smaller than the combination of orbital motion and earth's rotation, around 1.5%. That's going to be hard to notice.

Your "irrefutable" argument is anything but irrefutable. It's based on an error.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!

Why completley ignore everything I said?

I am not the one muddying the water, you have no idea how to calculate the moon's apparent speed.

Lets try it your way shall we?
You have 2 linear speeds.
What does this translate into in terms of ANGULAR speed?

Well, the moon is roughly 400 000 km away, with a max speed of 4400 km/hr.

That means the angular speed is roughly 4400/400000 radians per hour.
That works out to be 0.6 degrees per hour. But the moon appears to moves much faster than that, so you aren't even getting close to a working understanding.

Like I said, the important part is the ANGULAR velocity.
This is especially important as you aren't going up to the moon with a tape measure to measure how fast it is going. You are observing it from far away and observing how it appears to move through the sky, changing angle over time.

That is what causes the apparent motion of most things, the rotation of Earth.
What you are looking at is the residual after that angular velocity is taken away. But like I showed above, for a simple analysis, the difference is roughly 0.1%, not the BS you claim.

What you are basically doing is assuming Earth isn't spinning spouting a bunch of crap, and then concluding Earth isn't spinning, so effectively just going in circles.

You are the one muddying the waters here with your blatant misrepresentation of the facts.

Did you want a full mathematical derivation for the simple case of the moon orbiting around Earth's exact centre?
If not, can you point out anything wrong with my above claims? Or will I have to provide the full one either way?


And just like your other zig zag argument, it is entirely irrelavent if it is the moon moving and Earth stationary or Earth spinning with the moon moving, or the Earth moving with the moon stationary.

Instead, it is entirely to do with their path and distance.

THIS IS NOT AN ARGUMENT AGAINST A MOVING EARTH.
THIS IS AN ARGUMENT AGAINST A CLOSE MOON.

Again, a simple example:
Say the moon is orbiting over the tropic of cancer (which it does at times), in a perfect circle.
Now say we are viewing the moon from a position on the edge of the Arctic circle.

So lets start with the FE view.
For simplicity, I will ignore the angle of elevation as that will just change its apparent height and instead I will focus entirely on its apparent azimuthal velocity.
So you (the observer) are situation at a point on the arctic circle, some 23.43703 degrees from the North pole and thus roughly 2600 km.
Meanwhile, the moon is circling above the tropic, 66.56297 degrees from the north pole, and thus roughly 7400 km from it.
As the moon is following a path that has a radius of 7400 km, that puts its circumference at ~46000 km.
That puts its speed (nothing that it takes roughly 1 day to do a circle) at ~ 2000 km/hr.
This speed will be constant (as Earth isn't spinning in this retarded model).
Now then, when it is on its close path, it is a mere 4800 km away.
This puts its angular speed at 23 degrees per hour. So over your 5 minutes it would appear to move 5.8 degrees.
When it is furthest away it is 10000 km away. That puts its angular speed at 11 degrees per hour so in the 5 minutes it would appear to move 2.8 degrees.

So in your FE system with a close moon, it would be over double the rate.

The exact same applies if the moon is stationary and Earth is spinning or if Earth is spinning and the moon is moving.

Now lets compare that to the RE system.
Well, I already did that above.
You get a difference of 0.1%.

More importantly, it doesn't matter which is moving.

The math works out the same if Earth is stationary and the moon is moving around it, the moon is stationary with Earth spinning, or with both moving. All 3 are equivalent for the purposes of this "experiment"

The only thing this can distinguish between is a close moon or a far moon, and it indicates a far moon.

I don't have to be a great mathematician (like you) in order to prove that the earth is at rest.
No, you just need to have no idea what you are talking about and spout a bunch of shit which proves you know nothing about the topic at hand or are intentionally trying to deceive people.

But you have to be a great magician (like JackBlack) in order to prove that the earth is in motion.
No, it doesn't take any great deal of math or magic.
All it takes is a simple understanding of how relative motion (especially including circular paths) works. Something you either have no idea how to figure out or are blatantly misrepresenting.

You noticed well that it was kind of funny (from mathematical point of view, especially from magician's point of view) pointing out something so trivial, however, mathematical procedures (especially magician's procedures) demand clarity, especially if you are chronically drunk (as most of you (HC maniacs) are)...
We aren't the drunk ones.

What clarity do you need?

This argument is not a refutable argument, since this is an IRREFUTABLE argument (i shouldn't have pointed out such a trivial thing, again...lol)
Except it is refutable, as I did in the very first response.

Like I said before, stop dealing with BS linear speeds and instead focus on angular speeds.
Draw the system to show what is happening, or do you need me to do that for you?

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rabinoz

  • 26280
  • Real Earth Believer
At the edge of the Arctic circle there are many occasions in December (and even in June) when we can observe the moon (and measure the speed of it's motion) throughout substantial part of the day. There are even occasions when the moon at the edge of the Arctic circle doesn't set at all (throughout the day).
Rubbish!
Quote from: cikljamas
For example in a few days (9th June) the moon is going to rise above Fairbanks-Alaska (64,5 degr. N) at 23 h 38 min., and the moon is going to be visible above Fairbanks continually for more than 17 hours (the moon sets at 5 h 24 min. PM next day : 10th June)...
Really? Check again! the moon sets at 5 h 24 min AM next day: 10th June.
On June 9, 2107 the moon rises at 23:38 hrs (11:38 PM) and sets on 05:24 hrs at June 10, 2107 (5:24 AM).
Look at:

Fairbanks Alaska, Moon on 2017/06/10 at 05:25 hrs
So at midnight June 9, 2107 the moon is above the horizon, but sets again in only 5 hours and 24 minutes!
And look at the moon position at midday on June 10:

Fairbanks Alaska, Moon on 2017/06/10 at 12:00 hrs

Notice the notation "Position Moon under horizon"! Guess what that means.

But, I would like to know just how you calculate the moon's velocity from this stuff anyway, please explain!

As usual, your posts are so wrong that it is almost impossible to know how to attack them - like your stupid YoyTube videos.

Please run away and don't come back with stupid garbage like this!

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rabinoz

  • 26280
  • Real Earth Believer
What FE model can explain your "argument"?

This argument proves that the earth is at rest (no matter what shape of the earth turns out to be), this is going to be the best time for measuring non-existent speed's differences of moon's "apparent" westward motion within one polar arctic day (from Fairbanks) :

Still just as much total rubbish!

At the edge of the Arctic circle there are many occasions in December (and even in June) when we can observe the moon (and measure the speed of it's motion) throughout substantial part of the day. There are even occasions when the moon at the edge of the Arctic circle doesn't set at all (throughout the day).
Rubbish!
Quote from: cikljamas
For example in a few days (9th June) the moon is going to rise above Fairbanks-Alaska (64,5 degr. N) at 23 h 38 min., and the moon is going to be visible above Fairbanks continually for more than 17 hours (the moon sets at 5 h 24 min. PM next day : 10th June)...
Really? Check again! the moon sets at 5 h 24 min AM next day: 10th June.
On June 9, 2107 the moon rises at 23:38 hrs (11:38 PM) and sets on 05:24 hrs at June 10, 2107 (5:24 AM).

He already recognized that particular oops and presented the table for December '17 and December '21 in response. In the latter, the moon must be near its maximum extent north of the ecliptic while nearly full, because it is above the horizon for several days running.

His argument is still wrong, but at least he willingly acknowledged that error. That's better than you'll get from the likes of sandokhan and many others.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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rabinoz

  • 26280
  • Real Earth Believer
What FE model can explain your "argument"?

This argument proves that the earth is at rest (no matter what shape of the earth turns out to be), this is going to be the best time for measuring non-existent speed's differences of moon's "apparent" westward motion within one polar arctic day (from Fairbanks) :
You claim that "this argument proves that the earth is at rest". It does nothing of the sort.
To an observer on earth it makes not the slightest difference whether
          the moon orbits a stationary earth clockwise in 24 hrs and 50.5 mins,
          the earth rotates under a stationary moon once on 24 hrs and 50.5 mins or
          as actually happens, the earth rotates anti-clockwise once in 24 hours (relative to the sun)
                     and the moon orbits the earth anti-clockwise once in 29.5 days (relative to the sun).

There is no "'apparent westward motion"! At Fairbanks, the moon simply appears to circle around, but always appearing to move clockwise.

This site Astronomical Applications Dept, U.S. Naval Observatory will give the Altitude and Azimuth of the Moon for any city in the USA by name.

Here are the results for 2 hourly intervals for Fairbanks, Alaska on Dec 19, 2021.
                                                         
            Alt         Azm                   
hr:mn                (E of N)
00:00       49.5°       159.9°
02:00       49.5°       200.9°
04:00       41.7°       237.2°
06:00       30.0°       266.4°
08:00       17.9°       291.9°
10:00        7.9°       316.5°
12:00        1.8°       341.9°
14:00        0.9°         7.8°
16:00        4.9°        33.4°
18:00       13.6°        58.2°
20:00       25.2°        83.0°
22:00       37.4°       110.2°
00:00       47.3°       143.4°
         
And here are the compass points, if you can't work them out:

The moon obviously appears to circle around clockwise.

Here is the "side" of the earth-moon system, to scale first, as you always screw up scale in your zig-zag arguments!

Moon-Earth System - 24 hour moon Alaska
Done to scale like this, each part is too small to see properly, so here is the earth and moon separately.

Moon-Earth System - Earth

With the earth's tilt and the moon near the Northern extremity of its inclined orbit,
it can be seen above the horizon for a number of complete days.

It is expected to be at only 0.9° above the horizon at 2:00 PM on Dec 19, 2021.

These diagrams show the situation for midday, solar time,
when the moon would be just above the horizon in Fairbanks.



<------------------------- Around 384,000 km centre to centre distance --------------------->

Moon-Earth System - Moon
If you now imagine the earth rotating anti-clockwise about its slanting axis, to an observer at Fairbanks the moon would appear to orbit clockwise ( North -> NE -> E -> SE -> S -> SW -> W -> NW -> and back to -> N).

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cikljamas

  • 2061
  • Ex nihilo nihil fit
4400/2800 = 1,57
4400/60*5min. = 366,6
2800/60*5min. = 233,3
366,6/233,3 = 1,57 (also)
The difference between 4400 & 2800 (and between 366,6 & 233,3) = 46 %

The moon "apparently" moves across the sky (WESTWARDS) 0,5 degrees per 2 minutes. Moon's angular diameter = 0,5 degrees... So, the moon "apparently" moves westwards 1 angular diameter (0,5 degrees) per 2 minutes.

Since the difference between the speeds in above two different scenarios is 46 %, then we have to assume that the same directly proportional difference would be noticeable regarding the amount of the displacement of moon's angular diameter (as measured throughout certain period of time - let's say 5 min.), if above depicted different scenarios would really reflect what happens in our reality.

If such scenarios really existed, the difference of 46 % (regarding westward displacement of moon's angular diameter) would be quite substantial and very noticeable even with the naked eyes.

So, JackBlack, stop muddying the water!

Why completley ignore everything I said?

I am not the one muddying the water, you have no idea how to calculate the moon's apparent speed.

Lets try it your way shall we?
You have 2 linear speeds.
What does this translate into in terms of ANGULAR speed?

Well, the moon is roughly 400 000 km away, with a max speed of 4400 km/hr.

That means the angular speed is roughly 4400/400000 radians per hour.
That works out to be 0.6 degrees per hour. But the moon appears to moves much faster than that, so you aren't even getting close to a working understanding.

Like I said, the important part is the ANGULAR velocity.
This is especially important as you aren't going up to the moon with a tape measure to measure how fast it is going. You are observing it from far away and observing how it appears to move through the sky, changing angle over time.

That is what causes the apparent motion of most things, the rotation of Earth.
What you are looking at is the residual after that angular velocity is taken away. But like I showed above, for a simple analysis, the difference is roughly 0.1%, not the BS you claim.

Diameter of the earth at 65 degrees N (Fairbanks) = 5550 km
5550 * 3,14 = 17427/24 = 726/60 = 12,1 * 5 = 60 km
So, the earth rotates 60 km/5min
17427/60 = 290,45
360/290 = 1,24 degrees/5min (or 0,021 rad/5min)

The moon moves in it's orbit around the earth 0,041 degrees/5min

MIDNIGHT SCENARIO:

The full moon is above Fairbanks and Fairbanks is perfectly aligned with the moon and with the sun. The moon is right in front of us, the sun is right behind us.

The moon moves 0,041 degr. to the LEFT in it's huge orbit around the earth.
The earth moves 1,24 degr. to the LEFT in within much more smaller (but still giant) Arctic circle.

The consequence : "Apparent" motion of the moon = 2,5 apparent diameters (1,25 degrees) of the moon to the RIGHT!

What would happen if the earth moved 1,25 degrees to the RIGHT (instead of to the LEFT)?

"Apparent" motion of the moon would be equal to 2,5 apparent diameters of the moon to the LEFT.

If the earth were a spinning globe that is exactly what would happen 12 hours later (within NOON scenario).

So, how at NOON we can see the moon going to the RIGHT at all, since we move to the RIGHT, and the moon allegedly moves to the LEFT (although for negligible 0,041 degrees)?

Now, let's see how ANGULAR MOTION is of "great importance" within 5 minutes of our hypothetical observation of the moon's motion (around NOON and around MIDNIGHT) :

60 KM = 37,5 miles
37,5^2 ) 1406.25*8 = 11250/12/5280 = 0,17 miles/4 = 0,0425 miles = 0,068 km
Now, let's shrink our Arctic circle to one giant (but comprehensible) marry go round :
5550/100000 = 0,05 km = 50 m
Now, let's shrink our bulge in the middle of our 60 km to one super small (but comprehensible) length :
68m / 100000 = 0,68mm

So, if you imagine that we are spinning at the edge of one 50 m (in diameter) merry go round, during our circular motion which lasts 5 minutes we have to claim up 0,68mm (for the first 2,5 minutes of our circular motion), and then we have to descend 0,68mm (for the second 2,5 minutes long circular motion)!

So, shouldn't we consider such circular motion (for all intents and purposes) as a linear motion within this 5 minutes of observation of moon's motion?

If you have any doubts about that just watch this image and use your imagination while enlarging this circle to the certain (adequate to our case) extent :


"I can't breathe" George Floyd RIP

Diameter of the earth at 65 degrees N (Fairbanks) = 5550 km
5550 * 3,14 = 17427/24 = 726/60 = 12,1 * 5 = 60 km
So, the earth rotates 60 km/5min
17427/60 = 290,45
360/290 = 1,24 degrees/5min (or 0,021 rad/5min)
Why do you bother with any of those distances?

It is quite simple:
Earth rotates 360 degrees per 24 hours (technically slightly less, but irrelevant for this argument).
Thus its angular velocity=360 degrees/24 hours=15 degrees/hour.=1.25 degrees per 5 minutes.

You can do the same for the moon, with it having at orbital period of 29.53 days, and thus an angular velocity of 360degrees/29.53 days=0.042 degrees per 5 minutes.


The consequence : "Apparent" motion of the moon = 2,5 apparent diameters (1,25 degrees) of the moon to the RIGHT!
Yes, the moon apparently moves that small amount.

But you need to define what is left and right.
Again, left and right are linear terms. What you really have are clockwise (CW) and counter clockwise (CCW). They describe angular terms.

Lets say we are looking from above the north pole.
The Earth and the moon are both orbiting CCW.
The rotation of Earth will make everything appear to move CW, against the rotation of Earth, which makes the sun appear to rise in the east and set in the west, apparently circling around Earth in a clockwise direction.
This applies regardless of where you are and will make things appear to circle clockwise around the NP. That means if you are facing the object (and it is further out and you are in the north), it will appear to move to the right, all the time, regardless of if it is mid day or mid night or any other time.

So no, the moon won't reverse direction like you claim, just like the sun wouldn't.

You can also confirm this with a merry go round, or even a spinning chair, or a camera on a spinning disc.

What you are looking for is a slight change in the apparent angular speed, but you can't just use the numbers above to figure it out.

So, how at NOON we can see the moon going to the RIGHT at all, since we move to the RIGHT, and the moon allegedly moves to the LEFT (although for negligible 0,041 degrees)?
Exactly as explained above. Forget about the moon moving at all due to how slow it is. The main cause of the apparent motion is Earth's rotation. That makes it continue to appear to move against Earth's rotation, as is always observed, with it rising in the east and setting in the west.

Now, let's see how ANGULAR MOTION is of "great importance" within 5 minutes of our hypothetical observation of the moon's motion (around NOON and around MIDNIGHT) :
60 KM = 37,5 miles
37,5^2 ) 1406.25*8 = 11250/12/5280 = 0,17 miles/4 = 0,0425 miles = 0,068 km
Now, let's shrink our Arctic circle to one giant (but comprehensible) marry go round :
5550/100000 = 0,05 km = 50 m
Now, let's shrink our bulge in the middle of our 60 km to one super small (but comprehensible) length :
68m / 100000 = 0,68mm
So, if you imagine that we are spinning at the edge of one 50 m (in diameter) merry go round, during our circular motion which lasts 5 minutes we have to claim up 0,68mm (for the first 2,5 minutes of our circular motion), and then we have to descend 0,68mm (for the second 2,5 minutes long circular motion)!
So, shouldn't we consider such circular motion (for all intents and purposes) as a linear motion within this 5 minutes of observation of moon's motion?
If you have any doubts about that just watch this image and use your imagination while enlarging this circle to the certain (adequate to our case) extent :
No. It should be considered as angular motion, for the very reason you are pointing out. The distance moved is basically nothing. That is also the big problem with your zig-zag argument, the distance moved is nothing compared to how far away these objects is. The dominating factor is our turning.

Yes, for us, it is only that mere 0.68 mm, but what is the effect of that on the moon?
You are shrinking it by a factor of 100 000.
Well the moon is 400 000 km away.
That means the moon in your scaled down version is 4 km away.
Doing it the "common" way for those without great math skills, that puts its orbital circumference at 25 km.
That means 1 degree of that is roughly 7 m.

That is why the angular velocity is important, due to how little we are moving.

A better way to understand is to spin on the spot.
In this case your linear motion due to rotation is NOTHING.
So according to your reasoning, this means we should use that, treating it as linear motion.
And that should mean that everything appears to remain in the exact same position
But it doesn't. Instead, a slight turn can make far away objects appear to move a massive amount.

It isn't the linear motion or translation that makes it a big issue, it is that you are now looking in a different direction.
If you turn 1 degree, everything appears to turn 1 degree.

It is only with much closer objects does this start to fail.

*

cikljamas

  • 2061
  • Ex nihilo nihil fit
No. It should be considered as angular motion, for the very reason you are pointing out. The distance moved is basically nothing. That is also the big problem with your zig-zag argument, the distance moved is nothing compared to how far away these objects is. The dominating factor is our turning.

Yes, for us, it is only that mere 0.68 mm, but what is the effect of that on the moon?
You are shrinking it by a factor of 100 000.
Well the moon is 400 000 km away.
That means the moon in your scaled down version is 4 km away.
Doing it the "common" way for those without great math skills, that puts its orbital circumference at 25 km.
That means 1 degree of that is roughly 7 m.

That is why the angular velocity is important, due to how little we are moving.

So, while 4km distant moon (in our scaled down model) moved 7m to the LEFT we moved (on our merry go round) 0,5 meter to the LEFT (also), and expecting result of our 0,5 m long motion should be apparent translation of our moon to the RIGHT for 2,5 apparent moon's diameters. While we are traveling these 0,5 m long trip we have to surmount 0,68mm bulge (which is less than 1 fucking mm which is 25th part of one fucking inch), and such lateral motion you are still ready to call "circular motion" (due to the fact that we have to overcome a whole 25th part of one fucking inch by moving along 0,5 meters (out of 157 m of total circumference) long path of our 50m (in diameter) big merry go round)?

A better way to understand is to spin on the spot.

In this case your linear motion due to rotation is NOTHING.
So according to your reasoning, this means we should use that, treating it as linear motion.
And that should mean that everything appears to remain in the exact same position
But it doesn't. Instead, a slight turn can make far away objects appear to move a massive amount.

It isn't the linear motion or translation that makes it a big issue, it is that you are now looking in a different direction.
If you turn 1 degree, everything appears to turn 1 degree.

Try to verify this : move 0,5 m to the LEFT and see if 4 km distant object (with proportional dimensions to moon's alleged dimensions) would apparently translate 2,5 apparent moon's diameters to the RIGHT! You are perfectly aware that such experiment would prove that the moon is not so far away!!!

But even if we could achieve such result with our scaled down model, you would still have to verify THE CRUCIAL part of my argument : you should move to the RIGHT (Noon scenario) and see in which direction 2,5 apparent moon's diameters would move...


It is only with much closer objects does this start to fail.





"I can't breathe" George Floyd RIP

Enjoy yourselves, folks. I'm going to be on the road for a while and probably won't check in very often, if at all. See y'all in a few weeks!
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

No. It should be considered as angular motion, for the very reason you are pointing out. The distance moved is basically nothing. That is also the big problem with your zig-zag argument, the distance moved is nothing compared to how far away these objects is. The dominating factor is our turning.

Yes, for us, it is only that mere 0.68 mm, but what is the effect of that on the moon?
You are shrinking it by a factor of 100 000.
Well the moon is 400 000 km away.
That means the moon in your scaled down version is 4 km away.
Doing it the "common" way for those without great math skills, that puts its orbital circumference at 25 km.
That means 1 degree of that is roughly 7 m.

That is why the angular velocity is important, due to how little we are moving.
So, while 4km distant moon (in our scaled down model) moved 7m to the LEFT we moved (on our merry go round) 0,5 meter to the LEFT (also), and expecting result of our 0,5 m long motion should be apparent translation of our moon to the RIGHT for 2,5 apparent moon's diameters. While we are traveling these 0,5 m long trip we have to surmount 0,68mm bulge (which is less than 1 fucking mm which is 25th part of one fucking inch), and such lateral motion you are still ready to call "circular motion" (due to the fact that we have to overcome a whole 25th part of one fucking inch by moving along 0,5 meters (out of 157 m of total circumference) long path of our 50m (in diameter) big merry go round)?
No, that is not what I said at all.
However I did notice a mistake. That would have been 70 m.
The moon, in its 4km orbit, is moving at a rate of roughly 0.04 degrees per 5 minutes remember? The 70 m was for 1 degree, not even the 1.25 degrees.

Completely ignoring the rotation and instead just focusing on the translation, firstly, as you pointed out it can be approximated as a tangent and forget the in and out part.
Now then, we have our 50 m merry go round. I don't really care how accurate that is, I will leave it at that.
In 5 minutes we move 1.25 degrees around it. That means we would have moved ~ 1m. (closer to 1.09).
The moon, with its 4 km orbit, is moving 2.79 m.

That means the relative motion would be a maximum of 3.7 m and a minimum of 1.7 m (ignoring direction), and that is over 4 km.

If you were looking at buildings 4 km away, and just steps a few m to the left or right, do they appear to move? No.
That is because of just how small that is.
Using the maximum of 3.7 m, the angular displacement that causes is 0.05 degrees, for the 1.7 it is 0.02 degrees.

This linear motion is not causing the apparent motion of the moon.
Do you know what is? YOUR ROTATION. That is why it is important.
The motion, just due to translation would be a mere 0.05 degrees, at most.

But at the same time you have rotated 1.25 degrees.
You turning 1.25 degrees will make everything appear to move 1.25 degrees in the opposite direction.

Okay?
The translational motion of you and the moon combined will make it appear to move no more than 0.05 degrees. But your rotation will cause it to appear to move 1.25 degrees.

To put in another way:
You have moved 1.09 m (left or right, I don't give a shit).
The moon has moved 2.79 m, to the left.
But you are now looking at spot that is 70 m to the left of the moon.
What this means is that for your current view, the moon is roughly 70 m (if you like, again using simplified math to analyse it, you can say somewhere between 66.3 m and 68.3 m) to the right of where you are looking at.
That means the moon will appear to move to the right.


That is the primary thing which makes all celestial objects (i.e. everything except Earth) appear to move. The primary exceptions are close by meteors and artificial satellites if you wish to include them.


A better way to understand is to spin on the spot.

In this case your linear motion due to rotation is NOTHING.
So according to your reasoning, this means we should use that, treating it as linear motion.
And that should mean that everything appears to remain in the exact same position
But it doesn't. Instead, a slight turn can make far away objects appear to move a massive amount.

It isn't the linear motion or translation that makes it a big issue, it is that you are now looking in a different direction.
If you turn 1 degree, everything appears to turn 1 degree.

Try to verify this : move 0,5 m to the LEFT and see if 4 km distant object (with proportional dimensions to moon's alleged dimensions) would apparently translate 2,5 apparent moon's diameters to the RIGHT! You are perfectly aware that such experiment would prove that the moon is not so far away!!!
No, you are just helping to back me up.
This is exactly what I have said before, your little pathetic step to the left or right will have basically no impact on the apparent direction of this far away object.
But turning left or right will have a massive one.

So go and do it.
Go view a distant object (4km) and step even a few m to the left or right.
See if its position appears to change much.
Now, turn 1.25 degrees to the left or right and see what happens?

But even if we could achieve such result with our scaled down model, you would still have to verify THE CRUCIAL part of my argument : you should move to the RIGHT (Noon scenario) and see in which direction 2,5 apparent moon's diameters would move...
No, that is the most crucial flaw in your argument.
Moving to the left or right will have virtually no impact.
What will matter is the turning part.

Here is a simple experiment you can set up:
Get a turntable, where you can accurately control the angle.
Now if you like, for the moon, get either a much larger turn table, to mount the moon on (but make sure the small one can still turn independently or in a compensated way), or for the simplified version, just get a small linear track that you can move it along, or just move it yourself.

Now mount a camera on the turntable.
First, just as a control, mount in on the centre, then in a second run, mount it close to the moon (simulating the moon at closest point, so midnight for the full moon), then mount it far away
Now, set it in motion (or just take a before and after).
Notice what happens in all 3 cases?
The moon appears to move to the right.

Here is an actual scale drawing (using the equator to further emphasise the point):
https://cad.onshape.com/documents/896878716739e463043c15c5/w/738a02ab853e0a649e11fead/e/14f4cbf0e0fa6707e220bb0b

The one on the right is the initial condition, with the person close to the moon looking due south and the person far away looking due north, so both straight at the moon.
The one on the left is the final condition, after the 5 minutes. The moon is further along its orbit (to the left).
The person close to the moon is further along Earth (to the left), but the moon has moved an apparent 1.23 degrees to the right.
The person further away from the moon is further along Earth (to the right), but the moon has moved an apparent 1.19 degrees to the right.
Happy now?

This is one with the FE distances from before (with the moon turning 1.21 degrees to complete roughly 1 circle a day but go through a cycle once a month):
https://cad.onshape.com/documents/600d80bb7ae333fdbadf44fa/w/3bf7ef5d3de7e0031c3460d1/e/bce03ddfbf36eeab307c6195

Again, the one in the middle is the initial.
The one of the left is after 5 minutes.
Now, it is just the moon moving along its path (to the right).
For the person close to it, the moon has appeared to move 1.87 degrees to the right.
For the person far from it, the moon has appeared to move 0.90 degrees to the right.
See how it is much worse for the close moon.

As a bonus comparison, there is another one on the right.
This has this much smaller system rotating just like the real one.
So, the moon is further along its path (0.04 degrees, to the left).
The person close to the moon has rotated with Earth, to the left, but the moon appears 1.87 degrees to the right.
The person further away from the moon has rotated with Earth, to the right, and the moon appears 0.90 degrees to the right.

Notice something quite significant?
These angles are exactly the same as the small stationary Earth, only moon moving case.
This is because IT DOESN'T MATTER WHICH IS MOVING! ALL THAT MATTERS IS THE RELATIVE MOTION!
And that relative motion is the same with a stationary Earth and moving moon, a stationary moon and rotating Earth, or a rotating Earth with a moving moon (as long as they are matched).

So this can NEVER be an argument against a moving Earth.
All this is an argument for is the distance to the moon.

If the moon was close, like in the FE situation with it circling above the tropics, then over the 5 minute observation, at mid night you would expect an angular displacement of 1.87 degrees and at mid day it would be 0.90 degrees. That is over double for mid night.
If the moon was distant (like in reality), with it circling in its orbit, then you would expect it to move 1.23 degrees to the right at mid night and 1.19 degrees at mid day. That is a difference of less than 2%.

All your argument does is show the moon isn't close.
It doesn't show Earth is stationary. It CANNOT show Earth is stationary.
At best, you could observe this double speed at mid night relative to mid day and conclude the moon is close, but that isn't observed.

So just like your previous zig-zag argument, this one is a complete failure.

I'm not going to bother with your pictures (especially as it seems to have nothing to do with the argument at hand and instead is just ignoring the third dimension.
I have put in more than enough effort refuting your crap 3 times, pointing out what is wrong with it and showing how to do it correctly, with you just ignoring it.

Now are you going to give me the same courtesy and actually read what I have said and process it and either refute it rationally or accept it?

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rabinoz

  • 26280
  • Real Earth Believer
<< I'll leave all you Zigging and Zagging to YackBlack's tender mercies - for now! >>

But, here is a free present just for you! A visualisation of the sun and moon on the flat earth.



STOP SAYING THE EARTH IS FLAT-my interview with an international pilot(this will blow your mind, FLAT EARTH STREAM

Just look at the video from 1:22 to 8:50 - I do hope you like a nice Strauss Waltz.


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cikljamas

  • 2061
  • Ex nihilo nihil fit
1. JackBlack, so according to you modern science is not able to detect such (alleged) discrepancy regarding two (alleged) different speeds of moon's relative motion throughout ONE SINGLE polar day (4400 km/h vs 2800 km/h), even if the direction of earth's rotation wouldn't make any difference regarding the core of my ZIGZAG argument (which is : there would be unavoidable change in the direction of moon's apparent motion)???

2. According to you it doesn't matter weather the earth rotates to the RIGHT or to the LEFT? In both cases the moon would appear to move in one single direction? Then, you have to be able to sanely answer to this question :

Why the moon appears to move to the RIGHT all day long (throughout one single polar day)?

Because the earth turns to the LEFT (with respect to the moon), isn't that so?

Well, the earth turns to the LEFT (with respect to the moon), only first half of one polar day, during the second part of one polar day the earth turns to the RIGHT (with respect to the moon), so why the moon still appears to move to the RIGHT (westward) if now the earth turns to the RIGHT in relation to the moon?
« Last Edit: June 08, 2017, 03:31:48 AM by cikljamas »
"I can't breathe" George Floyd RIP

...
Well, the earth turns to the LEFT (with respect to the moon), only first half of one polar day, during the second part of one polar day the earth turns to the RIGHT (with respect to the moon), so why the moon still appears to move to the RIGHT (westward) if now the earth turns to the RIGHT in relation to the moon?

wow, you have really not a little bit of understanding of moving objects in a 3-dimensional room, especially if rotation is involved.

if i hear something like that, i always ask me: how does tie you shoe laces in the morning?

*

cikljamas

  • 2061
  • Ex nihilo nihil fit
...
Well, the earth turns to the LEFT (with respect to the moon), only first half of one polar day, during the second part of one polar day the earth turns to the RIGHT (with respect to the moon), so why the moon still appears to move to the RIGHT (westward) if now the earth turns to the RIGHT in relation to the moon?

wow, you have really not a little bit of understanding of moving objects in a 3-dimensional room, especially if rotation is involved.

if i hear something like that, i always ask me: how does tie you shoe laces in the morning?

What a marvelous refutation of my argument from a guy who knows how to tie shoe laces in the morning (i am not sure about the evening, given the consumed amount of drugs and alcohol during the day).. lol
"I can't breathe" George Floyd RIP

1. JackBlack, so according to you modern science is not able to detect such (alleged) discrepancy regarding two (alleged) different speeds of moon's relative motion throughout ONE SINGLE polar day (4400 km/h vs 2800 km/h), even if the direction of earth's rotation wouldn't make any difference regarding the core of my ZIGZAG argument (which is : there would be unavoidable change in the direction of moon's apparent motion)???
No. Modern science can measure such speeds. An unaided human cannot.
In fact, modern science has wonderful things, such as equatorial mount telescopes which can be motorised and programmed to remove the apparent motion due to Earth's rotation, to effectively track an object. Yes, there will still be parallax due to Earth's rotation and Earth's orbit, but these will be quite small for most objects.
The moon will be the most significantly effected by parallax due to Earth's rotation of the significant celestial objects, but that would be fairly small (and much greater with a much closer moon). The most easy to notice one effected by Earth's orbit would be the sun which appears to circle Earth once a year due to parallax.
There is no change in the moon's apparent direction, just the speed at which it appears to be moving.
It goes from 1.23 degrees per hour to 1.19 degrees per hour (at the equator)
If you wish to assert there is a change in apparent direction, then you need to prove it, going through my argument, showing what is wrong with it and correcting it.
I have pointed out what is wrong with yours, the fact that you are ignoring the rotation itself and instead trying to treat it as a translation.
The dominant factor in the change in apparent position of the moon is the fact that you look in a different direction (as a reference point anyway) after those 5 minutes of rotation.

But the other important fact I want to get through to you is that which one is moving doesn't actually matter, that you would get the exact same thing with a rotating Earth and moving moon as you would with a rotating Earth which is rotating at a slightly different speed with a stationary moon or a stationary Earth with a moon moving at a different rate to compensate.

All that matters is the relative motion and the distances.

So this cannot tell the difference between a stationary Earth and a rotating one.

What can produce a difference is the distance to the moon.


2. According to you it doesn't matter weather the earth rotates to the RIGHT or to the LEFT? In both cases the moon would appear to move in one single direction? Then, you have to be able to sanely answer to this question :
No. According to me, that slight translation due to the rotation is, for the most part, irrelevant. It is quite a small distance with viewing something much further away. This tiny translation does not significantly contribute to the moon's apparent motion.
It is a difference of 0.04 degrees out of 1.21 (from 1.19 to 1.23), and that is only if you could magically view it from the equator.
Viewing it from the arctic circle, with a radius of roughly 2500 km, that brings it down to between 1.202 to 1.218, or a difference of 0.016 out of 1.21)

Why the moon appears to move to the RIGHT all day long (throughout one single polar day)?
Because the earth turns to the LEFT (with respect to the moon), isn't that so?
Again, rotating left and right doesn't make much sense.
It is rotating CCW, that is what is important.
That means if you are looking towards the moon (from the northern hemisphere, preferably somewhere near the pole) and you will be turning such that you end up looking to the left.


Well, the earth turns to the LEFT (with respect to the moon), only first half of one polar day, during the second part of one polar day the earth turns to the RIGHT (with respect to the moon), so why the moon still appears to move to the RIGHT (westward) if now the earth turns to the RIGHT in relation to the moon?
See, this is the problem with using left or right.
Sure, the part of Earth you are on switching from moving left to moving right, but Earth is still rotating in the same direction, and it is still this rotation which is the main cause of the apparent motion of the moon.

If you don't believe me, you can try it yourself, even without anything fancy.
Draw a large circle, such that when in this circle you can view something quite a distance away (an example would be a 2.5 m radius circle viewing something 400 m away).
Now, stand on the near side of the circle facing the object, such that you are looking directly out from the circle.
Now, step to the side (going CCW around the centre, so now you are stepping to the left), and keep looking directly outwards from the circle, this means you are also turning around CCW (or to the left).
Notice how the distant object appears to be moving right out of your field of view?

Now try it with the other side, stand on the far side of the circle, viewing that same object. Now you are looking directly in towards the centre of the circle.
Now, just like before, step around the outside of the circle in a CCW direction, continually facing directly in.
Unlike before, you are now stepping to the right, however, like before, in order to continue looking directly in, you need to turn to the left (CCW), and just like before the object will appear to disappear out the right side of your vision.

You can also try to separate these 2 parts.
Just step sideways a small amount, without turning. Try both left and right if you like. The object stays pretty much right where it is.
Now try standing in the same spot and turning CCW (to the left). Notice how now the object appears to disappear out the right side of your vision?

This is the dominating factor, you turning to the left.
« Last Edit: June 08, 2017, 04:53:24 AM by JackBlack »