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Messages - Curiouser and Curiouser

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1
But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour.

But due to the earth's motion, the star does appear to rotate around the southern pole once per day?

Yes. Which is completely different than, and in no way can be described as "seeing that the apparent angular movement is about 15 degrees."

If star makes 360 degrees in 24 hours at constant speed, how many degrees it makes in one hour? :)

IF it makes 360 degrees in 24 hours (and I'll assume you're rounding a sidereal day up to 24 hours) then the rate would be 15 degrees per hour.  But only stars on the celestial equator move 360 degrees. Any other star moves by 360 * cos(celestial elevation).

Sigma Octantis or Polaris circle their respective celestial poles in little tiny circles once per day. They are NOT traversing an angular path of 360 degrees and are NOT moving at an apparent angular rate of 15 degrees/hour.

During solar day it was roughly 361 degrees per day.

How big is the difference between 15 and 15.042, please?

~~~~~

0.042 / 15 = 0.0028 = 0.28%

Congratulations. You've successfully determined the difference between a solar day and a sidereal day. Which has nothing whatsoever to do with the question or any of the presented arguments.

A star traces a full circle around the celestial pole in one sidereal day.
Its right ascension changes by 360 degrees.
Its apparent angular position changes by 360*cos(elevation from celestial equator) degrees.

Since you're fond of asking questions, let me return the favor.

I observe Polaris. One hour later I observe Polaris again. What is the apparent angular movement I have observed between the two times?

Is it 0.2 degrees?
Is it 15.0 degrees?

Good observation.
Instead of 24 hours for 361 degrees, it is 23 hours and 56 minutes for 360 degrees.

But when you divide those values you get the same result. :)

For the second time your reply has nothing whatsoever to do with the question or any of the presented arguments.

Observe Polaris. One hour later observe Polaris again. What is the apparent angular movement observed between the two times?

Is it 0.2 degrees?
Is it 15.0 degrees?
Is it something else?
Are you just going to reply with the difference between solar day and sidereal day again?

Please read my post again.
It is not about the difference between solar and sidereal day.

It is about the closeness to those 15 degrees aroud the Earth's axis
whether you apply your solar/sidereal difference or not.

I agree that it is not about the difference between solar and sidereal day. I never said, implied, or hinted that it was.

Please, though, answer this simplified version of the previous question, yes or no. The apparent angular movement of Polaris is 15 (+/-1) degrees per hour.

2
But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour.

But due to the earth's motion, the star does appear to rotate around the southern pole once per day?

Yes. Which is completely different than, and in no way can be described as "seeing that the apparent angular movement is about 15 degrees."

If star makes 360 degrees in 24 hours at constant speed, how many degrees it makes in one hour? :)

IF it makes 360 degrees in 24 hours (and I'll assume you're rounding a sidereal day up to 24 hours) then the rate would be 15 degrees per hour.  But only stars on the celestial equator move 360 degrees. Any other star moves by 360 * cos(celestial elevation).

Sigma Octantis or Polaris circle their respective celestial poles in little tiny circles once per day. They are NOT traversing an angular path of 360 degrees and are NOT moving at an apparent angular rate of 15 degrees/hour.

During solar day it was roughly 361 degrees per day.

How big is the difference between 15 and 15.042, please?

~~~~~

0.042 / 15 = 0.0028 = 0.28%

Congratulations. You've successfully determined the difference between a solar day and a sidereal day. Which has nothing whatsoever to do with the question or any of the presented arguments.

A star traces a full circle around the celestial pole in one sidereal day.
Its right ascension changes by 360 degrees.
Its apparent angular position changes by 360*cos(elevation from celestial equator) degrees.

Since you're fond of asking questions, let me return the favor.

I observe Polaris. One hour later I observe Polaris again. What is the apparent angular movement I have observed between the two times?

Is it 0.2 degrees?
Is it 15.0 degrees?

Good observation.
Instead of 24 hours for 361 degrees, it is 23 hours and 56 minutes for 360 degrees.

But when you divide those values you get the same result. :)

For the second time your reply has nothing whatsoever to do with the question or any of the presented arguments.

Observe Polaris. One hour later observe Polaris again. What is the apparent angular movement observed between the two times?

Is it 0.2 degrees?
Is it 15.0 degrees?
Is it something else?
Are you just going to reply with the difference between solar day and sidereal day again?

3
But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour.

But due to the earth's motion, the star does appear to rotate around the southern pole once per day?

Yes. Which is completely different than, and in no way can be described as "seeing that the apparent angular movement is about 15 degrees."

If star makes 360 degrees in 24 hours at constant speed, how many degrees it makes in one hour? :)

IF it makes 360 degrees in 24 hours (and I'll assume you're rounding a sidereal day up to 24 hours) then the rate would be 15 degrees per hour.  But only stars on the celestial equator move 360 degrees. Any other star moves by 360 * cos(celestial elevation).

Sigma Octantis or Polaris circle their respective celestial poles in little tiny circles once per day. They are NOT traversing an angular path of 360 degrees and are NOT moving at an apparent angular rate of 15 degrees/hour.

During solar day it was roughly 361 degrees per day.

How big is the difference between 15 and 15.042, please?

~~~~~

0.042 / 15 = 0.0028 = 0.28%

Congratulations. You've successfully determined the difference between a solar day and a sidereal day. Which has nothing whatsoever to do with the question or any of the presented arguments.

A star traces a full circle around the celestial pole in one sidereal day.
Its right ascension changes by 360 degrees.
Its apparent angular position changes by 360*cos(elevation from celestial equator) degrees.

Since you're fond of asking questions, let me return the favor.

I observe Polaris. One hour later I observe Polaris again. What is the apparent angular movement I have observed between the two times?

Is it 0.2 degrees?
Is it 15.0 degrees?

4
But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour.

But due to the earth's motion, the star does appear to rotate around the southern pole once per day?

Yes. Which is completely different than, and in no way can be described as "seeing that the apparent angular movement is about 15 degrees."

If star makes 360 degrees in 24 hours at constant speed, how many degrees it makes in one hour? :)

IF it makes 360 degrees in 24 hours (and I'll assume you're rounding a sidereal day up to 24 hours) then the rate would be 15 degrees per hour.  But only stars on the celestial equator move 360 degrees. Any other star moves by 360 * cos(celestial elevation).

Sigma Octantis or Polaris circle their respective celestial poles in little tiny circles once per day. They are NOT traversing an angular path of 360 degrees and are NOT moving at an apparent angular rate of 15 degrees/hour.

5
Flat Earth General / Re: Seeing the sun on the head is not real
« on: July 16, 2019, 12:19:48 PM »
Can you move a container of water without disturbing the waters' surface?  NO, you can't.

I most certainly can.

I just can't accelerate it.

Learn physics.

6
Out of interest, how did you measure it.  Something similar to a sundial or equatorial mount?

FWIW, Im most definitely not saying that a flat earth explains the apparent motion of the sun better than the real earth.

I just found the topic interesting, because I hadnt really thought this one through properly before.

Probably a stick in the ground. That appears to be the sophistication of many of the posters here. If at all. I suspect many of the posters claiming to have done "experiments" just make it up from their flawed understanding.

7
But how is it possible that I have actually measure this apparent movement of 15 degrees per hour? I measure this myself many times? How can that be possible?

It is possible because:
1) You do not understand the geometry of the positions of the earth and sun and how the sun moves across the sky with respect to the local earth coordinate system.
I do understand simple geometry. I studied mathematics and physics

Apparently not, because you said:
In the sphere model it is trivial: The earth rotated once full (360 degrees) in 24 hours (i.e. a day). That makes 15 degrees per hour. Simple.
This is not true.

Quote
2) You have measured incorrectly, and with insufficient error bars.

Could be a possibility, but with many repeated measurements it is easy to see that the deviation is less than 100%. That is sufficient to rule out flat earth

Your measurements are now bounded between 0 degrees/hour and 30 degrees/hour? This is the first time that you have attempted to put any tolerances on your values. (I, on the other hand, have specifically said that I am careful to try to use significant digits correctly in my values.) Given that, I will have to concede. Yes, the apparent angular motion of the sun across the sky is in fact between 0 and 30 degrees per hour.

Quote
3) You have not read and comprehended the explanations given in the previous (now this) topic.

Irrelevant. It is about the observation of the apparent motion of the sun. It is just about an observation. Independent of any explanation

And your observation, and the errors in your observation, were explained previously.

Quote
4) You have convinced yourself that a 360 degree earth rotation in 24 hours is equivalent to an apparent angular motion of 15 degrees/hour, and that has biased you into "fudging" your measurements to give the correct answer.

Incorrect. I am interested in reality, the truth, science, knowledge.

Then please explain why you have made measurements "many many times" and come to the wrong conclusion. Also why you characterize correct explanations as "irrelevant."

Quote
Even using predictions (like the NOAA Solar Calculator) confirm that the angular position of the sun does not agree with a 15 degree/hour apparent angular motion rate.

Irrelevant. If earth were flat the angular motion must be MUCH smaller close to a sunset/sunrise. An approximate measurement is enough.

Then why didn't you say that an approximate measurement was good enough weeks ago? This is the first time you have mentioned it. You said 15 degrees per hour. Since you have studied mathematics and physics, you would know that means 14.5 to 15.5 degrees per hour. I am telling you that is incorrect. I am also telling you that your "simple" explanation of 360 degrees/24 hours is incorrect.

8
I can't help you with your own reading comprehension and/or laziness. This is your topic. That you started. You are on the very page with the answer you seek.

Ok, then there is no answer to my question.

I just went through all the responses and comments. There is no answer!
Yes, there is. The answer is that your premise, the very assumption that your argument is built upon, is not correct, and to continue asserting it is like you have done is to ignore reality, which you are also doing when you say you haven't had this explained to you.

Typical distraction to the subject of gyroscopes is off topic.  This topic is about the apparent motion of the sun relative to an observer on Earth.  Nothing else.

Focus, people.

For what its worth I think the numbers should be 15 deg per hour during the equinox, gradually reduced to 13.75 deg per hour at the solstices.

I think we can just take the angular movement from equinoxes (360/24) and multiply by cosine of the offset (maximum 23.5 deg at solstices).

Can anyone confirm?

Bingo.

Ah thank you. That is what you mean.

So you think the sun does not move at an angular speed of 15 degrees per hour? Are you saying that premise is incorrect?
It does not.
Yes.

But how is it possible that I have actually measure this apparent movement of 15 degrees per hour? I measure this myself many times? How can that be possible?

It is possible because:
1) You do not understand the geometry of the positions of the earth and sun and how the sun moves across the sky with respect to the local earth coordinate system.
2) You have measured incorrectly, and with insufficient error bars.
3) You have not read and comprehended the explanations given in the previous (now this) topic.
4) You have convinced yourself that a 360 degree earth rotation in 24 hours is equivalent to an apparent angular motion of 15 degrees/hour, and that has biased you into "fudging" your measurements to give the correct answer.

Even using predictions (like the NOAA Solar Calculator) confirm that the angular position of the sun does not agree with a 15 degree/hour apparent angular motion rate.

9
I can't help you with your own reading comprehension and/or laziness. This is your topic. That you started. You are on the very page with the answer you seek.

Ok, then there is no answer to my question.

I just went through all the responses and comments. There is no answer!
Yes, there is. The answer is that your premise, the very assumption that your argument is built upon, is not correct, and to continue asserting it is like you have done is to ignore reality, which you are also doing when you say you haven't had this explained to you.

Typical distraction to the subject of gyroscopes is off topic.  This topic is about the apparent motion of the sun relative to an observer on Earth.  Nothing else.

Focus, people.

For what its worth I think the numbers should be 15 deg per hour during the equinox, gradually reduced to 13.75 deg per hour at the solstices.

I think we can just take the angular movement from equinoxes (360/24) and multiply by cosine of the offset (maximum 23.5 deg at solstices).

Can anyone confirm?

Bingo.

Ah thank you. That is what you mean.

So you think the sun does not move at an angular speed of 15 degrees per hour? Are you saying that premise is incorrect?
It does not.
Yes.

10
Flat Earth General / Re: Apparent motion of the sun
« on: July 16, 2019, 01:18:42 AM »
You already started a topic on this, and were already told that your premise was wrong.

https://www.theflatearthsociety.org/forum/index.php?topic=82127.0

11
Phew is debunked, Mr. Paper Tape and Tin Can.

I just measured the diameter of a precision steel rod (0.9980 +/- 0 0005 inches) with a set of Mitutoyo calipers that were calibrated 3 days ago. Placing stop blocks on a ground steel table, I rolled the steel rod two full revolutions, judging the start and stop points by the verticality of a scribe mark on the end of the rod with respect to a gage block. Using the calipers to then measure the distance between the stop blocks (and subtracting the diameter of the rod) gave the values of 2 x circumference from 6.263 to 6.289 over 6 measurements. This gives a measured value of C/D of between 3.13 and 3.15.

Placing the stop blocks 7.341 inches apart (the distance required if C/D was 3.1716), and rolling the steel rod between the blocks always resulted in the vertical scribe over-rotating by a few degrees.

Perhaps danang just sucks at measurement?

No. Danang definitely sucks at measurement. Repeatedly.

12
For reference, they used to be the third most powerful party until now and in one EU election they got around 10% iirc.

I read about them in Newsweek. Or was that the Asian Dawn Movement?

13
Flat Earth General / Re: Moon landing Hoax?
« on: July 13, 2019, 04:57:12 PM »
I have another mundane moon hoax question:

If we've never been to the Moon, then where did the retroreflectors come from?

Many moon hoax advocates only doubt manned landings, not uncrewed robotic missions. Several uncrewed craft landed on the moon in the '60s and "70s. The existence of retroreflectors that can be detected today is a poor proof of manned moon landings.

I agree, that goes for moon hoax advocates. But what about flat earth believer moon hoax advocates?  My experience is that the vast majority of the latter would in no way accept even uncrewed missions to the moon.

Robots landing on the moon in the early 70s and setting up retroreflectors is actually even MORE amazing.

So amazing, in fact, that it was done. In the '70s. By the Soviets. With an unmanned lunar rover.

https://www.space.com/amp/8295-lost-soviet-reflecting-device-rediscovered-moon.html

You should really recalibrate your amazement meter.

14
Flat Earth General / Re: Moon landing Hoax?
« on: July 13, 2019, 04:48:11 PM »
I have another mundane moon hoax question:

If we've never been to the Moon, then where did the retroreflectors come from?

Many moon hoax advocates only doubt manned landings, not uncrewed robotic missions. Several uncrewed craft landed on the moon in the '60s and "70s. The existence of retroreflectors that can be detected today is a poor proof of manned moon landings.

So, what to you propose is high proof of manned moon landings?

The cumulative self-consistent record, spanning all the fields over the decades involved is the proof.

A single piece of evidence that can be dismissed with a plausible alternative explanation is not a good proof.

15
Flat Earth General / Re: Moon landing Hoax?
« on: July 13, 2019, 11:11:11 AM »
I have another mundane moon hoax question:

If we've never been to the Moon, then where did the retroreflectors come from?

Many moon hoax advocates only doubt manned landings, not uncrewed robotic missions. Several uncrewed craft landed on the moon in the '60s and "70s. The existence of retroreflectors that can be detected today is a poor proof of manned moon landings.

16
Flat Earth General / Re: Cavendish experiment
« on: July 11, 2019, 11:43:52 PM »
I don't see the relevance of this experiment at all.

17
Where did you find out how high Hawking's IQ was?

Exactly. Yes, wise, where did you get this information from?

This topic belongs to Trump's negative politics. I to open this issue does not make me a part of the issue. I am definitely one of the furthest person to this thread hence opened it with all my courage. keep your personal problems with me or somebody else to the angry ranting.

"Why do you not support donald trump?" I gave answers. I can't help that you don't like them.

He were asking to you, not to me. Stop to do manipulation. Leading the question about Hawking's IQ isn't a talking about OP question.

Manipulation (n) Telling me something that's true that I don't want to believe. From the "Wise-ictionary. Definitions of My Flat Earth World."

18
This is a very advanced question for a flat earther.

When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour.

How can that be possible on a flat earth?

In the sphere model it is trivial: The earth rotated once full (360 degrees) in 24 hours (i.e. a day). That makes 15 degrees per hour. Simple.

But how can it work on a flat earth?

I disagree with your stated premise.

"When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour."

I disagree that the apparent angular movement is always the same.
I disagree that it is about 15 degree per hour.
I disagree that you have observed this ("Why do I see the sun move with 15 degrees per hour?") with the accuracy necessary to make this statement and are just taking someone else's word for it.

To refute this, provide observational information about the azimuth and elevation of the sun or the moon at pairs of times one hour apart giving the following:
Observed azimuth and elevation to an accuracy of at least one solar or lunar diameter
Location (Lat, Long)
Date
Time of each observation

For each date provide observations for object at low elevation (close to horizon) and high elevation (mid-day). Provide information for at least two different dates differing in time of year by at least 3 months.

Show that for each pair of observations an hour apart, the angular separation is the same, and is about 15 degrees.
The sundial is dependent on this fact.
Need I say more?
Yes, you do, because it is not a fact. See post above.

19
Typical distraction to the subject of gyroscopes is off topic.  This topic is about the apparent motion of the sun relative to an observer on Earth.  Nothing else.

Focus, people.

For what its worth I think the numbers should be 15 deg per hour during the equinox, gradually reduced to 13.75 deg per hour at the solstices.

I think we can just take the angular movement from equinoxes (360/24) and multiply by cosine of the offset (maximum 23.5 deg at solstices).

Can anyone confirm?

Bingo.

20
Here is an extreme example that may make you less confused.
No need because I'm not confused now.

But you are correct in that alex314's initial premise:
Quote
When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour.
is incorrect except at the poles.

No, your statement is incorrect. but I can see why you are confused.

I will address the nominal, geometric case without regards to atmospheric refraction, and references to the position of the sun are to the center of the sun. At the pole, at equinox, the sun will nominally trace the horizon. In this case the apparent angular motion of the sun is 360/degrees per solar day = 15.00 degrees/hour. (Easy to imagine, because the path of the sun is perpendicular to the Earth's axis of rotation.)

At the equator, the sun will trace a line from east to zenith to west in 12 hours = 15.00 degrees/hour. Again, easy to imagine and do the calculation in your head.

At any other latitude, while perhaps not as obvious, the path traced is at the same rate. You can use some simple geometry to convince yourself of this.

This situation changes when the sun is not at 90 degrees to the axis of the Earth's rotation. At the summer solstice, at the north pole, the sun traces a path in the sky at a constant elevation of ~23.5 degrees.

An object tracing a path in the sky at 23.5 degrees elevation in 24 hours is not moving at an angular rate of 15.00 degrees/hour. This should be obvious from the analogy to Sigma Octantis. Sigma Octantis at an elevation of 89 degrees is not moving at 15.00 degrees/hour; there must be a monotonic function that goes from object on the horizon moving at 15.00 degrees/hour to object at pole moving at 0.00 degrees/hour.

Or, if you're mathematically inclined, you can calculate the angular separation yourself from

https://books.google.com/books?id=MTGYxQyW998C&pg=PA66&lpg=PA66&dq=%22the+angle+between+two+celestial+objects%22&source=bl&ots=UMhTOyKjZG&sig=X3M3S3h7M-EHDsF6BNBJHNBIvZ4&hl=en&sa=X&ved=0ahUKEwi90-GVmtLUAhVs04MKHb0gBOYQ6AEIKzAB#v=onepage&q=%22the%20angle%20between%20two%20celestial%20objects%22&f=false

using either observed data or calculated solar position

https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html

You'll find that the solar rate of angular motion changes significantly. By no means is it always 15 degrees/hour.

Quote from: Curiouser and Curiouser
View Sigma Octantis, about a degree away from the celestial pole. Engage the equatorial mount drive on your telescope to keep Sigma Octantis centered in the telescope. This drive will rotate the telescope around the equatorial mount axis at a rate of approximately 15.0 degrees per hour.

But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour. The pointing of the telescope traced a portion of a cone with an apex angle of about 2 degrees. The angular rate at which Sigma Octantis moved with respect to the observer is a few tenths of a degree per hour, not 15 degrees per hour.
I have never mentioned "with respect to the observer". Everywhere I've been referring to rotation about the axis (gnomon) of a "properly aligned" sundial.

No. You butted in with that irrelevant observation, and have continued with it, even though you were told:
While this is true, it has nothing to do with the basic geometry that the apparent angular rate the sun and moon move is not always the same.
...
Whether or not a sundial has equally spaced tick marks in a particular orientation or unequally spaced tick marks in a different orientation, [snip] does not change the actual observed apparent position of the sun in the sky or the change in position versus time of the sun in the sky.
...
You're confusing uniform angular motion or measurement of a specific measuring device with the actual angular motion of an object in the sky.
Again, you're confusing uniform angular motion of a specific measuring device with the actual angular motion of an object in the sky.


Quote from: Curiouser and Curiouser
It should be obvious that this is the case because Sigma Octantis is not viewed by the observer at 90 degrees from the celestial pole.
Why is that even relevant?

For the geometrical reasons stated above.

Quote from: Curiouser and Curiouser
Similarly, because of the orientation of the earth's axis with respect to its plane of orbit, the sun is not always viewed by the observer at 90 degrees to the celestial pole, and changes throughout the year. Therefore the apparent angular motion rate changes as well, and is not constant.
The apparent angular rate of the sun about the earth's axis only varies slightly due to the earth's orbital ellipticity not due to the seasons.

What? Nothing about this discussion has anything to do with orbital ellipticity! Where did you get that? It has to do with the geometry of angles!

All the motion I have referred to about about the earth's axis of rotation and that is very nearly constant (unless you are bothered about a millisecond per day up or down). In earlier posts I tried to avoid this detail as it might lead to a flat ~ Globe slanging match.

No, I'm not concerned with a few milliseconds. I am concerned with an apparent angular rate of motion of a celestial object that is as slow as 13.8 degrees/hour rather than 15.0 degrees per hour. And I do try to take care to use significant digits in values I present.

The original post included
When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour.

In the sphere model it is trivial: The earth rotated once full (360 degrees) in 24 hours (i.e. a day). That makes 15 degrees per hour. Simple.

No. It's not. The observations and conclusions are wrong. It's a bad argument when you present "here's a simple fact" that is wrong. Which is why I asked for:

"To refute this, provide observational information about the azimuth and elevation of the sun or the moon at pairs of times one hour apart"

and

"Show that for each pair of observations an hour apart, the angular separation is the same, and is about 15 degrees."

Which was not done.

21
But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour.

But due to the earth's motion, the star does appear to rotate around the southern pole once per day?

Yes. Which is completely different than, and in no way can be described as "seeing that the apparent angular movement is about 15 degrees."

22

Also note that the 15 per hour is only uniform with a properly aligned equatorial sundial or similar instrument.
And this is one such similar instrument ...


Whether or not a sundial has equally spaced tick marks in a particular orientation or unequally spaced tick marks in a different orientation, as in

does not change the actual observed apparent position of the sun in the sky or the change in position versus time of the sun in the sky.

You're confusing uniform angular motion or measurement of a specific measuring device with the actual angular motion of an object in the sky.
If you want to learn a bit about vertical sundials, including the one you pictured, have a look in How To Make An Equatorial Sundial - With Photos


Your pictured sundial is not an equatorial ring or disc sundial but this is:

And anywhere in the Northern Hemisphere, including the equator, such a sundial can be aligned by pointing the gnomon directly at Polaris (its close enough).

I'm not confused by any such thing.

If the uniform angular velocity of a properly aligned equatorial ring or disc sundial is not caused by the uniform angular velocity of the light source about a parallel axis then please explain the cause.

You are quite confused.

Again, you're confusing uniform angular motion of a specific measuring device with the actual angular motion of an object in the sky.

Here is an extreme example that may make you less confused. Take your telescope with its equatorial mount drive system and go outside. Properly align your equatorial mount axis to the southern celestial pole. View Sigma Octantis, about a degree away from the celestial pole. Engage the equatorial mount drive on your telescope to keep Sigma Octantis centered in the telescope. This drive will rotate the telescope around the equatorial mount axis at a rate of approximately 15.0 degrees per hour.

But Sigma Octantis did not move angularly with respect to you at a rate of 15.0 degrees per hour. The pointing of the telescope traced a portion of a cone with an apex angle of about 2 degrees. The angular rate at which Sigma Octantis moved with respect to the observer is a few tenths of a degree per hour, not 15 degrees per hour.

It should be obvious that this is the case because Sigma Octantis is not viewed by the observer at 90 degrees from the celestial pole.

Similarly, because of the orientation of the earth's axis with respect to its plane of orbit, the sun is not always viewed by the observer at 90 degrees to the celestial pole, and changes throughout the year. Therefore the apparent angular motion rate changes as well, and is not constant.

23
Would the sun move at different rates near the horizon vs the top of the sky, according to basic geometry, can someone tell a noob?
The sun, moon and stars move a little slower very close to the horizon.
When right on the horizon they usually appear about 0.5 higher than their "geometric position" but that refraction is only half that at 2.5 above.
This is often seen making the sun seem squashed a little vertically just before it sets.

While this is true, it has nothing to do with the basic geometry that the apparent angular rate the sun and moon move is not always the same.


Also note that the 15 per hour is only uniform with a properly aligned equatorial sundial or similar instrument.
And this is one such similar instrument ...


Whether or not a sundial has equally spaced tick marks in a particular orientation or unequally spaced tick marks in a different orientation, as in

does not change the actual observed apparent position of the sun in the sky or the change in position versus time of the sun in the sky.

You're confusing uniform angular motion or measurement of a specific measuring device with the actual angular motion of an object in the sky.

24
I didn't call you stupid. I suggested you may be under-informed with respect to the context of the comment.

Back on topic, let's all take a moment to remember this is the thread where rab finally accepted FE! And I think C&C was waiting on some specific evidence supporting alex's claims in the OP.

What claim? I just wanted to know how flat earth explains the apparent motion of the sun of 15 degrees per hour. Thats all.

I did not make a claim.

You made a claim.


When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour.


That is a claim.

25
This is a very advanced question for a flat earther.

When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour.

How can that be possible on a flat earth?

In the sphere model it is trivial: The earth rotated once full (360 degrees) in 24 hours (i.e. a day). That makes 15 degrees per hour. Simple.

But how can it work on a flat earth?

I disagree with your stated premise.

"When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour."

I disagree that the apparent angular movement is always the same.
I disagree that it is about 15 degree per hour.
So you "disagree that the apparent angular movement is always the same" and  "disagree that it is about 15 degree per hour".
How then does a sundial like this keep quite good time with uniformly spaced hour lines?


Sundial time does drift slowly to 16 minutes ahead of clock time in November and  in the middle of February to 14 minutes behind.
But each hour the shadow moves very nearly 15.

Do you have reason to claim otherwise?

I would say that you and I have different criteria for "always the same," "about," "quite good," and "very nearly."

Shame on you for derailing the conversation with off-topic tangents. The topic is "Why do I [alex314] see the sun move with 15 degrees per hour?" My response is totally on topic with questions about what alex314 sees. Your response is totally off topic with questions about hardware not originally in the topic questions. You are a hypocrite for wandering off-topic while chastizing others for doing the same.

 ;D ;D ;D Also, why do you submit a picture of hardware that implies evidence that Australia doesn't exist? Hmmmmmm ...  ;D ;D ;D

26
This is a very advanced question for a flat earther.

When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour.

How can that be possible on a flat earth?

In the sphere model it is trivial: The earth rotated once full (360 degrees) in 24 hours (i.e. a day). That makes 15 degrees per hour. Simple.

But how can it work on a flat earth?

I disagree with your stated premise.

"When you observe the sun (or the moon), you can see that the apparent angular movement is always the same. It is about 15 degrees per hour. It means, the position of the sun and the moon changes by 15 degrees per hour."

I disagree that the apparent angular movement is always the same.
I disagree that it is about 15 degree per hour.
I disagree that you have observed this ("Why do I see the sun move with 15 degrees per hour?") with the accuracy necessary to make this statement and are just taking someone else's word for it.

To refute this, provide observational information about the azimuth and elevation of the sun or the moon at pairs of times one hour apart giving the following:
Observed azimuth and elevation to an accuracy of at least one solar or lunar diameter
Location (Lat, Long)
Date
Time of each observation

For each date provide observations for object at low elevation (close to horizon) and high elevation (mid-day). Provide information for at least two different dates differing in time of year by at least 3 months.

Show that for each pair of observations an hour apart, the angular separation is the same, and is about 15 degrees.

27
Flat Earth General / Re: Number one observation that earth is flat
« on: July 01, 2019, 12:04:14 PM »
Flat earthers only!

And ONLY response to this question. No discussion in this thread. Create a new thread when you want to discuss details.

Using "please" is a good way of making it sound like a respectful request and not a demand, and would go a long way towards improving the impression of your persona here.

28
The Lounge / Re: Motorcycle thread
« on: June 30, 2019, 05:02:23 PM »
Oh, forgot to say look at the Horizon as a whole not directly in front of the bike when riding, helps avoid dangers and cut turns.

And how far away is that horizon on a flat earth?

29
But you havent done that. Just do it and win your Nobel Prize.

So where is your Nobel Prize?

Since achieving a Nobel Prize seems to be your criteria for whether danang's arguments are valid, which Nobel Prize do you imagine danang would be eligible for if he was correct?

30
"I love 'phew' and not pi," says danang.
"Every other math expert can hang!"
But his brain is not well.
It has rot, and a smell
Which is much like the phew of his wang.

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