Why do I see the sun move with 15 degrees per hour?

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?

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.

Quote from: alex314 on June 30, 2019, 10:05:42 PM

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?

Quote from: Curiouser and Curiouser on July 01, 2019, 05:39:24 PM

Quote from: alex314 on June 30, 2019, 10:05:42 PM

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.

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

How then does a sundial like this keep quite good time with uniformly spaced hour lines?

Would you look at that. Another victory for FE!!

Quote from: rabinoz on July 01, 2019, 06:59:26 PM

How then does a sundial like this keep quite good time with uniformly spaced hour lines?

Would you look at that. Another victory for FE!!

That North Polar AEP map does have its uses even though Tom Bishop claims:

That's the mid-1800's model. The Zetetic  movement switched over to Bi-Polar models in the early 1900's upon direct discovery of the South Magnetic Pole.

Later unrelated offshoots continued to use and present the Monopole model.

Quote from: Curiouser and Curiouser on July 01, 2019, 05:39:24 PM

Quote from: alex314 on June 30, 2019, 10:05:42 PM

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?

Thank you for your support. The flat earth society believers especially thank you to explain the issue like a good way. With your support, the flat earth society will become great. Thank you again.

Quote from: rabinoz on July 01, 2019, 06:59:26 PM

How then does a sundial like this keep quite good time with uniformly spaced hour lines?

Would you look at that. Another victory for FE!!

Ah I see. It is not about truth for you, but to 'win' only.

I see

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?

Quote from: boydster on July 01, 2019, 09:34:53 PM

Quote from: rabinoz on July 01, 2019, 06:59:26 PM

How then does a sundial like this keep quite good time with uniformly spaced hour lines?

Would you look at that. Another victory for FE!!

Ah I see. It is not about truth for you, but to 'win' only.

I see

Click "Search." Type the following, including the quotes: "Another victory for FE!" Learn context.

Quote from: alex314 on July 01, 2019, 11:18:35 PM

Quote from: boydster on July 01, 2019, 09:34:53 PM

Quote from: rabinoz on July 01, 2019, 06:59:26 PM

How then does a sundial like this keep quite good time with uniformly spaced hour lines?

Would you look at that. Another victory for FE!!

Ah I see. It is not about truth for you, but to 'win' only.

I see

Click "Search." Type the following, including the quotes: "Another victory for FE!" Learn context.

LOL. I am not a brainless puppet. I can think!

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.

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.

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.

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 Amusing Planet: Campbell–Stokes Recorder: A Simple Device That Measures Sunshine

A Campbell–Stokes sunshine recorder outside Darwin Airport Meteorological Office, in Darwin, Australia. Photo credit: Bidgee/Wikimedia

Assuming it's the equinox (sun's over the equator) measure how far the sun moves across the sky in an hour from 45 degrees north.

Make the same measurement from 45 degrees south.

If the world is round both measurements will be the same. If the northern measurement is larger than the southern measurement then the world is flat.

Quote from: boydster on July 02, 2019, 04:22:32 AM

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.

Quote from: faded mike on July 01, 2019, 11:59:12 PM

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.

Actually, even easier proof of the shape of the world.

On the equinox when the sun's directly above the equator the sun rises at 90 degrees (due east) and sets at 270 degrees (due west).

This happens whether you are 45 degrees north, on the equator or 45 degrees south.

Therefore the earth is round.

Although this had been well observed throughout time, I now await claims that no one has ever observed this/I can't prove it/NASA are liars.

Quote from: rabinoz on July 02, 2019, 04:37:55 AM

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.

Quote from: Curiouser and Curiouser on July 02, 2019, 10:15:11 AM

Quote from: rabinoz on July 02, 2019, 04:37:55 AM

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.

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?

Quote from: Curiouser and Curiouser on July 02, 2019, 03:15:01 PM

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."

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:

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.

Take your telescope with its equatorial mount drive system and go outside. Properly align your equatorial mount axis to the southern celestial pole.

Sorry, but I don't have a telescope with an equatorial mount drive. I'd like to but can only dream.
But carry on. I'll try to imagine it.

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.

I knke that if you simply place a stick vertically in the ground it's shadow will only move at 15°/hour at the poles.

The axis of the sundial or of the "telescope with its equatorial mount drive system" is aligned parallel to the axis of rotation of the earth.
So the sundials shadow will move around the ring at 15°/hour and
provided it was aligned perfectly, the telescope will track Sigma Octantis rotating around its tiny 1° 2′ 37″ circle in once in one sidereal day or at 15.041°/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.

Why is that even relevant?

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.

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.

Quote from: turtles on July 02, 2019, 04:20:19 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 03:15:01 PM

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."

Ah right, yes, agreed.

Select one point in your backyard, or front yard, or your window, ...
Always use the same point.

Establish direct line between the point and the Sun at 8 am.
Establish direct line between the point and the Sun at 9 am.
What is the angle between the two lines?

What is the angle between 12 pm and 1 pm?
What is the angle between 6 pm and 7 pm?

Is it 15 degrees all three times?
If not, how different the angles are?

~~~~~

How to establish those lines?

Well, there are several ways.

One is to select a pole in your yard and at desired times stick a small tent stakes into the ground as the tip of the pole shadow moves.
Then you can tie some thread or thin rope from the pole tip to each stake.
Then just directly measure the angles between the threads.

Quote from: Curiouser and Curiouser on July 02, 2019, 03:15:01 PM

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.

I think C&C is right in principle, and I fully admit I’ve probably used this “fact” myself before without really thinking it through.

The example of a star trail near celestial poles is a good example.

However, I don’t think it’s dependent on latitude, only time of year.  At the equinoxes, the Sun should appear to trace a circular path, with us in center.  Latitude just changes the angle of the Sun’s apparent path.  Sun should appear to move 15 deg an hour everywhere.  I think.

At other times of the year, we are offset from the center of this imaginary circle, by a certain amount, peaking at the solstices at 23.5 deg offset.  So I believe the sun should move less per hour relative to us.  Although measured from one of the celestial poles, it’s still 15 deg per hour around that point.

I’ve not tried the maths on this yet though.  It’s not very obvious how to calculate it by thinking about the geometry in my head.  I need to sit down with pen and paper and scribble some diagrams to really picture it properly.  Until I do, consider this comment provisional.

I don’t think any of this affects the 15 deg intervals on a sundial though.

Someone let me know if that makes sense.

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?

A gyroscope stays in position for many hours, thus proving it is not the Earth moving. Sure the sun has moved, that's all.

Quote from: alex314 on June 30, 2019, 10:05:42 PM

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?

A gyroscope stays in position for many hours, thus proving it is not the Earth moving. Sure the sun has moved, that's all.

Evidence please!
And with a gyroscope large enough to have a drift rate low enough to measure the earth's rotation.
Little toy gyroscopes that Flat Earth YouTubers play with are hopeless at measuring the earth's rotation.

The "great" flat earther Bob Knodel manage to get the $20,000 needed (so he said) to buy a highly stable laser ring gyroscope to "prove that the earth was stationary".
Just search YouTube for "Bob Knodel proves the earth rotates" and look "what comes out of the woodwork".

The short version: Bob Knodel & His Ring Laser Gyroscope Experiment                                                  by: FlatEarth.ws

The longer version: Bob Knodel Proves the Globe using the Gyro of DOOM (for Flat Earth)                                                 by: WheresWa11y

And the Flat Earthers were not happy!

Bob Knodel of globebusters is a LIAR and a SATANIST
                                  and getting THE BOOT Beyond Flat Earth!
                                                by: Awake Souls
Don't dare tell the truth to these Flat Earthers that are so "far down the rabbit hole" they can't even see the light.

But if the earth is stationary please explain how marine gyro-compasses and gyro-theodilites can find precise true north?
Have a look at: Flat Earth General / Re: A gyroscope shows the Earth is stationary. « Message by rabinoz on July 08, 2018, 02:59:22 PM »

Please show that the results of the Ring Laser Gyroscope is consistent and not based on a algorithm like the experiment it is based on.

https://wiki.tfes.org/Ring_Laser_Gyroscope