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

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."

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

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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.

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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%

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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%

For the very tiny percentage of the visible stars on the celestial equator, that's all well and good, but that's not the sun (with two days per year being the exception), which is what the OP was asserting with reckless abandon, and it's not the vast majority of stars, either.

For the very tiny percentage of the visible stars on the celestial equator, that's all well and good, but that's not the sun (with two days per year being the exception), which is what the OP was asserting with reckless abandon, and it's not the vast majority of stars, either.

The topic question is "Why do I see the sun move with 15 degrees per hour?" and the sun appears to move 15° of longitude per hour.

Finding your Latitude and Longitude
Longitude
To find your longitude you just need to know how many hours apart you are from Greenwich, UK and a vertical stick to know when the sun is at its zenith over your present location.

It does not explicitly say "15° of longitude per hour" but that would have to be implied for the Wiki entry to be meaningful.

Maybe the OP could have stated it better than he did:

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.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
But how can it work on a flat earth?

But in any case, I see no problem explaining his "how can it work on a flat earth?".
The usual Flat Earth model explains that quite well with the sun moon circling 360° in 24 hours.

Sun dial doesn't exist.

On very north hemiplane at September equinox.. how come  the sun comes late for more than 1 hour? What about at March? Teens minutes late. >> Different from September's??

Globers, please repent 

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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?

Quote from: boydster on July 17, 2019, 04:43:36 AM

For the very tiny percentage of the visible stars on the celestial equator, that's all well and good, but that's not the sun (with two days per year being the exception), which is what the OP was asserting with reckless abandon, and it's not the vast majority of stars, either.

The topic question is "Why do I see the sun move with 15 degrees per hour?" and the sun appears to move 15° of longitude per hour.

As has already been covered in this thread, it does not appear to move 15° per hour most of the time. It varies throughout the year.

Quote from: Macarios on July 16, 2019, 08:51:37 PM

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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.

Quote from: Curiouser and Curiouser on July 17, 2019, 06:21:12 AM

Quote from: Macarios on July 16, 2019, 08:51:37 PM

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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?

Quote from: Macarios on July 17, 2019, 09:19:16 AM

Quote from: Curiouser and Curiouser on July 17, 2019, 06:21:12 AM

Quote from: Macarios on July 16, 2019, 08:51:37 PM

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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.

Quote from: Curiouser and Curiouser on July 17, 2019, 09:37:42 AM

Quote from: Macarios on July 17, 2019, 09:19:16 AM

Quote from: Curiouser and Curiouser on July 17, 2019, 06:21:12 AM

Quote from: Macarios on July 16, 2019, 08:51:37 PM

Quote from: Curiouser and Curiouser on July 16, 2019, 06:49:38 PM

Quote from: Macarios on July 16, 2019, 02:11:25 PM

Quote from: Curiouser and Curiouser on July 02, 2019, 05:43:45 PM

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."

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.

As has already been covered in this thread, it does not appear to move 15° per hour most of the time. It varies throughout the year.

It does not appear to move at very close to 15° per hour all of the time.

The variation in the rate of movement through the year is very slight and accumulates to only about 12 minutes fast or slow over months.

This still seems to be causing confusion among several posters who IMO generally seem to know their stuff.

Let’s try this thought experiment (or actual experiment if you have the equipment):

Imagine you are tracking a star 2 degs from the celestial pole  through a telescope with an equatorial mount.  Every hour, you turn the telescope approximately 15 degs, ignoring the sidereal day difference.

However this motion of the telescope is mainly just spinning it along its axis.  The angle the telescope is actually pointing at hardly changes.

It’s the same principle with the sun, to a much lesser extent.  So tracking the sun at 15 deg per hour around the Earth’s axis with an equatorial mount or even a sundial IS NOT THE SAME as 15 deg per hour change in position in the sky relative to an observer on Earth.  Except on the equinoxes. 

The difference isn’t much even at the solstices, but it is real.

Of course none of that changes that this is very hard (or impossible) to explain with standard flat earth “models”.  So C&C’s argument is perhaps a bit pedantic, but that’s still the argument everyone seems to be struggling with, and as far as I can tell, it’s correct.

Edit:  This is not about the difference between a solar day and a sidereal day, nor is it about the equation of time (although that will affect results slightly as well).

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.

Polaris surely follows the 360 degrees path around the celestial pole, no matter how close.
On that path it travels 15 degrees per hour.
At one moment it will be due east, and roughly 12 hours later it will be due west, 180 degrees away.

Alpha UMI Aa sits at the declination of 89° 15' 50.8" (89,264 degrees, or 0.736 degrees away from the celestial pole).
From the point of view of some random observer anywhere at the Earth's surface the total daily movement range will cover only 1.472 degrees.

During one day speed of the apparent movement is constant.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, let's try to predict the Sun's apparent traveling angles before we test them in some actual observation:

[spoiler='Sun Virtual Path Basic Data']

R: Rowbotham's model, Sun for equinoctial solar noon from 45 degrees north seen at 45 degrees elevation, which gives the height of 3110 miles = 5005 km
G: Globe model, Sun's distance from the ground 149 ± 3 million km
(the times are solar, the solar noon is at 12:00 pm)

1. England, Germany, Poland is mostly around 52 degrees north.
2. Boston, New York, Philadelphia, Washington (and Madrid, Rome, Istanbul, Beijing) are grouped around 40 degrees north.
3. Sydney, Canberra, Melbourne, Adelaide (and Santiago, Buenos Aires, Montevideo) are around 35 degrees south.

J: June solstice.
Sun's trajectory
JR: curve radius 7389 km,
     center of the curve 5005 km above the North pole,
     tangential speed 1934 km/h.
JG: curve radius 152 000 000 x cos(23.5) = 139 393 313 km,
      center of the curve 152 000 000 x sin(23.5) = 60 609 858 km above the center of the Earth,
      tangential speed 36 493 084 km/h.

S: September equinox.
Sun's trajectory
SR: curve radius 10 000 km,
      center of the curve 5005 km above the North pole
      tangential speed 2618 km/h.
SG: curve radius 149 000 000 km,
      center of the curve in the center of the Earth,
      tangential speed 39 008 109 km/h.

D: December solstice.
Sun's trajectory
DR: curve radius 12 611 km,
      center of the curve 5005 km above the North pole,
      tangential speed 3302 km/h.
DG: curve radius 146 000 000 x cos(23.5) = 133 890 771 km,
      center of the curve 146 000 000 x sin(23.5) = 58 217 364 km below the cenetr of the Earth,
      tangential speed 35 052 522 km/h.

(By now you already understood what would here be the meaning of, say, SR2: S - September Equinox, R - Rowbotham's model, 2 - observer is at 40 degrees north)

EDIT: Looks like "Spoiler" tag doesn't work on this forum.
[/spoiler]

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

JR1: 6 am / 6 pm - 10.45 degrees per hour, 12 pm - 18 degrees per hour
JG1: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour

JR2: 6 am / 6 pm - 8.35 degrees per hour, 12 pm - 20 degrees per hour
JG2: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour

JR3: 6 am / 6 pm - 3.75 degrees per hour, 12 pm - 16.5 degrees per hour
JG3: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour


SR1: 6 am / 6 pm - 11.5 degrees per hour, 12 pm - 19 degrees per hour
SG1: 6 am / 6 pm - 14.67 degrees per hour, 12 pm - 14.67 degrees per hour

SR2: 6 am / 6 pm - 10.4 degrees per hour, 12 pm - 21.3 degrees per hour
SG2: 6 am / 6 pm - 14.67 degrees per hour, 12 pm - 14.67 degrees per hour

SR3: 6 am / 6 pm - 4.9 degrees per hour, 12 pm - 22.5 degrees per hour
SG3: 6 am / 6 pm - 14.67 degrees per hour, 12 pm - 14.67 degrees per hour


DR1: 6 am / 6 pm - 12.4 degrees per hour, 12 pm - 18.9 degrees per hour
DG1: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour

DR2: 6 am / 6 pm - 18.7 degrees per hour, 12 pm - 21.3 degrees per hour
DG2: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour

DR3: 6 am / 6 pm - 6.5 degrees per hour, 12 pm - 32.5 degrees per hour
DG3: 6 am / 6 pm - 13.5 degrees per hour, 12 pm - 13.5 degrees per hour

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Obviously (without refraction) the apparent Sun's angular speed will vary much more in Rowbotham's model than in Globe model.

In Rowbotham's model it depends on location, and can be as low as 4 degrees in the morning (in Soth pole even lower),
and as high as 32 degrees (on Equator even higher).

In Globe model varies during a year, from 13.5 to 14.67 degrees, and during a day it remains practically constant, regardless the observer's location on the Earth.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, how fast is the Sun really traveling across the sky in the places where each of us lives?

Yes.

This is all C&C was trying to say.