<blockquote>
<p>At the equator, there is no geometric effect at all—the Sun's daily track above the horizon is always exactly half a circle.</p>
</blockquote>
<p>The issue is the definition of &quot;half a circle&quot;. What this refers to is how long the Sun (or any star for that matter) is above the horizon when the observer is at the equator. If you ignore atmospheric refraction and use the center of the Sun, then sunrise to sunset (or star rise to star set) is 12 hours long. 12 hours is half of 24 hours, so that is half a circle.</p>
<p>Let's take an extreme example. Polaris is approximately 0.7 degrees from the celestial north pole. Therefore, Polaris travels in a small circle (radius of 0.7 degrees) around the pole. It does not move very much. But Polaris is visible for 12 hours from rising to setting, so it is visible for exactly half a circle.</p>
<p>Another way to express this is by using the hour angle when the object rises or sets. The hour angle is the number of degrees (or hours of right ascension) between the object and the meridian. Specifically, the angle is measured between two planes:</p>
<ul>
<li>The plane of the meridian. The meridian is the lighter red line in the figures below, connecting due north (N), the zenith (Z), and due south (S).</li>
<li>The plane passing through celestial north, the object in question, and celestial south.</li>
</ul>
<p>When the observer is at the equator, all objects have an hour angle of 6 hours when rising.</p>
<p><a href="https://i.sstatic.net/ZBYp2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZBYp2.png" alt="Observer at the equator" /></a>
Figure 1: Observer at the equator. If the line of 0h Right Ascension (RA) is on the meridian (the lighter red line), then 6h RA is on the horizon all the way from due north (N) to east (E) to south (S). <em>Any object</em> that is rising has an hour angle of 6h. (This is true regardless of what line of RA is on the meridian.) Likewise, the hour angle is 6h when setting. 6h+6h=12h which is half of a full circle of 24h.</p>
<p>For observers north of the equator,</p>
<ul>
<li>Objects north of the equator have an hour angle larger than 6 hours when rising. Therefore, they are visible for more than &quot;half a circle&quot;.</li>
<li>Objects south of the equator have an hour angle smaller than 6 hours when rising. Therefore, they are visible for less than &quot;half a circle&quot;.</li>
</ul>
<p><a href="https://i.sstatic.net/7YKyQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7YKyQ.png" alt="Observer at +30 degrees latitude." /></a>
Figure 2: Observer at +30 degrees latitude. The blue lines are declination lines; the lighter blue line at 0 declination is the celestial equator. The hour angle of an object that is rising depends on the declination of the object. When the Sun is at +23 degrees declination, the hour angle is approximately 7h. When the Sun is at -23 degrees declination, the hour angle is approximately 5h. When the Sun is at 0 degrees declination, the hour angle is 6 hours, and this is true regardless of the observer's latitude (except at the poles).</p>
<p><strong>Time from rising to the meridian</strong>. There is a relationship between the hour angle and the time between when an object rises and when it reaches the meridian.</p>
<ul>
<li>For stars and planets (which move slowly day to day), consider these objects as being attached to the dome of the sky. Since the sky rotates through 24h RA in 23 hours 56 minutes of clock time, the clock time approximately equals the hour angle. (That is, 24h RA/23.93h clock time approximately equals 1.)</li>
<li>By definition of &quot;noon&quot;, the Sun appears to move 360 degrees (or 24h RA) around the sky in 24 hours of clock time. Thus, the time from rising to reaching the meridian equals the hour angle. At 0 latitude, Figure 1 shows that the Sun always rises 6 hours prior to noon, regardless of the date. At other latitudes, the time depends on the declination of the Sun, but the time always equals the hour angle.</li>
<li>For the Moon, the hour angle when rising is still shown by Figure 1 (always 6h) and Figure 2 (depends on the declination). But the Moon requires approximately 25 hour to rotate through 360 degrees (24h RA). Therefore, the time between moonrise and reaching the meridian is approximately 25/24=1.04 times longer than the hour angle.</li>
</ul>
<p>As mentioned at the beginning, sunrise (and moonrise) is based on when the limb of the Sun (or Moon) becomes visible and includes the effect of atmospheric refraction which makes the object appear to be higher in the sky. Also, observing from a high altitude where the horizon is more than 90 degrees from the zenith would affect the hour angle. All those types of effects are ignored in the previous discussion.</p>

CC BY-SA 4.0
2024-01-10T19:59:49.687Z
If Polaris travels so closely to the axis of the celestial north pole, then shouldn't it be visible 24 hours to observers in the Northern Hemisphere (although sunlight makes it practically invisible during daytime) and invisible to the Southern Hemisphere?
Jacob Miller
CC BY-SA 4.0
2024-01-10T20:12:23.060Z
At 0.7 degrees S latitude (and farther south), Polaris never rises. At the equator, Polaris is above the horizon for 12 hours. At 0.7 degrees N latitude (and farther north), Polaris is above the horizon for 24 hours a day.
JohnHoltz
CC BY-SA 4.0
2024-01-13T16:08:27.373Z
"When the observer is at the equator, all objects have an hour angle of 6 hours when rising": This is true for objects that don't orbit Earth (or at least orbital frequency is negligible compared to Earth's rotational frequency). Geo-synchronised sats for example are stationary for 24 hours. And the Moon ~ 6 hours 25 mins when rising. I know that you meant this regarding Sol and stars, but shouldn't "all objects" be better replaced by "relatively stationary objects"?
Jacob Miller
CC BY-SA 4.0
2024-01-14T00:36:12.500Z
"When the observer is at the equator, all objects have an hour angle of 6 hours when rising". This IS true for the Moon also (maybe if you ignore the horizontal parallax). The Right Ascension (RA) of the Moon minus the RA of the meridian is 6 hours. You are correct that in the case of the Moons, it takes longer than 6 hours for the Moon to reach the meridian because of its motion. Otherwise, the statement works out to be correct.
JohnHoltz
CC BY-SA 4.0
2024-01-14T04:39:24.853Z
But the Right Ascension (RA) of the Moon minus the RA of the meridian should be 25 minutes. This means that the Moon would reach the meridian about 25 minutes later than 6 hours. If the Right Ascension (RA) of the Moon minus the RA of the meridian is 6 hours then the moon will be visible for 18 hours because the moon will rise 6 hours earlier than the usual 6 hours quarter circle. :)
Jacob Miller
CC BY-SA 4.0
2024-01-14T12:30:25.813Z
Also one of the other things is altitude. More the altitude of the viewer more of the portion of the celestial sphere he can view, so objects even at the equator may be visible for more than 12 hours. Or at negative altitudes less of a field view.
Jacob Miller
CC BY-SA 4.0
2024-01-15T00:46:36.210Z
Can you kindly explain how can the Right Ascension (RA) of the Moon minus the RA of the meridian be 6 hours as you stated in the comment? :)
Jacob Miller
CC BY-SA 4.0
2024-01-15T03:49:47.437Z
It will require an image which will take me a few days to make. Basically, from the equator the horizon is the same as a line of RA, and the hour angle is the angle between the plane containing the meridian-observer and the plane containing the Earth's axis-rising object. The second plane coincides with the horizon.  The angle between the meridian and horizon is 90 degrees which equals 6 hours of right Ascension.
JohnHoltz
CC BY-SA 4.0
2024-01-15T03:57:31.100Z
I think you are confusing the hour angle with the time from rising to the meridian. For stars that is true (approximately). For the Sun that is almost true (because the Sun only moves 1/4 degree in 6 hours). For the Moon the time is different because of the Moon's large daily motion.
JohnHoltz
CC BY-SA 4.0
2024-01-16T04:05:52.220Z
Thank you very much for helping me understand.  :) So basically it is a measure of the angle of the diurnal arc of the celestial object in degrees expressed in hours. But, in your comment, why "almost" for the sun? The average time delay due to the apparent solar movement of 1 degree of Earth's rotation per day due to Earth's orbit around the sun is included in 24 hours. Without that, it is ~23 hours and 56 minutes which is called the stellar day.
Jacob Miller
CC BY-SA 4.0
2024-01-16T20:38:57.970Z
Yes on the diurnal angle. Good way to describe it. Since 6 hours RA equals 90 degrees, the sky needs to rotate 90+1/4 degrees for the Sun to move from rising to transit. At 360 degrees per 23 hours 56 minutes, that gives a sunrise to transit time of 6 hours (to 4 decimal places). I was not expecting that! Keep in mind this calc is based on theory (no atmospheric refraction, sunrise based on center of the Sun), so the real values will differ slightly. I will review my original answer, make adjustments, and add some figures to try to clarify based on the comments.
JohnHoltz
CC BY-SA 4.0
2024-01-23T23:07:20.323Z
This answer has many good points, but the argument about the visibility of Polaris is simply wrong. The North pole is visible for any place on or North of the equator for 24 hours a day, the whole year round. start stellarium, disable atmosphere, go fast forward, and watch :)
planetmaker
CC BY-SA 4.0
2024-01-24T04:21:20.287Z
@planetmaker Can I insert an image into a comment? Maybe not. Stellarium Online shows the altitude of Polaris at lower culmination as -0° 37' from latitude +0.01099°. That puts it below the theoretical horizon (as it should be). The "fake" horizon shown by the imagery is a few degrees too low compared to the theoretical horizon. (I think I am not that tall for the dip of the horizon to be a few degrees :-)
JohnHoltz
CC BY-SA 4.0
2024-01-24T07:17:31.280Z
Not in comments. But you are welcome and encouraged to amend your answer with additional explanation. Given refraction even - 0.7° will always be visible on the equator
planetmaker
CC BY-SA 4.0
2024-01-24T09:53:18.850Z
@JohnHoltz if I want to put an image in comments, I start an answer post that I will delete moments later. I then click the insert image button and generate a link for the image within stack exchange, then paste the link into my comment like this https://i.sstatic.net/fXKu6.png and then delete the dummy answer post.
uhoh
CC BY-SA 4.0
2024-01-29T06:02:19.750Z
@uhoh  Is it necessary to post the answer or is it enough to insert the image at the draft stage and discard the draft without posting the answer?
Jacob Miller
CC BY-SA 4.0
2024-01-29T07:29:08.003Z
@JacobMiller No, not necessary to post an answer. I use unposted answer areas as sandboxes all the time, mostly to make links for images just like I did here. There's always a small danger (especially when sandboxing a large passage or lots of MathJax formatting) that one accidentally clicks "Post Your Answer" by mistake, but that's not the end of the world (I've done it once or twice) you just edit, delete your work, and then type some simple blurb like "oops, I was sandboxing and accidentally hit post", save your edit, then click delete. But if you don't mistakenly click post, it goes away.
uhoh
CC BY-SA 4.0
2024-01-31T13:54:46.797Z
The "geometric effect" is the variation in the length of the Sun's track across the sky as a function of the season which is due to the Earth's axial tilt relative to its orbit around the Sun. This affects sun rising and setting times. As per this answer, for an observer at the equator, the diurnal arc of the sun throughout all the days of a year is 180$^{\circ}$ or an hour angle of 12 hours. So "At the equator, there is no geometric effect at all — the Sun's daily track above the horizon is always exactly half a circle." is a correct statement.
Jacob Miller
CC BY-SA 4.0
2024-02-01T06:36:43.600Z
RE: "This IS true for the Moon also (maybe if you ignore the horizontal parallax)." Isn't the parallax irrelevant for an unmoving observer?
Jacob Miller
CC BY-SA 4.0
2024-02-01T07:41:27.707Z
@uhoh I meant that in the context of the comment. Yes, the Moon appears to move faster at the observer's local meridian than at the horizon. But why shall it change the hour angle? It just makes the apparent angular velocity non-linear, But otherwise, the 12-hour diurnal arc for the equator should stay the same. :)
Jacob Miller
CC BY-SA 4.0
2024-02-01T10:49:46.867Z
@JacobMiller let's say that the moon has a circular orbit around a spherical Earth. Let's say the Moon's orbit's radius = 1 and the Earth's radius is 0.9. For a given spot on the Earth it will only be above the horizon for +/-28.5° of orbital motion, or ±1.7 hours with respect to its crossing of the local meridian, not ±6 hours. Now for the actual radius of Earth of 0.018 in this scheme, the effect is only a few minutes. Actually it has nothing to do with the Sun (sorry about that) it's strictly a finite size Earth/finite size lunar orbit thing, but I would still call it parallax.
uhoh
CC BY-SA 4.0
2024-02-01T10:51:04.953Z
@JacobMiller The Earth's radius has to shrink to zero before you get the ±6 hours.
uhoh
CC BY-SA 4.0
2024-02-03T06:53:21.010Z
@uhoh Thanks. I damned missed it. You are correct indeed. The earth below limits our viewing window of the "centre of the moon" because we are offset from the axis of rotation. The same effect for the centre of the Sun is likely negligible though because of the distance. I used "centre" because the comparison should be against the same horizontal distance from the edge of the object or centre. Otherwise, the angular size of the object counteracts this effect.
Jacob Miller
CC BY-SA 4.0
2024-02-03T11:35:21.093Z
Lunar parallax is quite important. And amateurs around the world have used it to contribute useful data about the lunar surface by doing precise transit timings. Also, the Moon is so close relative to the Earth radius that the Moon's angular diameter is slightly larger when it crosses the meridian compared to when it's rising. Which makes the Moon Illusion even more surprising. ;)
PM 2Ring
CC BY-SA 4.0
2024-02-04T00:22:34.970Z
@JacobMiller the R_earth = 0.9 example was already familiar to me because it corresponds to an artificial satellite in LEO at about 630 km altitude. Hanging out in [Space SE](https://space.stackexchange.com/) too much, I'm always thinking about satellites :-)
uhoh
CC BY-SA 4.0
2024-01-10T14:30:21.450Z
2024-01-23T18:14:45.603Z
2024-01-23T18:14:45.603Z
JohnHoltz
<p>This question is about an explanation of &quot;Sunrise and Sunset Times Near the Solstices&quot; on the official website of the US Naval Observatory.</p>
<p>It is mentioned in the explanation article that,</p>
<blockquote>
<p><a href="https://aa.usno.navy.mil/faq/rs_solstices#:%7E:text=At%20the%20equator%2C%20there%20is%20no%20geometric%20effect%20at%20all%E2%80%94the%20Sun%27s%20daily%20track%20above%20the%20horizon%20is%20always%20exactly%20half%20a%20circle.%20The%20clock%20effect%20completely%20determines%20when%20sunrise%20and%20sunset%20occur%20there" rel="nofollow noreferrer">At the equator, there is no geometric effect at all—the Sun's daily
track above the horizon is always exactly half a circle. The clock
effect completely determines when sunrise and sunset occur there.</a></p>
</blockquote>
<p>Now this seems inaccurate to me. Yes, the Geometric effect which occurs due to the seasonal decline of the sun is minimised at the equator relative to higher or lower latitudes.</p>
<ul>
<li>But, how can the <strong>geometric effect be eliminated at the equator</strong> as the sun as observed from the equator still <strong>travels from the Tropic of Cancer to the Tropic of Capricorn</strong> (<em>as this should also cause a geometric effect, although minimal relative to other latitudes, still significant relative to the clock effect which is the effect on the observed position of the sun due to the discrepancy between solar time and clock time</em>)?</li>
</ul>

CC BY-SA 4.0
2024-01-10T12:02:56.440Z
Excellent question. I will have to think more about it. Yet on the equator  the Sun always rises straight East, and sets straight West - only the position the Sun is at noon varies by season (and thus how the half circle the Sun moves on is arcing over the sky).
planetmaker
CC BY-SA 4.0
2024-01-10T12:45:45.953Z
@planetmaker It is not exactly a half-circle throughout the year. It is less than half a circle near winter and summer solstices although not as much a minor circle like at higher or lower latitudes. Rising straight east or setting straight west does not always represent a half circle. :)
Jacob Miller
CC BY-SA 4.0
2024-01-10T12:57:22.487Z
Any straight path on a sphere from exactly East to exactly West must be a half circle - what could it be otherwise? The rise and set positions of the Sun only changes during the year at lattitudes different to the equator.
planetmaker
CC BY-SA 4.0
2024-01-10T13:19:21.860Z
@planetmaker Indeed. My mistake. :) Yet It is not exactly a half-circle throughout the year because the Sun does not always rise straight East, and set straight West on the equator. It deviates from East-West, away from ~March 20 and ~September 23 equinoxes.
Jacob Miller
CC BY-SA 4.0
2024-01-10T15:02:25.360Z
The Sun's azimuth at sunrise & sunset certainly varies throughout the year, even at the equator. I have some diagrams here, https://astronomy.stackexchange.com/a/39685/16685 They were calculated using *very* simple Sun position equations, but they're ok as a first approximation. ;)
PM 2Ring
CC BY-SA 4.0
2024-01-10T18:29:47.697Z
@PM2Ring Those are some nice diagrams :) They represent considerable variation during the year even at lower latitudes.  Just to be sure am I right to consider it a bit of an oversight by the author to disregard, the geometrical effect as a contributing factor, for sunrise & sunset time along the Equator also?
Jacob Miller
CC BY-SA 4.0
2024-01-10T18:36:45.420Z
@Jacob Thanks! As the diagram shows, the apparent solar time (sundial time) of sunrise & sunset at the equator is exactly 6 o'clock, all year round.
PM 2Ring
CC BY-SA 4.0
2024-01-10T20:13:49.860Z
@PM2Ring I don't understand. What happens when the sun is apparently over the Tropics? Shouldn't this reduce daylight? There is almost no variation near the equator. This puzzles me. https://www.timeanddate.com/sun/brazil/macapa
Jacob Miller
CC BY-SA 4.0
2024-01-10T23:47:42.887Z
At the equator, the celestial poles are on the horizon, so the diurnal circles in a star trail photo are cut exactly in half. https://en.wikipedia.org/wiki/Star_trail
PM 2Ring
<observational-astronomy><positional-astronomy><sunrise><sunset><solstice>
Geometric effect on sun's position in the sky at the equator
Jacob Miller
# Geometric effect on sun's position in the sky at the equator

(Asked by: username_0 on 2024-01-09)

This question is about an explanation of "Sunrise and Sunset Times Near the Solstices" on the official website of the US Naval Observatory.

It is mentioned in the explanation article that,

> [At the equator, there is no geometric effect at all—the Sun's daily
> track above the horizon is always exactly half a circle. The clock
> effect completely determines when sunrise and sunset occur there.](https://aa.usno.navy.mil/faq/rs_solstices#:%7E:text=At%20the%20equator%2C%20there%20is%20no%20geometric%20effect%20at%20all%E2%80%94the%20Sun%27s%20daily%20track%20above%20the%20horizon%20is%20always%20exactly%20half%20a%20circle.%20The%20clock%20effect%20completely%20determines%20when%20sunrise%20and%20sunset%20occur%20there)

Now this seems inaccurate to me. Yes, the Geometric effect which occurs due to the seasonal decline of the sun is minimised at the equator relative to higher or lower latitudes.

* But, how can the **geometric effect be eliminated at the equator** as the sun as observed from the equator still **travels from the Tropic of Cancer to the Tropic of Capricorn** (*as this should also cause a geometric effect, although minimal relative to other latitudes, still significant relative to the clock effect which is the effect on the observed position of the sun due to the discrepancy between solar time and clock time*)?

**Tags:** observational-astronomy, positional-astronomy, sunrise, sunset, solstice

> **Comments:**
> **username_1**: Excellent question. I will have to think more about it. Yet on the equator  the Sun always rises straight East, and sets straight West - only the position the Sun is at noon varies by season (and thus how the half circle the Sun moves on is arcing over the sky).  
> **username_0**: @username_1 It is not exactly a half-circle throughout the year. It is less than half a circle near winter and summer solstices although not as much a minor circle like at higher or lower latitudes. Rising straight east or setting straight west does not always represent a half circle. :)  
> **username_1**: Any straight path on a sphere from exactly East to exactly West must be a half circle - what could it be otherwise? The rise and set positions of the Sun only changes during the year at lattitudes different to the equator.  
> **username_0**: @username_1 Indeed. My mistake. :) Yet It is not exactly a half-circle throughout the year because the Sun does not always rise straight East, and set straight West on the equator. It deviates from East-West, away from ~March 20 and ~September 23 equinoxes.  
> **username_2**: The Sun's azimuth at sunrise & sunset certainly varies throughout the year, even at the equator. I have some diagrams here, https://astronomy.stackexchange.com/a/39685/16685 They were calculated using *very* simple Sun position equations, but they're ok as a first approximation. ;)  
> **username_0**: @PM2Ring Those are some nice diagrams :) They represent considerable variation during the year even at lower latitudes.  Just to be sure am I right to consider it a bit of an oversight by the author to disregard, the geometrical effect as a contributing factor, for sunrise & sunset time along the Equator also?  
> **username_2**: @Jacob Thanks! As the diagram shows, the apparent solar time (sundial time) of sunrise & sunset at the equator is exactly 6 o'clock, all year round.  
> **username_0**: @PM2Ring I don't understand. What happens when the sun is apparently over the Tropics? Shouldn't this reduce daylight? There is almost no variation near the equator. This puzzles me. https://www.timeanddate.com/sun/brazil/macapa  
> **username_2**: At the equator, the celestial poles are on the horizon, so the diurnal circles in a star trail photo are cut exactly in half. https://en.wikipedia.org/wiki/Star_trail  

---

## 1 answers

### Answer by username_3 on 2024-01-10. Score: 6 | Accepted answer

> At the equator, there is no geometric effect at all—the Sun's daily track above the horizon is always exactly half a circle.

The issue is the definition of "half a circle". What this refers to is how long the Sun (or any star for that matter) is above the horizon when the observer is at the equator. If you ignore atmospheric refraction and use the center of the Sun, then sunrise to sunset (or star rise to star set) is 12 hours long. 12 hours is half of 24 hours, so that is half a circle.

Let's take an extreme example. Polaris is approximately 0.7 degrees from the celestial north pole. Therefore, Polaris travels in a small circle (radius of 0.7 degrees) around the pole. It does not move very much. But Polaris is visible for 12 hours from rising to setting, so it is visible for exactly half a circle.

Another way to express this is by using the hour angle when the object rises or sets. The hour angle is the number of degrees (or hours of right ascension) between the object and the meridian. Specifically, the angle is measured between two planes:

* The plane of the meridian. The meridian is the lighter red line in the figures below, connecting due north (N), the zenith (Z), and due south (S).
* The plane passing through celestial north, the object in question, and celestial south.

When the observer is at the equator, all objects have an hour angle of 6 hours when rising.

[![Observer at the equator](https://i.sstatic.net/ZBYp2.png)](https://i.sstatic.net/ZBYp2.png)
Figure 1: Observer at the equator. If the line of 0h Right Ascension (RA) is on the meridian (the lighter red line), then 6h RA is on the horizon all the way from due north (N) to east (E) to south (S). *Any object* that is rising has an hour angle of 6h. (This is true regardless of what line of RA is on the meridian.) Likewise, the hour angle is 6h when setting. 6h+6h=12h which is half of a full circle of 24h.

For observers north of the equator,

* Objects north of the equator have an hour angle larger than 6 hours when rising. Therefore, they are visible for more than "half a circle".
* Objects south of the equator have an hour angle smaller than 6 hours when rising. Therefore, they are visible for less than "half a circle".

[![Observer at +30 degrees latitude.](https://i.sstatic.net/7YKyQ.png)](https://i.sstatic.net/7YKyQ.png)
Figure 2: Observer at +30 degrees latitude. The blue lines are declination lines; the lighter blue line at 0 declination is the celestial equator. The hour angle of an object that is rising depends on the declination of the object. When the Sun is at +23 degrees declination, the hour angle is approximately 7h. When the Sun is at -23 degrees declination, the hour angle is approximately 5h. When the Sun is at 0 degrees declination, the hour angle is 6 hours, and this is true regardless of the observer's latitude (except at the poles).

**Time from rising to the meridian**. There is a relationship between the hour angle and the time between when an object rises and when it reaches the meridian.

* For stars and planets (which move slowly day to day), consider these objects as being attached to the dome of the sky. Since the sky rotates through 24h RA in 23 hours 56 minutes of clock time, the clock time approximately equals the hour angle. (That is, 24h RA/23.93h clock time approximately equals 1.)
* By definition of "noon", the Sun appears to move 360 degrees (or 24h RA) around the sky in 24 hours of clock time. Thus, the time from rising to reaching the meridian equals the hour angle. At 0 latitude, Figure 1 shows that the Sun always rises 6 hours prior to noon, regardless of the date. At other latitudes, the time depends on the declination of the Sun, but the time always equals the hour angle.
* For the Moon, the hour angle when rising is still shown by Figure 1 (always 6h) and Figure 2 (depends on the declination). But the Moon requires approximately 25 hour to rotate through 360 degrees (24h RA). Therefore, the time between moonrise and reaching the meridian is approximately 25/24=1.04 times longer than the hour angle.

As mentioned at the beginning, sunrise (and moonrise) is based on when the limb of the Sun (or Moon) becomes visible and includes the effect of atmospheric refraction which makes the object appear to be higher in the sky. Also, observing from a high altitude where the horizon is more than 90 degrees from the zenith would affect the hour angle. All those types of effects are ignored in the previous discussion.

> **Comments:**
> **username_0**: If Polaris travels so closely to the axis of the celestial north pole, then shouldn't it be visible 24 hours to observers in the Northern Hemisphere (although sunlight makes it practically invisible during daytime) and invisible to the Southern Hemisphere?  
> **username_3**: At 0.7 degrees S latitude (and farther south), Polaris never rises. At the equator, Polaris is above the horizon for 12 hours. At 0.7 degrees N latitude (and farther north), Polaris is above the horizon for 24 hours a day.  
> **username_0**: "When the observer is at the equator, all objects have an hour angle of 6 hours when rising": This is true for objects that don't orbit Earth (or at least orbital frequency is negligible compared to Earth's rotational frequency). Geo-synchronised sats for example are stationary for 24 hours. And the Moon ~ 6 hours 25 mins when rising. I know that you meant this regarding Sol and stars, but shouldn't "all objects" be better replaced by "relatively stationary objects"?  
> **username_3**: "When the observer is at the equator, all objects have an hour angle of 6 hours when rising". This IS true for the Moon also (maybe if you ignore the horizontal parallax). The Right Ascension (RA) of the Moon minus the RA of the meridian is 6 hours. You are correct that in the case of the Moons, it takes longer than 6 hours for the Moon to reach the meridian because of its motion. Otherwise, the statement works out to be correct.  
> **username_0**: But the Right Ascension (RA) of the Moon minus the RA of the meridian should be 25 minutes. This means that the Moon would reach the meridian about 25 minutes later than 6 hours. If the Right Ascension (RA) of the Moon minus the RA of the meridian is 6 hours then the moon will be visible for 18 hours because the moon will rise 6 hours earlier than the usual 6 hours quarter circle. :)  
> **username_0**: Also one of the other things is altitude. More the altitude of the viewer more of the portion of the celestial sphere he can view, so objects even at the equator may be visible for more than 12 hours. Or at negative altitudes less of a field view.  
> **username_0**: Can you kindly explain how can the Right Ascension (RA) of the Moon minus the RA of the meridian be 6 hours as you stated in the comment? :)  
> **username_3**: It will require an image which will take me a few days to make. Basically, from the equator the horizon is the same as a line of RA, and the hour angle is the angle between the plane containing the meridian-observer and the plane containing the Earth's axis-rising object. The second plane coincides with the horizon.  The angle between the meridian and horizon is 90 degrees which equals 6 hours of right Ascension.  
> **username_3**: I think you are confusing the hour angle with the time from rising to the meridian. For stars that is true (approximately). For the Sun that is almost true (because the Sun only moves 1/4 degree in 6 hours). For the Moon the time is different because of the Moon's large daily motion.  
> **username_0**: Thank you very much for helping me understand.  :) So basically it is a measure of the angle of the diurnal arc of the celestial object in degrees expressed in hours. But, in your comment, why "almost" for the sun? The average time delay due to the apparent solar movement of 1 degree of Earth's rotation per day due to Earth's orbit around the sun is included in 24 hours. Without that, it is ~23 hours and 56 minutes which is called the stellar day.  
> **username_3**: Yes on the diurnal angle. Good way to describe it. Since 6 hours RA equals 90 degrees, the sky needs to rotate 90+1/4 degrees for the Sun to move from rising to transit. At 360 degrees per 23 hours 56 minutes, that gives a sunrise to transit time of 6 hours (to 4 decimal places). I was not expecting that! Keep in mind this calc is based on theory (no atmospheric refraction, sunrise based on center of the Sun), so the real values will differ slightly. I will review my original answer, make adjustments, and add some figures to try to clarify based on the comments.  
> **username_1**: This answer has many good points, but the argument about the visibility of Polaris is simply wrong. The North pole is visible for any place on or North of the equator for 24 hours a day, the whole year round. start stellarium, disable atmosphere, go fast forward, and watch :)  
> **username_3**: @username_1 Can I insert an image into a comment? Maybe not. Stellarium Online shows the altitude of Polaris at lower culmination as -0° 37' from latitude +0.01099°. That puts it below the theoretical horizon (as it should be). The "fake" horizon shown by the imagery is a few degrees too low compared to the theoretical horizon. (I think I am not that tall for the dip of the horizon to be a few degrees :-)  
> **username_1**: Not in comments. But you are welcome and encouraged to amend your answer with additional explanation. Given refraction even - 0.7° will always be visible on the equator  
> **username_4**: @username_3 if I want to put an image in comments, I start an answer post that I will delete moments later. I then click the insert image button and generate a link for the image within stack exchange, then paste the link into my comment like this https://i.sstatic.net/fXKu6.png and then delete the dummy answer post.  
> **username_0**: @username_4  Is it necessary to post the answer or is it enough to insert the image at the draft stage and discard the draft without posting the answer?  
> **username_4**: @JacobMiller No, not necessary to post an answer. I use unposted answer areas as sandboxes all the time, mostly to make links for images just like I did here. There's always a small danger (especially when sandboxing a large passage or lots of MathJax formatting) that one accidentally clicks "Post Your Answer" by mistake, but that's not the end of the world (I've done it once or twice) you just edit, delete your work, and then type some simple blurb like "oops, I was sandboxing and accidentally hit post", save your edit, then click delete. But if you don't mistakenly click post, it goes away.  
> **username_0**: The "geometric effect" is the variation in the length of the Sun's track across the sky as a function of the season which is due to the Earth's axial tilt relative to its orbit around the Sun. This affects sun rising and setting times. As per this answer, for an observer at the equator, the diurnal arc of the sun throughout all the days of a year is 180$^{\circ}$ or an hour angle of 12 hours. So "At the equator, there is no geometric effect at all — the Sun's daily track above the horizon is always exactly half a circle." is a correct statement.  
> **username_0**: RE: "This IS true for the Moon also (maybe if you ignore the horizontal parallax)." Isn't the parallax irrelevant for an unmoving observer?  
> **username_0**: @username_4 I meant that in the context of the comment. Yes, the Moon appears to move faster at the observer's local meridian than at the horizon. But why shall it change the hour angle? It just makes the apparent angular velocity non-linear, But otherwise, the 12-hour diurnal arc for the equator should stay the same. :)  
> **username_4**: @JacobMiller let's say that the moon has a circular orbit around a spherical Earth. Let's say the Moon's orbit's radius = 1 and the Earth's radius is 0.9. For a given spot on the Earth it will only be above the horizon for +/-28.5° of orbital motion, or ±1.7 hours with respect to its crossing of the local meridian, not ±6 hours. Now for the actual radius of Earth of 0.018 in this scheme, the effect is only a few minutes. Actually it has nothing to do with the Sun (sorry about that) it's strictly a finite size Earth/finite size lunar orbit thing, but I would still call it parallax.  
> **username_4**: @JacobMiller The Earth's radius has to shrink to zero before you get the ±6 hours.  
> **username_0**: @username_4 Thanks. I damned missed it. You are correct indeed. The earth below limits our viewing window of the "centre of the moon" because we are offset from the axis of rotation. The same effect for the centre of the Sun is likely negligible though because of the distance. I used "centre" because the comparison should be against the same horizontal distance from the edge of the object or centre. Otherwise, the angular size of the object counteracts this effect.  
> **username_2**: Lunar parallax is quite important. And amateurs around the world have used it to contribute useful data about the lunar surface by doing precise transit timings. Also, the Moon is so close relative to the Earth radius that the Moon's angular diameter is slightly larger when it crosses the meridian compared to when it's rising. Which makes the Moon Illusion even more surprising. ;)  
> **username_4**: @JacobMiller the R_earth = 0.9 example was already familiar to me because it corresponds to an artificial satellite in LEO at about 630 km altitude. Hanging out in [Space SE](https://space.stackexchange.com/) too much, I'm always thinking about satellites :-)  

---

