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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.

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)?
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    $\begingroup$ 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). $\endgroup$ Jan 10 at 12:02
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    $\begingroup$ @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. :) $\endgroup$ Jan 10 at 12:45
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    $\begingroup$ 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. $\endgroup$ Jan 10 at 12:57
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    $\begingroup$ The Sun's azimuth at sunrise & sunset certainly varies throughout the year, even at the equator. I have some diagrams here, astronomy.stackexchange.com/a/39685/16685 They were calculated using very simple Sun position equations, but they're ok as a first approximation. ;) $\endgroup$
    – PM 2Ring
    Jan 10 at 15:02
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    $\begingroup$ At the equator, the celestial poles are on the horizon, so the diurnal circles in a star trail photo are cut exactly in half. en.wikipedia.org/wiki/Star_trail $\endgroup$
    – PM 2Ring
    Jan 10 at 23:47

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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 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. 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.

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    $\begingroup$ 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. $\endgroup$
    – JohnHoltz
    Jan 10 at 20:12
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    $\begingroup$ 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. $\endgroup$
    – JohnHoltz
    Jan 15 at 3:57
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    $\begingroup$ 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 $\endgroup$ Jan 24 at 7:17
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    $\begingroup$ @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. $\endgroup$
    – uhoh
    Jan 29 at 7:29
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    $\begingroup$ 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. ;) $\endgroup$
    – PM 2Ring
    Feb 3 at 11:35

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