<p>We must consider angular velocities of the Earth rotation and the orbital Moon motion relative to the stars. Then, we can calculate the angular velocity of the Moon relative to the Earth:</p>

<p><span class="math-container">$1-\frac{365}{27.32166\cdot 366}$</span> in sidereal days. Thus, the length of the tidal day is - in hours:</p>

<p><span class="math-container">$$\left(1-\frac{365}{27.32166\cdot 366}\right)^{-1}\cdot 24 \cdot \frac{ 365}{366} = 24.84631775 = \text{24h 50m 28.15s}$$</span> </p>

CC BY-SA 4.0
2019-03-01T22:31:59.427Z
2019-03-02T08:11:55.957Z
2019-03-02T08:11:55.957Z
Branislav
<p>A tidal day is the MEAN time between successive meridian transits of the Moon. (This is sometimes called called a lunar day, a term that should be deprecated for use in this sense, because it is also used for the length of a day on the Moon and should be reserved for the latter meaning to avoid ambiguity.) Its length exceeds a mean solar day of 24 hours by X minutes, which is also the (mean) time-length by which moonrise (or its meridian transit) is later each day than it was yesterday. (<a href="https://urldefense.proofpoint.com/v2/url?u=http-3A__oceanservice.noaa.gov_education_kits_tides_media_supp-5Ftide05.html&amp;d=AwMFaQ&amp;c=2X_btuPRWkGwRX26NHIotw&amp;r=uOwu8neHJUHgad2beBfZ56ZZjypo5ad7MWWuBxjkUfQ&amp;m=d_8Uoce1JnY95qK1NFicSrftfWFNbTC-imykJ52c384&amp;s=7TwwObW7gwrUwviHeMjRJVa_7PNTD3qBVVu7_oASD9c&amp;e=" rel="nofollow noreferrer">Here</a> is an animation by NOAA demonstrating this principle.)</p>

<p>The exact value of X is variously given in different places, ranging from about 48 to 52. However, I cannot find anywhere an authoritative source detailing the "correct" way to calculate X.</p>

<p>Does anyone know of an authoritative method of calculating X, and/or can anyone comment on the three methods given below.</p>

<p>(1) I have a personal theory that puts X at about 48.763 minutes, but I am uneasy about how correct the underlying logic of it is, which is this:</p>

<p>Consider two successive conjunctions (C1 and C2) of the MEAN sun and MEAN moon. At C1, the Sun's azimuth and the Moon's azimuth will, at ALL places on Earth, be the same, and, although this is no longer true for the rest of the lunation, it will once again be true at C2. This is analogous to two clocks: Clock 1 (the MEAN sun) keeps correct time, while clock 2 (the MEAN moon) loses a constant amount of time (X minutes) each day. Since the two clocks agree at C1 and at C2, clock 2 must have lost exactly 24 hours (1440 minutes) during the intervening synodic lunation, whose length (Ln) is 29.5305891203704 days. This suggests that X (the number of minutes that clock 2 loses per day) = (1440 / Ln) = 48.763.</p>

<p>However, I feel uneasy about this logic for this reason: If C1 occurs on April 1 at 06:00:00, UTC, and it is then sunrise at Tema, Ghana (on the prime meridian, near the equator), C2 will be on April 30 at 18:44:03, UTC, and that will not be sunrise at Tema; it will be dusk there, and it will be sunrise at east longitude 169°.</p>

<p>Does this negate the above logic suggesting that X = 24 hours / Ln?</p>

<p>(2) Another method for calculating X suggested to me is this:</p>

<p>It is based on the Moon's sidereal period (Ld), 27.321661 days. The MEAN angular distance along its orbit travelled by the Moon each day is (360° / Ld) = 13.17636° (which equates to a mean orbital speed of about 0.55° per hour.) So, each day, the Earth must rotate 360° plus an extra 13.17636° to point the same meridian back at the Moon again. The Earth's rotational speed is (360° / 24 hrs) = 15° per hour. At that rate, the extra time needed by the Earth to rotate those extra 13.17636° is [(13.17636/360) * 24] = 0.878424 hour = 52.7 minutes.</p>

<p>(3) I argue that even if method (2) is right in principle, it requires two amendments, because:</p>

<p>(a) The Earth rotates 360° in a sidereal day, which is nearly 4 minutes shorter than 24 hours, in which time the Earth rotates nearly 361°, or, more precisely, 360.98565°. (The fraction in the latter value is obtained from the average daily portion of the Earth's orbit: 360°/365.2422 days = 0.985647° per day.) (An animation demonstrating why this is so is <a href="http://moongazer.x10.mx/website/astronomy/downloads.htm#solarVsidereal" rel="nofollow noreferrer">here</a>.) So, the extra time needed by the Earth to rotate those extra 13.17636° is:</p>

<p>[(13.17636/360.98565)*24] = 0.8760 hours = 52.56 minutes.</p>

<p>(b) Furthermore, in those extra 52.56 minutes, the Moon has moved on in its orbit by nearly another half a degree. More precisely, it is (360/27.321661)*(52.56/1440) = 0.48094°. So the calculation should be:</p>

<p>[(13.17636+0.48094)/360.98565)*24] = 0.90800 hours = 54.48 minutes. </p>

<p>So we have three different methods of calculating X, yielding values of 48.763, 52.56 or 54.48 minutes.</p>

<p>Comments anyone? Either endorsements or criticisms of any of these three methods will be gratefully accepted.</p>

CC BY-SA 3.0
2016-11-20T15:24:05.493Z
The Moon's orbit is an ellipse, and it's tilted with respect to the earth's axis of rotation. That makes exact calculation difficult. Solar gravity mucks with things as well.
Wayfaring Stranger
<the-moon><positional-astronomy>
Length of a tidal day or by how much is the Moon late each day?
Moongazer
# Length of a tidal day or by how much is the Moon late each day?

(Asked by: username_0 on 2016-11-19)

A tidal day is the MEAN time between successive meridian transits of the Moon. (This is sometimes called called a lunar day, a term that should be deprecated for use in this sense, because it is also used for the length of a day on the Moon and should be reserved for the latter meaning to avoid ambiguity.) Its length exceeds a mean solar day of 24 hours by X minutes, which is also the (mean) time-length by which moonrise (or its meridian transit) is later each day than it was yesterday. ([Here](https://urldefense.proofpoint.com/v2/url?u=http-3A__oceanservice.noaa.gov_education_kits_tides_media_supp-5Ftide05.html&d=AwMFaQ&c=2X_btuPRWkGwRX26NHIotw&r=uOwu8neHJUHgad2beBfZ56ZZjypo5ad7MWWuBxjkUfQ&m=d_8Uoce1JnY95qK1NFicSrftfWFNbTC-imykJ52c384&s=7TwwObW7gwrUwviHeMjRJVa_7PNTD3qBVVu7_oASD9c&e=) is an animation by NOAA demonstrating this principle.)

The exact value of X is variously given in different places, ranging from about 48 to 52. However, I cannot find anywhere an authoritative source detailing the "correct" way to calculate X.

Does anyone know of an authoritative method of calculating X, and/or can anyone comment on the three methods given below.

(1) I have a personal theory that puts X at about 48.763 minutes, but I am uneasy about how correct the underlying logic of it is, which is this:

Consider two successive conjunctions (C1 and C2) of the MEAN sun and MEAN moon. At C1, the Sun's azimuth and the Moon's azimuth will, at ALL places on Earth, be the same, and, although this is no longer true for the rest of the lunation, it will once again be true at C2. This is analogous to two clocks: Clock 1 (the MEAN sun) keeps correct time, while clock 2 (the MEAN moon) loses a constant amount of time (X minutes) each day. Since the two clocks agree at C1 and at C2, clock 2 must have lost exactly 24 hours (1440 minutes) during the intervening synodic lunation, whose length (Ln) is 29.5305891203704 days. This suggests that X (the number of minutes that clock 2 loses per day) = (1440 / Ln) = 48.763.

However, I feel uneasy about this logic for this reason: If C1 occurs on April 1 at 06:00:00, UTC, and it is then sunrise at Tema, Ghana (on the prime meridian, near the equator), C2 will be on April 30 at 18:44:03, UTC, and that will not be sunrise at Tema; it will be dusk there, and it will be sunrise at east longitude 169°.

Does this negate the above logic suggesting that X = 24 hours / Ln?

(2) Another method for calculating X suggested to me is this:

It is based on the Moon's sidereal period (Ld), 27.321661 days. The MEAN angular distance along its orbit travelled by the Moon each day is (360° / Ld) = 13.17636° (which equates to a mean orbital speed of about 0.55° per hour.) So, each day, the Earth must rotate 360° plus an extra 13.17636° to point the same meridian back at the Moon again. The Earth's rotational speed is (360° / 24 hrs) = 15° per hour. At that rate, the extra time needed by the Earth to rotate those extra 13.17636° is [(13.17636/360) \* 24] = 0.878424 hour = 52.7 minutes.

(3) I argue that even if method (2) is right in principle, it requires two amendments, because:

(a) The Earth rotates 360° in a sidereal day, which is nearly 4 minutes shorter than 24 hours, in which time the Earth rotates nearly 361°, or, more precisely, 360.98565°. (The fraction in the latter value is obtained from the average daily portion of the Earth's orbit: 360°/365.2422 days = 0.985647° per day.) (An animation demonstrating why this is so is [here](http://username_0.x10.mx/website/astronomy/downloads.htm#solarVsidereal).) So, the extra time needed by the Earth to rotate those extra 13.17636° is:

[(13.17636/360.98565)\*24] = 0.8760 hours = 52.56 minutes.

(b) Furthermore, in those extra 52.56 minutes, the Moon has moved on in its orbit by nearly another half a degree. More precisely, it is (360/27.321661)\*(52.56/1440) = 0.48094°. So the calculation should be:

[(13.17636+0.48094)/360.98565)\*24] = 0.90800 hours = 54.48 minutes.

So we have three different methods of calculating X, yielding values of 48.763, 52.56 or 54.48 minutes.

Comments anyone? Either endorsements or criticisms of any of these three methods will be gratefully accepted.

**Tags:** the-moon, positional-astronomy

> **Comments:**
> **username_1**: The Moon's orbit is an ellipse, and it's tilted with respect to the earth's axis of rotation. That makes exact calculation difficult. Solar gravity mucks with things as well.  

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## 1 answers

### Answer by username_2 on 2019-03-01. Score: 1

We must consider angular velocities of the Earth rotation and the orbital Moon motion relative to the stars. Then, we can calculate the angular velocity of the Moon relative to the Earth:

$1-\frac{365}{27.32166\cdot 366}$ in sidereal days. Thus, the length of the tidal day is - in hours:

$$\left(1-\frac{365}{27.32166\cdot 366}\right)^{-1}\cdot 24 \cdot \frac{ 365}{366} = 24.84631775 = \text{24h 50m 28.15s}$$


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