Length of a tidal day

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Length of a tidal day

Moongazer
Dear calendar people,

I know that a number of members of this list have expertise in the field of positional astronomy, and I write this in the hope that someone can help me understand how to calculate the difference in length between a mean solar day (24 hours) and a tidal day (the mean time between successive culminations of the Moon).

I have a personal theory (detailed below) which yields a value of about 48 minutes, based on the Moon's synodic period. However, I have not seen this value given anywhere and, though my theory may be wrong, I have not been able to find anyone who can explain to me why it is wrong - i.e. what is the flaw in the underlying logic. Some of the values more usually given are 50, 51, and 52 minutes, and, while I understand how the last one is obtained, I believe that it should be 54 minutes for the reason given below.

(The value of 50 minutes is given by NOAA, together with a .gif animation on the above-linked webpage, showing why a tidal day is longer than a solar day.)

My specific questions are these:

Firstly, what is the flaw in the underlying reasoning behind the method suggested by my own theory, which is this:

The solar day and the tidal day are based on the apparent diurnal motions of the Sun and Moon. These are analogous to two clocks, which, at lunar conjunction, both show the same time. Clock 1 (the Sun) keeps correct time, while clock 2 (the Moon) loses a constant amount of time each day, until, at the next lunar conjunction, it again shows the same time as clock 1. This means that clock 2 has lost exactly 24 hours over the intervening period of 29.530589 days. Therefore, to find how much time (on average) the Moon falls behind the Sun per day, we should just divide 24 hours by the interval. So the calculations is:

24 hours = 1440 minutes.

1440 minutes / 29.530589 days = 48.762996 minutes per day.

My second question flows from an email exchange between myself and Professor Norman L Markworth of the Dept of Physics and Astronomy at Stephen F Austin State University, Texas, on whose website I found my most recent Google search result on this subject. It is an astronomy lab tutorial exercise headed, The Sidereal Period of the Moon, which says:

        From night to night the moon moves roughly 13° toward the east. This causes the moon to rise roughly 52 minutes later each day.

I wrote to them querying this, from which ensued an email exchange excerpted below.

Can anyone on this list either confirm or show fault with my responses to the professor?

My responses are preceded by {MG-- and followed by --MG}
(The breakup into numbered points of the professor's reply is my own.)

Hope someone can help shed some light on this,

Moongazer.

{MG--

In a statement of the form, "the Moon will rise (or set or culminate) x minutes later each day," there are a couple of implied qualifying clauses that are not mentioned explicitly because they are assumed to be self-evident. However, for the purposes of this discussion, it would be well to explicate them:

The words "later, each day," mean later, on an ordinary clock showing (mean) solar time, and "each day" means a mean solar day of 24 hours, during which the Moon travels approximately 13.1763° along its orbit and the Earth rotates almost 361°. The more precise value used below for the Earth's daily rotation is 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.)

Please see my further comments (in red) interspersed with the text of your reply.

Points 1 and 2 of Prof Markworth's reply explains how the Moon, in its eastward orbital motion, travels an angular distance of approximately 13.18° per day around our skies.

--MG}

[3]  This means that the Earth has to turn an extra 13.18 degrees so that tonight's moon is in the same sky position as last night’s moon. If last night the Moon was just at the eastern horizon, tonight at the same time it is 13.18 degrees below the eastern horizon.

[4]  Now, how long does it take the good ol' Earth to turn 13.18 degrees? Well, in 24 hours, it turns 360 degrees, so in (13.18/360) x 24 = 0.88 hours or 52.7 minutes, the sky rotates the extra 13.18 degrees.

{MG--

[4a]  Not so. In 24 hours the Earth turns nearly 361°, so the calculation should be as follows. (As it turns out, even with additional arithmetic precision, the difference is not significant to this discussion, but, for pedagogic reasons, it would have been preferable to put it as:)

(13.17636/360.98565) * 24 = 0.8760 hours = 52.56 min

[4b]  Furthermore, since this time (52.56 minutes) is nearly an hour, in which time the Moon will have moved on in its orbit almost half a degree further, we should factor in the additional orbital motion of the Moon during that time, which is: (360/27.321661) * (52.56/1440) = 0.48093°, so the calculation should be:

13.65729/360.98565) * 24 = 0.90800 hours = 54.48 min

--MG}

[5]  Now, why do we use the time of the Sidereal month instead of the Synodic month which is 29.53 days? Because we are interested in the Moon's position relative to the background sky (Sidereal) not whether it is in the same orientation with respect to the Sun and Earth.

{MG--

[5a]  If we accept point [5], then (no argument with [6], but) what then is the point of taking the average of the two as suggested in point [7]? If the Moon's synodic orientation is irrelevant, then we should take no account of it at all.

--MG}

[6]  A Synodic month separates one New Moon from the next New Moon, or any corresponding similar lunar phases on any two cycles. If we were to use the Synodic month, we would get a lunar shift of 12.1 degrees per day, and the Moon would rise 48.7 minutes later each night.

[7]  The average of the two is 50.7 minutes. The difference between the two is 4 minutes, which is just the amount that the Sun has moved to the East in ITS motion along the sky. This emphasizes that Sidereal time does not depend on the location of the Sun, but Synodic does!
 
 Dr. Norm Markworth
Department of Physics and Astronomy
Stephen F Austin State University
Box 13044
Nacogdoches, TX 75962

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Re: Length of a tidal day

Helios
Dear Moongazer, calendar people,

1 ) there is the lunar or tidal day between meridan transits. This is 1 day 50 minutes 28 seconds = 1 / [ 1 - ( 1 / M ) ]

2 ) there is the excess to one solar day of one lunar day. This is 50 minutes 28 seconds = 1 / [ M - 1 ]

3 ) there is the extent to which the moon is late after 1 day. This is 48 minutes 46 seconds = 1 / M

Now we can calculate the angle that the Moon travels in one day.
( 360° )*[ 1 - ( 1 / M ) ] = 347.81°
In one hour, the moon travels
( 15° )*[ 1 - ( 1 / M ) ] = 14.49°
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Re: Length of a tidal day

Moongazer
Dear Helios and calendar people,

Thank you, Helios, for your reply, but I am having trouble understanding it.

(I am a hobbyist student of positional astronomy (see my website on this subject at tiny.cc/astronomy), but my knowledge of mathematics is very poor.)

Can you please clarify:

(a)  In point 2 (I start with the easier one) you give a function: 1 / [ M - 1 ]. What does it do, what is its derivation, and what is the value of M there?

And the same goes for the function in point 1 (the previous paragraph) of: 1 / [ 1 - ( 1 / M ) ].

(b)  In the last paragraph, you present two calculations for the mean angular distance travelled by the Moon in one day and in one hour. They cannot be referring to the Moon's real orbital motion eastward, for the values of those are, respectively, about 13.18° and 0.5°. It seems to me that you are referring there to the Moon's apparent diurnal motion westward. If so, that would explain the value 15° in the last line.

However, I am similarly handicapped there as in relation to your points (1) and (2) by my lack of understanding of those mysterious functions containing the term M, whose value I do not know. M appears in five places. Does it have the same value everywhere? And what should it be? (I apologise for my mathematical ignorance.)

(c)  In point 3, you say that the Moon is late each day by 48 minutes, 46 seconds. This seems to support my method of [24 hrs / (Moon's mean synodic period)], based on the analogy of the two clocks that agree at lunar conjunctions, the slow one losing exactly 24 hours in each synodic lunation. But how do we reconcile this with the value (50 min, 28 sec) you give in point 2 for the excess of a tidal day over a solar day? Wouldn't the excess be the same as the amount of time by which the Moon is late each day?

Moongazer
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Re: Length of a tidal day

Moongazer
In reply to this post by Helios
Dear Helios and Calendar people,

P.S. to my earlier reply to Helios:

My apologies again. I've just noticed something I should not have missed earlier - that it is possible to obtain the value of M from the equation in point 3. Doesn't mean I understand it all, but at least it gives me a starting point to work from. I will work backwards from point 3, using the value of M obtained from there and see where that gets me.

Moongazer
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Re: Length of a tidal day

Moongazer
In reply to this post by Helios
Dear Helios and calendar people,

P.P.S. to my earlier reply to Helios:

This is what I have done with your equations (but I'm not sure I went about this the right way.)

At first I thought "1" was a numeric literal, but that got me nowhere, so I then assumed it meant one day or 86,400 seconds.

Then, starting from the value in (3) of 48 min, 46 sec, or 2,926 seconds, I got:

86400/M = 2926, so:
M = 86400/2926 = 29.528366370471633629528366370472,
which I recognised as the length in days of a synodic lunation.

I then plugged that value for M into (2) and, working in days, I get
1 / (M - 1) = 0.03505283082157318446462371516879 days,
which, as you wrote, equates to 50 mins, 28 secs.

Then, going to your last paragraph, and plugging the same value for M into those equations, I get:
347.8083333 ...° and 14.492° (just as you wrote) as the daily and hourly angular distance of the Moon's apparent diurnal motion westward.

BUT:

(a) I have no understanding of why those "magic formulae" work. If you can explain it, I would very much like to know this.

(b) It seems to me that if, indeed, point 3 was your starting point, you started from my theory that to obtain the amount by which the Moon is late each day, one should divide 24 hrs by M, based on the analogy of the two clocks that agree at lunar conjunctions, the slow one losing exactly 24 hours in the interval between conjunctions. This does indeed yield a value of 48 mins, 46 seconds. Is there any general acceptance that this theory of mine is a correct assumption to proceed from? And, if it is, then,

(c) This begs the question: why does the amount by which the Moon is late each day differ from the excess of a tidal day above a 24-hour day? They should be the same, shouldn't they?

Moongazer.
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Re: Length of a tidal day

Karl Palmen
In reply to this post by Helios
Dear Moongazer, Helios and Calendar People

Helios has not defined his variables nor explained to derivation of his formulae.

Helios has used the same 1/( 1 - 1/x) formula as he has used for the nodetide, with just a different value for x and the unmentioned time unit.

-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Helios
Sent: 13 November 2016 01:00
To: [hidden email]
Subject: Re: Length of a tidal day

Dear Moongazer, calendar people,

1 ) there is the lunar or tidal day between meridan transits. This is 1 day
50 minutes 28 seconds = 1 / [ 1 - ( 1 / M ) ]

KARL REPLIES: I think M is the number of days in a synodic month and also the numerator is one day.
I prefer to have the synodic month M in any unit of time and this introduces D for the length of the day. We then have

D /( 1 - D/M ) =
1/( 1/D - 1/M )

For D=1, we get Helios's formula from either of these.

Now we see that if the lunar day is L then

1/L = 1/D - 1/M
1/D = 1/L + 1/M

These reciprocals may be thought of as the count rates of their respective cycles
1/D counts 1 per day
1/M counts 1 per synodic month and
1/L counts 1 per lunar day.
The count of days equals the count of lunar days plus the count of synodic months.

I hope this explains the derivation of Helios's formula.


My calculation makes it 24 hours 50.472... minutes = 24 hours 50 minutes 28.33 seconds.

Also 1/L = 1/D - 1/M implies
M/L = M/D - 1
The synodic month in lunar days is exactly one less than the synodic month in days.


HELIOS CONTINUED
2 ) there is the excess to one solar day of one lunar day. This is 50
minutes 28 seconds = 1 / [ M - 1 ]

KARL REPLIES:
D/( 1 - D/M) - D =
D*M/( M - D ) - D =
M/( M/D - 1 ) - D =
(M - D*(M/D - 1)/(M/D - 1) =
(M - M + D)/(M/D - 1) =
D/(M/D - 1)

Noting M/L = M/D - 1
We have D*L/M


HELIOS CONTINUED:
3 ) there is the extent to which the moon is late after 1 day. This is 48
minutes 46 seconds = 1 / M

KARL REPLIES:
This is D*D/M = D/(M/D)

It refers to the angle by which the moon is behind after one day measured in units of D/360°.
This is not really a time interval at all, but an angle expressed in time units.

This I reckon to be 0.033863... days = 12.19 degrees.


HELIOS CONTINUED:
Now we can calculate the angle that the Moon travels in one day.
( 360° )*[ 1 - ( 1 / M ) ] = 347.81°
In one hour, the moon travels
( 15° )*[ 1 - ( 1 / M ) ] = 14.49°

KARL REPLIES:
Helios effectively multiplies 360 degrees by D/L, then by H/L, where H is an hour.
Note that 347.81° Helios calculated added to the 12.19 degrees I calculated is exactly 360 degrees.
For the one hour of moon travel, Helios could add one more decimal place to get 14.492° .

I hope this helps.

Karl

16(03(16





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Re: Length of a tidal day

Irv Bromberg
Dear Moongazer, Helios and Kalendarists:

I don't understand Moongazer's logic for the calculation of the length of the tidal day.

In desktop astronomy programs like Starry Night or Stellarium, Moon returns to almost the same place in the sky about every 24h 50m 30s. This should be the length of the tidal day. In these programs this timing interval can be used to depict the "lunalemma", see actual photograph here:

http://epod.usra.edu/blog/2014/05/lunar-analemma.html

Stellarium is free and available for almost any computing platform, see:

http://www.stellarium.org/

The length of the tidal day is the (mean) interval between successive culminations of Moon -- when it appears to cross the local celestial Meridian from east to west (meridian transit). So you need an algorithm to reckon when this happens for some specified locale -- it will probably best and simplest to choose the Prime Meridian at the Equator.

-- Irv Bromberg, University of Toronto, Canada

http://www.sym454.org/lunar/
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Re: Length of a tidal day

Karl Palmen
In reply to this post by Moongazer
Dear Moongazer and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Moongazer
Sent: 13 November 2016 12:27
To: [hidden email]
Subject: Re: Length of a tidal day


(c) This begs the question: why does the amount by which the Moon is late
each day differ from the excess of a tidal day above a 24-hour day? They
should be the same, shouldn't they?

KARL REPLIES: The supposed lateness of the moon is not a time interval at all, but an angle of about 12.19 degrees, which is how far the moon is behind after 24 hours. The real lateness is the 50 minutes 28.33 seconds calculated earlier. This I expect to equal the average daily delay of the tides, but not the actual delay, because the sun also influences the tides and may cause them to occur earlier or later (by a few minutes I expect).

Karl

16(03(16
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Re: Length of a tidal day

Irv Bromberg
In reply to this post by Irv Bromberg
From: East Carolina University Calendar discussion List [[hidden email]] on behalf of Irv Bromberg [[hidden email]]
Sent: Tuesday, November 15, 2016 09:16

I don't understand Moongazer's logic for the calculation of the length of the tidal day.


On further pondering what Moongazer did, I suspect that the mistake is that in effect he started with a lunar culmination for a specific locale, but ended with a lunar culmination for a different locale.

-- Irv Bromberg, University of Toronto, Canada

http://www.sym454.org/lunar/


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Re: Length of a tidal day

Karl Palmen
In reply to this post by Moongazer
Dear Moongazer and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Moongazer
Sent: 13 November 2016 12:27
To: [hidden email]
Subject: Re: Length of a tidal day

Dear Helios and calendar people,


(b) It seems to me that if, indeed, point 3 was your starting point, you
started from my theory that to obtain the amount by which the Moon is late
each day, one should divide 24 hrs by M, based on the analogy of the two
clocks that agree at lunar conjunctions, the slow one losing exactly 24
hours in the interval between conjunctions. This does indeed yield a value
of 48 mins, 46 seconds. Is there any general acceptance that this theory of
mine is a correct assumption to proceed from? And, if it is, then,

KARL REPLIES: The slow clock would be 48 mins 46 seconds behind after 24 hours of the normal clock, but 50 mins 28 seconds behind after 24 of its own hours.

Karl

16(03(16

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Re: Length of a tidal day

Phil De Rosa
In reply to this post by Karl Palmen
Thank you Brian.  I got them so I think it's my Hotmail address that's at
fault.

Phil

-----Original Message-----
From: Karl Palmen
Sent: Tuesday, November 15, 2016 6:20 AM
To: [hidden email]
Subject: Re: Length of a tidal day

Dear Moongazer and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[hidden email]] On Behalf Of Moongazer
Sent: 13 November 2016 12:27
To: [hidden email]
Subject: Re: Length of a tidal day


(c) This begs the question: why does the amount by which the Moon is late
each day differ from the excess of a tidal day above a 24-hour day? They
should be the same, shouldn't they?

KARL REPLIES: The supposed lateness of the moon is not a time interval at
all, but an angle of about 12.19 degrees, which is how far the moon is
behind after 24 hours. The real lateness is the 50 minutes 28.33 seconds
calculated earlier. This I expect to equal the average daily delay of the
tides, but not the actual delay, because the sun also influences the tides
and may cause them to occur earlier or later (by a few minutes I expect).

Karl

16(03(16
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Re: Length of a tidal day

Karl Palmen
In reply to this post by Moongazer
Dear Moongazer and Calendar People

A thought of a simpler way to explain this:

MOONGAZER SAID:

24 hours = 1440 minutes.

1440 minutes / 29.530589 days = 48.762996 minutes per day.


KARL REPLIES:

The correct answer is

1440 minutes / 28.530589 days = 50.472144 minutes per day.

One needs to divide the 24 hours by the number of tidal days in a synodic month, which is one less than the number of days in a synodic month.

Converting to minutes and seconds, one gets 50 minutes 28.3 seconds as reckoned before.


28.530589 of these periods make the 24 hours added to 28.530589 days to get 28.530589 tidal days equal to one synodic month.


Karl

16(04(06

-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Moongazer
Sent: 12 November 2016 04:52
To: [hidden email]
Subject: Length of a tidal day

Dear calendar people,

I know that a number of members of this list have expertise in the field of
positional astronomy, and I write this in the hope that someone can help me
understand how to calculate the difference in length between a mean solar
day (24 hours) and a tidal day (the mean time between successive
culminations of the Moon).

I have a personal theory (detailed below) which yields a value of about 48
minutes, based on the Moon's synodic period. However, I have not seen this
value given anywhere and, though my theory may be wrong, I have not been
able to find anyone who can explain to me why it is wrong - i.e. what is the
flaw in the underlying logic. Some of the values more usually given are 50,
51, and 52 minutes, and, while I understand how the last one is obtained, I
believe that it should be 54 minutes for the reason given below.

(The value of 50 minutes is given by  NOAA
<http://oceanservice.noaa.gov/education/kits/tides/media/supp_tide05.html>
, together with a .gif animation on the above-linked webpage, showing why a
tidal day is longer than a solar day.)

My specific questions are these:

Firstly, what is the flaw in the underlying reasoning behind the method
suggested by my own theory, which is this:

The solar day and the tidal day are based on the apparent diurnal motions of
the Sun and Moon. These are analogous to two clocks, which, at lunar
conjunction, both show the same time. Clock 1 (the Sun) keeps correct time,
while clock 2 (the Moon) loses a constant amount of time each day, until, at
the next lunar conjunction, it again shows the same time as clock 1. This
means that clock 2 has lost exactly 24 hours over the intervening period of
29.530589 days. Therefore, to find how much time (on average) the Moon falls
behind the Sun per day, we should just divide 24 hours by the interval. So
the calculations is:

24 hours = 1440 minutes.

1440 minutes / 29.530589 days = 48.762996 minutes per day.


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