I referred to the instruction manual of my new watch
that I got from Walmart last week and noted it have a moon phase indicator. It
says the moon age is calculated by
29.53 * (L / 360) = d
in which L is the moon angular distance and d means moon
age in days
the moon angular distance may be the same term as the
lunar ecliptic longitude.
If I were to use that for the days of the lunar month of
the lunisolar or lunar calendar...
floor(29.53... * (L / 360)) + 1 = d
However, if I were to calculate the solar days using the
solar ecliptic longitude...
floor(365.2422... * (S / 360)) = D
in which S means the solar ecliptic longitude, and D
means the day of the solar year.

On 2009 Apr 22, at 18:25 , ELITE 3000 wrote:
> I referred to the instruction manual of my new watch that I got from > Walmart last week and noted it have a moon phase indicator. It says > the moon age is calculated by > 29.53 * (L / 360) = d Irv replies: The given value of 29.53 days is only 29d 12h 43m 12.0s, which is about 51 seconds too short. The actual lunar cycle varies from about 29d 6h 30m to 29d 20h in length (longer cycles near Earth orbital perihelion, shorter cycles near Earth orbital aphelion), so such a fixed length calculation can be wildly inaccurate. The proper way to calculate the lunar age is to determine the actual number of days and fraction of a day elapsed since the most recent actual lunar conjunction, properly accounting for the time zone and daylight saving time where applicable. However, if the watch indicates the result on a small dial then the eye might not notice the inaccuracy of their simplistic fixed length cycle calculation.  Irv Bromberg, Toronto, Canada <http://www.sym454.org/lunar/> 
Hi all,
For most people it is absolutely sufficient to have an approximation of the moon age as an integral number of days. For that purpose the "29.53 * (L / 360) = d" formula is absolutely appropriate. "L" should be the (easy to visually estimate) angular distance (elongation of course) between Moon and Sun, but when the moon is waning it must be 360 minus the visual angle. Else it would give the days left until next new moon. Of course, for more accuracy Irv is right, but I guess most walmartwatchbuyers don't care about a moon age accurate to the second. At the moment I am sending THIS mail (20090423 10:00:00 UTC) the moon age is 27.82919 days (according to Meeus' moon phase algorithm, see www.henkreints.nl/astro). The angle between moon and sun is 22°43'34", so a visual estimate might be 20, resulting in 340 to be used in the walmart formula, which then yields 27.89, so not bad at all. _________________________________________________ Kind regards / met vriendelijke groeten, Henk Reints Oorspronkelijke tekst Irv Bromberg > On 2009 Apr 22, at 18:25 , ELITE 3000 wrote: >> I referred to the instruction manual of my new watch that I got from >> Walmart last week and noted it have a moon phase indicator. It says >> the moon age is calculated by >> 29.53 * (L / 360) = d > > Irv replies: The given value of 29.53 days is only 29d 12h 43m 12.0s, > which is about 51 seconds too short. The actual lunar cycle varies > from about 29d 6h 30m to 29d 20h in length (longer cycles near Earth > orbital perihelion, shorter cycles near Earth orbital aphelion), so > such a fixed length calculation can be wildly inaccurate. > > The proper way to calculate the lunar age is to determine the actual > number of days and fraction of a day elapsed since the most recent > actual lunar conjunction, properly accounting for the time zone and > daylight saving time where applicable. > > However, if the watch indicates the result on a small dial then the > eye might not notice the inaccuracy of their simplistic fixed length > cycle calculation. > >  Irv Bromberg, Toronto, Canada > > <http://www.sym454.org/lunar/> > 
In reply to this post by Ryan Provost2
What is S?
Victor On Wed, Apr 22, 2009 at 5:25 PM, ELITE 3000 <[hidden email]> wrote: > I referred to the instruction manual of my new watch that I got from Walmart > last week and noted it have a moon phase indicator. It says the moon age is > calculated by > > 29.53 * (L / 360) = d > > in which L is the moon angular distance and d means moon age in days > > the moon angular distance may be the same term as the lunar ecliptic > longitude. > > If I were to use that for the days of the lunar month of the lunisolar or > lunar calendar... > > floor(29.53... * (L / 360)) + 1 = d > > However, if I were to calculate the solar days using the solar ecliptic > longitude... > > floor(365.2422... * (S / 360)) = D > > in which S means the solar ecliptic longitude, and D means the day of the > solar year. 
On 2009 Apr 23, at 10:52 , Victor Engel wrote:
Irv replies: S is the letter after R and before T, although I've been thinking about proposing an alphabet reform that will make the alphabet symmetrical! Seriously, though, Ryan adequately defined it in his original message and I quote: However, if I were to calculate the solar days using the solar ecliptic longitude... floor(365.2422... * (S / 360)) = D in which S means the solar ecliptic longitude, and D means the day of the solar year. Note that although the WalMart lunar age expression could be in error by many hours, the above proposed solar cycle age expression, which will approximate the age of the solar cycle relative to the northward equinox, can be in error by several days, as Earth moves faster near perihelion and slower near aphelion. Again, it would be best to compute the number of days and fraction of day elapsed since the equinox up to the present moment, accounting also for the time zone and daylight saving time if applicable. Is there any need to know the solar cycle age? Why should the northward equinox be the starting point? (This is implicit in directly using the ecliptic solar longitude as in Ryan's expression quoted above, because that starts from zero at the northward equinox.) Perhaps the mean year length constant used in the expression should be the one that has a stable calendar season at or near the desired starting point? 
In reply to this post by Victor Engel
Dear CalndrL, I'm trying to find out if anyone has developed or is developing an 'extensible' date object  most likely this would be in XML Schema, to be interoperable between many softwares and platforms? For example, a 'simple' date object might look something like this: event: Death of John Doe day: april 23 year: 1981 weekday: thursday note: Obituary in NYT A more robust and complex date object could link to other date objects (birth, marriage  or include these as 'subevents'). Could handle ambiguous or multiple records (3 different sources claiming 3 different dates  London Times says it was April 19). Other fields could help with leap year conversions, various calendars, etc. Thus an event called 'Death of John Doe' could contain a small amount of unambiguous data, or a large amount of conflicting data. More notefields could be added, or more links to other events, even more weekdays? XML can be designed to handle these kinds of addon or multiple field elements. Thank you, Richard Ellsberry 
Hi Richard, There is no one standard for historical metadata but probably the closest thing to what you've described would be HEML: Historical Markup Language. More info here http://heml.mta.ca/hemlcocoon/

In reply to this post by Irv Bromberg
the number 29.53 is an APPROXIMATE number of days (NOT the exact number of
days) of of the lunation.  From: "Irv Bromberg" <[hidden email]> Sent: Wednesday, April 22, 2009 6:48 PM To: <[hidden email]> Subject: Re: Moon age calculations > On 2009 Apr 22, at 18:25 , ELITE 3000 wrote: >> I referred to the instruction manual of my new watch that I got from >> Walmart last week and noted it have a moon phase indicator. It says the >> moon age is calculated by >> 29.53 * (L / 360) = d > > Irv replies: The given value of 29.53 days is only 29d 12h 43m 12.0s, > which is about 51 seconds too short. The actual lunar cycle varies from > about 29d 6h 30m to 29d 20h in length (longer cycles near Earth orbital > perihelion, shorter cycles near Earth orbital aphelion), so such a fixed > length calculation can be wildly inaccurate. > > The proper way to calculate the lunar age is to determine the actual > number of days and fraction of a day elapsed since the most recent actual > lunar conjunction, properly accounting for the time zone and daylight > saving time where applicable. > > However, if the watch indicates the result on a small dial then the eye > might not notice the inaccuracy of their simplistic fixed length cycle > calculation. > >  Irv Bromberg, Toronto, Canada > > <http://www.sym454.org/lunar/> > > 
In reply to this post by Victor Engel
for calculating the solar day?
floor(365.2422... * (S / 360)) = D S means the solar ecliptic longitude <===== D means the day of the solar year.  From: "Victor Engel" <[hidden email]> Sent: Thursday, April 23, 2009 10:52 AM To: <[hidden email]> Subject: Re: Moon age calculations > What is S? > > Victor > > On Wed, Apr 22, 2009 at 5:25 PM, ELITE 3000 <[hidden email]> wrote: >> I referred to the instruction manual of my new watch that I got from >> Walmart >> last week and noted it have a moon phase indicator. It says the moon age >> is >> calculated by >> >> 29.53 * (L / 360) = d >> >> in which L is the moon angular distance and d means moon age in days >> >> the moon angular distance may be the same term as the lunar ecliptic >> longitude. >> >> If I were to use that for the days of the lunar month of the lunisolar or >> lunar calendar... >> >> floor(29.53... * (L / 360)) + 1 = d >> >> However, if I were to calculate the solar days using the solar ecliptic >> longitude... >> >> floor(365.2422... * (S / 360)) = D >> >> in which S means the solar ecliptic longitude, and D means the day of the >> solar year. > > 
In reply to this post by Ryan Provost2
On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote:
> the number 29.53 is an APPROXIMATE number of days (NOT the exact number of > days) of of the lunation. Well, it'd be hard to define an exact value. The classical value is exactly 29 days + 12 hours + 793/1080 hour, which works out to about 29.530594 days. The modern mean is usually given as 29.530589 days; the difference is less than half a second. Rounding to 29.53 days still gets you within a minute of the correct mean. Past a certain point, it doesn't matter how accurate your approximation to the mean is, because any given individual lunation can be up to 15 hours off the mean value. So really, you're only getting an approximation to the nearest whole day, which still might be off by one.  Mark J. Reed <[hidden email]> 
Dear Mark and Calendar People
Original Message From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Mark J. Reed Sent: 23 April 2009 22:47 To: [hidden email] Subject: Re: Moon age calculations On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: > the number 29.53 is an APPROXIMATE number of days (NOT the exact number of > days) of of the lunation. Well, it'd be hard to define an exact value. The classical value is exactly 29 days + 12 hours + 793/1080 hour, which works out to about 29.530594 days. The modern mean is usually given as 29.530589 days; the difference is less than half a second. Rounding to 29.53 days still gets you within a minute of the correct mean. Past a certain point, it doesn't matter how accurate your approximation to the mean is, because any given individual lunation can be up to 15 hours off the mean value. So really, you're only getting an approximation to the nearest whole day, which still might be off by one. KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) as an approximation or even 30 for a quick reckoning in the head. Karl 10(07(29 till noon 
Well, the error is of course cumulative. 29.53 is accurate to about a
day per century; 29.5 to about a day per three years. 30 is off by a day every 2 to 3 months. On 4/24/09, Palmen, KEV (Karl) <[hidden email]> wrote: > Dear Mark and Calendar People > > Original Message > From: East Carolina University Calendar discussion List > [mailto:[hidden email]] On Behalf Of Mark J. Reed > Sent: 23 April 2009 22:47 > To: [hidden email] > Subject: Re: Moon age calculations > > On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: >> the number 29.53 is an APPROXIMATE number of days (NOT the exact number >> of >> days) of of the lunation. > > Well, it'd be hard to define an exact value. The classical value is > exactly 29 days + 12 hours + 793/1080 hour, which works out to about > 29.530594 days. The modern mean is usually given as 29.530589 days; > the difference is less than half a second. Rounding to 29.53 days > still gets you within a minute of the correct mean. > > Past a certain point, it doesn't matter how accurate your > approximation to the mean is, because any given individual lunation > can be up to 15 hours off the mean value. So really, you're only > getting an approximation to the nearest whole day, which still might > be off by one. > > KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) > as an approximation or even 30 for a quick reckoning in the head. > > Karl > > 10(07(29 till noon > >  Sent from my mobile device Mark J. Reed <[hidden email]> 
Dear Mark
I thought the formula was to calculate the moon given its phase as its angular position relative to the sun and so the age is reset to 0 each new moon, so preventing any error from accumulating over the months. Karl 10(07(30 Original Message From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Mark J. Reed Sent: 24 April 2009 14:34 To: [hidden email] Subject: Re: Moon age calculations Well, the error is of course cumulative. 29.53 is accurate to about a day per century; 29.5 to about a day per three years. 30 is off by a day every 2 to 3 months. On 4/24/09, Palmen, KEV (Karl) <[hidden email]> wrote: > Dear Mark and Calendar People > > Original Message > From: East Carolina University Calendar discussion List > [mailto:[hidden email]] On Behalf Of Mark J. Reed > Sent: 23 April 2009 22:47 > To: [hidden email] > Subject: Re: Moon age calculations > > On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: >> the number 29.53 is an APPROXIMATE number of days (NOT the exact number >> of >> days) of of the lunation. > > Well, it'd be hard to define an exact value. The classical value is > exactly 29 days + 12 hours + 793/1080 hour, which works out to about > 29.530594 days. The modern mean is usually given as 29.530589 days; > the difference is less than half a second. Rounding to 29.53 days > still gets you within a minute of the correct mean. > > Past a certain point, it doesn't matter how accurate your > approximation to the mean is, because any given individual lunation > can be up to 15 hours off the mean value. So really, you're only > getting an approximation to the nearest whole day, which still might > be off by one. > > KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) > as an approximation or even 30 for a quick reckoning in the head. > > Karl > > 10(07(29 till noon > >  Sent from my mobile device Mark J. Reed <[hidden email]> 
Dear Karl and Mark,
That was my impression. Perhaps this is a good time to question how the watch calculates the phase (from which it derives the age). Victor On Fri, Apr 24, 2009 at 8:37 AM, Palmen, KEV (Karl) <[hidden email]> wrote: > Dear Mark > > I thought the formula was to calculate the moon given its phase as its angular position relative to the sun and so the age is reset to 0 each new moon, so preventing any error from accumulating over the months. > > Karl > > 10(07(30 > > Original Message > From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Mark J. Reed > Sent: 24 April 2009 14:34 > To: [hidden email] > Subject: Re: Moon age calculations > > Well, the error is of course cumulative. 29.53 is accurate to about a > day per century; 29.5 to about a day per three years. 30 is off by a > day every 2 to 3 months. > > > On 4/24/09, Palmen, KEV (Karl) <[hidden email]> wrote: >> Dear Mark and Calendar People >> >> Original Message >> From: East Carolina University Calendar discussion List >> [mailto:[hidden email]] On Behalf Of Mark J. Reed >> Sent: 23 April 2009 22:47 >> To: [hidden email] >> Subject: Re: Moon age calculations >> >> On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: >>> the number 29.53 is an APPROXIMATE number of days (NOT the exact number >>> of >>> days) of of the lunation. >> >> Well, it'd be hard to define an exact value. The classical value is >> exactly 29 days + 12 hours + 793/1080 hour, which works out to about >> 29.530594 days. The modern mean is usually given as 29.530589 days; >> the difference is less than half a second. Rounding to 29.53 days >> still gets you within a minute of the correct mean. >> >> Past a certain point, it doesn't matter how accurate your >> approximation to the mean is, because any given individual lunation >> can be up to 15 hours off the mean value. So really, you're only >> getting an approximation to the nearest whole day, which still might >> be off by one. >> >> KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) >> as an approximation or even 30 for a quick reckoning in the head. >> >> Karl >> >> 10(07(29 till noon >> >> > >  > Sent from my mobile device > > Mark J. Reed <[hidden email]> > > 
Good point; I veered off of the original topic into mean lunation use
in general. I would like to know how the watch determine's the moon's longitude... On Fri, Apr 24, 2009 at 9:58 AM, Victor Engel <[hidden email]> wrote: > Dear Karl and Mark, > > That was my impression. Perhaps this is a good time to question how > the watch calculates the phase (from which it derives the age). > > Victor > > On Fri, Apr 24, 2009 at 8:37 AM, Palmen, KEV (Karl) > <[hidden email]> wrote: >> Dear Mark >> >> I thought the formula was to calculate the moon given its phase as its angular position relative to the sun and so the age is reset to 0 each new moon, so preventing any error from accumulating over the months. >> >> Karl >> >> 10(07(30 >> >> Original Message >> From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Mark J. Reed >> Sent: 24 April 2009 14:34 >> To: [hidden email] >> Subject: Re: Moon age calculations >> >> Well, the error is of course cumulative. 29.53 is accurate to about a >> day per century; 29.5 to about a day per three years. 30 is off by a >> day every 2 to 3 months. >> >> >> On 4/24/09, Palmen, KEV (Karl) <[hidden email]> wrote: >>> Dear Mark and Calendar People >>> >>> Original Message >>> From: East Carolina University Calendar discussion List >>> [mailto:[hidden email]] On Behalf Of Mark J. Reed >>> Sent: 23 April 2009 22:47 >>> To: [hidden email] >>> Subject: Re: Moon age calculations >>> >>> On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: >>>> the number 29.53 is an APPROXIMATE number of days (NOT the exact number >>>> of >>>> days) of of the lunation. >>> >>> Well, it'd be hard to define an exact value. The classical value is >>> exactly 29 days + 12 hours + 793/1080 hour, which works out to about >>> 29.530594 days. The modern mean is usually given as 29.530589 days; >>> the difference is less than half a second. Rounding to 29.53 days >>> still gets you within a minute of the correct mean. >>> >>> Past a certain point, it doesn't matter how accurate your >>> approximation to the mean is, because any given individual lunation >>> can be up to 15 hours off the mean value. So really, you're only >>> getting an approximation to the nearest whole day, which still might >>> be off by one. >>> >>> KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) >>> as an approximation or even 30 for a quick reckoning in the head. >>> >>> Karl >>> >>> 10(07(29 till noon >>> >>> >> >>  >> Sent from my mobile device >> >> Mark J. Reed <[hidden email]> >> >> > >  Mark J. Reed <[hidden email]> 
In reply to this post by Victor Engel
Sirs:
>29 days + 12 hours + 793/1080 hour..... =29^{d}.530594135802469 135802469 135802..=29^{d} 12^{h} 44^{m} 3s.33.. (ancient Indian, I believe, used 29^{d} 12^{h} 44^{m} 2^{s}.8) = 29^{d}.530587962963; while modern value of 29.530589 =29^{d} 12^{h} 44^{m} 2^{s}.8896 >KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) Thanks Karl, sir. Brij Bhushan Vij Today: (MJD 2454947)/1361+D125W1705 (G. Friday, 2009 April 24H15:96 (decimal) EST Aa Nau Bhadra Kritvo Yantu Vishwatah Rg Veda Jan:31; Feb:29; Mar:31; Apr:30; May:31; Jun:30 Jul:30; Aug:31; Sep:30; Oct:31; Nov:30; Dec:30 (365th day of Year is World Day) My Profile:http://www.brijvij.com/bbv_2colvipBrief.pdf HOME PAGE: http://www.brijvij.com/ ******As per Kali VGRhymeCalendaar***** "Koi bhi cheshtha vayarth nahin hoti, purshaarth karne mein hai" Contact # 001 (201) 6758548 > Date: Fri, 24 Apr 2009 08:58:01 0500 > From: [hidden email] > Subject: Re: Moon age calculations > To: [hidden email] > > Dear Karl and Mark, > > That was my impression. Perhaps this is a good time to question how > the watch calculates the phase (from which it derives the age). > > Victor > > On Fri, Apr 24, 2009 at 8:37 AM, Palmen, KEV (Karl) > <[hidden email]> wrote: > > Dear Mark > > > > I thought the formula was to calculate the moon given its phase as its angular position relative to the sun and so the age is reset to 0 each new moon, so preventing any error from accumulating over the months. > > > > Karl > > > > 10(07(30 > > > > Original Message > > From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Mark J. Reed > > Sent: 24 April 2009 14:34 > > To: [hidden email] > > Subject: Re: Moon age calculations > > > > Well, the error is of course cumulative. 29.53 is accurate to about a > > day per century; 29.5 to about a day per three years. 30 is off by a > > day every 2 to 3 months. > > > > > > On 4/24/09, Palmen, KEV (Karl) <[hidden email]> wrote: > >> Dear Mark and Calendar People > >> > >> Original Message > >> From: East Carolina University Calendar discussion List > >> [mailto:[hidden email]] On Behalf Of Mark J. Reed > >> Sent: 23 April 2009 22:47 > >> To: [hidden email] > >> Subject: Re: Moon age calculations > >> > >> On Thu, Apr 23, 2009 at 2:25 PM, ELITE 3000 <[hidden email]> wrote: > >>> the number 29.53 is an APPROXIMATE number of days (NOT the exact number > >>> of > >>> days) of of the lunation. > >> > >> Well, it'd be hard to define an exact value. The classical value is > >> exactly 29 days + 12 hours + 793/1080 hour, which works out to about > >> 29.530594 days. The modern mean is usually given as 29.530589 days; > >> the difference is less than half a second. Rounding to 29.53 days > >> still gets you within a minute of the correct mean. > >> > >> Past a certain point, it doesn't matter how accurate your > >> approximation to the mean is, because any given individual lunation > >> can be up to 15 hours off the mean value. So really, you're only > >> getting an approximation to the nearest whole day, which still might > >> be off by one. > >> > >> KARLSAYS: Given that is so, you might as well use 29.5 (rather than 29.53) > >> as an approximation or even 30 for a quick reckoning in the head. > >> > >> Karl > >> > >> 10(07(29 till noon > >> > >> > > > >  > > Sent from my mobile device > > > > Mark J. Reed <[hidden email]> > > > > > Rediscover Hotmail®: Now available on your iPhone or BlackBerry Check it out. 
Free forum by Nabble  Edit this page 