"The Calendar" by David Ewing Duncan: Numerous Errors?

I have recently read "The Calendar" by David Ewing Duncan. It is an
excellent book, and one that often had me going back over the pages to
check or compare what I was currently reading with what I had read
earlier. In doing so, however, I found several discrepancies that I
couldn't explain. The question is, are these errors, or have I simply
misunderstood?

My copy is a paperback edition published by Fourth Estate, London, in
1999. I can only quote page numbers from my book, so my apologies if
your page numbers are different. In the order that they appear in the
book, my queries are as follows;

1. Calendar Index, page vi - The year as amended by Pope Gregory XIII
(the Gregorian calendar): 365 days, 5 hours, 48 minutes, 20 seconds.
Shouldn't this be 365 days, 5 hours, 49 minutes, 12 seconds?

2. Calendar Index, page vi - Length of time the Gregorian calendar is
off from the true solar year: 25.96768 seconds. Deducting this from
the figure I have assumed in item 1 above, the true solar year would
be 365 days, 5 hours, 48 minutes, 46.03232 seconds. Why be so
precise, and then get it wrong? According to some sources, this
figure corresponds to the 1840's, not 1900, 1996, 1999 or 2000 (all
years considered to be current at various locations in the book).

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?

4. Timeline, page xxi, Year 1100 - The year by Omar Khayyam is given
as 365d 5h 49m 12s (and the same again on page 278), but on page 190
it is given as 365.24219858156 days. This converts to 365d 5h 48m
45.96s, so why the difference? I believe that the first figure is
actually the Gregorian year. The second figure is 99.99999% of that
given for the atomic year.

5. Time Stands Still, page 97 - In the table, 6 Kalends April is given
as 26th March, whereas other sources give this as 27th March. Which
is correct?

6. Time Stands Still, page 98. The formula 22 + 11 + 33 - 30 + 3,
should read 22 + 11 = 33 - 30 = 3?

7. The Strange Journey of 365.242199, page 154 - He (Aryabhata)
estimates the length of the solar year at 365.3586805 days, some 2
hours 47 minutes and 44 seconds off from the true year in Aryabhata's
era, which equalled 365.244583 days. The footnote says that this is
about 7 seconds shorter than our current year. Only three errors
here! These are (i) 365.3586805 days is 2 hours 47 minutes and 44
seconds off our era, not the year in 499; (ii) 365.244583 days equals
365 days, 5 hours, 52 minutes, 12 seconds, which is actually 3 minutes
26 seconds longer than 365.242199 days; and (iii) In 499, the year
would have been about 7 seconds longer, not shorter, than our current
year.

8. From the House of Wisdom to Darkest Europe, page 190 - Omar Khayyam
appears to have calculated the year to within 4 seconds of what it was
in 1079. Duncan doesn't make this point, but just brushes it of as
'overly precise'. Any comments?

9. From the House of Wisdom to Darkest Europe, page 190-191 - Ulugh
Beg gave a measurement for the length of the year that came to 365
days, 5 hours, 49 minutes and 15 seconds, just 25 seconds too long. I
make this figure about 29 seconds longer than our time, and 27 seconds
longer than in 1440. Where does the 25 seconds come from?

10. Solving the Riddle of Time, page 277-278 - The table contains two
differences from what is given elsewhere in the book. The
measurements for Omar Khayyam and the Gregorian calendar are not as
quoted on pages 190 and 277 respectively. See items 4 and 1 above for
further details.

My conclusions? It's a great book, that probably took Duncan many
years to research. However, next time, he should get somebody to
check his mathematics!

Regards,



Philip Clarke

Apparently you're not the only one who has issues with the accuracy of
Duncan's Calendar. The following quote can be found at:
http://www.friesian.org/week.htm
"The usefulness of Duncan's book is compromised by many, many errors,
sometimes egregious ones. His treatment of luni-solar calendars is a
disaster."

Have you written him regarding your observations? His contact information
is on Literati.Net at http://literati.net/Duncan/DuncanContacts.htm

"Philip Clarke" wrote...

I have recently read "The Calendar" by David Ewing Duncan. It is an
excellent book, and one that often had me going back over the pages to
check or compare what I was currently reading with what I had read
earlier. In doing so, however, I found several discrepancies that I
couldn't explain. The question is, are these errors, or have I simply
misunderstood?

I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Philip Clarke wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?
--
Tom Rankin - Programmer by day, amateur astronomer by night!
Mid-Hudson Astronomy Association - http://mhaa.whodeanie.com
Views and Brews - http://viewsandbrews.com

When replying, remove the capital letters from my email address.

Tom,

Thanks for your response. Yes, I'm sure you're right that rounding
has taken place.

On page 322, Duncan defines the year as 31,556,925.9747 seconds (the
standard up until 1967). Multiplying this by the Cesium rate gives
290,091,200,278,565,436 oscillations. What I don't understand,
though, is why this is rounded UP to 290,091,200,500,000,000. I would
have been quite happy with 290,091,200,278,000,000, though.

Regards,


Philip Clarke


Tom Rankin wrote in message . ..
I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Philip Clarke wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?

On Sun, 25 Jul 2004 12:44:45 GMT, Tom Rankin
wrote:

I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

Maybe it is the other way around: They choosed the length of the
second so that it gives a nice round figure for the caesium
oscillations?


It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Philip Clarke wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?

JRS: In article , dated
Sun, 25 Jul 2004 12:44:45, seen in news:sci.astro.amateur, Tom Rankin
posted :
Philip Clarke wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?
I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Please put responses after (trimmed) quotes.

The figure is not for oscillations of caesium; it is for the frequency
of radiation corresponding with a hyperfine transition of caesium.

More importantly, the figure is exact; it defines the SI second :

"La seconde est la durée de 9 192 631 770 périodes de la radiation
correspondant à la transition entre les deux niveaux hyperfins de l'état
fondamental de l'atome de cesium 133 (CGPM 13, 1967, Resolution 1). "


One cannot speak of the length of the year without defining which year -
Gregorian Calendar, UTC, Tropical, Sidereal, Anomalistic, Eclipse, etc.

There appear to be 290,091,439,519,565,040 oscillations in an average
UTC year disregarding leap-seconds - EXACTLY. For an ordinary year,
289,898,835,498,720,000; for a leap year, 290,693,078,883,648,000; also
EXACTLY.


Philip :

Q1, yes, ... 12 seconds; easily worked out.

Does the information in Duncan suffice to pin down, with certainty and
allowing for days starting early, the exact date and time in the
Gregorian or Julian Calendars of the first moment of Year 1 Month 1 Day
1 of the Islamic Calendar (disregarding variation of local time with
longitude)?

I considered the book to be quite good, but prefer

URL:http://www.users.zetnet.co.uk/egrichards/ E.G.Richards,
"Mapping Time - The Calendar and its History",
OUP 1998, ISBN 0-19-850413-6 (now 2nd Edn, and pbk.).
Book & site include algorithms; site has errata.

I've not seen the second edition; one does need to deal with the errata,
though they are mostly typos.


See URL:http://www.merlyn.demon.co.uk/datelinx.htm#PM .

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - w. FAQish topics, links, acronyms
PAS EXE etc : URL:http://www.merlyn.demon.co.uk/programs/ - see 00index.htm
Dates - miscdate.htm moredate.htm js-dates.htm pas-time.htm critdate.htm etc.

Dr John Stockton wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?
I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Please put responses after (trimmed) quotes.

Sorry; I've just re-read the Posting Style Guide.


The figure is not for oscillations of caesium; it is for the frequency
of radiation corresponding with a hyperfine transition of caesium.

Noted; for bevity I will use the term "caesium frequency".


More importantly, the figure is exact; it defines the SI second :

"La seconde est la durée de 9 192 631 770 périodes de la radiation
correspondant à la transition entre les deux niveaux hyperfins de l'état
fondamental de l'atome de cesium 133 (CGPM 13, 1967, Resolution 1). "

Oui, je comprends.


One cannot speak of the length of the year without defining which year -
Gregorian Calendar, UTC, Tropical, Sidereal, Anomalistic, Eclipse, etc.

Agreed. Unfortunately, I wasn't too clear here. I was actually
looking to verify the accuracy of the mean solar year that I had
originally referred to in the previous question.


There appear to be 290,091,439,519,565,040 oscillations in an average
UTC year disregarding leap-seconds - EXACTLY. For an ordinary year,
289,898,835,498,720,000; for a leap year, 290,693,078,883,648,000; also
EXACTLY.

I can see that these figures relate to 365.2425, 365 and 366 days,
respectively. Isn't the first figure the Gregorian year, or is this
the same as the UTC year?

Since I couldn't verify the exact length of the mean solar year, I was
hoping to approach the problem from another angle. Of course, I could
have multiplied the "caesium frequency" by 365 days, 5 hours, 48
minutes, 46.03232 seconds, but it was this duration that I was trying
to prove. Initially, I was struck by an internet article which
described how the second was originally defined as 1 / 31,556,925.975
of a year, and then after Scientists complained that this was not
accurate enough, they re-defined the second as 1 / 31,556,925.9747 of
a year. By definition, a (mean solar?) year is 31,556,925.9747
seconds in duratin. The question is, if 0.003 of a second made a
difference to Scientists in 1967, then why was the "caesium frequency"
of the current year only defined to the nearest 0.5 seconds? I have
looked at formulae by Newcomb and others without success.


Philip :

Q1, yes, ... 12 seconds; easily worked out.

I thought so, thanks.


Does the information in Duncan suffice to pin down, with certainty and
allowing for days starting early, the exact date and time in the
Gregorian or Julian Calendars of the first moment of Year 1 Month 1 Day
1 of the Islamic Calendar (disregarding variation of local time with
longitude)?

I'm sorry, but I don't really understand this question.


I considered the book to be quite good, but prefer

URL:http://www.users.zetnet.co.uk/egrichards/ E.G.Richards,
"Mapping Time - The Calendar and its History",
OUP 1998, ISBN 0-19-850413-6 (now 2nd Edn, and pbk.).
Book & site include algorithms; site has errata.

I've not seen the second edition; one does need to deal with the errata,
though they are mostly typos.

Thanks, I will try to get a copy.

Many thanks for your reponse.

Regards,


Philip Clarke

JRS: In article ,
dated Wed, 28 Jul 2004 16:36:33, seen in news:sci.astro.amateur, Philip
Clarke posted :
Dr John Stockton wrote:

There appear to be 290,091,439,519,565,040 oscillations in an average
UTC year disregarding leap-seconds - EXACTLY. For an ordinary year,
289,898,835,498,720,000; for a leap year, 290,693,078,883,648,000; also
EXACTLY.

I can see that these figures relate to 365.2425, 365 and 366 days,
respectively. Isn't the first figure the Gregorian year, or is this
the same as the UTC year?

Yes; no, except if leap-seconds are disregarded. The difference between
Greg & UTC year varies according to the antics of our globe. UTC has
never been less than Greg since current rules started, has certainly
never exceeded it by more than two seconds, and AFAIR never by two, but
ICBW.


Does the information in Duncan suffice to pin down, with certainty and
allowing for days starting early, the exact date and time in the
Gregorian or Julian Calendars of the first moment of Year 1 Month 1 Day
1 of the Islamic Calendar (disregarding variation of local time with
longitude)?

I'm sorry, but I don't really understand this question.

Well, anyone else is free to answer it; or indeed to pin down the exact
date ant time in any other trustworthy and cite-able manner. Anyone
care to ask their Imam?

P.S. Actually, since writing that, I've borrowed a Duncan - which does
not address the point; it says little about Islamic calendars.

--
© John Stockton, Surrey, UK. Turnpike v4.00 MIME. ©
Web URL:http://www.merlyn.demon.co.uk/ - w. FAQish topics, links, acronyms
PAS EXE etc : URL:http://www.merlyn.demon.co.uk/programs/ - see 00index.htm
Dates - miscdate.htm moredate.htm js-dates.htm pas-time.htm critdate.htm etc.

(Philip Clarke) wrote in message . com...
Dr John Stockton wrote:

3. Calendar Index, page vi - The year as measured in oscillations of
atomic cesium: 290,091,200,500,000,000. If Cesium oscillates
9,192,631,770 times per second, then there are 794,243,384,928,000
oscillations in a day. Is the year really an exact multiple of
500,000,000 oscillations?
I can take a shot at #3.

Since the oscillation of Cesium is not known to 18 places, there is most
likely a round off taking place here.

It is interesting though, that this seems to imply that the length of
the year is only defined to roughly .05 seconds.

Please put responses after (trimmed) quotes.

Sorry; I've just re-read the Posting Style Guide.


The figure is not for oscillations of caesium; it is for the frequency
of radiation corresponding with a hyperfine transition of caesium.

Noted; for bevity I will use the term "caesium frequency".


More importantly, the figure is exact; it defines the SI second :

"La seconde est la durée de 9 192 631 770 périodes de la radiation
correspondant à la transition entre les deux niveaux hyperfins de l'état
fondamental de l'atome de cesium 133 (CGPM 13, 1967, Resolution 1). "

Oui, je comprends.


One cannot speak of the length of the year without defining which year -
Gregorian Calendar, UTC, Tropical, Sidereal, Anomalistic, Eclipse, etc.

Agreed. Unfortunately, I wasn't too clear here. I was actually
looking to verify the accuracy of the mean solar year that I had
originally referred to in the previous question.


There appear to be 290,091,439,519,565,040 oscillations in an average
UTC year disregarding leap-seconds - EXACTLY. For an ordinary year,
289,898,835,498,720,000; for a leap year, 290,693,078,883,648,000; also
EXACTLY.

I can see that these figures relate to 365.2425, 365 and 366 days,
respectively. Isn't the first figure the Gregorian year, or is this
the same as the UTC year?

Since I couldn't verify the exact length of the mean solar year, I was
hoping to approach the problem from another angle. Of course, I could
have multiplied the "caesium frequency" by 365 days, 5 hours, 48
minutes, 46.03232 seconds, but it was this duration that I was trying
to prove.

You will have to rely on your own commonsense in the absense of any
worthwhile response and you need not go too far before you recognise
why contemporary explanations are misleading and ineffective.

To determine the annual orbital cycle as 365 day 5 hours 49 min you
are required to determine the equable 24 hour day First whereby the
equable hour,minute and second are determined as subdivisions of the
24 hour day.

If you agree that it is not possible to calculate the annual cycle
without first determining the equable 24 hour day you will be half way
to developing a far better appreceation of the calendar system as an
offshoot of the principles which determine the equable day.

Using the Sun as a reference for the motions of the Earth,the combined
constant axial and variable orbital motion generates a change from
one rotation for a given longitude meridian to the next complete
rotation.The brilliance of our ancestors was to equalise the variation
by adding and subtracting appropriate minutes and seconds to
longitudinal noon to smooth out the variations and facilitate the
seamless transition from one 24 hour day to the next 24 hour
day,Monday into Tuesday,ect.

This correction is known as the Equation of Time.

In 1677,Flamsteed altered the 24 hour/360 degree equivalency and
linked the rotation of the Earth directly to stellar circumpolar
motion giving the value for rotation through 360 degrees as the
sidereal 23 hours 56 min 04 sec.He screwed up the ancient exquisite
reasoning that benefited humanity with the 24 hour day and
subsequently the calendar year based on that equable day.

The other responses you received so far indicate that in the 21st
century,men are incapable of telling what the rotation rate of the
Earth is through 360 degrees choosing as their value the sidereal
value.The correct value is and always will be 24 hours insofar as the
Equation of Time correction is the exquisite equalising method to
screen out all the variables introduced by variable orbital
motion,finite light distance or indeed variations in the rotation rate
of the Earth itself to facilitate the difference of 1 degree rotation
per 4 min, 15 degrees per 1 hour and ultimately 360 degrees per 24
hours.









Initially, I was struck by an internet article which
described how the second was originally defined as 1 / 31,556,925.975
of a year, and then after Scientists complained that this was not
accurate enough, they re-defined the second as 1 / 31,556,925.9747 of
a year. By definition, a (mean solar?) year is 31,556,925.9747
seconds in duratin. The question is, if 0.003 of a second made a
difference to Scientists in 1967, then why was the "caesium frequency"
of the current year only defined to the nearest 0.5 seconds? I have
looked at formulae by Newcomb and others without success.


Philip :

Q1, yes, ... 12 seconds; easily worked out.

I thought so, thanks.


Does the information in Duncan suffice to pin down, with certainty and
allowing for days starting early, the exact date and time in the
Gregorian or Julian Calendars of the first moment of Year 1 Month 1 Day
1 of the Islamic Calendar (disregarding variation of local time with
longitude)?

I'm sorry, but I don't really understand this question.


I considered the book to be quite good, but prefer

URL:http://www.users.zetnet.co.uk/egrichards/ E.G.Richards,
"Mapping Time - The Calendar and its History",
OUP 1998, ISBN 0-19-850413-6 (now 2nd Edn, and pbk.).
Book & site include algorithms; site has errata.

I've not seen the second edition; one does need to deal with the errata,
though they are mostly typos.

Thanks, I will try to get a copy.

Many thanks for your reponse.

Regards,


Philip Clarke

Philip Clarke wrote:

Since I couldn't verify the exact length of the mean solar year,

I'm not sure this is what you're getting at, but the length of the mean
tropical year isn't constant. It's getting shorter, and the measurement
is getting more accurate. 31556925.9747 sec is the canonical length of
a specific year (1900). The J2000 mean tropical year is 31556925.187488
sec.

By definition, a (mean solar?) year is 31,556,925.9747 seconds

It's not defined as a number of seconds. The year 1900 was in effect
defined to be this length, but in general, the mean tropical year is the
mean amount of time it takes the sun to return to a given longitude on
the ecliptic, and the ecliptic changes over time.

[...] why was the "caesium frequency" of the current year only
defined to the nearest 0.5 seconds? I have looked at formulae by
Newcomb and others without success.

The "caesium frequency" was just rounded off, possibly because the
length of a year isn't constant. I don't think there's any deeper
explanation than that.

Giving this number to greater precision would have implied a specific
measurement of a particular year, and it seems clear from the context
you've quoted that the number was merely an approximation derived from
the current definition of the SI second.

If you're hoping to use a higher precision version of this number to
to get "the" length of a year, you've been misled.

The Newcomb formula gives the longitude of the sun on a given date.

L = L0 + 129602768".13 C + 1".089 C^2 + ...

where L0 = 279° 41' 48".04 is the longitude of the sun on 1 Jan 1900 at
12:00Z, and C is the number of Julian centuries since that date. (A
Julian century is exactly 36525 days.)

Hopefully somebody will correct me if I'm wrong about this, since I'm
not an expert on how the definition of the ephemeris second was derived,
but I believe a version of Newcomb's formula with only the linear term
was used,

L = L0 + 129602768".13 C

and following this truncated version of the formula, the length of the
tropical year beginning 1 Jan 1900 12:00Z is just the value of C for
which L = L0 + 360°. Solving for C,

C = 1296000" / 129602768".13
= 0.0099997864142842054588143772543... Julian centuries
= 365.24219878173060438319512921348... days
= 31556925.974741524218708059164045... seconds

There's no physical justification for the extra precision--the year 1900
wasn't "really" precisely this long. The value 31556925.9747 arises
simply from rounding at the number of significant digits in the formula.
It also depends on ignoring the higher order terms. If those are added,
the number differs by on the order of 0.003 sec.

- Ernie http://home.comcast.net/~erniew