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"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 |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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. |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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? |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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. |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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 |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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. |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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"The Calendar" by David Ewing Duncan: Numerous Errors?
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 |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
In message , Oriel36
writes The same old BS. 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. But the day is defined as the time between meridian passages of the sun, and it's divided up into hours, minutes, and seconds. In Civil Time the day is assumed to be of constant length. What's so difficult about that idea? 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. Why not post the Flamsteed statement, or post a link to it? I'd be interested to hear from someone who actually knows what they are talking about, but AFAICS Flamsteed was responsible for the idea of the equation of time. BTW, either you have a virus or you are deliberately sending me unsolicited emails. -- What have they got to hide? Release the full Beagle 2 report. Remove spam and invalid from address to reply. |
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"The Calendar" by David Ewing Duncan: Numerous Errors?
Jonathan Silverlight wrote in message ...
In message , Oriel36 writes The same old BS. To be fair to you,the mistake is a really old one and precedes the gravitational agenda (which makes use of the mistake) by a decade.The first professional astronomer was Flamsteed who set out to prove that the Earth rotates constantly on its axis through the motion of the fixed stars thereby making use of this observation for determining planetary longitude. Flamsteed was incorrect. 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. But the day is defined as the time between meridian passages of the sun, and it's divided up into hours, minutes, and seconds. In Civil Time the day is assumed to be of constant length. What's so difficult about that idea? 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. Why not post the Flamsteed statement, or post a link to it? I did many times - .... our clocks kept so good a correspondence with the Heavens that I doubt it not but they would prove the revolutions of the Earth to be isochronical" http://www-gap.dcs.st-and.ac.uk/~his...ongitude2.html A direct consequence of that statement can be found in any website refering to the sidereal value - http://hypertextbook.com/facts/1999/JennyChen.shtml I'd be interested to hear from someone who actually knows what they are talking about, but AFAICS Flamsteed was responsible for the idea of the equation of time. You won't find any,there is just myself. You can however appeal to those who are familiar with the equable 24 hour clock day,the Equation of Time correction from the noon determination and how one 24 hour day elapses seamlessly into the next 24 hour day by the 'Equation of Time' method but then you will encounter those who are aware that the precise value for axial rotation through 360 degrees,is and always will be 24 hours. BTW, either you have a virus or you are deliberately sending me unsolicited emails. |
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