Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

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Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Karl Palmen

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year). Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 

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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day

How about February 28th?

Here's the fallacy behind the error:

From February 28, 1904 to February 28, 2096 is 192 years.

Suppose that there weren't a leapday in 2096.

Surely then March 1 2096 is so close to February 28, 2096, that you can still call it 192 years, without important error, for the purpose of judging how much the  +.24219 days per year positive displacement has displaced the calendar during those 192 years.

I mean, without a leapday in 2096, surely the displacement of the calendar doesn't suddently take a whole-day jump at any point. The relation between date & ecliptic-longitude is an objective observable fact, not subject to different answers depending on our assumptions. So I assumed that there's no reason to call it different between February 28 & March 1, if there were no leapday that year. It seemed justifiable to say that, from 1904's leap-insertion-point to 2096's leap-insertion-point is 192 years.

So, 192 X .2425 - 49 = -2.44

If the example of January 10th 1904 to January 10th 2097 is mentioned, then the answer could be, "Sure, that's a longer duration, so of course it results in a different answer. Looking at dates 193 years apart, instead of 192 years, encompassing the same number of leapdays, will of course result in less overall negative net displacement."

Of course there is a leapday in 2096, right between February 28 & March 1. So that's one of the 49 leapdays in the interval of interest.

Of course there must be a fallacy there somewhere. That's how I got the incorrect answer of -2.44 days.

I just wanted to tell how I got the incorrect -2.44 days answer.


(Replying farther down) :

, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year). Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:


Yes, but what interests me most is the largest displacement between one extreme & the next extreme. That occurs during the negatively-changing half of the 400-year cycle.
 
Michael Ossipoff


 

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 


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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff
Typo:

I said:

"Surely then March 1 2096 is so close to February 28, 2096, that you can still call it 192 years, without important error, for the purpose of judging how much the  +.24219 days per year positive displacement has displaced the calendar [with respect to the mean year] during those 192 years."

I meant +.2425, not +.24219

Michael Ossipoff

On Tue, Jan 10, 2017 at 5:18 PM, Michael Ossipoff <[hidden email]> wrote:


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day

How about February 28th?

Here's the fallacy behind the error:

From February 28, 1904 to February 28, 2096 is 192 years.

Suppose that there weren't a leapday in 2096.

Surely then March 1 2096 is so close to February 28, 2096, that you can still call it 192 years, without important error, for the purpose of judging how much the  +.24219 days per year positive displacement has displaced the calendar during those 192 years.

I mean, without a leapday in 2096, surely the displacement of the calendar doesn't suddently take a whole-day jump at any point. The relation between date & ecliptic-longitude is an objective observable fact, not subject to different answers depending on our assumptions. So I assumed that there's no reason to call it different between February 28 & March 1, if there were no leapday that year. It seemed justifiable to say that, from 1904's leap-insertion-point to 2096's leap-insertion-point is 192 years.

So, 192 X .2425 - 49 = -2.44

If the example of January 10th 1904 to January 10th 2097 is mentioned, then the answer could be, "Sure, that's a longer duration, so of course it results in a different answer. Looking at dates 193 years apart, instead of 192 years, encompassing the same number of leapdays, will of course result in less overall negative net displacement."

Of course there is a leapday in 2096, right between February 28 & March 1. So that's one of the 49 leapdays in the interval of interest.

Of course there must be a fallacy there somewhere. That's how I got the incorrect answer of -2.44 days.

I just wanted to tell how I got the incorrect -2.44 days answer.


(Replying farther down) :

, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year). Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:


Yes, but what interests me most is the largest displacement between one extreme & the next extreme. That occurs during the negatively-changing half of the 400-year cycle.
 
Michael Ossipoff


 

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 



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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Irv Bromberg
In reply to this post by Karl Palmen
From: East Carolina University Calendar discussion List [[hidden email]] on behalf of Karl Palmen [[hidden email]]
Sent: Tuesday, January 10, 2017 11:03

Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).


Irv replies: Yes, my mistake in relying on my lousy memory instead of looking it up. It is shown on page 5 of my jitter charts collection here:

http://home.primus.ca/~kalendis/solar/jitter-charts-leap-day.pdf

and the jitter range is given there as 2+79/400 days, which, as Karl wrote, is 1/400 day less than the 2.2 days that I mistakenly quoted earlier.
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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff
In reply to this post by Karl Palmen


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

Karl:

As you said in another thread (which I'll probably reply to later), displacement isn't really so simple. It's true that it's a difference or change in the relation between date and solar ecliptic longitude, but that doesn't define it completely. There are different ways that difference or change can be measured/defined.

What makes calendar leapyear-rules such a uniquely confusing subject is that  (as I mentioned once before) the matter of leapyears & displacements is a combination of physical fact, and conventional fictitious things.

Yes of course displacement is most meaningfully & usefully defined between times that have the same date in different years. And, by that measure, the Gregorian jitter-range is about 2.2 days

But, for better or worse, there's a conventional assumption that the calendar's displacement is the same over a period that you've reasonably named a "displacement-year" (DY), and that the calendar's displacement changes abruptly at the end of a displacement year, when (if the ending year isn't a leapyear) it increases (with respect to a reference year of length Y) by an amount equal to Y minus the length of that particular displacement-year.

You made the reasonable suggestion that the DY should begin at the leap-insertion-point (LIP). That's the most convenient and logical place to start the DY.  Bu that assumption, the Gregorian jitter-range again is about 2.2 days.

But, convenient & logical though it may be, that isn't the only possible place to start the DY. If the DY starts at January 1, then doesn't that mean that the Gregorian's jitter-displacement range can reach 2.44?

I was interested in the maximum value that the Gregorian jitter-range could have, by any interpretation or measure.

Another thing: This conventional notion of a DY in which the calendar-displacement is (defined as) constant, and at the end of which the calendar-displacement abruptly changes by an amount equal to Y minus the DY's length....Isn't it only just a convenient conventional notion, without physical basis?

Maybe arguably the calendar-displacament, is, by one possible definition, continually changing. The date and the ecliptic longitude are measured in two completely different units (days vs degrees). but the ongoing change in the date and in the ecliptic longitude can be described with a common unit: Percent completion of a (calendar vs tropical) year, from some arbitrarily-chosen starting-point.

Regardless of what starting-point is chosen, while the calendar's completion-percentage is increasing by 1/365, the Gregorian average year's completion-percent is increasing by 1/365.2425

So,by that interpretation, the common calendar year's completion percentage is constantly gaining on that of the MTY, in terms of completion-percentage from some arbitrary unnamed starting-point.

How about as reckoned from just barely before leapday 1904 and just barely after leaday 2096? That's barely over 192 years, and that steady gain of (common-lengsth) calendar year over MTY would reach about 192 X .2425

...if there were no leapdays in that interval. But, with the 49 leapyears, the net change in the relation between the two years' completion-percentages amounts to -2.44 days' worth.

I'm just saying that the logical and convenient start of the DY at the LIP isn't the only possible interpretation. The DY could be started elsewhere, like at January 1st.

And, in fact, the displacement could be defined as the calendar-year's constantly steadily increasing gain in completion-percentage with respect to that of the calendar's mean-year.   ...instead of assuming a DY during which calendar-displacement is constant, and at the end of which calendar-displacement abruptly increases by the Y minus the DY's length.

Maybe your interpretation is the logical and convenient one, but there are other interpretations, and there probably isn't any reason why one interpretation is objectively more valid than another.

By the way, just as it makes the most sense to say that the DY begins at the LIP, wouldn't it also make more sense to start the calendar year at the equinox at which the ecliptic-longitude numbering begins?

Of course that conflicts with the widely-agreed feeling that the year begins in midwinter, and so i'm not advocating changing the beginning of the calendar year to the March equinox. I just mention it as something that, in a way, would make more sense. (but, in another way, wouldn't). I realize that there have been periods in history during which the calendar year was regarded as starting at the March equinox.

Michael Ossipoff















 

Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 


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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff
In my post of a few minutes ago, there were several places where i said "MTY" when I meant "Gregorian average year".

Every instance, in that post, where I said "MTY", I meant "Gregorian average year".

...because the topic was the Gregorian jitter-range.


Michael Ossipoff

On Sat, Jan 14, 2017 at 5:04 PM, Michael Ossipoff <[hidden email]> wrote:


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

Karl:

As you said in another thread (which I'll probably reply to later), displacement isn't really so simple. It's true that it's a difference or change in the relation between date and solar ecliptic longitude, but that doesn't define it completely. There are different ways that difference or change can be measured/defined.

What makes calendar leapyear-rules such a uniquely confusing subject is that  (as I mentioned once before) the matter of leapyears & displacements is a combination of physical fact, and conventional fictitious things.

Yes of course displacement is most meaningfully & usefully defined between times that have the same date in different years. And, by that measure, the Gregorian jitter-range is about 2.2 days

But, for better or worse, there's a conventional assumption that the calendar's displacement is the same over a period that you've reasonably named a "displacement-year" (DY), and that the calendar's displacement changes abruptly at the end of a displacement year, when (if the ending year isn't a leapyear) it increases (with respect to a reference year of length Y) by an amount equal to Y minus the length of that particular displacement-year.

You made the reasonable suggestion that the DY should begin at the leap-insertion-point (LIP). That's the most convenient and logical place to start the DY.  Bu that assumption, the Gregorian jitter-range again is about 2.2 days.

But, convenient & logical though it may be, that isn't the only possible place to start the DY. If the DY starts at January 1, then doesn't that mean that the Gregorian's jitter-displacement range can reach 2.44?

I was interested in the maximum value that the Gregorian jitter-range could have, by any interpretation or measure.

Another thing: This conventional notion of a DY in which the calendar-displacement is (defined as) constant, and at the end of which the calendar-displacement abruptly changes by an amount equal to Y minus the DY's length....Isn't it only just a convenient conventional notion, without physical basis?

Maybe arguably the calendar-displacament, is, by one possible definition, continually changing. The date and the ecliptic longitude are measured in two completely different units (days vs degrees). but the ongoing change in the date and in the ecliptic longitude can be described with a common unit: Percent completion of a (calendar vs tropical) year, from some arbitrarily-chosen starting-point.

Regardless of what starting-point is chosen, while the calendar's completion-percentage is increasing by 1/365, the Gregorian average year's completion-percent is increasing by 1/365.2425

So,by that interpretation, the common calendar year's completion percentage is constantly gaining on that of the MTY, in terms of completion-percentage from some arbitrary unnamed starting-point.

How about as reckoned from just barely before leapday 1904 and just barely after leaday 2096? That's barely over 192 years, and that steady gain of (common-lengsth) calendar year over MTY would reach about 192 X .2425

...if there were no leapdays in that interval. But, with the 49 leapyears, the net change in the relation between the two years' completion-percentages amounts to -2.44 days' worth.

I'm just saying that the logical and convenient start of the DY at the LIP isn't the only possible interpretation. The DY could be started elsewhere, like at January 1st.

And, in fact, the displacement could be defined as the calendar-year's constantly steadily increasing gain in completion-percentage with respect to that of the calendar's mean-year.   ...instead of assuming a DY during which calendar-displacement is constant, and at the end of which calendar-displacement abruptly increases by the Y minus the DY's length.

Maybe your interpretation is the logical and convenient one, but there are other interpretations, and there probably isn't any reason why one interpretation is objectively more valid than another.

By the way, just as it makes the most sense to say that the DY begins at the LIP, wouldn't it also make more sense to start the calendar year at the equinox at which the ecliptic-longitude numbering begins?

Of course that conflicts with the widely-agreed feeling that the year begins in midwinter, and so i'm not advocating changing the beginning of the calendar year to the March equinox. I just mention it as something that, in a way, would make more sense. (but, in another way, wouldn't). I realize that there have been periods in history during which the calendar year was regarded as starting at the March equinox.

Michael Ossipoff















 

Irv was wrong about the jitter being exactly 2.2 days. It is exactly 2.1975 days (perhaps, something led him to add 1/400).

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 



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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff
In reply to this post by Karl Palmen


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  


Karl said:

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

Right, so what's the Gregorian calendar's displacement (disregarding leapdays), with respect to the Gregorian average year, from February 28th, 11:59 p.m., 1904 to February 28, 11:59 p.m. 2096?

I'd say that's 192 X .2425 = 46.56

Karl said:

Jitter is only meaningful when comparing  the same date in different years

[endquote]

Is it meaningful to speak of the calendar's displacement (with respect to some mean year like 365.2425) of a date-&-time being about the same as that of as time that's 2 minutes earlier?

Remember that the +.2425 annual displacement with respect to the Gregorian mean year that occurs in an abrupt jump isn't what's really happening. The only displacement that really occurs in an abrupt jump is that caused by a leapday.

So on Februrary 28th, 11:59 p.m, 2096, the calendar is (disregarding leapdays) displaced 46.56 days, with respect to February 28th, 11:59, 1904.

Ok, then how about March 1, 12:01 a.m., 2096? How much is the calendar then displaced with respect to February 28th, 11:59 p.m., 1904?  Less than it was 2 minutes previous? If so, then explain how & why that could be. (Remember, we're disregarding leapdays).

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

Is that error the source of the 2.2 answer for the Gregorian jitter-range?

I'd said:

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error. 

[endquote]

I suggest that it's best to avoid ad hominem language.  ...because, in that way, we avoid saying things that we later won't be proud of, when we're wrong.

State your claim, say why you believe it to be so, but it's best to refrain from characterizing the other person, or what they said, just in case you might be mistaken.

Michael Ossipoff














 

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 


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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff

A wong wording in the message that I posted a few minutes ago::

I said:

" Less than it was 2 minutes previous? If so, then explain how & why that could be. (Remember, we're disregarding leapdays)."

I meant:

".2425 days different from what it was 2 minutes previous?"




On Sun, Jan 15, 2017 at 10:24 AM, Michael Ossipoff <[hidden email]> wrote:


On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  


Karl said:

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

Right, so what's the Gregorian calendar's displacement (disregarding leapdays), with respect to the Gregorian average year, from February 28th, 11:59 p.m., 1904 to February 28, 11:59 p.m. 2096?

I'd say that's 192 X .2425 = 46.56

Karl said:

Jitter is only meaningful when comparing  the same date in different years

[endquote]

Is it meaningful to speak of the calendar's displacement (with respect to some mean year like 365.2425) of a date-&-time being about the same as that of as time that's 2 minutes earlier?

Remember that the +.2425 annual displacement with respect to the Gregorian mean year that occurs in an abrupt jump isn't what's really happening. The only displacement that really occurs in an abrupt jump is that caused by a leapday.

So on Februrary 28th, 11:59 p.m, 2096, the calendar is (disregarding leapdays) displaced 46.56 days, with respect to February 28th, 11:59, 1904.

Ok, then how about March 1, 12:01 a.m., 2096? How much is the calendar then displaced with respect to February 28th, 11:59 p.m., 1904?  Less than it was 2 minutes previous? If so, then explain how & why that could be. (Remember, we're disregarding leapdays).

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

Is that error the source of the 2.2 answer for the Gregorian jitter-range?

I'd said:

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error. 

[endquote]

I suggest that it's best to avoid ad hominem language.  ...because, in that way, we avoid saying things that we later won't be proud of, when we're wrong.

State your claim, say why you believe it to be so, but it's best to refrain from characterizing the other person, or what they said, just in case you might be mistaken.

Michael Ossipoff














 

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13




 



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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Karl Palmen
In reply to this post by Michael Ossipoff

Dear Michael and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 14 January 2017 22:04
To: [hidden email]
Subject: Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

 

 

 

On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 28 December 2016 04:32
To: [hidden email]
Subject: Re: New Henry-Hanke calendar/website, old shortcomings

 



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

Karl:

As you said in another thread (which I'll probably reply to later), displacement isn't really so simple. It's true that it's a difference or change in the relation between date and solar ecliptic longitude, but that doesn't define it completely. There are different ways that difference or change can be measured/defined.

KARL REPLIES: Yes it is not so simple. There are at least two types. There is the type that Michael calls “Calculated Displacement”, which is used to define an arithmetic minimum displacement calendar. There is also “Actual Displacement” which is the displacement of actual solar ecliptic longitudes (for ideal positions in the calendar year). I had to add the part in () for it to be meaningful to me. The expression “change in the relation between date and solar ecliptic longitude” is hopelessly vague on its own. The ideal positions are arbitrary and need to be defined for year-round accuracy to be meaningful.

 

What makes calendar leapyear-rules such a uniquely confusing subject is that  (as I mentioned once before) the matter of leapyears & displacements is a combination of physical fact, and conventional fictitious things.

Yes of course displacement is most meaningfully & usefully defined between times that have the same date in different years. And, by that measure, the Gregorian jitter-range is about 2.2 days

But, for better or worse, there's a conventional assumption that the calendar's displacement is the same over a period that you've reasonably named a "displacement-year" (DY), and that the calendar's displacement changes abruptly at the end of a displacement year, when (if the ending year isn't a leapyear) it increases (with respect to a reference year of length Y) by an amount equal to Y minus the length of that particular displacement-year.

You made the reasonable suggestion that the DY should begin at the leap-insertion-point (LIP). That's the most convenient and logical place to start the DY.  Bu that assumption, the Gregorian jitter-range again is about 2.2 days.

But, convenient & logical though it may be, that isn't the only possible place to start the DY. If the DY starts at January 1, then doesn't that mean that the Gregorian's jitter-displacement range can reach 2.44?

KARL REPLIES: When measuring the jitter of a calendar, one must use the displacement year of that calendar, which for Gregorian begins March 1.

 

I was interested in the maximum value that the Gregorian jitter-range could have, by any interpretation or measure.

 

Another thing: This conventional notion of a DY in which the calendar-displacement is (defined as) constant, and at the end of which the calendar-displacement abruptly changes by an amount equal to Y minus the DY's length....Isn't it only just a convenient conventional notion, without physical basis?

KARL REPLIES: Yes. For a calendar with a leap day or leap week inserted at one place in the year, it is sufficient to consider the displacement of one date such as January 1 to define such as calendar and minimise its displacement.

Extension to other dates is merely a convention and does not lead to a more accurate calendar, but it may be useful in defining year round accuracy.

However if you choose ANY convention that does not give exactly the same displacement value to two dates with no leap day/week insertion point (LIP) between them, then the resulting displacements will be different from the calculated displacements of a minimum displacement calendar and will in general have a larger magnitude.

 

Maybe arguably the calendar-displacament, is, by one possible definition, continually changing. The date and the ecliptic longitude are measured in two completely different units (days vs degrees). but the ongoing change in the date and in the ecliptic longitude can be described with a common unit: Percent completion of a (calendar vs tropical) year, from some arbitrarily-chosen starting-point.

KARL REPLIES: Percent completion of a calendar year not fixed relative to a calendar date. It is different in a leap year than in a common year. It is therefore not an appropriate measure. An appropriate measure is time elapsed since the start of the displacement year. Then each date (including time of day) in a year has exactly the same time since start of the displacement year. Starting the displacement year just after the insertion point of the leap day or week (LIP) is necessary for this.

This time (elapsed since the start of the displacement year) may be converted into angle by dividing by the calendar mean year (Y) and multiplying by 360 degrees. Such an angle measure may be used to define the ideal solar ecliptic longitudes (ISEL), by adding an arbitrary angle to it, corresponding to the ISEL just after the LIP.

I see from a later note that instead of dividing by the length of the mean calendar year (Y), Michael might divide by the length of the common year. This would squeeze all the ISEL into a common year, get the continuously changing calculated displacements (as CSEL – ISEL), but also increase displacements both calculated and actual.

I hope Michael will read, think about and respond to my note in which I define the ideal solar ecliptic longitude (ISEL), calculated solar ecliptic longitude (CESL) and actual solar ecliptic longitude (ASEL).

Karl

16(05(19

 

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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Karl Palmen
In reply to this post by Michael Ossipoff

Dear Michael and Calendar People

 

MICHAEL SAID:

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

KARL REPLIES: It’s no error. It arises from the fact that the mean time (over a Gregorian 400-year cycle) that March 1, 12:01 am occurs after February 28th, 11:59 pm is 2 minutes + 0.2425 day.

One may think of each LIP ideally containing a mean leap day/week equal in duration the a day/week multiplied by the proportion of years that are leap years. Then the sudden jump in displacement arises from a deviation from this ideal.

I then think more about this ideal year and I think what Michael is doing is to stretch the common year to fill the duration of the calendar mean year to eliminate to gaps at the LIPs. This would result in a calculated displacement that changes at a constant rate every non-leap day and goes down 1 each leap day or leap week. This displacement would have a range that extends outside of (-0.5, +0.5] as in his original proposal for minimum displacement calendar for when the LIP is not at the end of the year. Here I use units of day for a leap day calendar and week for a leap week calendar.

I realise that the space, in which this ideal year lies, is the space of ideal solar ecliptic longitudes (ISEL) and by stretching the common year, he squeezes all the ISEL into a common year. Whereas there would otherwise be a few ISEL that would occur either in the next displacement year or in the leap day/week.  These few ISEL are an inconvenience eliminated by the squeezing of the ISEL into a common year. The price of this resulting convenience is an increase in both the calculated and actual displacements.

Karl

16(05(19

 

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 15 January 2017 15:25
To: [hidden email]
Subject: Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

 

 

 

On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

 

Karl said:

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

 

Right, so what's the Gregorian calendar's displacement (disregarding leapdays), with respect to the Gregorian average year, from February 28th, 11:59 p.m., 1904 to February 28, 11:59 p.m. 2096?

I'd say that's 192 X .2425 = 46.56

Karl said:

Jitter is only meaningful when comparing  the same date in different years

[endquote]

Is it meaningful to speak of the calendar's displacement (with respect to some mean year like 365.2425) of a date-&-time being about the same as that of as time that's 2 minutes earlier?

Remember that the +.2425 annual displacement with respect to the Gregorian mean year that occurs in an abrupt jump isn't what's really happening. The only displacement that really occurs in an abrupt jump is that caused by a leapday.

So on Februrary 28th, 11:59 p.m, 2096, the calendar is (disregarding leapdays) displaced 46.56 days, with respect to February 28th, 11:59, 1904.

Ok, then how about March 1, 12:01 a.m., 2096? How much is the calendar then displaced with respect to February 28th, 11:59 p.m., 1904?  Less than it was 2 minutes previous? If so, then explain how & why that could be. (Remember, we're disregarding leapdays).

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

Is that error the source of the 2.2 answer for the Gregorian jitter-range?

I'd said:

 

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error. 

[endquote]

I suggest that it's best to avoid ad hominem language.  ...because, in that way, we avoid saying things that we later won't be proud of, when we're wrong.

State your claim, say why you believe it to be so, but it's best to refrain from characterizing the other person, or what they said, just in case you might be mistaken.

Michael Ossipoff

 

 



 

 

 

 



 

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13



 

 

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Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

Michael Ossipoff
About having the LIP other than at the end of the year:

Originally, when I posted my Minimum-Displacement Calendar (multi-version) proposal last spring & summer, I wanted the nonfixed Roman version to be so true to tradition that its LIP is at the end of February 28th.

When you (Karl) pointed out the complexity that that would bring in, as regards the choice of Dzero, realized that having the LIP other than at the end of the year wouldn't be a good idea, even for a nonfixed Roman version.

Anyway, I've realized also that a multi-version proposal isn't such a good idea either. And so now my Minimum-Displacement Calendar proposal consists only of the 30,30,31 quarters, in the leapweek fixed version with the Minmum-Displacement leapyear rule.

When it's only one version, that allowed me to leave out a lot of definitions and terms that were complicating the proposal.

My only concession to multichoice is that I offer Dzero values of 0 or -.6288   ...with -.6288 as the recommendation.  ...with an epoch of Gregorian January 2nd, 2017.

I've posted my 1-version Minimum-Displacement Calendar proposal at the Calendar Wiki, and it's shown at the recent-changes page, but (like the recently-posted Hanke-Henry proposal) it hasn't yet appeared at the Proposed Calendars page.

About displacement & jitter-calculation, it's more & more obvious that displacement isn't a simple topic.

As I was saying, it's a complex mix of difficultly-separated physical fact & conventional fiction.

If I don't have an LIP other than at the end of the year, I hope that gets rid of some of the contradictions and incompatibilities.

We're using different definitions, and thereby getting different answers (2.2 & 2.44) for the Gregorian jitter-range.

Those definitions seem a matter of choice, not hard fact. Maybe my choices & definitions are contrary to accepted convention.

You mentioned that the notion of a constantly-increasing calendar-displacement can result in a calculation of greater max displacement, when the LIP isn't at year-end. But now I don't propose a calendar with LIP not at year-end.

But isn't it undeniable that, as the Roman-Gregorian common-year undergoes 1/365 of a year-completion, the Gregorian mean-year undergoes 1/365.2425 of a year-completion, and the MTY undergoes 1/365.24217 of a year-completion?

(A wikipedia page said that the MTY currently has about 365.24217 mean solar days.)

So isn't the Roman-Gregorian common year steadily, constantly gaining on those other 2 years every day, every minute?

Sure, leapyear-rules, including my Minimum-Displacement leapyear-rule, at the end of a common-year, add (Y-365) days  to D. But that's just adding what has been accumulating all year, isn't it?

Of course the LIP is the time to find out what value the displacement has reached, for the purpose of determining if a leapweek is needed. That's a good reason for calculating the new D value at that time, at the LIP.


Michael Ossipoff

On Mon, Jan 16, 2017 at 8:08 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael and Calendar People

 

MICHAEL SAID:

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

KARL REPLIES: It’s no error. It arises from the fact that the mean time (over a Gregorian 400-year cycle) that March 1, 12:01 am occurs after February 28th, 11:59 pm is 2 minutes + 0.2425 day.

One may think of each LIP ideally containing a mean leap day/week equal in duration the a day/week multiplied by the proportion of years that are leap years. Then the sudden jump in displacement arises from a deviation from this ideal.

I then think more about this ideal year and I think what Michael is doing is to stretch the common year to fill the duration of the calendar mean year to eliminate to gaps at the LIPs. This would result in a calculated displacement that changes at a constant rate every non-leap day and goes down 1 each leap day or leap week. This displacement would have a range that extends outside of (-0.5, +0.5] as in his original proposal for minimum displacement calendar for when the LIP is not at the end of the year. Here I use units of day for a leap day calendar and week for a leap week calendar.

I realise that the space, in which this ideal year lies, is the space of ideal solar ecliptic longitudes (ISEL) and by stretching the common year, he squeezes all the ISEL into a common year. Whereas there would otherwise be a few ISEL that would occur either in the next displacement year or in the leap day/week.  These few ISEL are an inconvenience eliminated by the squeezing of the ISEL into a common year. The price of this resulting convenience is an increase in both the calculated and actual displacements.

Karl

16(05(19

 

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Michael Ossipoff
Sent: 15 January 2017 15:25
To: [hidden email]
Subject: Re: Jitter Calculation RE: New Henry-Hanke calendar/website, old shortcomings

 

 

 

On Tue, Jan 10, 2017 at 11:03 AM, Karl Palmen <[hidden email]> wrote:

Dear Michael, Irv and Calendar People

 

From: East Carolina University Calendar discussion List [mailto:[hidden email]



Of course it's true that you (Irv) and Karl have been around this subject a lot more than I have, and that I'm an amateur, and relatively a beginner.

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error.  

 

Karl said:

Jitter is only meaningful when comparing  the same date in different years. One can choose any date besides a leap day, for example January 10. Then we can compare January 10, 1904 with January 10, 2097. The interval is 193 years with 49 leap days, which is 2.1975 days more than the average of 193*0.2425 leap days. This applies for any starting date from March 1, 1903 to February 28, 1904 (in the same displacement year).

 

Right, so what's the Gregorian calendar's displacement (disregarding leapdays), with respect to the Gregorian average year, from February 28th, 11:59 p.m., 1904 to February 28, 11:59 p.m. 2096?

I'd say that's 192 X .2425 = 46.56

Karl said:

Jitter is only meaningful when comparing  the same date in different years

[endquote]

Is it meaningful to speak of the calendar's displacement (with respect to some mean year like 365.2425) of a date-&-time being about the same as that of as time that's 2 minutes earlier?

Remember that the +.2425 annual displacement with respect to the Gregorian mean year that occurs in an abrupt jump isn't what's really happening. The only displacement that really occurs in an abrupt jump is that caused by a leapday.

So on Februrary 28th, 11:59 p.m, 2096, the calendar is (disregarding leapdays) displaced 46.56 days, with respect to February 28th, 11:59, 1904.

Ok, then how about March 1, 12:01 a.m., 2096? How much is the calendar then displaced with respect to February 28th, 11:59 p.m., 1904?  Less than it was 2 minutes previous? If so, then explain how & why that could be. (Remember, we're disregarding leapdays).

It's customary to calculate the amount of displacement that the annual +(Y-365) calendar displacement would add up to, without any leapdays, and then subtract the leapdays:

192 X .2425 - 49 = -2.44

I guess it's easy to start believing in the factual reality of  an annual abrupt +(Y-365) increase in calendar displacement, at the LIP, though that of course isn't really what's happening.

That could lead you to believe that the calendar-displacement at March 1, 12:01 a.m., 2096 is  .2425 days more than it was on February 28th, 11:59 p.m. (disregarding leapday).

But that would be an error.

Is that error the source of the 2.2 answer for the Gregorian jitter-range?

I'd said:

 

So, this might be another of my amateur-beginner's errors--but, for the Gregorian jitter-range, I get 2.44 days, not 2.2

KARL REPLIES:  It is an amateur beginner’s error. 

[endquote]

I suggest that it's best to avoid ad hominem language.  ...because, in that way, we avoid saying things that we later won't be proud of, when we're wrong.

State your claim, say why you believe it to be so, but it's best to refrain from characterizing the other person, or what they said, just in case you might be mistaken.

Michael Ossipoff

 

 



 

 

 

 



 

Michael has expressed a desire to add calendar drift to this jitter. Suppose it’s a constant drift relative to a mean year of 365.24219 days. Then it is not sufficient to look at just one interval between the greatest displacements in the 400-year cycle, one must look at both intervals. So if comparing January 10, one must not only look at (Jan 10, 1904 – Jan 10, 2097) but must also look at (Jan 10, 1797 – Jan 10, 1904) or (Jan 10, 2097 – Jan 10, 2304).  When comparing with the calendar mean year of 365.2425 days these two results are equal in magnitude differing only in sign and that’s why it is then sufficient to look at one interval. But when one adds drift this changes and we get:

(Jan 10, 1904 – Jan 10, 2097): +2.1975 + (193*(0.00031)) = +2.25733 days

(Jan 10, 2097 – Jan 10, 2304): -2.1975 + (207*(0.00031)) = -2.13333 days.

Karl

16(05(13