Well,
I better review the 5515year cycle. (http://www.nabble.com/5515YearLuniSolarCycleto20632776r0.html)
It says that there are 2031 leap lunisolar years and 1336 leap solar
years but I calculated the leap years in the leap week (fixedweek)
calendar in Excel*, and there are 978 leap fixedweek years. (correct the
numbers if I'm wrong.). Is there any more accurate year cycles than
this?
*I used excel for this.
Row A is the year
Row B is
Floor((365+(1336/5515))*A/7)Floor(365+(1336/5515))*(A1)/7)
Row C is B52 (1 = leap year, 0 = non leap
year)

Dear Calendar People The 5515year cycle was discovered by Helios (unless you know of
an earlier discoverer) and has the following special property: (1) months can be made up of a repetition of one lunar
calendar cycle listed in "Lunar Calendar Cycles less than 1885
Months" http://www.hermetic.ch/cal_stud/lunarcal/lunacycl.htm
, (2) has fewer than 10,000 years and (3) has a mean year between 365.2416 and 365.2428
days. Very few cycles have this property. One other is the 725year
cycle of 183 49month cycles, which has barely enough accuracy to qualify. This is what makes the 5515year cycle special. The months of the 5515year cycle can be arranged into a lunar
calendar with a 1749month cycle. These months can be arranged into a Helios
cycle. Thirty nine of these 1749month cycles can be made into 5515 lunisolar years
by having 1023 leap months in these years. These too can be arranged into a
Helios cycle. The 5515year cycle can be made in my lunisolar spread sheets http://www.thelight.com/cal/kp_Lunisolar_xls.html
, by setting A=5515 (number of years), B=2031 (number of leap months) and C=1071
(number of days in excess of 354 per 12 month year and 384 per 13 month year).
It shows a mean year of 365.24225 days and a mean month of 29.5305889 days. The
spreadsheet also shows that the 5515year cycle has the equivalent to 1336 leap
days. Here are some less important properties of the 5515year cycle: The total number of months in the 5515year cycle is 39*1749=68211=99*689.
Hence the months could be grouped into 689 Octaeterides (of 99 months each),
each of which could be divided into 8 octaeteris years to give a total of 5512 octaeteris
years. These octaeteris years would drift through the seasons three times every
5515year cycle or once every 13 cycles of 1749 months. If we were to have a
solar calendar for the 5515year cycle and divide the year into 13 solar
months, then the octaeteris year would drift one of these solar months each lunar
1749month cycle. The 5515year cycle does not have a whole number of weeks so
cannot be used by a leap week calendar ( (5515+1336) mod 7 =5). So the 978 leap
week figure must be wrong. It would provide the equivalent to 7*9785515 = 1331
leap days which is 5 days short of the 1336 required. However, the 5515year cycle could be used by a leap trecena
calendar, where a trecena is a 13day period. Then we’d have
(5515+1336)/13 = 527 leap trecenas in each 5515year cycle. See http://en.wikipedia.org/wiki/Trecena
about the trecena. Below, I mark figures confirmed correct in green and
incorrect figures in red. Also the formula for Row B does (and hence row C) does not repeat
every 5515 years, but will repeat in 7*5515=38605 years. Karl 10(06(10 From: East Carolina University Calendar
discussion List [mailto:[hidden email]] On Behalf Of ELITE
3000 Well, I better review the 5515year cycle. (http://www.nabble.com/5515YearLuniSolarCycleto20632776r0.html)
It says that there are 2031 leap lunisolar
years and 1336 leap solar years but I
calculated the leap years in the leap week (fixedweek) calendar in Excel*, and
there are 978 leap fixedweek years.
(correct the numbers if I'm wrong.). Is there any more accurate year cycles
than this? *I
used excel for this. Row A
is the year Row B
is Floor((365+(1336/5515))*A/7)Floor(365+(1336/5515))*(A1)/7) Does not repeat every
5515 years! Row C
is B52 (1 = leap year, 0 = non leap year)
Scanned by iCritical. 
So, I would extend the cycle to 38,605 (7 x 5,515)
leap week years and noted that there are 6851 leap years. So under the
Symmetry454 year cycle, there are 52 leap years in the 293 year cycle, so what's
the difference between the 52/293 and the 6851/38605 patterns, and which one is
more accurate?
by the way, if we were to keep the 5,515 year
pattern for the leap week calendar over 7 cycles, a leap year should be
added at the 7th cycle at the 5,515th year (every 38,605th year) to keep up with
the solar calendar.
BTW: I would like to add the 6851/38605 pattern to
the Kalendis program.
And I saw that the current developmental version
of Kalendis (v9.707[1427]) (Revised on CE2009.02.19Th Gregorian) doesn't have
the International fixed 13month and the world calendar! I would like the
feature back in Kalendis in the future version so I would like to experiment
with the World, the international fixed 13month and symmetry454 and 010 under
the 6851/38605 pattern and the offset from the 52/293 and to convert to the
Gregorian dates. (and I'll use the EliteKalkulator (separate from Kalendis) to
convert the dates from Gregorian to the EMTS date
elements)
From: [hidden email]
Sent: Friday, March 06, 2009 7:49 AM
To: [hidden email]
Subject: Re: 5515year cycle review Dear
Calendar People The
5515year cycle was discovered by Helios (unless you know of an earlier
discoverer) and has the following special property: (1) months can be made up of a repetition of one lunar
calendar cycle listed in "Lunar Calendar Cycles less than 1885
Months" http://www.hermetic.ch/cal_stud/lunarcal/lunacycl.htm
, (2) has fewer than 10,000 years and (3) has a mean year between 365.2416 and 365.2428
days. Very
few cycles have this property. One other is the 725year cycle of 183 49month
cycles, which has barely enough accuracy to qualify. This
is what makes the 5515year cycle special. The
months of the 5515year cycle can be arranged into a lunar calendar with a
1749month cycle. These months can be arranged into a Helios cycle. Thirty nine
of these 1749month cycles can be made into 5515 lunisolar years by having 1023
leap months in these years. These too can be arranged into a Helios
cycle. The
5515year cycle can be made in my lunisolar spread sheets http://www.thelight.com/cal/kp_Lunisolar_xls.html
, by setting A=5515 (number of years), B=2031 (number of leap months) and C=1071
(number of days in excess of 354 per 12 month year and 384 per 13 month year).
It shows a mean year of 365.24225 days and a mean month of 29.5305889 days. The
spreadsheet also shows that the 5515year cycle has the equivalent to 1336 leap
days. Here
are some less important properties of the 5515year cycle: The
total number of months in the 5515year cycle is 39*1749=68211=99*689. Hence the
months could be grouped into 689 Octaeterides (of 99 months each), each of which
could be divided into 8 octaeteris years to give a total of 5512 octaeteris
years. These octaeteris years would drift through the seasons three times every
5515year cycle or once every 13 cycles of 1749 months. If we were to have a
solar calendar for the 5515year cycle and divide the year into 13 solar months,
then the octaeteris year would drift one of these solar months each lunar
1749month cycle. The
5515year cycle does not have a whole number of weeks so cannot be used by a
leap week calendar ( (5515+1336) mod 7 =5). So the 978 leap week figure must be
wrong. It would provide the equivalent to 7*9785515 = 1331 leap days which is 5
days short of the 1336 required. However,
the 5515year cycle could be used by a leap trecena calendar, where a trecena is
a 13day period. Then wed have (5515+1336)/13 = 527 leap trecenas in each
5515year cycle. See http://en.wikipedia.org/wiki/Trecena
about the trecena. Below, I mark figures confirmed correct in green and incorrect figures in red. Also the formula for Row B does (and hence row C) does not repeat every 5515 years, but will repeat in 7*5515=38605 years.
Karl 10(06(10 From: East
Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of ELITE 3000 Well, I better review the 5515year cycle.
(http://www.nabble.com/5515YearLuniSolarCycleto20632776r0.html)
It says that there are 2031 leap lunisolar
years and 1336 leap solar years but I
calculated the leap years in the leap week (fixedweek) calendar in Excel*, and
there are 978 leap fixedweek years.
(correct the numbers if I'm wrong.). Is there any more accurate year cycles than
this? *I
used excel for this. Row A
is the year Row B
is Floor((365+(1336/5515))*A/7)Floor(365+(1336/5515))*(A1)/7) Does not repeat every
5515 years! Row C
is B52 (1 = leap year, 0 = non leap
year)
Scanned by iCritical. 
In reply to this post by Karl Palmen  UKRI STFC
Just to follow up on that, I created a PDF file
(508 KB) from a Microsoft Excel spreadsheet of the entire 7x5515 year pattern
for the leapweek calendar, and the 5515year lunisolar and solar year pattern
and to determine if the year is a leap year or not. Included is the spreadsheet
itself in the Microsoft Office 972003 (.xls, 5.14 MB) and Microsoft Office 2007
(.xlsx, 1.49 MB) formats.
From: [hidden email]
Sent: Friday, March 06, 2009 7:49 AM
To: [hidden email]
Subject: Re: 5515year cycle review Dear
Calendar People The
5515year cycle was discovered by Helios (unless you know of an earlier
discoverer) and has the following special property: (1) months can be made up of a repetition of one lunar
calendar cycle listed in "Lunar Calendar Cycles less than 1885
Months" http://www.hermetic.ch/cal_stud/lunarcal/lunacycl.htm
, (2) has fewer than 10,000 years and (3) has a mean year between 365.2416 and 365.2428
days. Very
few cycles have this property. One other is the 725year cycle of 183 49month
cycles, which has barely enough accuracy to qualify. This
is what makes the 5515year cycle special. The
months of the 5515year cycle can be arranged into a lunar calendar with a
1749month cycle. These months can be arranged into a Helios cycle. Thirty nine
of these 1749month cycles can be made into 5515 lunisolar years by having 1023
leap months in these years. These too can be arranged into a Helios
cycle. The
5515year cycle can be made in my lunisolar spread sheets http://www.thelight.com/cal/kp_Lunisolar_xls.html
, by setting A=5515 (number of years), B=2031 (number of leap months) and C=1071
(number of days in excess of 354 per 12 month year and 384 per 13 month year).
It shows a mean year of 365.24225 days and a mean month of 29.5305889 days. The
spreadsheet also shows that the 5515year cycle has the equivalent to 1336 leap
days. Here
are some less important properties of the 5515year cycle: The
total number of months in the 5515year cycle is 39*1749=68211=99*689. Hence the
months could be grouped into 689 Octaeterides (of 99 months each), each of which
could be divided into 8 octaeteris years to give a total of 5512 octaeteris
years. These octaeteris years would drift through the seasons three times every
5515year cycle or once every 13 cycles of 1749 months. If we were to have a
solar calendar for the 5515year cycle and divide the year into 13 solar months,
then the octaeteris year would drift one of these solar months each lunar
1749month cycle. The
5515year cycle does not have a whole number of weeks so cannot be used by a
leap week calendar ( (5515+1336) mod 7 =5). So the 978 leap week figure must be
wrong. It would provide the equivalent to 7*9785515 = 1331 leap days which is 5
days short of the 1336 required. However,
the 5515year cycle could be used by a leap trecena calendar, where a trecena is
a 13day period. Then wed have (5515+1336)/13 = 527 leap trecenas in each
5515year cycle. See http://en.wikipedia.org/wiki/Trecena
about the trecena. Below,
I mark figures confirmed correct in green
and incorrect figures in red.
Also the formula for Row B does (and hence row C) does not repeat every 5515
years, but will repeat in 7*5515=38605 years. Karl 10(06(10 From: East
Carolina University Calendar discussion List [mailto:[hidden email]]
On Behalf Of ELITE 3000 Well, I better review the 5515year cycle.
(http://www.nabble.com/5515YearLuniSolarCycleto20632776r0.html)
It says that there are 2031 leap lunisolar
years and 1336 leap solar years but I
calculated the leap years in the leap week (fixedweek) calendar in Excel*, and
there are 978 leap fixedweek years.
(correct the numbers if I'm wrong.). Is there any more accurate year cycles than
this? *I
used excel for this. Row A
is the year Row B
is Floor((365+(1336/5515))*A/7)Floor(365+(1336/5515))*(A1)/7) Does not repeat every
5515 years! Row C
is B52 (1 = leap year, 0 = non leap
year)
Scanned by iCritical. 7x5515 leap week, 5515 solar and 5515 lunisolar.xls (7M) Download Attachment 7x5515 leap week, 5515 solar and 5515 lunisolar.xlsx (2M) Download Attachment 7x5515 leap week, 5515 solar and 5515 lunisolar.pdf (696K) Download Attachment 
Ryan, a.k.a. Elite 3000 recently sent an email of size circa 10 megabytes.
He did this by attaching PDF file (508 KB) Microsoft Office 972003 (.xls, 5.14 MB) Office 2007 (.xlsx, 1.49 MB) formats. The c. 7.13 MB actually expands in email because email is basically a 7bit byte world thus a special method is required to transmit 8bit binary data. This is what caused an already large email to further expand. Ryan meant well but his email could have been made smaller by splitting it up into three emails, compressing (.zip) the files, and sending fewer files. The three files all represent the same data. Some people have still only dial up access. In some countries, people pay according to bandwidth use. With emails of this size, sometimes the recipient not only can not receive the email, the large email never completes and thus blocks the receipt of subsequent emails. The Office 2007 Excel file compresses 39% to 948KB. The Office 2007 likely was not required since Office 2007 can read the older formats. The Excel older (pre 2007) format compresses 78% to 1170KB. The PDF file compresses 50% to 256KB. Regards, Gerry (Lowry) P.S.: Ryan, thank you for contributing. 
Op 7mrt2009, om 12:11 heeft gerry_lowry (alliston ontario canada)
het volgende geschreven: > Ryan, a.k.a. Elite 3000 recently sent an email of size circa 10 > megabytes. ... > With emails of this size, sometimes the recipient not only can not > receive the email, > the large email never completes and thus blocks the receipt of > subsequent emails. Moreover, some providers block oversized emails, or otherwise the mailing list often does not allow them; for example I never received such an email by Elite3000 . It is good practice to make sizeable files available for download from some website, and just post the link.  Tom Peters 
Tom Peters wrote, in part "It is good practice to make sizeable files available
for download from some website, and just post the link." There are ways to do this even if one does not have their own website. For example, "Google Documents":  Create and share your work online  Upload from and save to your desktop  Edit anytime, from anywhere  Pick who can access your documents  Share changes in real time  Files are stored securely online  It's FREE! g. 
I'm sorry. I didn't know about the filters to block oversized emails.
 From: "gerry_lowry (alliston ontario canada)" <[hidden email]> Sent: Saturday, March 07, 2009 2:39 PM To: <[hidden email]> Subject: Re: using Ryan Provost (Elite 3000) as an example of good intentions (was: 5515year cycle review) > Tom Peters wrote, in part "It is good practice to make sizeable files > available > for download from some website, > and just post the link." > > There are ways to do this even if one does not have their own website. > > For example, "Google Documents": >  Create and share your work online >  Upload from and save to your desktop >  Edit anytime, from anywhere >  Pick who can access your documents >  Share changes in real time >  Files are stored securely online >  It's FREE! > > g. > > 
In reply to this post by Ryan Provost2
On 2009.03.06, at 15:59 , ELITE 3000 wrote:
Longterm accuracy depends on the calendar mean year relative to the astronomical target mean year, not the number of years per cycle. If the astronomical target is the northward equinox, then the present era mean year target is about 365 days 5 hours 49 minutes. The 293year cycle mean year is slightly shorter than that = 365+^{71}/_{293} days = 365 days 5h 48 minutes 56+^{152}/_{293} seconds, but due to the tidal slowing of the Earth rotation rate it is better to be a few seconds too short. In the present era the calendar season for the 293year cycle is at about 4.1° of solar longitude (March 25), very close to the northward equinox. It is a few degrees after the equinox because the mean year of the 293year cycle is slightly too short. The 38605year cycle has an even shorter mean year of 365+^{1336}/_{5515 days = 365 days 5 hours 48 minutes 50+290/1103 seconds. This is too short for the northward equinoctial year, but what is your target anyway?} The 5515year leap cycle has a calendar season at about 10.8° of solar longitude (April 1st). It is an extra several degrees after the equinox because of its shorter mean year. If selection of this long cycle has anything to do with the lunar cycle, that is not a good idea, because the mean lunar cycle is getting progressively shorter in terms of mean solar days, so it is futile to use any fixed cycle for more than about a millennium, plus or minus, depending on how much drift is deemed tolerable.
That will not happen. Given the cyclic and progressive changes in the lengths of the equinoctial and solstitial mean years, there is no need for such long leap cycles. The longest cycles in Kalendis are <1000 years, except for the 559/3150 cycle, which is only there because in order to make a leap week mean year equivalent to the Revised Julian calendar, it is necessary to use 7 x 450 = 3150 years. It so happens that I've been thinking about removing support for the 3150year cycle anyway, because it can't be made symmetrical, and the distribution of its leap intervals is too long and complex.
The public version of Kalendis never supported the original leap day versions of the 13Month calendar nor The World Calendar because I was unable to devise a satisfactory method to display the null days in the monthly calendar view. Previous versions did support leap week variants of those calendars, and they still exist in my developmental version, but I suppressed them in the version that I sent you. You can obtain the same leap week variant of The World Calendar within the Symmetry window by choosing "Experiment" mode, then choosing the 30+31+30 structure, then click on the word "Quarters" until it changes to 31+30+30, then choose the desired leap rule. Of course I can simply switch back on the availability of The World Calendar / 13Month Calendar window, but I don't see the point in that. I suppressed them because I don't support those calendar reform proposals. 
In reply to this post by Ryan Provost2
Dear Ryan and Calendar People I see no point whatsoever in making a leap week calendar have a
mean year of 365 1336/5515 days! I emphasised that the virtue of the 5515year cycle is that it can
be realised by a lunar calendar. This lunar calendar could be like my yerm
lunar calendar, except that every other 52yerm cycle would be extended to 55
yerms so creating a 107yerm cycle of 1749 months (about 141.41 years) of which
39 form a 5515year cycle. I’ve listed all leap week calendar cycles whose
mean year is between 365.241936 days and 365.2425 days and have fewer
than 1000 years at http://www.hermetic.ch/cal_stud/palmen/lweek1.htm#cycle
. I consider them all to be accurate, because they have calendar seasons. Of these, the one that has a mean year closest to 365 1336/5155
days (about 365.242248 days) is the 355year cycle with 63 leap weeks (63/355). The accuracy that the mean year of 365 1336/5515 days would
perish well before the 38,605 yearcycle would be completed, which is far
longer than any rulebased calendar can be expected to last. If you want your leap week cycle to have a whole number of lunar
months, you may choose a cycle from http://www.thelight.com/cal/Lunisolar7.html
. The number of leap weeks is (Years+Leap)/7. Karl 10(06(12 till noon From: East Carolina University Calendar
discussion List [mailto:[hidden email]] On Behalf Of ELITE
3000 So, I
would extend the cycle to 38,605 (7 x 5,515) leap week years and noted that
there are 6851 leap years. So under the Symmetry454 year cycle, there are 52
leap years in the 293 year cycle, so what's the difference between the 52/293
and the 6851/38605 patterns, and which one is more accurate? by
the way, if we were to keep the 5,515 year pattern for the leap week
calendar over 7 cycles, a leap year should be added at the 7th cycle at the
5,515th year (every 38,605th year) to keep up with the solar calendar. BTW:
I would like to add the 6851/38605 pattern to the Kalendis program. And I
saw that the current developmental version of Kalendis (v9.707[1427]) (Revised
on CE2009.02.19Th Gregorian) doesn't have the International fixed 13month and
the world calendar! I would like the feature back in Kalendis in the future
version so I would like to experiment with the World, the international fixed
13month and symmetry454 and 010 under the 6851/38605 pattern and the offset
from the 52/293 and to convert to the Gregorian dates. (and I'll use the
EliteKalkulator (separate from Kalendis) to convert the dates from Gregorian to
the EMTS
date elements) From: [hidden email] Sent: Friday, March 06,
2009 7:49 AM To: [hidden email]
Subject: Re: 5515year cycle
review Dear Calendar People The 5515year cycle was discovered by Helios (unless you know of
an earlier discoverer) and has the following special property: (1) months can be made up of a repetition of one lunar
calendar cycle listed in "Lunar Calendar Cycles less than 1885
Months" http://www.hermetic.ch/cal_stud/lunarcal/lunacycl.htm
, (2) has fewer than 10,000 years and (3) has a mean year between 365.2416 and 365.2428
days. Very few cycles have this property. One other is the 725year
cycle of 183 49month cycles, which has barely enough accuracy to qualify. This is what makes the 5515year cycle special. The months of the 5515year cycle can be arranged into a lunar
calendar with a 1749month cycle. These months can be arranged into a Helios
cycle. Thirty nine of these 1749month cycles can be made into 5515 lunisolar
years by having 1023 leap months in these years. These too can be arranged into
a Helios cycle. The 5515year cycle can be made in my lunisolar spread sheets http://www.thelight.com/cal/kp_Lunisolar_xls.html
, by setting A=5515 (number of years), B=2031 (number of leap months) and
C=1071 (number of days in excess of 354 per 12 month year and 384 per 13 month
year). It shows a mean year of 365.24225 days and a mean month of 29.5305889
days. The spreadsheet also shows that the 5515year cycle has the equivalent to
1336 leap days. Here are some less important properties of the 5515year cycle: The total number of months in the 5515year cycle is 39*1749=68211=99*689.
Hence the months could be grouped into 689 Octaeterides (of 99 months each),
each of which could be divided into 8 octaeteris years to give a total of 5512
octaeteris years. These octaeteris years would drift through the seasons three
times every 5515year cycle or once every 13 cycles of 1749 months. If we were
to have a solar calendar for the 5515year cycle and divide the year into 13
solar months, then the octaeteris year would drift one of these solar months
each lunar 1749month cycle. The 5515year cycle does not have a whole number of weeks so
cannot be used by a leap week calendar ( (5515+1336) mod 7 =5). So the 978 leap
week figure must be wrong. It would provide the equivalent to 7*9785515 = 1331
leap days which is 5 days short of the 1336 required. However, the 5515year cycle could be used by a leap trecena
calendar, where a trecena is a 13day period. Then we’d have
(5515+1336)/13 = 527 leap trecenas in each 5515year cycle. See http://en.wikipedia.org/wiki/Trecena
about the trecena. Below, I mark figures confirmed correct in green and
incorrect figures in red. Also the formula for Row B does (and hence row C) does not
repeat every 5515 years, but will repeat in 7*5515=38605 years. Karl 10(06(10 From: East Carolina University Calendar
discussion List [mailto:[hidden email]] On Behalf Of ELITE
3000 Well, I better review the 5515year cycle. (http://www.nabble.com/5515YearLuniSolarCycleto20632776r0.html)
It says that there are 2031 leap lunisolar
years and 1336 leap solar years but I
calculated the leap years in the leap week (fixedweek) calendar in Excel*, and
there are 978 leap fixedweek years.
(correct the numbers if I'm wrong.). Is there any more accurate year cycles
than this? *I
used excel for this. Row A
is the year Row B
is Floor((365+(1336/5515))*A/7)Floor(365+(1336/5515))*(A1)/7) Does not repeat every
5515 years! Row C
is B52 (1 = leap year, 0 = non leap year)
Scanned by iCritical. 
In reply to this post by gerry lowry +1 705 4297550 wasaga beach ontario canada
I was a victim of the excessive size of this attachment. To help
others, I'll describe how I bypassed the problem. My Internet Service Provider (ISP) uses Google Gmail (1 GB of storage), but I link via a dialup connection (I'm planning to upgrade). Several tries to download my email all were terminated by the server when this message was being received because of excessive time elapsed (that never happens when downloading other large Internet files), but my connection to the Internet was not terminated. I tried several filters in my email reader, Outlook Express, including one that was supposed to delete any message at the server that had an attachment. No filter was successful. I then accessed my email account at my ISP via a browser (Internet Explorer 7). This allowed me to see all messages waiting on the server (all old messages that had been read and deleted via Outlook Express were still there!). Only one of the recent messages had an attachment, so I summarily deleted it. I then successfully downloaded all remaining messages via Outlook Express (no filter), including a few older messages with 'small' attachments that were under 100 KB each. Joe Kress  Original Message  From: "gerry_lowry (alliston ontario canada)" <[hidden email]> To: <[hidden email]> Sent: Saturday, March 07, 2009 7:11 AM Subject: using Ryan Provost (Elite 3000) as an example of good intentions (was: 5515year cycle review) > Ryan, a.k.a. Elite 3000 recently sent an email of size circa 10 > megabytes. > > He did this by attaching > PDF file (508 KB) > Microsoft Office 972003 (.xls, 5.14 MB) > Office 2007 (.xlsx, 1.49 MB) formats. > > The c. 7.13 MB actually expands in email because email is > basically a 7bit byte world thus a special method is required > to transmit 8bit binary data. This is what caused an already > large email to further expand. > > Ryan meant well but his email could have been made smaller by > splitting it up into three emails, compressing (.zip) the > files, and sending fewer files. > > The three files all represent the same data. > > Some people have still only dial up access. In some countries, > people pay according to bandwidth use. > > With emails of this size, sometimes the recipient not only can > not receive the email, the large email never completes and > thus blocks the receipt of subsequent emails. > > The Office 2007 Excel file compresses 39% to 948KB. > The Office 2007 likely was not required since Office 2007 can > read the older formats. > > The Excel older (pre 2007) format compresses 78% to 1170KB. > > The PDF file compresses 50% to 256KB. > > > Regards, > Gerry (Lowry) > > P.S.: Ryan, thank you for contributing. > 
Joe, I'm guessing you used what's commonly called a "webmail" interface.
If that fails, as it might depending on the particular server side webmail software that happens to be in use, it's possible to use the program "telnet" and a command line interface. After connecting with "telnet", one would list the emails which would basically display each message number with its size, example: 3370 18786 3371 244498 3372 86039 3373 10262846 <==== Ryan's message 3374 4837 3375 3969 DELE 3373 <===== delete message using its message number Regards, Gerry (Lowry) 
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