Talk:Hiver Calendar

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OK, I've looked at CT AM7 and the Hiver calendar is totally screwed. The book says Glea rotates about its own axis once in "about 22 hours", a HF year is 143 cycles = 179 standard days. I'm going to have to work this out on a spreadsheet to get this to work. Gruffty Hivers: If you tolerate us, then your children will be next... 13:09, 3 June 2008 (UTC)

The Basics[edit]

OK: 1 Hiver day = 22 standard hours (according to CT AM7). On that basis:

  • 11 standard days (of 24 standard hours each) equals 12 Hiver days (of 22 standard hours each) exactly;
  • 264 standard hours (or 11 standard days) equals 12 Hiver days (of 22 standard hours each) exactly.

That's my starting point for sorting Teh Hivorz Kalenda Mesz out. Gruffty Hivers: If you tolerate us, then your children will be next... 13:29, 3 June 2008 (UTC)

Ignore all of the above, I have changed tack to look at the HF year. Gruffty Hivers: If you tolerate us, then your children will be next... 13:10, 4 June 2008 (UTC)

The Year[edit]

  • A Hiver year = 179 standard days = 4,296 standard hours;
  • 4,296 standard hours divided by 22 standard hours (per Hiver day) = 195.2727 Hiver days (of 22 standard hours each).
  • A Hiver "day" is termed a "cycle" by the Hivers: 1 cycle = 1 day = 22 standard hours.
  • 195.2727 Hiver days/cycles (of 22 standard hours each) = 1 Hiver year = 179 standard days = 4,296 standard hours.

However.... CT AM7 p. 15 states that Glea orbits New Primary every 358 days (i.e. 2 * 179 standard days). Thus the Hivers get 2 years for the price of one. Gruffty Hivers: If you tolerate us, then your children will be next... 13:50, 3 June 2008 (UTC)

Ignore all of the above, I have changed tack to look at the HF year. Gruffty Hivers: If you tolerate us, then your children will be next... 13:11, 4 June 2008 (UTC)

Slippage[edit]

It looks like Hiver years slip by a year for every 52 standard years. Each standard year = 2 Hiver years + 168 hours (7 standard days). So:

  • 1 standard year = 2 HF years + 1 week
  • 2 standard years = 4 HF years + 2 weeks
  • 3 standard years = 6 HF years + 3 weeks

and so on. Gruffty Hivers: If you tolerate us, then your children will be next... 11:52, 4 June 2008 (UTC)

Ignore all of the above, I have changed tack to look at the HF year. Gruffty Hivers: If you tolerate us, then your children will be next... 13:05, 4 June 2008 (UTC)

Uh Oh[edit]

OK, so I've punched in the numbers and....it doesn't work:

  • 22 standard hours * 143 cycles = 3146 standard hours;
  • 179 standard days * 24 standard hours = 4296 standard hours;
  • Difference = 1150 standard hours between standard hours for 24 hour days and standard hours for 22 hour days.

Option 1[edit]

Increase the number of cycles in a HF year, but keep a HF day/cycle at 22 standard hours:

  • 22 standard hours * 195 cycles = 4296 standard hours;
  • 179 standard days * 24 standard hours = 4296 standard hours;
  • Difference = +52 cycles against 143 originally.

Option 2[edit]

Increase the number of standard hours in a HF day/cycle, but keep 143 days/cycles per HF year:

  • 30.04195804 standard hours * 143 cycles = 4296 standard hours;
  • 179 standard days * 24 standard hours = 4296 standard hours;
  • Difference = +8.041958042 standard hours per HF day against 22 originally.

Better see what Marc says about all of this... Gruffty Hivers: If you tolerate us, then your children will be next... 13:04, 4 June 2008 (UTC)

Tom Rux Calculations[edit]

Note From Tom Rux The Hiver year in my calculations = 143 cycles. Snrdog 03:54, 21 September 2008 (UTC)

Here is the result of my attempt to verify the Hiver calendar information as presented in CT: AM 7 and conversion formulas presented on the conversion chart of the Date Conversion article.

1. The length of the Hiver year as presented in CT: AM 7 of 179 days does not support the 2 dates that have been converted from Imperial to HF. The Date Conversion article provided a third date which also is not supported by the published Hiver information.

1A. I used the formula HF year = (Imperial year - 410) X Constant where the Constant is equal to 2.0391. The Constant was determined by

round(365 Imperial/179 Hive,10) = round(2.039106145,4) = 2.0391.
(Imperial year - 410) X 2.0391
CT:AM 7 page 2 date: 1111 Imperial = 1401 HF
My calculations have the date as 1111 Imperial = 1429 HF
CT:AM 7 page 17 date: -4698 Imperial = -10217 HF
My calculations have the date as -4698 Imperial = -10416 HF
Date Conversion Article page 2 date: 1105 Imperial = 1389 HF
My calculations have the date as 1105 Imperial = 1417 HF

1B. The Date Conversion article uses a Constant of 2 which returns dates that are ± 1 year from the three dates.

(Imperial year - 410) X 2
CT:AM 7 page 2 date: 1111 Imperial = 1401 HF
My calculations have the date as 1111 Imperial = 1402 HF
CT:AM 7 page 17 date: -4698 Imperial = -10217 HF
My calculations have the date as -4698 Imperial = -10216 HF
Date Conversion Article page 2 date: 1105 Imperial = 1389 HF
My calculations have the date as 1105 Imperial = 1390 HF

1C. I used both dates from CT: AM 7 (pages 2 and 17) and the date from the Date Conversion Hiver section to determine a Constant: To determine the Constant I reworked the Imperial to Hiver formula to: Constant = round(HF date/(Imperial date - 410),4)

1C1. CT:AM 7 page 2 Constant = 1.9986

Constant = round(1401/(1111 - 410),5)
Constant = round(1401/701,5)
Constant = round(1.99857,4) = 1.9986

1C2. CT:AM 7 page 17 Constant = 2.0002

Constant = round(-10217/(-4698 - 410),5)
Constant = round(-10217/-5108,5)
Constant = round(2.00020,4) = 2.0002

1C3. Date converion article page 2 Constant = 1.9986

Constant = round(1389/(11105- 410),5)
Constant = round(1389/695,5)
Constant = round(1.998556,4) = 1.9986

1C1 & 1C3: (Imperial year - 410) X 1.9986

CT:AM 7 page 2 date: 1111 Imperial = 1401 HF
My calculations shows that the dates match
CT:AM 7 page 17 date: -4698 Imperial = -10217 HF
My calculations have the date as -4698 Imperial = -10209 HF
Date Conversion Article page 2 date: 1105 Imperial = 1389 HF
My calculations shows that the dates match

1C2: (Imperial year - 410) X 2.0002

CT:AM 7 page 2 date: 1111 Imperial = 1401 HF
My calculations have the date as 1111 Imperial = 1402 HF
CT:AM 7 page 17 date: -4698 Imperial = -10217 HF
My calculations shows the dates match
Date Conversion Article page 2 date: 1105 Imperial = 1389 HF
My calculations have the date as 1105 Imperial = 1390 HF

Calculations 1B and 1C2 have the least impact on the published dates.

2. Since the 179 day length does not appear to support the listed dates I did more calculations to determine the HF days per year, day length, hours in a total day, and the hours in an interval.

2A. HF in Imperial days per year:

1111 and 1105 Imperial Constant = 1.9986, Imperial year = 365
Days per year = round(365/1.9986,3) = round(182.631,0) = 183 days
-4698 Imperial Constant 2.0002, Imperial year 365
Days per year = round(365/2.0002,3) = round(182.482,0) = 182 days

Date Conversion constant of 2 = round(365/2,1) = round(182.5,0) = 183 days.

2B. HF in Imperial Day length = Imperial days/143 HF days

Date conversion article constant 2, 1111, and 1105: round(183/143,4) = round(1.2797,2) = 1.28

-4698: round(182/143,4) = round(1.2727,2) = 1.27

2C. Hours

Date conversion article constant 2, 1111, and 1105:

24 x 1.28 = 30.72 hours

-4698: 24 x 1.27 = 30.48 hours

2D. Hiver intervals:

Date conversion article constant 2, 1111, and 1105:

round(30.72/6,4) = round(5.1200,2) = 5.12 hours

-4698: round(30.48/6,4) = round(5.0800,2) = 5.08 hours

3. Conclusions:

Formula Constant of 2 has the same error of 1 year for all listed OTU dates.

The Hiver timekeeping units appear to be:

Year = 182.5 Imperial standard days
Day = 1.275 Imperial standard day
Hours in day = 30.6 Imperial standard hours
Interval = 5.1 Imperial standard hours

The only change is the length of the year which changed from 179 to 182.5.

Tom Rux - tmr0195@comcast.net