Timeline for Leap year formula to arbitrary precision
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2022 at 18:42 | vote | accept | Michael Engelhardt | ||
| S Oct 29, 2022 at 18:42 | vote | accept | Michael Engelhardt | ||
| Oct 29, 2022 at 18:42 | |||||
| S Oct 29, 2022 at 18:42 | vote | accept | Michael Engelhardt | ||
| S Oct 29, 2022 at 18:42 | |||||
| Oct 29, 2022 at 18:35 | vote | accept | Michael Engelhardt | ||
| S Oct 29, 2022 at 18:42 | |||||
| Oct 26, 2022 at 15:59 | answer | added | Michael Lugo | timeline score: 5 | |
| Oct 26, 2022 at 14:54 | comment | added | Jeppe Stig Nielsen | Yes, over the geological eras, the actual number a calendar needs to approximate in order to keep the "natural" day (day-night cycle as seen far from Earth's poles) and the "natural" year (summer-winter cycle as seen far from the equator) in sync will vary a lot. This is mostly because of tidal braking of Earth's rotation (other effects exist as well). For example, there is going to be an era when that number is $365.2000$ which calls for a simple leap-year rule. Much later, it will be $365.0000$, and you can have a calendar with no leap years. At $364.0000$, it is exactly 52 "weeks". | |
| Oct 26, 2022 at 12:47 | comment | added | Henry | The average Gregorian day of $365.2425$ days differs from the current figure for the mean tropical year of $365.24219$ mean solar days by so little that by the time they are more than a full day apart both the length of a tropical year and the length of a solar day (measured in seconds) will have changed, in a not totally predictable way. Indeed your $365.242375$ and my $365.24219$ are already different. | |
| Oct 26, 2022 at 11:17 | history | became hot network question | |||
| Oct 26, 2022 at 4:48 | answer | added | Anthony Quas | timeline score: 8 | |
| Oct 26, 2022 at 3:56 | comment | added | Michael Engelhardt | @LSpice - I don't know of such an instance. That's an interesting additional question - if the answer to my original question is yes, then is the greedy algorithm always good enough, or are there cases where one has to be more sophisticated ... | |
| Oct 26, 2022 at 3:46 | comment | added | LSpice | Ha, right! I was trying to come up with a counterexample for the greedy algorithm, and somehow convinced myself that the greedy algorithm had to pick $a_2 = 4$. Do you know an instance where the greedy algorithm (choose the smallest possible $a_n$ at each stage that is on the right side of $r$) fails? | |
| Oct 26, 2022 at 3:40 | review | Close votes | |||
| Oct 27, 2022 at 6:35 | |||||
| Oct 26, 2022 at 3:38 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
added 28 characters in body
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| Oct 26, 2022 at 3:36 | comment | added | Michael Engelhardt | @LSpice - Well, that's $0.45 = 1/2 - 1/20$ ... | |
| Oct 26, 2022 at 3:31 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
added 139 characters in body
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| Oct 26, 2022 at 3:29 | comment | added | LSpice | Wouldn't something like $\frac1 2 - \frac1 4 + \frac1 5$ be impossible to hit in this way? | |
| Oct 26, 2022 at 3:17 | history | asked | Michael Engelhardt | CC BY-SA 4.0 |