A study into Kerbin leap years

[RA3236]

So I was randomly thinking about a timeline, and it occurred to me that I wanted to do a study into leap years (don't ask how that makes sense). So here it is:

The Guide to Kerbin Leap Years

(I sooo wanted to put a reference to The Hitchhikers Guide to the Galaxy here. And I almost did!)

Contents:

1. A brief revisit to Earth-Kerbin times

2. Calculating when Kerbin leap years occur

A brief revisit to Earth-Kerbin times

We can basically sum this up in a table:

Kerbin-Earth time relationships

Standard Unit	Earth Time	Kerbin Time
Seconds (s)	1000ms	1000ms
Minutes (min)	60s	60s
Hours (h)	60min	60min
Days (d)	24h	6h
Weeks (wk)	7d	10d
Months (mth)	4 — 4 3/7 wk	3wk - 3wk + 3d
Years (yr)	12 mth in a standard calendar, 365.24 d	9,203,544.6s, 426.090028d

A quick explanation into these timetables:

A Kerbin second, minute and hour is exactly the same as an Earth second, minute and hour. A Kerbin Day is 6 hours long, contrasting Earth's 24 hour day. There are 365.24 days in an Earth Year, which is much more than Kerbin's 426.090028 days. A Kerbin week would therefore become 10 days and a standard Kerbin month would be about 3 weeks long and a year would contain 14 months.

Before we get into the next section we should outline when leap years occur on Earth. A leap year on Earth is on every year that is divisible by 4, on every year that is divisible by 400 and every 3600 or so years (this is to make up for yet another day gained other the course of time).

Calculating when Kerbin leap years occur

Method One:

  Reveal hidden contents

Kerbin leap years are something of an interesting concept. There's a long way of doing this, but it would bore a lot of people, so Ill outline it for you guys.

In order to find when to add an extra day, we need to find when the addition of the decimal point after '426' reaches one. So, without further ado:

.090028*12=1.080336.

Very interesting, I'm sure. So this originally means every 12 years. Horray.

We aren't done yet, as theres still a small fraction after that one. Thankfully, this is solved by adding that number to itself until it causes the day number to increase by two.

1.080336*13=14.044368.

This means again, every 12*13=156 years we need another leap day. Except the year 156 will be a double leap year, as it will have two extra days (as it is a multiple of 12).

14.044369*23=323.020464.

I think you get the picture. This means every 156*23=3588 years is a triple leap year, meaning three extra days. Ill do this two more times.

323.020464*49=15,828.002736.

This means again every 3588*49=175,812 years will be a quadruple leap year. Actually, we'll stop here because I doubt anyone will want to go past the year 3588 anyways.

We can fix this a little. We'll make it so that every n*12 year is a leap year, n*156+1 is a leap year, n*3588+2 is a leap year and every n*175812+3 year is a leap year. I'm done for today.

This means that the year 2,109,747 is a leap year.

Method Two:

  Reveal hidden contents

This method was suggested by @FancyMouse.

This time we are going to use continued fractions. Most of the maths for the continued fraction is below, but ill sum it up at the end.

426.090028*500000=213,045,014

213,045,014/500000=426 r 45014

500000/45014=11 r 4846

45014/4846=9 r 1400

4846/1400=3 r 646

1400/646=2 r 108

646/108=5 r 106

108/106=1 r 2

106/2= 53

== 106522507/250000 = 426 22507/250000

The first equation sums up that multiplying 426.090028 creates a whole number. We can use that same multiple to find how much of a remainder stood there afterwards. The second equation does this. After that, it's pretty clear what's going to happen. We divide the previous examples divisor by the remainder and get the numbers for the continued fraction. The last equation is in fact the value of 426.090028 in fractional format.

How would This change our previous answer?

It really doesn't. This shows that for every two hundred and fifty thousand years we have on Kerbin, we need approximately 22507 years to be leap years.

How can we find the exact values for this?

Well:

250000/22507=11.1, 11 r 2423.

This means that for every eleven years, at least one of them must be a leap year.

We then expand on this by dividing 250000 by 2423:

250000/2423=103.2, 103 r 431

So every eleven years and 103 years.

250000/431=580 r 20

Every 11, 103 and 580 years...

250000/20=12500

So that's for every 11, 103, 580 and 12500 years. I think we're done now.

Notice how I did not put 10 down as the first answer, because I wanted exact answers and not to arrange them into a nice order.

Any suggestions or changes for this info is welcomed, and moderators are free to move this thread. If you have an opinion suggesting that the weeks or months are wrong, make a separate thread or shut up. None of that in this thread. Also, if there's some random equation to make the maths here easier to understand, please inform me, as I hardly know maths compared to the rest of the community.

Edited December 31, 2016 by RA3236
Method two

[FancyMouse]

There is actually a math theory that can give you the most accurate approximation when one wants to restrict how big the denominator is. The design of the leap year interval could very well use the result from the continued fraction expansion of 0.090028 and you may come up with something better.

[RA3236]

  On 12/31/2016 at 10:20 AM, FancyMouse said:

There is actually a math theory that can give you the most accurate approximation when one wants to restrict how big the denominator is. The design of the leap year interval could very well use the result from the continued fraction expansion of 0.090028 and you may come up with something better.

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Ill have a look at it soon  currently an hour to New Years  thanks

Edited December 31, 2016 by RA3236

[RA3236]

@FancyMouse my second method is in, using continued fractions.

[msasterisk]

Isn't a Kerbin month (AKA munth) about 6 days, based on the Mün's orbital period?  Where did you get 10 days?

[RA3236]

  On 1/1/2017 at 3:49 PM, msasterisk said:

Isn't a Kerbin month (AKA munth) about 6 days, based on the Mün's orbital period?  Where did you get 10 days?

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Actually, I got 30-34 days for a month. If you try getting 6 day months, there'd by half a bazillion (no maths required ) in a year, and would be simply impractical. Of course, it would be better to base the minths off Minmus, with an orbital period of 49 days, but 30 divides somewhat well (minus a few decimals) into 426 and I didn't want to end up changing a heap of stuff about it.

Its all up to what you prefer, really.

[msasterisk]

I like minths! Also, hehe, I'm bad at tables it seems, I mixed up weeks and months.

[tseitsei89]

I find this information completely useless but very intriguing nevertheless. So the best kind of information  Well done good sir

[RA3236]

  On 1/2/2017 at 4:39 AM, tseitsei89 said:

I find this information completely useless but very intriguing nevertheless. So the best kind of information  Well done good sir

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Well, the non-average KSP Fan writer may use this  thanks