Lemniscata and Kepler intro
This is a brief intro about Kepler’s laws for planetary motion and the Lemniscata curve.
I’ll update and explain better later, but now I’ll give you some indications so that you can calculate this yourselves.
Taking the four dates for zero shift from true and average mid-day:
24th December, 15 April, 14 June, and 31st August, you should make four lines intercepting in a center for those lines, but instead of putting them in a cross like disposition, you should do the following:
Opposite to 24th December in the same line but in opposite sense of direction as viewed from the center where the lines intercept each other, you should put 21st June.
With a set of squares rules you can divide the area into angles of 30 degrees, being each degree one day in difference (remember the legal year!! 🙂 ). By doing so, the area divided equally by these two dates, will be equally divided into portions of thirty degrees from any of the previous two dates, being every area encompassed by 30 degrees equivalent (with a 5º 15′ of arc in an entire year of difference) to one month.
I did a drawing but it is not well drawn, so I’ll upload it when I have it well measured and have scanned it with my father’s printer, and sent it to me from his laptop (which, by the way, I gave to him some years ago).
I’ll continue tomorrow.
[December 27th 2015, some hours later, update]
So, now you should have one straight line dividing an area whose limits are the 24th December, on one side, and the 21st of June on the opposite side.
The divisions, in 30 arc degrees (sets of squares are 90º, 45º, 45º, for one whose name in Spanish is “escuadra”, and 90º, 60º, 30º, for the other whose name in Spanish is “cartabón”, I’ll call them respectively isosceles, and escalene in this post) can be made easily by putting one side of the isosceles rule on the line whose limits are the two dates I said, and taking the escalene so that its longest side on the right angle is touching the line and the isosceles rule.
By doing this recursively putting care in making the intercepting point coincide (this point would be the position for the Sun), for all lines drawn like this, you will divide the paper on which you are making this drawing into twelve circular sectors, equally distributed.
[Edited and corrected in May the 23rd 2016:]
[WRONG: Kepler found that velocity for the Earth’s path around the Sun is not constant, being faster when it is near the Sun (perihelion -> perimetral Sun), than when it is far from the Sun (afelion -> afar from the Sun)].
[This below is correct].
Kepler found that velocity for the Earth’s path around the Sun is not constant, being faster when it is far from the Sun (afelion -> afar from the Sun) than when it is near the Sun (perihelion -> perimetral Sun). This happens in winter in the North hemisphere, and in summer in the South hemisphere (if you don’t have this concept clear…, don’t worry, trust me, it is so, I’ll think of an olive-like analogy later… 🙂 ).
But months don’t have thirty days, and year don’t have 360 days. Don’t worry about this, because this is a humanity scale made division for days and year, of which stars, planets, satellites, and, generally speaking, astra, have their own scales for movement into real three-dimensional space.
Just think about this (to make it simple):
Taking the Lemniscata curve you will see it has an eight like shape and goes not round as a circle, I made some indications on the image for which I provided the link above. Just look at this image, and keep reading below.
This curve crosses itself in one point which is out of the vertical axis, a little on the right of it, the two dates in a year which are in this interception point are April the 13th, and August the 29th.
By measuring the angle between these other two dates in the shortest path between the two lines (you should draw, because nither of these fall into the twelve areas division lines), you will find this angle has a value of 139 degrees of arc circumference, measured this way, and 360 – 139 = 221 º of arc circumference if you go the other way around.
In degrees them both are odds, but in arcmins they become even, because one degree is 60 arcmins so, if you want to divide those values, in halves, you’d have 110º 30′ in the long path middle, and 69º 30′ in the short path, between these to dates April the 13th, and August the 29th.
Those dates are also two degrees (days) far from 15th of April, and 31st of August.
You can think they are on opposite sides for the lines of 31st August, and 15th April, being in one case a little (2º) on the right, and in the other a little (2º) on the left, but you are not considering the sense of movement.
[I made the drawing putting 24th December on the left and 21st June on the right, if you did the other way, then think on it reversed to the references I am giving].
In fact, bearing in mind the trajectory for the planet…
- Which one?
- … I don’t know, I forgot…
… those dates are both before, two days before (2º before), the 31st August, and 15th April zero deviation between true and average mid-day.
You could think of those dates (31st August, and 15th April zero deviation between true and average mid-day) as the two values for a semi-minor axis of an ellipse, but you’re wrong. It is not so (I’ll explain why later).
The thing is, the Earth changes its rotation axis angle orientation (from negative to positive, and the other way around) respect the plane of its path in those two dates, and this is what makes the values for clockmins change from being subtracted, to being added, in order to make mid-day values for true and average coincide.
And today’s update goes this far, so you will have time to think about it yourselves, and I have time to put a little order in my disordered house, which needs a bit of attention, in order not to be engulfed by monster-furballs 🙂
[January the 3rd 2016 update]
By doing all I previously told in this post, you should have twelve lines equally distributed around a circle, and two more lines for August the 29th, and Aprile the 13th.
Now let’s draw November the 4th, and the equally distant date from December the 24th, that’s February the 11th.
November the 4th and February the 11th are the two dates in a year where the clockmins to subtract or add respectively take their maximun values. You make one degree to coincide with one day, and there are 94º | days between one another.
I know what you are thinking, I said to start with December the 24th on one side and June the 21st on the other, and you think I’m wrong, because these are not opossite taking one degree for each day apart. Yeah, I know, but I did this on purpose, because the Earth’s orbit is not a perfect circle, and in these two dates is when one of the orthonormal values for the components of movement pass from positive to negative, being the zero values in these two opposite days.
It is an inaccurate fast way to balance the difference between the 360 + 5,25 days in a year, and the 360 degrees in a circumference. And I started with these two, because these two are limit values, such as those of 15th April, and 31st August on the other axis for movement.
So, to balance equally those 5 and a quarter days to make this calculus fit, you take 2, 1, 1, and 1.25, (I’ll fine-tune later on) in each of the extreme values, those corresponding to the Earth changing from going towards, to going apart from the Sun, and going upwards, to going downwards the line joining the previous two places in its orbit, and you’ll have the Lemniscata elliptically distributed around the Sun.
Being the intuitive idea that the Earth lets herself wander around two, one, one and a quarter, and one days while changing its sense of direction. And all this tripping around is what makes the true year differ from the 360 legal year 5.25 days.
I’ll go on (or not 🙂 ) in a number (II) for this post, in forthcoming dates.