Unexplained Equation

I want to find the two exact centuries in Earth's history/future where the tropical year is/was prophesized to be 365.19000 si days long and 365.26000. I previously worked it out by hand using a table at https://www.hermetic.ch/cal_stud/cal_art.html#Variation and know they should be around 140,000 BC and 1,126,000 AD. But I want to use the equation at https://www.astro.uni.torun.pl/~kb/Papers/JRASC/Tropic.htm also at https://scienceworld.wolfram.com/astronomy/TropicalYear.html * The length of an average tropical year depends on where in the earth's orbit you consider the start/end point. It's a circular elipse obviously so it sounds like bullshit but its true for some reason. I don't care but if asked put december solstice/jan 1 * Si days of 86400 seconds just assume * don't care about equation methodology warning labels The problem I have is I want to use the equation. They just hand to people and think to myself "what fuarrrkin website or calculator in creation is it supposed to be used on".

>The length of an average tropical year depends on where in the earth's orbit you consider the start/end point. It's a circular elipse obviously so it sounds like bullshit but its true for some reason. Of course, do you thing the slowing down of earth happens in large portions at the end of each year? Of course not, it happens constantly, so the lenght of the year depends on when.do you start to count that year because it gets longer every second. >what fuarrrkin website or calculator in creation is it supposed to be used on". It's a cubic equation, you have to make it equal the number you are searching and solve it for T. It can be solved on paper with pen but wolframalpha does it for you: write "365.2421896698-00000615359T-7.29E-10T^2+2.64E-10T^3 = 365.19000" into wolframalpha.com's search filed and it will tell you all kinds of things, at the bottom there will be the solutions, 3 of them (a cubic equation can.have 1, 2 or 3 solutions.). One of those is a large negative, the other a large positive number, the third is a tiny number close to zero. The equation is almost symmetrical because the constant part is a number very close to the result, so if we reduce it to zero, the constant part is a very small number The three solutions: 4.82795*10^7 -4.82794*10^7 8.48117*10^-8 According to the second equation, the solution of T has to be multiplied by 36525, then 2451545 added to it, to get JD the three JDs for the three solutions of T equal: 1 763 411 189 045 -1 763 402 633 455 2 451 545 The third one can be discarded, all of its value comes from the constant part of the second equation, it gives the same value for almost everything. Now play with the two other values: If you type into duckduckgo "1 763 411 189 045 Julian date to Gregorian date" all calculators say it's not valid, and AI hallucinates it's either today's date or 31th december this year, but of course it's not. If you divide it by 365.2, and substract 4713 from it (the gregorian year of the start of the julian calendar), you get year 4 828 613 000 For the other solution: (you have to add 4713 this time): 4 828 589 573 BC This obviously can't be a solution because it is 100 million years older than the Earth itself. So it's either the first one or i fuarrrked up something.

>>13314 Well, saying it's almost symmetrical is retarded, i'm sorry, i wanted to say, its solutions are two points almost the opposite of each other, and a point that is almost zero.