Discussion of Note 369(8)

Very interesting idea! This is one way of solving the problem when theta1 = theta1 (theta, phi, chi) and r = r(theta, phi, chi). This would mean that the nutations and precessions of the earth would depend on where it is in its orbit around the sun. The coordinate transformation can be worked out, and I could do this today. The result will be very complicated but the computer could probably deal with it. The nutations and precessions of the earth as a freely rotating symmetric top have already been worked out in a previous note and written up in Section 3 of UFT368, already showing signs of being a popular paper. The computer results revealed nutations and precessions which would mean that there would be warm and cold spells in the climate.

To: EMyrone@aol.com
Sent: 03/02/2017 16:45:03 GMT Standard Time
Subj: Re: 369(8): General Theory of the Gyroscope and Milankovitch Cycles

I understand this as follows: In the case of Milankowitch theory the earth gyro is freely floating in the earth orbit, therefore there is no U term “m g h cos(theta)”. In addition to the three Euler angles we need the two plane polar coordinates of the mass point in the orbit around the sun: r and theta_1. There is probably a coupling betrween theta_1 (and possibly r) and the Euler angle coordinates. Instead of setting up the kinetic energy in separate terms one has to use the standard Lagrange method. This is to set up the complete coordinate transformation

(X, Y, Z) –> (r, theta_1, theta, phi, psi)

and compute the Lagrangian

L = 1/2 m (X^2 + Y^2 + Z^2) – U(r).

This then will contain all couplings.

Horst

Am 03.02.2017 um 12:35 schrieb EMyrone:

The gyro in the presence of a general external torque (4), generating the potential energy (2), is given by solving the lagrangian (14). If there is no interaction between the rotation of the gyroscope and the translation of its fixed point, the problem is solved by Eqs. (21) to (24), in which (24) is independent. This means that there is simple force on the gyro’s point, giving the external torque (4). However if U is a function of r and theta, Eqs. (21), (22), (24) and (26) must be solved simultaneously, giving a lot more information. The Milankovitch cycles are given by solving Eqs. (29) to (31), and the orbital motion in a plane by solving Eqs. (34) to (36). If for some reason condition (37) holds, then the Milankovitch cycles are affected in a way to be determined by the Maxima code. So I will proceed to write up Sections 1 and 2 of UFT369.


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