This is excellent progress, the problem has almost been set up and solved, a remarkable achievement because it is a five variable problem. I used the definitions given in Eqs (10.99) of Marion and Thornton, who used a theta rotation generator (10.93). This is different from my theta rotation generator, which I took from another reference. There is freedom of choice in these definitions. I recommend using the self consistent Marion Thornton results. The elliptical orbit of the centre of mass has been obtained correctly, so the program is working up to that point. I will look in to incorporating moments of inertia into the statement of the problem. In the first instance another mass m sub 1 can be placed at the end of the vector r sub 1, just as a simple model. The general method used by Marion and Thornton is given by Eqs. (10.6a) and (10.6b). The kinetic energy is divided into the translational part, which the code has evaluated correctly to give an ellipse, and the rotational part:

T(rot) = (1/2) sigma over i and j I sub ij omega sub i omega sub j

where I is the moment of inertia tensor. For principal moments of inertia this reduces to

T(rot) = (1/2) sigma over i I sub i omega sub i squared.

So I can proceed like this in the next note. The angular velocities are given in terms of the Euler angles in Eq. (10.102). The problem is almost solved and I will write out these suggestions in the next note, UFT370(1), a paper that can be dedicated to the Milankowitch cycles. Note that the usual method given by Marion and Thornton means that there is no dependence on the Euler angles of r and theta1. So the problem simplifies to one two variable problem (r and theta1, translational motion), and one three variable problem (r, theta and chi) for rotational motion. The complete problem, however, consists of finding a dependence of theta1 and r on theta, phi and chi. This is why the Cartesian method of Note 369(9) was used, so it is well worth trying to iron out the teething problems.

EMyrone

Sent: 06/02/2017 23:30:26 GMT Standard Time

Subj: Re: Discussion of 369(9)I programmed the Lagrangian evaluation of note 9. My first result is that the parameters A,B,C,… in eq.(8) come out differently from the matrix multiplication (6), see bottom of first page of the protocol.

For simplicity I used

r12 = r13 = 0

so that only r11 remains effective. Continuing with these parameters, the lagrangian remains manageable. However there are problems with the numerical solution. I separated the gravitational and rotational part for test. The gravitational part gives an ellipse, see attached figures. The rotational part (in the protocol) gives Lagrange equations with denominator zero, and, at least in the phi part, numerator zero, see eq. o103 on page 9. Anything goes wrong here, either I have an error in the caculation, or the model with bold r and bold r1 is not suitable. I wonder if the rigid body model has been applied correctly, r1 points to a mass element which is not present in the calculation. I would expect that any type of moment of inertia has to enter the calculation.

Horst

Am 06.02.2017 um 08:19 schrieb EMyrone:

Agreed with the first remark. They fix the position of a point on the earth’s surface relative to its centre of mass, and this also answers the second point, because they define the principal moments of inertia (Marion and Thornton chapter ten of the third edition). Note 369(9) was a development of your earlier suggestion of using Cartesian coordinates in the lab frame to develop all the possible interactions. Introducing a rotational lagrangian is superfluous because rotational motion is already included through the motion of vector r sub 1. As you can see this was initially defined in the frame (1, 2, 3) of the principal moments of inertia of the gyroscope (the earth) and then transformed into (X, Y, Z). The Euler transformation is inevitably very complicated but the idea is simple. The great advantage of this method, if it can be crunched out, is that it accounts for all possible motions with five Lagrange variables, r, theta1, theta, phi, chi.

To: EMyrone

Sent: 05/02/2017 15:46:41 GMT Standard Time

Subj: Re: 369(9): General Theory of the Milankovitch CyclesI have some difficulties of understanding:

1. The constants r11, r12, r13 describe the position of an arbitrary point in the rotating body. Can these be chosen arbitrarily so that we can observe the motion of a point at the earth surface for example?

2. In the Lagrangian (11) there are no rotational energy terms. Therefore the Lagrange equations will not contain any moments of inertia. I guess that these terms of kinetic energy are to be added and you omitted them for clarity.

Horst

Am 04.02.2017 um 13:15 schrieb EMyrone:

This general theory shows that five simultaneous Euler Lagrange equations in five Lagrange variables must be solved in general. All the motions are inter related, and are made up of the nutations and precessions of an asymmetric top in orbit around the sun. So the Milankowitch cycles are very intricate. It may be possible to solve this problem numerically using Maxima’s code for simultaneous differential equations. The nutations and precessions depend on the position of the orbit. For example they are different at perihelion and aphelion. The earth is approximately a symmetric top and one of its precessions results in the equinox precession developed in UFT119 with the complementary method of the gravitomagnetic field. So if possible, and when Horst has time, graphics of these intricate motions would be full of interest.