369(8): General Theory of the Gyroscope and Milankovitch Cycles

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.

a369thpapernotes8.pdf

Advertisements
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: