This will be the first time that this problem has been correctly solved, so your coding is full of interest, in particular the lab frame translational motion of the point of the gyro can be computed. If the upward Z axis force generated by the spin of the gyro is the same as the gravitational force downward, the point will float. So Laithwaite would feel no weight if he is holding a floating point. The reason why this two hundred and fifty year old problem has not been solved before is that it is exceedingly intricate, and code is needed to solve four simultaneous differential equations in four variables: r, theta, phi and chi.
To: EMyrone@aol.com
Sent: 28/01/2017 08:54:57 GMT Standard Time
Subj: Re: 369(1): Complete Analytical Mechanics of the GyroscopeThis is a model with a free floating foot point of the gyro. It seems that the reference to the earth radius and gravitational earth potential is not necessary, it is easier to refer to any fixed point at the earth surface from where R is measured. This avoids appearance of G amd M in the Lagrange equations. I will set up this version today afternoon by computer.
Horst
Am 27.01.2017 um 14:39 schrieb EMyrone:
The geometry is defined in Figure(1), and in general the gyroscope’s point is allowed to move with respect to the centre of the earth. So the point can move up or down. The analytical problem reduces to the simultaneous solution of four differential equations, (17) to (19) and (32). The first three are the same as in UFT368, for the pure rotational motion of the gyro. They are supplemneted by Eq. (32) for the motion of R, where R is he distance between the point of hte gyro and the centre of the Earth. Eq. (32) is the necessary link between rotational and translation motions of the gyro. In the replicated Laithwaite experiment the point of the gyro (the common origins of (1, 2, 3) and (X, Y, Z)) is held at the height of Laithwaite’s arm.