To address the first point the problem can be defined so that the origins of (X, Y, Z) and (1, 2, 3) are the same, r = h cos theta is in the positive Z axis and R(t) is the variable height above a lab bench in the minus Z axis. I will sketch this out in the next note. This configuration means that the foot of the gyro is allowed to slide up and down the Z axis. This configuration is in fact that of Note 349(1). In Figure (1) of that note I displaced r = h cos theta and R to the right for clarity. In an experiment, the foot of the gyro can be attached to a stand with a low friction collar of some kind that allows the foot to move up and down a stand. Will the gyro fall to the lab bench or will it not? In order to define the constants of motion in the presence of an external lab frame force or torque, the lagrangian must include the potential energy due to the external torque, and then solved in theta, phi, chi and r. I am not quite clear as to what is meant by an additional constraint. Probably it means that the three simultaneous equations in the three Euler angles are solved as before, and the origin is moved in some way. However, in my opinion the complete solution of the problem is as in Note 349(1), with four Lagrange variables, theta, phi, chi and R1 and four simultaneous differential equations. This is because that set of equations covers all possibilities – it is a complete solution. I agree that R can be reinterpreted as the height above a lab bench instead of as the radius of the earth. The solution must be able to show whether or not a gyro fixed horizontally to a stand with a low friction collar will fall to the lab bench. Does it need another upward force to stop it falling? If the foot of the gyro is fixed permanently to the stand, so that teh foot cannot move, it will move around and its motion is as described in your very interesting graphics. The stand and gyro can be placed on a balance, so that the total weight can be measured, first with a static, non spinning gyro, then with a spinning gyro. I will calculate the effect of an upward force in the next note.
To: EMyrone@aol.com
Sent: 28/01/2017 18:06:43 GMT Standard Time
Subj: Comments: Re: Discussion of 369(1): Complete Analytical Mechanics of the GyroscopeSome comments:
Using a moving fixed point would normally mean that the angles have to be measured from the coordinate origin of the lab. If the origin of the spherical coordinate system is static in the lab (which is the usual proceeding in applying Lagrange theory) then the expressions for the kinetc energy will change and there will be a coupling between R and other coordinates. This is not the case if the coordinate origin of the x-y plane is shifted with R as we did before.
Can we be sure that the constants of motion remain valid if the gyro is moved by any other form of external force/torque? A constant L_psi means a constant rotation speed around body axis x_3. This could change if external forces are working.
Instead of a new coordinate R, we could use an additional constraint R(t). This would describe what Laithwaite did. I guess that the enforced shift leads to an enhanced nutation of the gyro so it will follow and overshoot the shift in the next cycle of nutation. Then the gyro would go upwards without external force application.
Horst
Am 28.01.2017 um 10:12 schrieb EMyrone:
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
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.