Many thanks again, and congratulations in turn! This rigorous solution shows that the spinning of the gyroscope cannot balance the force of gravitation without an extra upward force. The elliptic spiral is particularly interesting, the centre of mass follows this path as it is pulled downwards by the force of gravity. A balancing upward force modelled by m = m1 allows the gyro to float, so if held above the lab bench it would feel weightless. This is the observation by Laithwaite, who was perfectly right. This effect can be used for heavy engineering, for example to reduce drag in railway systems. That requires the skills of a practical engineer. All this shows the power and elegance of the Euler Lagrange equations.

To: EMyrone@aol.com

Sent: 29/01/2017 16:28:01 GMT Standard Time

Subj: New Results for 369(1)I calculated the correct equations, see attachment. theta_dot_dot and R_dot_dot are given by %i28 on page 5. Interestingly, theta_dot_dot is identical to the former version Eq.(19) in the note, except that the term mgh has vanished, it has moved in the R_dot_dot equation which is more complicated now. The numerical solution shows a nearly free falling mass, there is a small oscillation in the velocity (page 9). Most interesting is the space curve of the centre of mass (page 10). It is an elliptic spiral now, the mass falling about 300 meters in 8 sec, mass is 10 kg, rotating relatively slowly (otherwise it is difficult to see the main effects).

We have no complete precession any more. I saw this already in a test calculation for paper 368 with m-m1=0 as you proposed. This approach (without R variable) described the effect very appropriate, congatulations! I will write up section 3 of paper 368 now, then all this will become more transparent.

Horst

Am 29.01.2017 um 14:36 schrieb EMyrone:

Many thanks, the space curves of the omega vector look particularly interesting and all your rotational results can be used for UFT368. I have just sent over the calculations for a gyro fixed to a lab stand a distance R above the bench. The force in the lab frame (X, Y, Z) is F = mg (frame (X Y, Z)). This is always equal to the force in the moving frame (1, 2, 3). This means that the acceleration vector in Cartesian coordinates is the same as the acceleration vector in spherical polar coordinates. In vector notation this means:

F = m (dv / dt) (X, Y, Z) = m ((dv / dt + omega x v) (frame (1,2,3)) = mg (X, Y, Z)

In plane polars this leads to results such as:

r double dot – r theta dot squared = mg = – MG / r squared

which is the Leibnitz equation. On the left hand the acceleration is expressed in the moving frame (1, 2, 3) (plane polar coordinates). On the right hand side the acceleration is expressed in the lab frame (Cartesian coordinates. The gyro is the same type of problem. So all is OK.

To: EMyrone

Sent: 29/01/2017 11:10:24 GMT Standard Time

Subj: Re: First Computer Results for 369(1)I believe I forgot the attachments, here they are. I made an error in the kinetic energies, it has been corrected now.

Eq. (28) in the note is not a suitable equation for numerical solution because the changes in R_1 are extremey small, this is a “stiff” diff. equation compared to (17-19). Eq.(28) includes the changes in potenial energy that are extremely small, it is better for numerical stability to use the approach

R_1 dotdot = -g

and to measure R from the work bench. The variable R (not R_1) is the additional Lagrange variable. This gives the Translational kinetic energy

The appearance of theta leads to a change in the Lagrange equation for theta. Therefore the angular part is affected. It could be that L_phi is no more a constant of motion. the theta equation (%i8) is more complicatd now.

I let the equations with constants (section 5) of motion as before so that you can see the graphs I described in my previous email. However these are not the right results.

Another point: These Lagrange equations hold for a moving frame of angles (moving with R). It seems to me that we have to use a frame fixed to the lab for obtaining meaningful results.

Horst

PS: I updated the version of the Maxima input program and now have some difficulties with copying graphics to email and with pdf output.

Am 29.01.2017 um 09:44 schrieb EMyrone:

These are all very interesting results, in general the set of Lagrange variables most suitable to the problem should be chosen. There is some freedom of choice of Lagrange variables as you know, but also some rules. The very important important advance is that the code is available for solution of the problem. this is code for the solution of complicated simultaneous differential equations. The angles are measured in the same way as in Problem 10.10 of Marion and Thornton, but the foot of the gyro is allowed to move. The reason for this is that the centre of mass of the gyro can move upwards, giving lift. For some reason I cannot see the graphics, can you resend them? I cannot see the protocol but your method seems to be OK. To check this, it should be equivalent to the simultaneous solution of Eqs. (17) to (19) and (32). Eq. (32) was set up deliberately to find an equation that involves both rotation and translation (i.e it contains both R and theta). As you infer, R can be interpreted as the height of the point of the gyro above a laboratory bench. The overall aim is to find whether the gyro can be elevated, i.e. to find whether its point can move upwards. Although I cannot see them, the graphics look to be very interesting, especially the space curve of the centre. In order to get a complete solution can Eqs. (17) to (19) and (32) be solved simyltaneously with R interpreted as the height above the laboratory bench? The nutations and rotations / precessions are also very interesting. This is probably the first time in 250 years that this problem has been completely solved, so it should be addressed in as many useful ways as are relevant, with different Lagrange variables and so on. This code is very useful and is the beginning of a new phase of research. I can suggest an experiment: fix the point of the gyro to a stand a distance R above the laboratory frame, so it is horizontal as in the Laithwaite experiment, and observe its motion. Does its centre of mass rise above the point at which its foot is fixed to the stand, R above the lab bench?

To: EMyrone

Sent: 28/01/2017 17:46:26 GMT Standard Time

Subj: Re: Discussion of 369(1): Complete Analytical Mechanics of the GyroscopeIf R in eq.(27) is the earth radius, it is in very good approximation:

h << R

so that R is practically not altered. Therefore R can better be measured from any point in the lab. To my understanding the angles are measured from the foot point of the gyro as before. We therefore obtain the additional equation

R dot dot = -g.

This is decoupled from the other gyro equations, i.e. the gyro moves in free fall.

I added this equation (in Hamilton form) to the set of equations:

r dot = v

v dot = -gThe results are in the protocol. First I re-computed the Lagrange equations directly, then I used the easier form with constants of motion. Due to a pdf printing problem, the formulas are not readable, but you can look at the graphics:

1. theta dot(t), theta(t) — nutation

2. phi(t), psi(t) — precession and rotation of body

3. v(t), R(t) — free fall solution4. omega components in 1-2-3 frame and modulus omega (omega_3=const)

5. space curve of the centre of mass with nutation/precession

6. space curve of omega vector in 1-2-3 frame. This is motion in a plane (rosetta) since omega_3=const.I will produce similar graphics for paper 368 in better quality.

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.