Discussion of Note 369(4)

Many thanks to the Co President!

The Cartan spin connection of the spherical polar coordinate system

It is pleasing that this new era in physics, and unification, increases clarity across the board – including in terms of mathematical structure. You certainly deserve medals in all of the sciences and mathematics. Newton must be smiling!!

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369(4) : The Cartan spin connection of the spherical polar coordinate system

This is given by the 3 x 3 matrix in Eq. (28), it is an antisymmetric rotation matrix in three dimensions. It can be used to describe a three dimensional orbit, and in general can be used for any problem in dynamics. This proves for the first time that the spherical polar coordinate system is a special case of Cartan geometry. Note that theta and phi are the same as the theta and phi Euler angles. Any curvilinear coordinate system is a special case of Cartan geometry.

a369thpapernotes4.pdf

Discussion of Note 369(3)

Yes it was first explained in UFT119 using field equations and the gravitomagnetic field. That method should be equivalent to the kinematic method using the Euler equations. The link between the two methods is the spin connection. I would say that the two methods are complementary, and forging a link between them would be very interesting. The first thing to do is to show that the usual expressions for velocity and acceleration in spherical polars (or any valid coordinates system) are examples of the covariant derivative of Cartan geometry. As you know this has already been done in plane polars in recent papers. In UFT368 the calculations look to be on the classical level, but they are also generally covariant because the derivative in the moving frame is a generally covariant derivative.

To: EMyrone@aol.com
Sent: 31/01/2017 10:21:40 GMT Standard Time
Subj: Re: Discussion of Note 369(3)

Wasn’t the equinox explained by other mechanisms, for example gravitomagnetic field? There seems to be some ambiguity.

Horst

Am 30.01.2017 um 18:05 schrieb EMyrone:

Many thanks, many things can be done now with the new Maxima code for solving simultaneous differential equations. Provided that the problem is defined precisely, the code can solve the equations no matter how complicated. The trajectories of the freely rotating symmetric top are theta(t), phi(t) and chi(t), and solutions can also be obtained for the freely rotating asymmetric top. Then, gravitational interaction between m and M is coded in, taking care to use the same frame of reference self consistently. This results in very complicated algebra, but complication is no problem. The Milankovich cycles should come out of this code. This could not have been done in the time of Euler and Lagrange because of the complexity. They could use intelligent approximations of course.

To: EMyrone
Sent: 30/01/2017 14:29:27 GMT Standard Time
Subj: Re: New Results for 369(1)

Excellent, congratulations both. As you say, yet another first rigorous analysis.

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External torque on gyro

This looks like a good method, and with the code, ideas like this are of immediate interest. The general or canonical method of defining torque is

Tq = (dL / dt) (X, Y, Z) = (dL / dt + omega x L)(1, 2, 3)

leading to Eqs. (10.120) of Marion and Thornton, the Euler equations with torque. Here L is angular momentum. These are Eqs. (2) to (4) of UFT368, and the code can be used to build up any torque from the Euler angle trajectories of UFT368.

To: EMyrone@aol.com
Sent: 31/01/2017 09:58:53 GMT Standard Time
Subj: External torque on gyro

To model an external torque around the Z axis, we could add this in the
Lagrange equation for phi. Since this equation is of first order, I
suggest to make an additional time derivative. Then we can write:

phi dot dot = … + Tq/I_phi

where I_phi is the momentum of inertia aroud the Z axis. Perhaps this
can be approximated by

I_phi approx. (X^2+Y^2)*m

What is the “canonical” method to introduce a torque? It must appear in
the kinetic energy then, before applying the Lagrange mechanism.

Horst

Daily Report Sunday 29/1/17

The equivalent of 67,835 printed pages was downloaded (247.328 megabytes) from 2135 downloaded memory files and 468 distinct visits each averaging 3.4 memory pages and 9 minutes, printed pages to hits ratio of 31.77, top ten referring domains 2,205,078, w 1447, main spiders Google, MSN and Yahoo. Collected ECE2 1141, Top ten 1036, Collected Evans / Morris 957(est), Collected scientometrics 632, Barddoniaeth 434, Principles of ECE 272, Collected Eckardt / Lindstrom 149, Autobiography volumes one and two 145, PECE 131, Collected Proofs 122, Evans Equations 108, UFT88 98, ECE2 78, Engineering Model 64 (est), CEFE 47, Self charging inverter 38, Llais 31, UFT321 30, UFT311 28, UFT313 26, UFT314 23, UFT315 18, UFT316 19, UFT317 24, UFT318 15, UFT319 29, UFT320 19, UFT322 25, UFT323 24, UFT324 20, UFT325 22, UFT326 24, UFT327 17, UFT328 27, UFT329 23, UFT330 17, UFT331 18, UFT332 22, UFT333 19, UFT334 17, UFT335 23, UFT336 20, UFT337 12, UFT338 19, UFT339 12, UFT340 15, UFT341 16, UFT342 14, UFT343 25, UFT344 22, UFT345 19, UFT346 17, UFT347 19, UFT348 19, UFT349 21, UFT351 20, UFT352 32, UFT353 14, UFT354 38, UFT355 33, UFT356 28, UFT357 15, UFT358 20, UFT359 16, UFT360 10, UFT361 10, UFT362 14, UFT363 22, UFT364 28, UFT365 24, UFT366 78, UFT367 31, UFT368 6 to date in January 2017. Informatics University of Kiel UFT section extensive; Princeton University my page; Bezeq Data Centre Israel extensive download; Pakistan Education and Research Network general; University of Durham UFT42. Intense interest all sectors, updated usage file attached for January 2017.

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Discussion of Note 369(3)

Many thanks, many things can be done now with the new Maxima code for solving simultaneous differential equations. Provided that the problem is defined precisely, the code can solve the equations no matter how complicated. The trajectories of the freely rotating symmetric top are theta(t), phi(t) and chi(t), and solutions can also be obtained for the freely rotating asymmetric top. Then, gravitational interaction between m and M is coded in, taking care to use the same frame of reference self consistently. This results in very complicated algebra, but complication is no problem. The Milankovich cycles should come out of this code. This could not have been done in the time of Euler and Lagrange because of the complexity. They could use intelligent approximations of course.

To: EMyrone@aol.com
Sent: 30/01/2017 14:29:27 GMT Standard Time
Subj: Re: New Results for 369(1)

Excellent, congratulations both. As you say, yet another first rigorous analysis.

Sent from my Samsung device

Discussion of 369(1)

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 Gyroscope

Some 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 Gyroscope

This 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.

369(3): General Lagrangian Theory of the Earth’s Precessions and Nutations

The general theory is very intricate, but can be solved by solving Eqs. (6) to (8) and (19) simultaneously to give the trajectories r(t), theta(t), phi(t) and chi(t). These trajectories are related to the Milankowich cycles, so this kind of theory could predict the climate changes due to the Earth’s precessions and nutations in the gravitational fields of the sun, moon and planets. It is clear from inspection that the Milankowich cycles are very intricate, as is well known. This may be the first complete solution of the problem.

a369thpapernotes3.pdf

AW: First Computer Results for 369(1)

OK thanks again!

To: EMyrone@aol.com
Sent: 30/01/2017 08:39:57 GMT Standard Time
Subj: AW: First Computer Results for 369(1)

Concerning my latest results i sent over, there is an inconsistency in the XY scaling of the centre of mass motion. I will check this today.
Horst

Von meinem Samsung Gerät gesendet.

New Results for 369(1)

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 Gyroscope

If 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 = -g

The 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 solution

4. 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 Gyroscope

This 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.

369(1b).pdf