To: EMyrone@aol.com

Sent: 25/02/2017 18:14:35 GMT Standard Time

Subj: Re: Further Discussion of Note 371(3) and 371(2)I have calculated the Lagrange equation (17) for r_1. It depends on the omega’s via (11-13). The omega’s depend on time so their time derivatives have to be evaluated. Omitting them gives a bit simpler expressions but seems not to be justified, see protocol file 371(2).pdf

When the full angular dependence (6-8) is introduced, the Lagrange equation for r_1,2,3 becomes extremely complicated, extending over more than one page in the protocol 371(2a).pdf. If the Euler angles are treated as Lagrange variables too, the equations have additionally to be resolved according to the second derivatives which was also done. Then there is a coupling of second derivatives across all equations. Trying to bring them into canonical form fails. Either the equations are too complicated or they are not resolvable. I guess that the latter is the case.

Horst

Am 25.02.2017 um 10:54 schrieb EMyrone:

The point mass M is placed at the origin and the point mass m is at (r1, r2, r3). The motion of the axes e1, e2, and e3 is defined with the dynamics of the Euler angles through the spin connection and equations (6) to (8). This gives six equations in six unknowns, Eqs. (17) to (22). That is an exactly determined problem therefore. The three degrees of freedom are the three dimensions of three dimensional space. Only three coordinates are being used, e sub 1, e sub 2 and e sub 3, and six Lagrange variables, r1, r2, r3, theta, phi and chi. The Euler equations for a rigid object are not being used, because there are no moments of inertia being used. Have you tried running these six simultaneous equations through Maxima, to find how the system behaves? If it gives reasonable results all looks OK. In spherical polar coordinates the mass M is at the origin and the mass m is at vector r. Solutions in the spherical polar system are also full of interest and much less complicated. It is also possible to use your idea of a unit vector by adapting the gyroscope with one point fixed. There are many interesting things to work on adn they will all create a lot of interest. It is best to work with spherical polar coordinates by Ockham’s Razor, but the Euler angles give a very large amount of new information. The important thing is the ability of Maxima so solve very complicated sets of simultaneous differential equations.

To: EMyrone

Sent: 25/02/2017 09:09:39 GMT Standard Time

Subj: Re: Discussion of Note 371(3) and 371(2)Your argument on time-dependence of the (e1, e2, e3) frame is correct. However it is not possble to describe a probelm with 3 degreees of freedom by 6 coordinates, at least not in Lagrange theory. Then you obtain an underdetermined system of equations, there is no unique solution.

A second point is to strictly discern if a masspoint or a rigid body is considered. You cannot use the Lagrange equations for a rigid body and apply it to a masspoint because then you have too many coordinates.

On the other hand it is possible to use Eulerian angles for masspoints. However you have already 3 coordinates so you cannot introduce additional translations. This type of application seems to be restricted to pure rotations on a unit sphere. The situation is different for a rigid body again.Horst

Am 25.02.2017 um 09:53 schrieb EMyrone:

It is a good idea to use the gyro with one point fixed for orbital theory, the mass M is at the fixed point of the gyro, and mass m is separated by a distance r from M. Unlike problem 10.10 of Marion and Thornton, however, the distance r is not constant. I will look in to this and go back to the basics of the derivation of the Euler equation from variational calculus, Marion and Thornton chapter five (Euler 1744). Howevber, I think that all is OK for the following reasons. It is true that the Euler angles relating frame (i, j, k) and (e1, e2, e3) are constants by definition, provided that frame (e1, e2, e3) always has the same orientation with respect to frame (i, j, k) and provided that the two frames are static. Then theta, phi and chi, being constants, cannot be used as variables. I agree about this point. However in Note 371(2), frame (e1, e2, e3) is moving with respect to (i. j. k), and so theta, phi and chi are also moving. This is because the Cartesain frame (i, jk, k) is static by definition, but frame (e1, e2, e3) is dynamic, i.e. e1, e2 and e3 depend on time, but i, j, and k do not depend on time. Similarly in spherical polars, (i, j, k) static, but (e sub r, e sub theta, e sub phi) is time dependent so r, theta and phi all depend on time. The lagrangian (1) of Note 371(2) is true for any definition of v, and Eq. (10) of that note is true for any definition of the spin connection (e.g. plane polar, spherical polar, Eulerian, and any curvilinear coordinate system in three dimensions). So Eqs. (11) to (16) of that note are correct. So it is correct to set up the lagrangian (16) using the Lagrange variables r1(t), r2(t), r3(t), theta(t), phi(t) and chi(t). It could also be set up with plane polar or spherical polar coordinates. We have already correctly solved those problems using the lagrangian method, The fundamental property of the spin connection is to show how the axes themselves move and it is valid to re express the angles of the plane polar and spherical polar coordinates as Eulerian angles. In the orbital problem, the Eulerian angles are all time dependent. So they vary in this sense, and can be used as Lagrange variables. The Euler variable x of chapter five of Marion and Thornton is t, x = t. To sum up, it is true that the Euler angles are constants when viewed as angles defining the orientation of a static (1, 2, 3) with respect to a static (X, Y, Z), but in Eq. (10) of Note 371(2), the components of the spin conenction are defined in terms of time dependent Euler angles, which are therefore Lagrange variables. This fact can be seen from Eqs. (6) to (8) of the note, in which appear phi dot, theta dot, and chi dot. These are in general non zero, i.e. they are all time dependent angular velocities.

To: EMyrone

Sent: 24/02/2017 18:35:19 GMT Standard Time

Subj: Re: Note 371(3) : Definition of Reference FramesThanks, this clarifies the subject. I think we must be careful in applying Lagrange theory. A mass point in 3D is described by 3 variables that are either [X. Y. Z] or [r1, r2,

r3]. The Eulerian angles describe the coordinate transformation between both frames of reference bold [i, j, k] and bold [e1, e2, e3]. The Eulerian angles are the sam for all points [r1, r2, r3]. So they cannot be subject to variation in the Lagrange mechanism. The degree of freedom must be the same in both frames, otherwise we are not dealing with generalized coordinates and Lagrange theory cannot be applied.A Lagrange approach in Eulerian coordinates can be made if we describe the motion of a mass point that is described as a unit vector in the [e1, e2, e3] frame. Then [r1, r2, r3] is fixed and the Eulerian angles can indeed be used for Lagrange variation. This is done for the gyro with one point fixed.

I calculated the transformation matrix A of the note and its inverse. The order of rotations is different from that in M&T.

Horst

Am 24.02.2017 um 15:11 schrieb EMyrone:

In this note the reference frames used in Note 371(2) are defined. The mass m orbits a mass M situated at the origin. The spherical polar coordinates are defined, and frame (X, Y, Z) is rotated into frame (1, 2, 3) with the matrix of Euler angles as in Eq. (11). The inverse of this matrix can be used to define r1, r2, and r3 in terms of X, Y, Z. In plane polar coordinates the orbit is a conic section with M at one focus as is well known.The planar elliptical orbit does not precess, but using spherical polar coordinates and Eulerian angles the orbit is no longer planar and precesses.

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To: EMyrone@aol.com

Sent: 25/02/2017 16:33:44 GMT Standard Time

Subj: Re: The Basics of the Euler Lagrange TheoryAgreed, the method of Lagrangian multipliers is more basic and can for example be applied when functions are not continuously differentiable. I will try a solution of eqs. (17-19) of note 371(2).

Horst

Am 25.02.2017 um 12:47 schrieb EMyrone:

I reviewed these basics as described by Marion and Thornton. The number of proper Lagrange variables is equal to the number of degrees of freedom, but it is also possible to use the method of undetermined multipliers when the number of Lagrange variables is greater than the number of degrees of freedom. However, a simple solution to the problem is to solve Eqs. (17) to (19) of Note 371(2) simultaneously, to give r1, r2 and r3 in terms of theta, phi and chi. These are the required orbits. There are three dimensions (degree of freedom) and three proper Lagrange variables, r1, r2, and r3. So there are three differential equations in three unknowns, an exactly determined problem. The orbits are r1(theta, phi, chi), r2(theta, phi, chi) and r3(theta, phi, chi). Finally use

r squared = r1 squared + r2 squared + r3 squared

to find r(theta, phi, chi) and its precessions. In the planar limit it should reduce to a conic section without precession.

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To: burleigh.personal@gmail.com, emyrone@aol.com

Sent: 25/02/2017 23:25:33 GMT Standard Time

Subj: Full Spanish version of UFT 369 (3 sections)Hello Dave,

Please find enclosed the full version of the Spanish UFT 369 paper, for posting.

Thanks.Regards,

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r squared = r1 squared + r2 squared + r3 squared

to find r(theta, phi, chi) and its precessions. In the planar limit it should reduce to a conic section without precession.

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To: EMyrone@aol.com

Sent: 25/02/2017 09:26:44 GMT Standard Time

Subj: Re: FOR POSTING: Section 3 of paper 369Many thanks for your eulogising comments. The method of numerical solution of the complete set of gyro equations is indeed very powerfull and applicable for cases I never thougth of.

Horst

Am 25.02.2017 um 10:18 schrieb EMyrone:

These results are full of interest and will attract a large readership. I can see that through the scientometrics. These graphics and analyses are clearly thought out as usual, and the results are interpreted so that the readership will be able to understand them without ploughing through the maths if they want to concentrate on engineering essentials. The results also show the great power of the Maxima code, controlled by the skill and experience of co author Dr Horst Eckardt. A huge amount of new information comes out of a problem that up to a few week ago was only vaguely understood, half understood and misinterpreted. Having understood the problem at last, all kinds of engineering solutions become possible, so we can engineer new kinds of railway systems for example with reduced drag. In the aerospace industry these results can eb used to enginner new kinds of mechanism that will allow aircraft to lift off, helping the wings lift the plane. These mechanisms can also be used in vehicles of all kinds to reduce drag. This is an outstanding section. It shows what can be done with the elegant mathematics of the eighteenth century enlightenment. A large number of possibilities has suddenly emerged.

Sent: 24/02/2017 19:02:41 GMT Standard Time

Subj: Section 3 of paper 369I finished section 3 of paper 369. I gave several examples for gyro

motion, including application of an external torque. I could not verify

the Shipov experiment (lifting a spinning top by a torque in Z

direction). This seems to be an effect of different origin, maybe a

fluid dynamics effect of spacetime, although it is astonishing that it

is of macroscopic order.Horst

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To: EMyrone@aol.com

Sent: 25/02/2017 09:09:39 GMT Standard Time

Subj: Re: Discussion of Note 371(3) and 371(2)Your argument on time-dependence of the (e1, e2, e3) frame is correct. However it is not possble to describe a probelm with 3 degreees of freedom by 6 coordinates, at least not in Lagrange theory. Then you obtain an underdetermined system of equations, there is no unique solution.

A second point is to strictly discern if a masspoint or a rigid body is considered. You cannot use the Lagrange equations for a rigid body and apply it to a masspoint because then you have too many coordinates.

On the other hand it is possible to use Eulerian angles for masspoints. However you have already 3 coordinates so you cannot introduce additional translations. This type of application seems to be restricted to pure rotations on a unit sphere. The situation is different for a rigid body again.Horst

Am 25.02.2017 um 09:53 schrieb EMyrone:

It is a good idea to use the gyro with one point fixed for orbital theory, the mass M is at the fixed point of the gyro, and mass m is separated by a distance r from M. Unlike problem 10.10 of Marion and Thornton, however, the distance r is not constant. I will look in to this and go back to the basics of the derivation of the Euler equation from variational calculus, Marion and Thornton chapter five (Euler 1744). Howevber, I think that all is OK for the following reasons. It is true that the Euler angles relating frame (i, j, k) and (e1, e2, e3) are constants by definition, provided that frame (e1, e2, e3) always has the same orientation with respect to frame (i, j, k) and provided that the two frames are static. Then theta, phi and chi, being constants, cannot be used as variables. I agree about this point. However in Note 371(2), frame (e1, e2, e3) is moving with respect to (i. j. k), and so theta, phi and chi are also moving. This is because the Cartesain frame (i, jk, k) is static by definition, but frame (e1, e2, e3) is dynamic, i.e. e1, e2 and e3 depend on time, but i, j, and k do not depend on time. Similarly in spherical polars, (i, j, k) static, but (e sub r, e sub theta, e sub phi) is time dependent so r, theta and phi all depend on time. The lagrangian (1) of Note 371(2) is true for any definition of v, and Eq. (10) of that note is true for any definition of the spin connection (e.g. plane polar, spherical polar, Eulerian, and any curvilinear coordinate system in three dimensions). So Eqs. (11) to (16) of that note are correct. So it is correct to set up the lagrangian (16) using the Lagrange variables r1(t), r2(t), r3(t), theta(t), phi(t) and chi(t). It could also be set up with plane polar or spherical polar coordinates. We have already correctly solved those problems using the lagrangian method, The fundamental property of the spin connection is to show how the axes themselves move and it is valid to re express the angles of the plane polar and spherical polar coordinates as Eulerian angles. In the orbital problem, the Eulerian angles are all time dependent. So they vary in this sense, and can be used as Lagrange variables. The Euler variable x of chapter five of Marion and Thornton is t, x = t. To sum up, it is true that the Euler angles are constants when viewed as angles defining the orientation of a static (1, 2, 3) with respect to a static (X, Y, Z), but in Eq. (10) of Note 371(2), the components of the spin conenction are defined in terms of time dependent Euler angles, which are therefore Lagrange variables. This fact can be seen from Eqs. (6) to (8) of the note, in which appear phi dot, theta dot, and chi dot. These are in general non zero, i.e. they are all time dependent angular velocities.

To: EMyrone

Sent: 24/02/2017 18:35:19 GMT Standard Time

Subj: Re: Note 371(3) : Definition of Reference FramesThanks, this clarifies the subject. I think we must be careful in applying Lagrange theory. A mass point in 3D is described by 3 variables that are either [X. Y. Z] or [r1, r2, r3]. The Eulerian angles describe the coordinate transformation between both frames of reference bold [i, j, k] and bold [e1, e2, e3]. The Eulerian angles are the sam for all points [r1, r2, r3]. So they cannot be subject to variation in the Lagrange mechanism. The degree of freedom must be the same in both frames, otherwise we are not dealing with generalized coordinates and Lagrange theory cannot be applied.

A Lagrange approach in Eulerian coordinates can be made if we describe the motion of a mass point that is described as a unit vector in the [e1, e2, e3] frame. Then [r1, r2, r3] is fixed and the Eulerian angles can indeed be used for Lagrange variation. This is done for the gyro with one point fixed.

I calculated the transformation matrix A of the note and its inverse. The order of rotations is different from that in M&T.

Horst

Am 24.02.2017 um 15:11 schrieb EMyrone:

In this note the reference frames used in Note 371(2) are defined. The mass m orbits a mass M situated at the origin. The spherical polar coordinates are defined, and frame (X, Y, Z) is rotated into frame (1, 2, 3) with the matrix of Euler angles as in Eq. (11). The inverse of this matrix can be used to define r1, r2, and r3 in terms of X, Y, Z. In plane polar coordinates the orbit is a conic section with M at one focus as is well known.The planar elliptical orbit does not precess, but using spherical polar coordinates and Eulerian angles the orbit is no longer planar and precesses.

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To: EMyrone@aol.com

Sent: 25/02/2017 07:21:27 GMT Standard Time

Subj: PS: Re: Note 371(3) : Definition of Reference FramesPS: a solution can be to consider a masspoint with 3 internal degrees of freedom, corresponding to the Euler angles. In addition there are 3 degrees of freedom for external motion (cartesian or spherical coordinate system). The internal degrees of freedom represent a spin resp. spin moment. It is not useful to introduce further translational coordinates arbitrarily on the surface of a body or so, because this does not conform to the Lagrange concept. The latter is made for masspoints only.

Horst

Am 24.02.2017 um 19:34 schrieb Horst Eckardt:

Thanks, this clarifies the subject. I think we must be careful in applying Lagrange theory. A mass point in 3D is described by 3 variables that are either [X. Y. Z] or [r1, r2, r3]. The Eulerian angles describe the coordinate transformation between both frames of reference bold [i, j, k] and bold [e1, e2, e3]. The Eulerian angles are the sam for all points [r1,

r2, r3]. So they cannot be subject to variation in the Lagrange mechanism. The degree of freedom must be the same in both frames, otherwise we are not dealing with generalized coordinates and Lagrange theory cannot be applied.A Lagrange approach in Eulerian coordinates can be made if we describe the motion of a mass point that is described as a unit vector in the [e1, e2, e3] frame. Then [r1, r2, r3] is fixed and the Eulerian angles can indeed be used for Lagrange variation. This is done for the gyro with one point fixed.

I calculated the transformation matrix A of the note and its inverse. The order of rotations is different from that in M&T.

Horst

Am 24.02.2017 um 15:11 schrieb EMyrone:

In this note the reference frames used in Note 371(2) are defined. The mass m orbits a mass M situated at the origin. The spherical polar coordinates are defined, and frame (X, Y, Z) is rotated into frame (1, 2, 3) with the matrix of Euler angles as in Eq. (11). The inverse of this matrix can be used to define r1, r2, and r3 in terms of X, Y, Z. In plane polar coordinates the orbit is a conic section with M at one focus as is well known.The planar elliptical orbit does not precess, but using spherical polar coordinates and Eulerian angles the orbit is no longer planar and precesses.

]]>

Sent: 24/02/2017 19:02:41 GMT Standard Time

Subj: Section 3 of paper 369I finished section 3 of paper 369. I gave several examples for gyro

motion, including application of an external torque. I could not verify

the Shipov experiment (lifting a spinning top by a torque in Z

direction). This seems to be an effect of different origin, maybe a

fluid dynamics effect of spacetime, although it is astonishing that it

is of macroscopic order.Horst

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