Archive for January, 2017

First Computer Results for 369(1)

Monday, January 30th, 2017

Agreed with this, it will be very interesting to plot R(t), the motion of the point of the gyro on the lab stand in the Z axis. This experiment is easy to carry out at some later stage.

To: EMyrone@aol.com
Sent: 29/01/2017 14:49:00 GMT Standard Time
Subj: Re: First Computer Results for 369(1)

I think we now near the completion. In eq.(14) a factor of h is missing, to be added later. We agree that the Lagrange variables are theta, phi, psi, R. R_1 is an auxiliary variable only. The equations to be solved simultaneously are (8-10) and (14). Eq. (16) is a special condition which can be evaluated a posteriori. It cannot replace (14) because then we would have 4 equations for only 3 variables.

There are still two points to be resolved. Due to (4), the kinetic energy contains mixed R and theta terms. This leads to a more complicated expression for theta dot dot (eq. 10). See eq. E1a (after %i8) in the protocol.

This equation contains R dot dot, and the equation for R dot dot (E4a) contains theta dot dot. So we cannot use this form directly for the solver, we have to resolve first the two equations for theta dot dot and R dot dot, as we did in paper 325. Then we can proceed as usual. I will do the linear equation solving by Maxima.

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.

Daily Report Saturday 28/1/17

Monday, January 30th, 2017

The equivalent of 54,632 printed pages was downloaded (199.189 megabytes) from 2115 downloaded memory files (hits) and 410 distinct visits each averaging 3.9 memory pages and 6 minutes, main spiders Google, MSN and Yahoo, top ten referring domains 2,220,454, w 1447. Collected ECE2 1094, Top ten 1013, Evans / Morris 924 (est), Collected scientometrics 603, Barddoniaeth 413, Principles of ECE 260, F3(Sp) 176, Eckardt / Lindstrom 139, Autobiography volumes one and two 137, PECE 129, Evans Equations 110, Collected Proofs 119, UFT88 97, Engineering Model 59, CEFE 46, Self charging inverter 37, PLENR 32, Llais 30, UFT321 29, UFT311 27, UFT313 25, UFT314 22, UFT315 17, UFT316 18, UFT317 23, UFT318 14, UFT319 28, UFT320 17, UFT322 24, UFT323 23, UFT324 19, UFT325 21, UFT326 22, UFT327 16, UFT328 26, UFT329 21, UFT330 15, UFT331 17, UFT332 21, UFT333 18,UFT334 15, UFT335 22, UFT336 19, UFT337 11, UFT338 18, UFT339 10, UFT340 14, UFT341 15, UFT342 13, UFT343 24, UFT344 21, UFT345 17, UFT346 16, UFT347 18, UFT348 18, UFT349 20, UFT351 17, UFT352 31, UFT353 13, UFT354 36, UFT355 32, UFT356 27, UFT357 14, UFT358 18, UFT359 15, UFT360 9, UFT361 9, UFT362 13, UFT363 21, UFT364 27, UFT365 23, UFT366 77, UFT367 30, UFT368 4 to date in January 2017. Informatics University of Kiel UFT extensive; University of Leon Spain UFT199; Ateneo de Manila University Philippines general; Maribor University Slovenia general. Intense interest all sectors, updated usage file attached for January 2017.

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First Computer Results for 369(1)

Sunday, January 29th, 2017

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@aol.com
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(1a).pdf

369(2): Motion of a Gyroscope Attached to a Stand

Sunday, January 29th, 2017

The geometry is defined in Fig (1), and two cases considered: 1) the foot of the gyro is attached to the stand firmly, so it cannot move up and down the Z axis, and 2) the foot is allowed to move up and down the Z axis. In case (1), (Laithwaite experiment), the weightless condition is given by Eq. (20). In case (1), Eqs. (8) to (10) and 16) must be solved simultaneously. In case (2) Eqs. (8) to (10) and (14) must be solved simultaneously.

a369thpapernotes2.pdf

First Computer Results for 369(1)

Sunday, January 29th, 2017

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@aol.com
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.

Lagrangian Methods in Electrodynamics

Sunday, January 29th, 2017

Agreed, it is possible to derive the Proca equation from a lagrangian as in Ryder “Quantum Field Theory” in which lagrangian methods are used in various ways. They are the standard model methods, highly developed, but very abstract. In dynamics (Marion and Thornton) the covariant derivative of any vector F in three dimensions is:

DF / dt (XYZ frame) = (dF / dt + omega x F) (123 frame)

This equation leads directly to the Euler equations. It can be generalized to the covariant derivative of Cartan geometry for an vector V sup a in any dimension so the angular velocity omega bold is related to the Cartan spin connection.

To: EMyrone@aol.com
Sent: 28/01/2017 15:53:48 GMT Standard Time
Subj: Re: Continuing with UFT369

This is very interesting, perhaps this opens a new method for solving eclectromagnetic problems by the Lagrange method.

Horst

Am 28.01.2017 um 10:37 schrieb EMyrone:

The next steps will be to show that the Euler equations can be derived from Cartan geometry using well defined spin connections. So in unified field theory there are also Euler equations of electromagnetism for example.

Daily Report 27/1/17

Sunday, January 29th, 2017

The equivalent of 55,578 printed pages was downloaded (292.639 megabytes) from 1787 downloaded memory files (hits) and 446 distinct visits each averaging 3.2 memory pages and 6 minutes, printed pages to hits ratio of 31.10, main spiders Google, MSN and Yahoo, top ten referrring domains 2,204,979 w 1447. Collected ECE2 1047, Top ten items 992, Evans / Morris 891, Collected scientometrics 570, Barddoniaeth 406, Principles of ECE 257, F3(Sp) 173, Collected Eckardt / Lindstrom 135, Autobiography volumes one and two 131, PECE 129, Collected Proofs 117, Evans Equations 109, UFT88 93, ECE2 77, Engineering Model 57, CEFE 35, Self charging inverter 35, PLENR 32, Llais 28, UFT321 28, UFT311 27, UFT313 25, UFT314 21, UFT315 16, UFT316 16, UFT317 22, UFT318 14, UFT319 27, UFT320 17, UFT322 24, UFT323 23, UFT324 19, UFT325 20, UFT326 20, UFT327 16, UFT328 25, UFT329 20, UFT330 15, UFT331 16, UFT332 19, UFT333 17, UFT334 15, UFT335 20, UFT336 17, UFT337 11, UFT338 16, UFT339 9, UFT340 13, UFT341 14, UFT342 12, UFT343 22, UFT344 20, UFT345 16, UFT346 16, UFT347 18, UFT348 17, UFT349 19, UFT351 17, UFT352 27, UFT353 11, UFT354 35, UFT355 31, UFT356 26, UFT357 13, UFT358 17, UFT359 14, UFT360 9, UFT361 9, UFT362 13, UFT363 20, UFT364 27, UFT365 23, UFT366 77, UFT367 30, UFT368 1 to date in January 2017. City of Winnipeg UFT papers, Deusu search engine Poetry, Informatics University of Kiel UFT papers extensive; Mathematics University of California Berkeley UFT87; University of Warwick Autobiography volume two, AIAS staff. Intense interest all sectors, updated usage file attached for January 2017.

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This server could not verify that you are authorized to access the document requested. Either you supplied the wrong credentials (e.g., bad password), or your browser doesn’t understand how to supply the credentials required.

Additionally, a 401 Unauthorized error was encountered while trying to use an ErrorDocument to handle the request.

Discussion of 369(1): Complete Analytical Mechanics of the Gyroscope

Saturday, January 28th, 2017

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 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.

UFT368 Sections 1 and 2, and Background Notes Posted

Saturday, January 28th, 2017

Many thanks again, much appreciated as ever! UFT368 will now appear off www.aias.us and www.upitec.org in the reading rooms of the copyright libraries in Britain and Ireland (www.webarchive.org.uk) and also on the Wayback Machine (www.archive.org). So it is permanently archived for conservation and the new generation. It is of course permanently accessible to anyone with a computer

To: EMyrone@aol.com
Sent: 28/01/2017 05:33:32 GMT Standard Time
Subj: Re: FOR POSTING: UFT368 Sections 1 and 2, and Background Notes

Posted today

Dave

On 1/26/2017 3:27 AM, EMyrone wrote:

This is UFT368 Sections 1 and 2 on the analytical mechanics of the gyroscope. The Laithwaite condition is defined in equation (41).

Daily Report 26/1/17

Saturday, January 28th, 2017

The equivalent of 72,511 printed pages was downloaded, (264.374 megabytes) from 2010 downloaded memory files (hits) and 475 distinct visits, each averaging 2.9 memory pages and 6 minutes, printed pages to hits ratio of 36.07, top ten referrals total 2,203,869, w 1447, main spides Google, MSN and Yahoo. Collected ECE2 1013, Top ten 969, Evans / Morris 858(est), Collected scientometrics 535(est), Barddoniaeth 398, Principles of ECE 251, F3(Sp) 171, Autobiography volumes one and two 130, Eckardt / Lindstrom papers 130, PECE 127, Collected proofs 109, Evans Equations 106, UFT88 89, ECE2 77, Engineering Model 53, CEFE 45, Self charging inverter 34, PLENR 31, Llais 28, UFT321 28, UFT311 26, UFT313 24, UFT314 20, UFT315 15, UFT316 16, UFT317 20, UFT318 14, UFT319 27, UFT320 17, UFT322 24, UFT323 23, UFT324 18, UFT325 20, UFT326 20, UFT327 15, UFT328 22, UFT329 20, UFT330 14, UFT331 16, UFT332 17, UFT333 17, UFT334 14, UFT335 19, UFT336 16, UFT337 11, UFT338 16, UFT339 9, UFT340 13, UFT341 14, UFT342 11, UFT343 21, UFT344 19, UFT345 16, UFT346 15, UFT347 17, UFT348 16, UFT349 19, UFT351 17, UFT352 26, UFT353 11, UFT354 32, UFT355 30, UFT356 24, UFT357 12, UFT358 17, UFT359 14, UFT360 9, UFT361 9, UFT362 13, UFT363 20, UFT364 27, UFT365 22, UFT366 73, UFT367 28 to date in January 2017. City of Winnipeg UFT Section; University of Quebec Trois Rivieres UFT367; Eset Corporation Home Page; Deusu search engine photographs, filtered statistics; Leibniz University Hannover UFT26, UFT260; Informatics University of Kiel UFT Section extensive; Physics University of Wuerzburg UFT238b; University of New Hampshire Foundations of Physics leaflet; Complutense University Madrid Spanish Section; Llyfrgell Genedlaethol Cymru AIAS Staff. Intense interest all sectors, updated usage file attached for January 2017.

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