Quote from: aero on 07/02/2015 12:36 AMThe higher density views are uploaded.https://drive.google.com/folderview?id=0B1XizxEfB23tfkVzeXVub2NpTm5fanZTTTdrLXNiT3VHaV9FYnB6TVpDUmJsWjRQbEUwdE0&usp=sharingThese Meep runs were made at resolution = 250 which is 2.5 times higher than previously uploaded views. These are the 14 final time slices of a 32 cycle run. 14 images for each view, separated by 0.1 cycle of the drive frequency 2.45 GHz. The Gaussian noise bandwidth of 2.45 GHz * .025 was used which seems reasonable for a magnetron. The 58 mm dipole antenna was located parallel to and 1/4 wavelength from the small end plate, excited with the Ez field component.Included are two models of the 10.2 inch NSF-1701 cavity, one using copper and the other using Perfect metal.I expect questions.Looking forward to seeing somebody make movies from these higher density runs !

The higher density views are uploaded.https://drive.google.com/folderview?id=0B1XizxEfB23tfkVzeXVub2NpTm5fanZTTTdrLXNiT3VHaV9FYnB6TVpDUmJsWjRQbEUwdE0&usp=sharingThese Meep runs were made at resolution = 250 which is 2.5 times higher than previously uploaded views. These are the 14 final time slices of a 32 cycle run. 14 images for each view, separated by 0.1 cycle of the drive frequency 2.45 GHz. The Gaussian noise bandwidth of 2.45 GHz * .025 was used which seems reasonable for a magnetron. The 58 mm dipole antenna was located parallel to and 1/4 wavelength from the small end plate, excited with the Ez field component.Included are two models of the 10.2 inch NSF-1701 cavity, one using copper and the other using Perfect metal.I expect questions.

Quote from: zellerium on 07/01/2015 12:47 AMHow hot can a magnetron get before it is 'damaged'? I imagine all internal parts are expanding, I wonder how long it takes for parts to plastically deform.I don't know how magnetrons are constructed. They can probably withstand a temperature above 100 C without getting damaged

How hot can a magnetron get before it is 'damaged'? I imagine all internal parts are expanding, I wonder how long it takes for parts to plastically deform.

Quote from: Rodal on 07/02/2015 12:16 AMQuote from: WarpTech on 07/02/2015 12:11 AMQuote from: aero on 07/01/2015 08:33 PM...Yes, a very big help from zxcvb, Thank you again, and welcome!Aero, what is the output power of the antenna in the simulation? The numbers are much larger than they were, but they are still much, much lower than I would expect under a resonant condition. ToddTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.Sorry, I must've missed that detail. How do I scale it to V/m and A/m then? I'm looking for some peak and average values so I can estimate the copper losses.Todd

Quote from: WarpTech on 07/02/2015 12:11 AMQuote from: aero on 07/01/2015 08:33 PM...Yes, a very big help from zxcvb, Thank you again, and welcome!Aero, what is the output power of the antenna in the simulation? The numbers are much larger than they were, but they are still much, much lower than I would expect under a resonant condition. ToddTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.

Quote from: aero on 07/01/2015 08:33 PM...Yes, a very big help from zxcvb, Thank you again, and welcome!Aero, what is the output power of the antenna in the simulation? The numbers are much larger than they were, but they are still much, much lower than I would expect under a resonant condition. Todd

...Yes, a very big help from zxcvb, Thank you again, and welcome!

Quote from: WarpTech on 07/02/2015 01:12 AMQuote from: Rodal on 07/02/2015 12:16 AMQuote from: WarpTech on 07/02/2015 12:11 AMQuote from: aero on 07/01/2015 08:33 PM...Yes, a very big help from zxcvb, Thank you again, and welcome!Aero, what is the output power of the antenna in the simulation? The numbers are much larger than they were, but they are still much, much lower than I would expect under a resonant condition. ToddTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.Sorry, I must've missed that detail. How do I scale it to V/m and A/m then? I'm looking for some peak and average values so I can estimate the copper losses.ToddHere: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot.

Quote from: WarpTech on 07/02/2015 01:12 AMQuote from: Rodal on 07/02/2015 12:16 AMTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.Sorry, I must've missed that detail. How do I scale it to V/m and A/m then? I'm looking for some peak and average values so I can estimate the copper losses.ToddHere: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot. And by comparing the data from the two runs, you may have a shot. The Perfect metal is lossless as I understand it. Everything that I set in the control file were identical so, except for adjustments that Meep may make internally (I don't know about anything in particular) the runs data should be comparable.

Quote from: Rodal on 07/02/2015 12:16 AMTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.Sorry, I must've missed that detail. How do I scale it to V/m and A/m then? I'm looking for some peak and average values so I can estimate the copper losses.Todd

Todd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.

Quote from: Rodal on 07/02/2015 12:57 AMQuote from: aero on 07/02/2015 12:36 AMThe higher density views are uploaded.https://drive.google.com/folderview?id=0B1XizxEfB23tfkVzeXVub2NpTm5fanZTTTdrLXNiT3VHaV9FYnB6TVpDUmJsWjRQbEUwdE0&usp=sharingThese Meep runs were made at resolution = 250 which is 2.5 times higher than previously uploaded views. These are the 14 final time slices of a 32 cycle run. 14 images for each view, separated by 0.1 cycle of the drive frequency 2.45 GHz. The Gaussian noise bandwidth of 2.45 GHz * .025 was used which seems reasonable for a magnetron. The 58 mm dipole antenna was located parallel to and 1/4 wavelength from the small end plate, excited with the Ez field component.Included are two models of the 10.2 inch NSF-1701 cavity, one using copper and the other using Perfect metal.I expect questions.Looking forward to seeing somebody make movies from these higher density runs !On it.

Quote from: aero on 07/02/2015 01:49 AMQuote from: WarpTech on 07/02/2015 01:12 AMQuote from: Rodal on 07/02/2015 12:16 AMQuote from: WarpTech on 07/02/2015 12:11 AMQuote from: aero on 07/01/2015 08:33 PM...Yes, a very big help from zxcvb, Thank you again, and welcome!Aero, what is the output power of the antenna in the simulation? The numbers are much larger than they were, but they are still much, much lower than I would expect under a resonant condition. ToddTodd, are you taking into account that these fields are in Meep dimensionless units ( http://ab-initio.mit.edu/wiki/index.php/Meep_Introduction#Units_in_Meep )? (not normal SI dimensional units)Also, the value of the field at the antenna is higher than shown, I had to clip the values of the antenna so that the electric fields can be seen.Sorry, I must've missed that detail. How do I scale it to V/m and A/m then? I'm looking for some peak and average values so I can estimate the copper losses.ToddHere: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot.So the electric field has to be multiplied by to get Volts/meterand H is already in Amps/meter (H does not need any transformation)where does the scale factor, a=0.3, enter the picture?Into the denominator , which is in meters?So really to get the Electric field one has to multiply by 3767.3/3 = 1255.77 Volts/meter and the Magnetic field H has to be multiplied by 1/0.3 = 3.33333 Amps/meter

...I think aero adjusted the time, the distances and the frequency with the scale factor (0.3). One therefore needs to be very careful with conversions to SI.

Be careful about reading health books. You may die of a misprint.

Quote from: deuteragenie on 07/02/2015 08:43 AM...I think aero adjusted the time, the distances and the frequency with the scale factor (0.3). One therefore needs to be very careful with conversions to SI.Quote from: Mark TwainBe careful about reading health books. You may die of a misprint.Taking everything into account, the transformation factor for the Electric Field becomes so that, as remarked by Todd, what one needs to know is whether aero used Io=1 or some other value. The value of a is automatically taken into account in the denominator of Ditto for the magnetizing field H, where the transformation factor is : We need aero to disclose what value he used for Io.

All mushrooms can be eaten; some of them will kill you.

Quote from: aero on 07/02/2015 12:36 AMThe higher density views are uploaded.https://drive.google.com/folderview?id=0B1XizxEfB23tfkVzeXVub2NpTm5fanZTTTdrLXNiT3VHaV9FYnB6TVpDUmJsWjRQbEUwdE0&usp=sharingThese Meep runs were made at resolution = 250 which is 2.5 times higher than previously uploaded views. These are the 14 final time slices of a 32 cycle run. 14 images for each view, separated by 0.1 cycle of the drive frequency 2.45 GHz. The Gaussian noise bandwidth of 2.45 GHz * .025 was used which seems reasonable for a magnetron. The 58 mm dipole antenna was located parallel to and 1/4 wavelength from the small end plate, excited with the Ez field component.Included are two models of the 10.2 inch NSF-1701 cavity, one using copper and the other using Perfect metal.I expect questions.Trying to understand the terminology being used. In traditional Finite Difference terminology (von Neumann, Ricthmyer, Lax, Hildebrand, and other classic authors, etc.) the terms "finite difference gridpoint" or "finite difference meshpoint" and "finite difference grid" or "finite difference mesh" is used to characterize the geometrical points at which the Finite Difference scheme is implemented. Von Neumann, Lax, etc., did not use the terminology "pixel" or "resolution" to refer to the Finite Difference mesh.In some of the Meep literature I see graphic display terms being used like "pixel" instead of "grid point". And now the term "resolution" as in "display resolution".I would rather use the traditional, mathematical terms rather than these graphic terms, because using graphic terms is apt for confusion. Confusion between the graphic display post-processing of Finite Difference solutions and the Finite Difference mesh used to obtain the solution.For example, let's say that we have a Finite Difference mesh in the x y plane with 100 FD grid points in both the x and y directions, equally spaced.For graphic displays one has many alternatives to display the FD solution, for example:1) display the field variables at every FD grid point (giving 100x100 pixels)2) display the field variables at every other FD gridpoint (giving 50x50 pixels)3) use spline (or other form of interpolaton) to display variables in between FD gridpoints in addition to displaying field variables at the gridpoints (giving much more than >>100x100 pixels)which shows that one should distinguish between the graphic display pixels and resolution and the Finite Difference mesh, because they are not necessarily the same. And in my experience with codes they have not been the same.QUESTION: is what is being displayed in the Meep graphs obtained at every FD grid point, and that's why display terminology is used instead of FD classic terminology?

...Lattice (http://ab-initio.mit.edu/wiki/index.php/Meep_Reference#lattice) specifies the size of the computational cell, whereas resolution specifies the computational grid resolution, in pixels per distance unit. If the lattice is 100 x 100 and the resolution 10, it results in 1000 x 1000 pixel images.

Quote from: deuteragenie on 07/02/2015 02:49 PM...Lattice (http://ab-initio.mit.edu/wiki/index.php/Meep_Reference#lattice) specifies the size of the computational cell, whereas resolution specifies the computational grid resolution, in pixels per distance unit. If the lattice is 100 x 100 and the resolution 10, it results in 1000 x 1000 pixel images.But is pixel used only in its proper form:pixel = the smallest element of an image that can be individually processed in a video display system.Or is it being used in some of the Meep literature to mean "Finite Difference Grid Point (or Mesh point)" ?which would be the wrong use of the word pixel, since pixel is a graphic term that should not be confused with the FD gridpoint, since they are not necessarily the same

..From reading the manuals, it appears that it is used adequately id est to represent one graphical unit. Maybe what is causing confusion is that the lattice could be 1 x 1, and resolution 100, giving images 100 x 100 pixels, which depending on the distance of 1 lattice unit can represent 1 mm, 1 meter or 1 km.

For CYLINDRICAL simulations with |m| > 1, compute more accurate fields near the origin r = 0 at the expense of requiring a smaller Courant factor. Empirically, when this option is set to true, a Courant factor of roughly min[0.5,1 / ( | m | + 0.5)] (or smaller) seems to be needed. The default is false, in which case the Dr, Dz, and Br fields within |m| pixels of the origin are forced to zero, which usually ensures stability with the default Courant factor of 0.5, at the expense of slowing convergence of the fields near r = 0.

Quote from: Rodal on 07/02/2015 01:57 AMQuote from: aero on 07/02/2015 01:49 AMsnip ...Here: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot.So the electric field has to be multiplied by to get Volts/meterand H is already in Amps/meter (H does not need any transformation)where does the scale factor, a=0.3, enter the picture?Into the denominator , which is in meters?So really to get the Electric field one has to multiply by 3767.3/3 = 1255.77 Volts/meter and the Magnetic field H has to be multiplied by 1/0.3 = 3.33333 Amps/meter I think aero adjusted the time, the distances and the frequency with the scale factor (0.3). One therefore needs to be very careful with conversions to SI.

Quote from: aero on 07/02/2015 01:49 AMsnip ...Here: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot.So the electric field has to be multiplied by to get Volts/meterand H is already in Amps/meter (H does not need any transformation)where does the scale factor, a=0.3, enter the picture?Into the denominator , which is in meters?So really to get the Electric field one has to multiply by 3767.3/3 = 1255.77 Volts/meter and the Magnetic field H has to be multiplied by 1/0.3 = 3.33333 Amps/meter

snip ...Here: http://meepunits.wikia.com/wiki/Meep_unit_transformation_WikiWith that and knowing that my scale factor, a, is 0.3, take your best shot.