-
#720 by SeeShells on 25 Dec, 2015 04:48
-
#721 by mwvp on 25 Dec, 2015 10:06
-
Ho Ho Ho! Merry Christmas!
I'll share a gift with you, FWIW...
https://en.wikipedia.org/wiki/Fizeau_experiment
"...In 1895, Hendrik Lorentz predicted the existence of an extra term due to dispersion:[S 3]:15–20"
What's that about? The Fresnel Drag coefficient, corrected for dispersion - the last term gamma/n * dn/dgamma. Of course the velocity of light isn't changing according to Relativity, but the frequency and momentum.
Now although this applies to a dielectric, perhaps it just as well applies to a sort of geometric inversion of a dielectric; a hollow reflector - a frustrum?
The implication is that since for empty space (n=1) there's no doppler shift; an accelerating, dispersive frustrum will cause a doppler shift, but if you have a dielectric accelerating along with, and inside a frustrum, it will reduce the doppler shift. However, that may be offset if the dielectric increases the cavity Q, increases the loss at the base or reduces loss at the apex of the frustrum.
-
#722 by SeeShells on 25 Dec, 2015 12:18
-
The copper sheeting in on the inside on the outside are the composite materials.
Carbon Fiber reinforced Plastic (CFRP) have a much lower conductivity than full metal and therefore higher Ohmic losses. The resin matrix material also have a tang-delta. Carbon Fiber at the inner surfave of the cavity will decrease the Q!
Bottom 3.17mm plate is 2.5mm thick of Alumina Ceramic, the top plate is 10mm. Bonded .032" .80mm O2 Free Copper....
Thanks for the link. Coming into the discussion around the 5th of November, I missed a lot.
My comments on the warm ends and cool side wall was more about the cool walls and her ceramics. Not so much surprise... I think her frustum walls are 1/4" copper while the ends are thinner and bonded to ceramic disks. (Assuming the epoxy was JB Weld.., from the pic, but I think she just called it magic...paste?)
Secret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
Ihavehad a ceramic coffee cup that when heated in a microwave would get so hot you could not touch it by anything other than the handle, while another ceramic cup would just be warm to touch. Coffee in both at or above boiling... Coffee in the hot cup and the cup itself, would cool down very quickly.., while the other would remain hot longer.
If the ceramic was exposed to microwaves inside of the cavity they might get hotter but the ceramic plates are on the outside.
When Shell gets the frustum back in working order and starts generating data, it would be important to know just what the thermal conductivity of her ceramic end plates are, when evaluating thermal data.
http://www.azom.com/properties.aspx?ArticleID=52
do we know what particular ceramic material Shell used?
Aluminum nitride and silicon carbide thermal conductivity is not that bad (about 1/2 that of copper), they are used in applications that emit heat. Zirconia's thermal conductivity is significantly lower than metals (about 1/20 that of copper), hence used for kiln walls, for example.
Also what matters for the EM Drive is the thermal diffusivity, rather than the thermal conductivity alone, because of the transient nature of the problem, hence the mass density of the ceramic is also important.
See above link
I don't remember her mentioning specifics, but then I only really read through thread 5 rather quickly. Gave up on the shear volume of the past threads, so what I know from the past is sketchy and based on very limited searches.
From the pictures it does look maybe 3/4 to an inch thick???
The next frustum I'm using curved endplates and will be bonding two layers of carbon fiber to it ~ 6-10mm , First layer will be a highly thermally conductive mix and the second for strength. It's critical to keep the endplates from warping from the high energy modes.
I'll still be using the Quartz rod through the center, polished on both ends and mirrored on one for Laser Interferometer testing through the center of the Quartz rod so I can measure any time displacements.
Still will be using the tune chamber on the top but not as long and the one I have.
Merry Christmas ALL!
Shell
There are several people here that have more knowledge in doing a Laser Interferometer testing than me. I'll need to ask, could I only mirror one half of one end to measure a pulse going down and back and just through to see if the directional travel times differ?
I have a lot of experience with this point and many experimental data.
The first plot shows a conical metallic cavity where the small end plate is made of CFRP(large fiber density), the second the same resonator but with a copper plate at the small end. These are real measurement data, not FEM simulations.
The coupling is not optimal but is shows the difference. Resonant mode is TE011.
Shell -
#723 by OnlyMe on 25 Dec, 2015 13:48
-
...
Thanks for the link. Coming into the discussion around the 5th of November, I missed a lot.
My comments on the warm ends and cool side wall was more about the cool walls and her ceramics. Not so much surprise... I think her frustum walls are 1/4" copper while the ends are thinner and bonded to ceramic disks. (Assuming the epoxy was JB Weld.., from the pic, but I think she just called it magic...paste?)
Secret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
Ihavehad a ceramic coffee cup that when heated in a microwave would get so hot you could not touch it by anything other than the handle, while another ceramic cup would just be warm to touch. Coffee in both at or above boiling... Coffee in the hot cup and the cup itself, would cool down very quickly.., while the other would remain hot longer.
If the ceramic was exposed to microwaves inside of the cavity they might get hotter but the ceramic plates are on the outside.
When Shell gets the frustum back in working order and starts generating data, it would be important to know just what the thermal conductivity of her ceramic end plates are, when evaluating thermal data.
http://www.azom.com/properties.aspx?ArticleID=52
Shell,
On the first point I guess I have graduated from vivid imagination to hallucination. Where I came up with 1/4" is not even worth chasing down.
The issue with the two coffee cups was not that one heats up in the microwave, I never put an empty cup in the microwave.., it was that with the water for coffee being heated to microwave boiling, one cup was too hot to touch and both the cup and hot water cooled off pretty quickly. Too fast to enjoy the hot coffee... While the other cup only became warm to touch and the water/coffee in it, remained hot longer. One was a good heat conductor and the other not so much... Where not so much is better in the case of a coffee cup..., if you like hot coffee.
-
#724 by Rodal on 25 Dec, 2015 14:55
-
Dr. Rodal:
If both the small end plate and the large end plate of Shell's frustum of a cone are Alumina ceramic on the outside, this may (**), unfortunately, going to make it more difficult to immediately interpret the thermal camera data concerning what mode was really excited in Shell's truncated cone, due to the lower thermal diffusivity of Alumina (as warned by user OnlyMe ). A transient thermal analysis may be necessary to interpret the thermal camera data (**).
I'm pretty sure SeeShell's large end plate is also copper backed ceramic (see update #4 images on her gofundme page).
...
The thermal camera infrared will not be measuring the copper temperature but the lower, thermally diffused, temperature of the ceramic (*). If the Alumina was thick enough, two-dimensional diffusion across the surface of the Alumina may diffuse the actual temperature boundaries necessary to properly identify the mode shape.
What is the thickness of the Alumina exterior on the outside of the large end plate in Shell's test ?
PS: I understood fromQuoteBottom 3.17mm plate is 2.5mm thick of Alumina Ceramic, the top plate is 10mm. Bonded .032" .80mm O2 Free Copper.
that the small plate Alumina was 2.5 mm thick.
Was the top plate Alumina 10 mm thick ? and the top plate copper was 0.8 mm thick ?
If the Alumina on the top end plate was 10mm thick (or similarly 10mm - 0.8mm =9.2 mm thick), that is 4 (10mm/2.5mm=4, or 9.2/2.5=3.7) times thicker than the Alumina on the small plate.
Recall that the Fourier time (https://en.wikipedia.org/wiki/Fourier_number ) goes like the square of the thickness,
(where alpha is the thermal diffusivity)
so this means that for diffussion time purposes, the Alumina on the top plate effectively diffuses 4^2 = 16 times (or (9.2/2.5)^2= 14 times) more slowly than the Alumina on the small end plate (for the same Fourier number).
_________
(*)
The thermal conductivity of Alumina (Aluminum Oxide) ceramic ranges from 12 to 39 W/m K) (SI units) compared to copper's 385.0 W/(m K) so, it is about 1/10 to 1/30 that of copper.
Similarly, the thermal diffusivity of sintered Alumina is about 1/10 that of copper:
thermal diffusivity of sintered Alumina 0.111× 10^(−4) m^2/s
thermal diffusivity of copper 1.11 × 10^(−4) m^2/s
(**) I haven't run any numbers (except for these dimensional analysis calculations), so can't tell at the moment
. Have to wait for more data to be confirmed...
-
#725 by OnlyMe on 25 Dec, 2015 15:36
-
Dr. Rodal:
If both the small end plate and the large end plate of Shell's frustum of a cone are Alumina ceramic on the outside, this is unfortunately going to make it more difficult to immediately interpret the thermal camera data concerning what mode was really excited in her truncated cone, due to the lower thermal diffusivity of Alumina.
I'm pretty sure SeeShell's large end plate is also copper backed ceramic (see update #4 images on her gofundme page).
...
The thermal camera infrared will not be measuring the copper temperature but the lower, thermally diffused, temperature of the ceramic (*). If the Alumina was thick enough, two-dimensional diffusion across the surface of the Alumina may diffuse the actual temperature boundaries necessary to properly identify the mode shape.
What is the thickness of the Alumina exterior on the outside of the large end plate in Shell's test ?
PS: I understood fromQuoteBottom 3.17mm plate is 2.5mm thick of Alumina Ceramic, the top plate is 10mm. Bonded .032" .80mm O2 Free Copper.
that the small plate Alumina was 2.5 mm thick.
Was the top plate Alumina 10 mm thick ? and the top plate copper was 0.8 mm thick ?
If the Alumina on the top end plate was 10mm thick (or similarly 10mm - 0.8mm =9.2 mm thick), that is 5 times thicker than the Alumina on the small plate.
Recall that the Fourier time (https://en.wikipedia.org/wiki/Fourier_number ) goes like the square of the thickness,
(where alpha is the thermal diffusivity)
so this means that for diffussion time purpose, the Alumina on the top plate effectively diffuses 5^2 = 25 times (or (9.2/2)^2= 21 times) more slowly than the Alumina on the small end plate (for the same Fourier number).
_________
(*)
The thermal conductivity of Alumina (Aluminum Oxide) ceramic ranges from 12 to 39 W/m K) (SI units) compared to copper's 385.0 W/(m K) so, it is about 1/10 to 1/30 that of copper.
1.11 × 10^(−4) m^2/s
One of Shell's replies (below), to my imaginative guesswork was, - bold text added by myself for emphasis.QuoteSecret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
I thought there was some reference in Thread 4, that set Shells frustum wall thickness at 1/4 inch O2 free copper? But at this point I am not sure that I was not dreaming that! -
#726 by Rodal on 25 Dec, 2015 15:40
-
...
Are the side copper walls of Shell's experiment with a frustum of a cone just bare copper with no ceramic on the outside ?
One of Shell's replies (below), to my imaginative guesswork was, - bold text added by myself for emphasis.QuoteSecret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
While the end plates have Alumina ceramic on the outside, with the large plate Alumina being about 4 times thicker than the Alumina on the small end plate? -
#727 by OnlyMe on 25 Dec, 2015 15:49
-
...
Are the side copper walls of Shell's experiment with a frustum of a cone just bare copper with no ceramic on the outside ?
One of Shell's replies (below), to my imaginative guesswork was, - bold text added by myself for emphasis.QuoteSecret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
While the end plates have Alumina ceramic on the outside, with the large plate Alumina being about 5 times thicker than the Alumina on the small end plate?
Yes, only the endplates are ceramic backed. I had assumed that was because they were thin and the frustum wall 1/4 inch, but again I am beginning to get confused... I also thought the ceramic at both endplates was the same thickness???
Even where the endplates are ceramic backed.., to avoid buckling and being thermally deformed, the fact that her magnetron is removed from the frustum, I would have expected that though thermally detecting the exact mode might be difficult, addressing external thermal convection should be easier.
I have always believed it is more important to start out trying to recreate the claimed thrust.., above thermal and systemic noise, before spending too much time, effort and money, trying to figure out just where that thrust comes from. IOW better to catch the rabbit, before trying to cook it. -
#728 by Rodal on 25 Dec, 2015 16:06
-
...
Are the side copper walls of Shell's experiment with a frustum of a cone just bare copper with no ceramic on the outside ?
One of Shell's replies (below), to my imaginative guesswork was, - bold text added by myself for emphasis.QuoteSecret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
While the end plates have Alumina ceramic on the outside, with the large plate Alumina being about 5 times thicker than the Alumina on the small end plate?
Yes, only the endplates are ceramic backed. I had assumed that was because they were thin and the frustum wall 1/4 inch, but again I am beginning to get confused... I also thought the ceramic at both endplates was the same thickness???
Even where the endplates are ceramic backed.., to avoid buckling and being thermally deformed, the fact that her magnetron is removed from the frustum, I would have expected that though thermally detecting the exact mode might be difficult, addressing external thermal convection should be easier.
...
Once these tests are confirmed and verified in their present form, the experiments could be re-run in the future, for further verification and comparison with 1) the whole EM Drive externally insulated (including insulation of the lateral conical walls) and/or 2) heating the EM drive to a similar extent but without feeding microwaves into the cavity, as has been proposed elsewhere. -
#729 by OnlyMe on 25 Dec, 2015 16:18
-
...
Are the side copper walls of Shell's experiment with a frustum of a cone just bare copper with no ceramic on the outside ?
One of Shell's replies (below), to my imaginative guesswork was, - bold text added by myself for emphasis.QuoteSecret Squirrel Sauce. Frustum walls and sides are .032" O2 free copper sheeting water jet cut.
I used JB Weld thinned the paste with Acetone, mounted the ceramic plate onto a old 78 record player platter, poured the thinned mixture onto the center while spinning. It spun out assuring a very thin uniform coating of JB Weld on the ceramic plate. Laid the .032" copper onto it then sandwiched it with a heavy uniform weight. Left it to bond for 48 hours ~74F.
While the end plates have Alumina ceramic on the outside, with the large plate Alumina being about 5 times thicker than the Alumina on the small end plate?
Yes, only the endplates are ceramic backed. I had assumed that was because they were thin and the frustum wall 1/4 inch, but again I am beginning to get confused... I also thought the ceramic at both endplates was the same thickness???
Even where the endplates are ceramic backed.., to avoid buckling and being thermally deformed, the fact that her magnetron is removed from the frustum, I would have expected that though thermally detecting the exact mode might be difficult, addressing external thermal convection should be easier.
...
The experiments could be re-run in the future, for further verification and comparison with 1) the whole EM Drive externally insulated (including insulation of the lateral conical walls) and/or 2) heating the EM drive to a similar extent but without feeding microwaves into the cavity, as has been proposed elsewhere.
#1 was my intent in suggesting the castable forge material as a casing for the whole frustum. The casting material would act as a thermal barrier, possibly even resulting a more uniform 3D thermal radiation when fully heated. It would concentrate heat in the metal frustum, but being ridged might counter any associated thermal distortion... Before even considering that kind of set up it would be best to see what real heat Shell's data turns up, when we start getting data. Too much don't insulate.
#2 I think should be doable with Shell's design because once you know how much heat is generated when it is in operation, it would be easy to set up a matching heat source — no microwaves. Again her magnetron is out of the thermal picture of the frustum.
-
#730 by aero on 25 Dec, 2015 17:38
-
Regarding the 1/4 inch thick copper. Only place I know that uses 1/4 inch thick is my (and perhaps other's) meep model. It is modelled as being that thick solely for computational convenience and has no other relationship to anyone's actual hardware, that I know of.
aero -
#731 by OnlyMe on 25 Dec, 2015 17:43
-
Regarding the 1/4 inch thick copper. Only place I know that uses 1/4 inch thick is my (and perhaps other's) meep model. It is modelled as being that thick solely for computational convenience and has no other relationship to anyone's actual hardware, that I know of.
aero
Well, like I said I apparently have a vivid imagination....
Off now for a few last minutes Santa drops, out to warm up the sled and wake up the rain deer.
Merry Christmas, to all. -
#732 by ThereIWas3 on 25 Dec, 2015 17:46
-
Everyone recalls from your microwave oven Owner's Manual (you read that, right? ) that you should not operate the oven with it empty. If the RF energy is not absorbed in heating the food, it reflects back to the magnetron, which can then overheat.
An EmDrive looks to me a lot like an empty microwave oven. Where is all the energy going? Clearly some goes into heating the walls of the frustrum, and we are going to meausre that. If there is a force generated, some energy has to go into that, and exactly how much that is will be interesting.
But some is being reflected back. A measurement of VSWR could let us calculate that, so can figure how much net energy goes into the frustrum to do something there. -
#733 by RFPlumber on 25 Dec, 2015 19:29
-
Everyone recalls from your microwave oven Owner's Manual (you read that, right?
) that you should not operate the oven with it empty. If the RF energy is not absorbed in heating the food, it reflects back to the magnetron, which can then overheat.
An EmDrive looks to me a lot like an empty microwave oven. Where is all the energy going? Clearly some goes into heating the walls of the frustrum, and we are going to meausre that. If there is a force generated, some energy has to go into that, and exactly how much that is will be interesting.
But some is being reflected back. A measurement of VSWR could let us calculate that, so can figure how much net energy goes into the frustrum to do something there.
Microwave oven is likely not a (well) matched cavity. EmDrive should be. Then nearly all of the RF power from magnetron will be absorbed and dissipated by cavity walls. If EmDrive cavity is not properly matched then any reference to magnetron power is basically useless (as a large % of that power could be reflecting back and just overheating the magnetron itself).
EDIT
Actually I am not sure if it will be "nearly all of the RF power" absorbed or only "nearly all of the _50%_ of the RF power" absorbed (the other 50% absorbed by the magnetron). Because the best power ratio one can hope to transfer to a critically coupled cavity is 1/2... A 800W rated magnetron - is it 800W of RF power delivered to a particular load or 800W RF power produced or 800W AC power consumed (with 70% conversion efficiency to RF)? -
#734 by Rodal on 25 Dec, 2015 19:56
-
Everyone recalls from your microwave oven Owner's Manual (you read that, right?
) that you should not operate the oven with it empty. If the RF energy is not absorbed in heating the food, it reflects back to the magnetron, which can then overheat.
An EmDrive looks to me a lot like an empty microwave oven. Where is all the energy going? Clearly some goes into heating the walls of the frustrum, and we are going to meausre that. If there is a force generated, some energy has to go into that, and exactly how much that is will be interesting.
But some is being reflected back. A measurement of VSWR could let us calculate that, so can figure how much net energy goes into the frustrum to do something there.
And the "figure of merit" for EM Drives has been proposed by original proponents as the measured force divided by the InputPower. This has generated controversy concerning the consequences for conservation of energy if self-acceleration of the EM Drive would be proportional to the InputPower (as per the now classic discussions on Conservation of Energy by Frobnicat, DeltaMass and WarpTech, in these threads).
But as you point out, comparisons made with this "figure of merit" also raise questions as to how much of that power went into heating the walls (and convecting away as natural convection) and back to the RF source (the coupling is never perfect), and therefore how valid are comparisons made on the basis of the measured force divided by the InputPower, since for most experiments the amount of power dissipated into heat is not reported and VSWR measurements are not always made available. -
#735 by ThereIWas3 on 25 Dec, 2015 20:07
-
The classic way to measure actual energy into a device is to immerse it in water and measure the temperature rise. Luckily, SeeShells' coax-fed design permits this, beacause all the electronic devices are some feet away, and the frustrum is airtight.
Using a thermal camera is a good first approximation, but the more precise method may be required eventually. I do not see a problem with meauring the energy this way, as long as you are not trying to collect force data on the same run. -
#736 by aero on 25 Dec, 2015 20:10
-
The waveguides into the frustum are located vertically into the large end 180 from each other. The antennas were 1/4 WL simple copper wire located in the waveguides.
When you say "vertically", is that with the frustrum lying on its side, so the feedlines come through the side walls at right angles to the wall? And how far from the large end? Or are they entering through the large end? I which case, how far from the centerline?
The RF is phased to match in the center of the frustum ~3wl.
What the waveguides see is the frustum as a "load dump" as the Q increases past and through it's peak to collapse, the frustum acts as a RF load dump or also called simply a load. By creating a slightly mismatched phasing in the waveguides I can force a dump at a defined time into the frustum. Then letting the shape of the frustum do what it does best and related to its shape. That's to create a pulsed push or force or call it what you want, in plain talk it allows the frustum shape to do what it does best in dumping it's high energy in a focused way.
Lasers do the same thing but in this case I can force it, I can tune it to happen just when it needs to.
Make sense?
Shell
added: In other words I'm controlling the decays of the high energy Q instead of single waveguides or antennas into the frustum which can cause rotational modes even through one cycle.
Or I can set the phasing between the two anywhere and also tune the cavity lengths. Makes for a lot of testing but I think the data we'll gain is worth it.
This sentence: "The RF is phased to match in the center of the frustum ~3wl." has confused me from day one.
My question: If the phase is matched at the sources and the two sources are equidistant from the center, then isn't the RF phase matched at the center? And isn't that what Shell wants?
Does the splitter on the coax feed introduce a phase mismatch? -
#737 by RFPlumber on 25 Dec, 2015 20:14
-
Progress update on my pendulum-based test. While I cannot yet report on whether EmDrive is producing thrust or not (congrats to Shell!), I can at least report that a 50 Ohm dummy RF load is definitely not producing any
. This is a good solid start; rumor has it that the only thing one needs to change now in order to obtain thrust is to use a frustum-shaped cavity instead of a dummy load… We shall see.
...
That is an impressive build! Could you post the CSVs you've generated so far, if you've saved them? It would be nice to have a little program read in a run and say "this data set shows non-zero thrust with probability X" before there is any actual frustum data to overfit to, and I'd like to take a stab at that.
Thank you. CSVs for 2 test runs attached...
Btw, both this request and some other feedback (by tellmeagain and others) made me take another look at these null test runs and, specifically, on the RF on period. A very unscientific and absolutely statistically insignificant examination of individual mid-point values both inside and outside the range suggests there is indeed a null force present, likely on the order of 20..40 uN. Collecting more runs, allowing for a longer RF on period and then applying some formal statistics to it will very likely show this force. Those 10A currents do indeed come with a cost. Still, I hope that if any thrust is ever observed on this build, it will be on the order of at least 100+ uN, and so it would be visible without resorting to statistical methods, just like the force from the electrostatic plates is.
-
#738 by ThereIWas3 on 25 Dec, 2015 20:59
-
My question: If the phase is matched at the sources and the two sources are equidistant from the center, then isn't the RF phase matched at the center?
My meep modelling has been assuming that the signals arriving at the frustrum antennas are exactly in phase with each other and therefore I do not bother modelling anything before that (toward the magnetron). Equal phase splitters are certainly possible and as long as the pieces of coax are cut to the same length it should not be a problem. -
#739 by ThereIWas3 on 25 Dec, 2015 22:25
-
I was listening to an episode of the NPR radio program "Science Friday" and the subject was "Failure in Science". One of the participants was a physics teacher. She has this written on the whiteboard in her classroom:Quote
Failure is not an Option. It is a Requirement.
The panel discussed things like the Michelson-Morely Experiment which, though not proving what it set out to prove, by that very failure set the stage for the concepts of Relativity and Quantum Mechanics. Every failure in science moves us closer to the truth, provided there is good data.
. Have to wait for more data to be confirmed...