Really simple
Quote from: Monomorphic on 11/10/2016 02:11 pmQuote from: Peter Lauwer on 11/10/2016 01:43 pmThe TM modes seem to have the highest Q for this cavity. Does that make sense? The loop is having its field in the axial direction of the cavity.Thanks,PeterThat is because you're using side wall injection. You are going to have a very difficult time (just like NASA did) trying to excite the TE01x modes. For example, I was able to locate TE012 at ~4.407Ghz, but it is VERY weak with the antenna mounted to the side and is very close to a TM mode. If you move the loop to the center things clear up greatly. Also, FEKO says your antenna wire diameter of 2mm is too much. The ratio of wire radius and length are not ideal. You may need to switch to a thinner wire.Thanks a lot for the advice and the sim, Monomorphic. I will make a new cavity with 1 mm wire. But, uhh..., how about the supply wires if you move the loop to the center? In a sim it is easy to do, but you will have strong influence of the wires from the side or endplate (either coax or two copper wires).How about placing the loop on one of the endplates, halfway the center and edge, and opening (B-field) facing the center? The loop is not placed at a maximum then, that's true. I will try it out.Peter
Quote from: Peter Lauwer on 11/10/2016 01:43 pmThe TM modes seem to have the highest Q for this cavity. Does that make sense? The loop is having its field in the axial direction of the cavity.Thanks,PeterThat is because you're using side wall injection. You are going to have a very difficult time (just like NASA did) trying to excite the TE01x modes. For example, I was able to locate TE012 at ~4.407Ghz, but it is VERY weak with the antenna mounted to the side and is very close to a TM mode. If you move the loop to the center things clear up greatly. Also, FEKO says your antenna wire diameter of 2mm is too much. The ratio of wire radius and length are not ideal. You may need to switch to a thinner wire.
The TM modes seem to have the highest Q for this cavity. Does that make sense? The loop is having its field in the axial direction of the cavity.Thanks,Peter
What is the impulse response of your antenna? In other words, if you apply a very short current pulse (ns), what does the ringing feedback signal look like? Is it a nice uniform sine wave at the resonant frequency of the cavity, or is there harmonic distortion from overlapping modes? Just curious. I was thinking about doing exactly what you have done, just to see what comes out and how I could design more powerful, ruggedized driving circuit to power it. I'm looking for oscillograms, before I start making some of my own.
Quote from: zellerium on 11/09/2016 05:51 pmQuote from: Peter Lauwer on 11/09/2016 04:35 pmMy first test cavity (cylinder), measured with a network analyzer (Agilent E8364B). Length = 130 mm, r = 49 mm. Coupling loop, diameter 15 mm, at 1/4 of length (32.5 mm), 2 mm silvered copper wire.Aim was to see what Q can be reached with untreated (unpolished) copper pipe and plates and this coupling (and to get some experience with copper soldering). And whether I can identify the modes (not yet).In the picture 2-3 GHz, 6 peaks can be seen:2.013 GHz Q= low2.136 GHz Q=~6002.358 GHz Q=~23k2.381 GHz Q=low2.642 GHz Q= not calculated (order 100-1000)2.917 GHz Q= not calculated (order 100-1000)Cheers,PeterI ran a quick HFSS eigenmode sim on it without the loop and the resonant frequency seems to correlate very well. I'm not sure what mode you had ~ 2GHz, the TE111 should be the first one. Maybe an effect of the antenna. Hope this helpsKurt[edit: each field plot is normalized to the same scale]Hi Kurt,Can you please also run a simulation 3-4 GHz? It will be interesting to identify more modes. Is there a TE011?3.249 GHz TM012? 3.381 GHz 3.747 GHz TM110? Q=12k3.938 GHz TM111? Q=20kThe TM modes seem to have the highest Q for this cavity. Does that make sense? The loop is having its field in the axial direction of the cavity.Thanks,Peter
Quote from: Peter Lauwer on 11/09/2016 04:35 pmMy first test cavity (cylinder), measured with a network analyzer (Agilent E8364B). Length = 130 mm, r = 49 mm. Coupling loop, diameter 15 mm, at 1/4 of length (32.5 mm), 2 mm silvered copper wire.Aim was to see what Q can be reached with untreated (unpolished) copper pipe and plates and this coupling (and to get some experience with copper soldering). And whether I can identify the modes (not yet).In the picture 2-3 GHz, 6 peaks can be seen:2.013 GHz Q= low2.136 GHz Q=~6002.358 GHz Q=~23k2.381 GHz Q=low2.642 GHz Q= not calculated (order 100-1000)2.917 GHz Q= not calculated (order 100-1000)Cheers,PeterI ran a quick HFSS eigenmode sim on it without the loop and the resonant frequency seems to correlate very well. I'm not sure what mode you had ~ 2GHz, the TE111 should be the first one. Maybe an effect of the antenna. Hope this helpsKurt[edit: each field plot is normalized to the same scale]
My first test cavity (cylinder), measured with a network analyzer (Agilent E8364B). Length = 130 mm, r = 49 mm. Coupling loop, diameter 15 mm, at 1/4 of length (32.5 mm), 2 mm silvered copper wire.Aim was to see what Q can be reached with untreated (unpolished) copper pipe and plates and this coupling (and to get some experience with copper soldering). And whether I can identify the modes (not yet).In the picture 2-3 GHz, 6 peaks can be seen:2.013 GHz Q= low2.136 GHz Q=~6002.358 GHz Q=~23k2.381 GHz Q=low2.642 GHz Q= not calculated (order 100-1000)2.917 GHz Q= not calculated (order 100-1000)Cheers,Peter
Quote from: Gs on 11/09/2016 02:17 amReally simpleNo, net momentum is conserved at each collision. It's not 'lost' as heat. It's always balanced. Sorry.
Quote from: Bob012345 on 11/10/2016 04:58 pmQuote from: Gs on 11/09/2016 02:17 amReally simpleNo, net momentum is conserved at each collision. It's not 'lost' as heat. It's always balanced. Sorry. I was unable to open the PDF so I am unsure just what you were responding to. That said on the face of it your statement is false, because heat is a measure of the (more or less randomized) momentum of a group of involved atoms and molecules.
Quote from: OnlyMe on 11/10/2016 10:03 pmQuote from: Bob012345 on 11/10/2016 04:58 pmQuote from: Gs on 11/09/2016 02:17 amReally simpleNo, net momentum is conserved at each collision. It's not 'lost' as heat. It's always balanced. Sorry. I was unable to open the PDF so I am unsure just what you were responding to. That said on the face of it your statement is false, because heat is a measure of the (more or less randomized) momentum of a group of involved atoms and molecules.Wrong. Heat is a randomized motion with an average momentum of zero. Heat does not carry, nor transfer momentum.
Strike a metal plate several times with a hammer and you will find that both the metal plate and hammer heat up. Since the only interaction between the two is the impact transferring momentum, a portion of the kinetic energy (momentum) is converted to thermal energy.
Quote from: Peter Lauwer on 11/10/2016 02:42 pmQuote from: Monomorphic on 11/10/2016 02:11 pmQuote from: Peter Lauwer on 11/10/2016 01:43 pmThe TM modes seem to have the highest Q for this cavity. Does that make sense? The loop is having its field in the axial direction of the cavity.Thanks,PeterThat is because you're using side wall injection. You are going to have a very difficult time (just like NASA did) trying to excite the TE01x modes. For example, I was able to locate TE012 at ~4.407Ghz, but it is VERY weak with the antenna mounted to the side and is very close to a TM mode. If you move the loop to the center things clear up greatly. Also, FEKO says your antenna wire diameter of 2mm is too much. The ratio of wire radius and length are not ideal. You may need to switch to a thinner wire.Thanks a lot for the advice and the sim, Monomorphic. I will make a new cavity with 1 mm wire. But, uhh..., how about the supply wires if you move the loop to the center? In a sim it is easy to do, but you will have strong influence of the wires from the side or endplate (either coax or two copper wires).How about placing the loop on one of the endplates, halfway the center and edge, and opening (B-field) facing the center? The loop is not placed at a maximum then, that's true. I will try it out.PeterAdded some sketch to the image.
I thought the circumference of a loop antenna should ideally be equal to wavelength.
Quote from: OnlyMe on 11/11/2016 02:08 amStrike a metal plate several times with a hammer and you will find that both the metal plate and hammer heat up. Since the only interaction between the two is the impact transferring momentum, a portion of the kinetic energy (momentum) is converted to thermal energy. What do you means by Kinetic energy (momentum) ?It sounds to me like writing acceleration(mass)Kinetic energy and momentum are very different.Kinetic energy is easily converted into thermal energy by any impact.Momentum is never in GR. The General relativity is constructed by this rule (and some others) Momentum is conserved, even in perfect inelastic collision.Momentum converted into heat needs totally new physics.
To answer the question you began with, "What do you means by Kinetic energy (momentum) ?. In the larger context "momentum" is a specialized subset of the "kinetic energy" of an object. Both of them are dependent on the motion of mass, one in a more limited sense than the other.
When you speak of momentum as defined classically being conserved even in inelastic colisions, and that heat and momentum are different and unrelated observables, you are expressing model and theory based conclusions.
There are many ways that electrical/electromagnetic energy can be used to generate kinetic energy/momentum.., and momentum can then be used to generate both heat and electrical/electromagnetic energy. As long as the energy balances (CoE is satisfied) CoM is conserved.
Quote from: Monomorphic on 11/11/2016 12:38 pmI thought the circumference of a loop antenna should ideally be equal to wavelength. The Eagleworks guys use much smaller loops: 13.5 mm (circumference ~42 mm ) for 1.94 GHz. Wavelength ~155 mm, so a little bigger than 1/4 lambda.
Quote from: OnlyMe on 11/11/2016 02:08 amTo answer the question you began with, "What do you means by Kinetic energy (momentum) ?. In the larger context "momentum" is a specialized subset of the "kinetic energy" of an object. Both of them are dependent on the motion of mass, one in a more limited sense than the other.The big difference is that Momentum can be negative, when Kinetic Energy can't.That is why, when you have a big system, isolated from other systems, and composed of several subsystems, you can give momentum to some subsystems. But the sum of the momentum given is zero. For a positive momentum given to any small subsystem, there is a negative momentum given to some others, and the sum of all the gained momentum is still zero, in standard physics.Quote from: OnlyMe on 11/11/2016 02:08 amWhen you speak of momentum as defined classically being conserved even in inelastic colisions, and that heat and momentum are different and unrelated observables, you are expressing model and theory based conclusions.Yes, that is why I had clearly written in GR. My point is not to say that conversion from momentum to heat is impossible, but that it is not compatible with GR.Quote from: OnlyMeThere are many ways that electrical/electromagnetic energy can be used to generate kinetic energy/momentum.., and momentum can then be used to generate both heat and electrical/electromagnetic energy. As long as the energy balances (CoE is satisfied) CoM is conserved.What you call CoM is not the standard CoM. The standard CoM means that an isolated system always keep the same momentum. No momentum can be converted into heat according to CoM.You may develop a new theory where Momentum can be converted into heat. But in that case CoM is not satisfied. Conservation of Momentum means that the momentum can't be converted into energy.
In GR, which again is a geometric model describing the tidal effects of gravitation, momentum as a component of the math, is not an observable. It is a descriptive component of the geometric model.Momentum as discussed in terms of the anomalous thrust and EmDrives is an observable... Even if it has not yet been conclusively observed in published data.
Quote from: OnlyMe on 11/11/2016 05:07 pmIn GR, which again is a geometric model describing the tidal effects of gravitation, momentum as a component of the math, is not an observable. It is a descriptive component of the geometric model.Momentum as discussed in terms of the anomalous thrust and EmDrives is an observable... Even if it has not yet been conclusively observed in published data.Here is something that I find really confusing... when we see a small object hit the Earth, we observe that its momentum has changed from non-zero to zero. On the other hand, we know that the total momentum of the system "earth + object" has not changed, so the Earth must have gained some momentum. Obviously this cannot be measured, but we know it must be true (otherwise CoM would be violated). Now when we look at "anomalous thrust" devices (such as EmDrive), we observe that they apparently gain momentum. It would be logical to assume that this is caused by interaction with some other object (or objects) which are gaining the opposite momentum, but we don't know the mechanism of this interaction yet (woodward effect and interaction with other objects in the universe is one hypothesis). Why do some people choose to throw away CoM and assume that it's not "pushing against" (interacting with) anything?