Quote from: Rodal on 05/17/2015 07:51 pmQuote from: Notsosureofit on 05/17/2015 07:44 pm@RODALIt seems (to me) that that is quite correct, but that does not mean that that is for max force since it neglects frequency. I can see how that might be for max laser sideband generation w/o constraining the frequency. (ie very high radial modes)Thanks. But where does the frequency enter into the equation for L? all I see is a, b, r2bar, r1bar, lo geometrical parameters. It seems that the electromagnetic field Power only enters through the parameter (U_{o})^{2} which is built inside the length "lo", so as we said, we need a number for "lo" to make any more progress.See Marco's Eq. 19, 20 and 28 in http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830137It is the choice of the resonant mode that fixes l0 and this depends on frequency (indirectly). You should check Egan link for some numbers to put in.

Quote from: Notsosureofit on 05/17/2015 07:44 pm@RODALIt seems (to me) that that is quite correct, but that does not mean that that is for max force since it neglects frequency. I can see how that might be for max laser sideband generation w/o constraining the frequency. (ie very high radial modes)Thanks. But where does the frequency enter into the equation for L? all I see is a, b, r2bar, r1bar, lo geometrical parameters. It seems that the electromagnetic field Power only enters through the parameter (U_{o})^{2} which is built inside the length "lo", so as we said, we need a number for "lo" to make any more progress.See Marco's Eq. 19, 20 and 28 in http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=830137

@RODALIt seems (to me) that that is quite correct, but that does not mean that that is for max force since it neglects frequency. I can see how that might be for max laser sideband generation w/o constraining the frequency. (ie very high radial modes)

Quote from: txdrive on 05/17/2015 06:41 pmQuote from: TheTraveller on 05/17/2015 06:25 pmQuote from: txdrive on 05/17/2015 05:52 pmQuote from: TheTraveller on 05/17/2015 05:35 pmQuote from: txdrive on 05/17/2015 05:13 pmWith regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.Last line say it all:QuoteIf the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?It's a geometric optics approximation. Doesn't hold precisely.For example suppose you got a beam of coherent light, 5mm across, 500nm wavelength, in space. It's not going all parallel to it's original direction, it's spreading, to about 1m across at 10 000m from the source (Makes a fuzzy "Airy disk" pattern). edit: The Poynting vector on the sides of a perfect cylinder over this beam, is not parallel to the axis. It's this Poynting vector that matters.Built and used many optical telescopes, including a 12 inch, Schmidt Cassegrain. Diffraction and I got to be good enemies.There is no natural spread for the beam inside the cavity. It is controlled by the guide and cutoff wavelengths determined by the cavity diameter it travels through

Quote from: TheTraveller on 05/17/2015 06:25 pmQuote from: txdrive on 05/17/2015 05:52 pmQuote from: TheTraveller on 05/17/2015 05:35 pmQuote from: txdrive on 05/17/2015 05:13 pmWith regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.Last line say it all:QuoteIf the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?It's a geometric optics approximation. Doesn't hold precisely.For example suppose you got a beam of coherent light, 5mm across, 500nm wavelength, in space. It's not going all parallel to it's original direction, it's spreading, to about 1m across at 10 000m from the source (Makes a fuzzy "Airy disk" pattern). edit: The Poynting vector on the sides of a perfect cylinder over this beam, is not parallel to the axis. It's this Poynting vector that matters.

Quote from: txdrive on 05/17/2015 05:52 pmQuote from: TheTraveller on 05/17/2015 05:35 pmQuote from: txdrive on 05/17/2015 05:13 pmWith regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.Last line say it all:QuoteIf the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.We were discussing radiation pressure generated from EM waves according to Maxwell's equations (attached). This pressure is apparently subject to cosine angle loss and thus when the EM wave moves along the frustum side walls and is not bouncing off the frustum spherical end plates, there is no Maxwell radiation pressure generated on the frustum side walls.Or did I not understand what the cosine loss factor is for in the lower of the 2 equations?

Quote from: TheTraveller on 05/17/2015 05:35 pmQuote from: txdrive on 05/17/2015 05:13 pmWith regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.Last line say it all:QuoteIf the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.You're ignoring diffraction. If the frustum side walls were not there, the radiation would leak out.edit: write an equation of form E(position, time) and H(position, time) . Maybe it would be clearer for a cylinder with flat end pieces? You can't just take geometric optics approximation from Wikipedia and apply it to microwaves in a reasonable sized cavity... it won't be correct then.

Quote from: txdrive on 05/17/2015 05:13 pmWith regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.Last line say it all:QuoteIf the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.The effective radiation source is the vertex of the frustum. The frustum side wall are aligned to, pointing at the vertex. Therefore the radiant pressures on the side walls are zero.

With regards to the forces on the side of the cavity, all I can say is, write the field equations for your fields (what is the H and E at a given point in space), then I can calculate the forces on the boundaries and/or show that your fields violate Maxwell's equations. When it is a picture I don't know what exactly you're talking about.

If the absorbing surface is planar at an angle a to the radiation source, the intensity across the surface will be reduced.

and by bouncing off the spherical end plates, which as I see it orient the EM waves so they travel / slide along the side walls at a 0 bounce radiant cosine angle.

Test 03 Success. I have thrust.I modified the setup and now i weight the frustum. the precision is much better.I will make a modifications to be able to adjust the cavity length to achieve the resonance, so i should have more thrust then.

Quote from: Iulian Berca on 05/17/2015 11:20 pmTest 03 Success. I have thrust.I modified the setup and now i weight the frustum. the precision is much better.I will make a modifications to be able to adjust the cavity length to achieve the resonance, so i should have more thrust then.Do you have the weight of the setup, so that we can calculate the newtons of thrust from this?

....Dear Jose,Thank you a lot for your effort to this question. Also thanks to @notsureofit for his help. You are right. I have checked this with Maple and optimized the function L(r) as a function of r1, r2 and U0 also. And yes, r1=r2 is the solution! I have other checks to do and my notebook just keeps on freezing for a computation of these but I hope to complete all in a few days. This means that Harold White experiments could represent a great leap beyond in experimental general relativity. This was my initial hope. As a theoretical physicist please don't ask me about applications!Give me a few days for further checks and I will update my draft, with due acknowledgements.Regards.

Test 03 Success. I have thrust.

Quote from: Iulian Berca on 05/17/2015 11:20 pmTest 03 Success. I have thrust.I'm new to this thread, but doesn't Shawyer's theory paper say that it is supposed to be Big-Endian? That is, that a larger force is applied on the inner face of the big end than on the inner face of the small end, so that if it were flying free in space it would accelerate big end first?Is Iulian's Little-Endian negative reading consistent with other theory or experimental results?~Kirk

Quote from: kdhilliard on 05/18/2015 01:17 amQuote from: Iulian Berca on 05/17/2015 11:20 pmTest 03 Success. I have thrust.I'm new to this thread, but doesn't Shawyer's theory paper say that it is supposed to be Big-Endian? That is, that a larger force is applied on the inner face of the big end than on the inner face of the small end, so that if it were flying free in space it would accelerate big end first?Is Iulian's Little-Endian negative reading consistent with other theory or experimental results?~KirkWatch Shawyers dynamic test. http://emdrive.com/dynamictests.htmlMoves toward small end. So both tests generate device movement in the same direction. Toward the small end.

Quote from: TheTraveller on 05/18/2015 01:27 amQuote from: kdhilliard on 05/18/2015 01:17 amQuote from: Iulian Berca on 05/17/2015 11:20 pmTest 03 Success. I have thrust.I'm new to this thread, but doesn't Shawyer's theory paper say that it is supposed to be Big-Endian? That is, that a larger force is applied on the inner face of the big end than on the inner face of the small end, so that if it were flying free in space it would accelerate big end first?Is Iulian's Little-Endian negative reading consistent with other theory or experimental results?~KirkWatch Shawyers dynamic test. http://emdrive.com/dynamictests.htmlMoves toward small end. So both tests generate device movement in the same direction. Toward the small end.Also interesting, and just to remind folks, it appears that Iulian has the waveguide directly inside the fulstrum ending some what near the centre of the cavity. Sawyer, et al all seemed to use an antenna of some type that was close to one side of the cavity.

Quote from: demofsky on 05/18/2015 01:48 amQuote from: TheTraveller on 05/18/2015 01:27 amQuote from: kdhilliard on 05/18/2015 01:17 amQuote from: Iulian Berca on 05/17/2015 11:20 pmTest 03 Success. I have thrust.I'm new to this thread, but doesn't Shawyer's theory paper say that it is supposed to be Big-Endian? That is, that a larger force is applied on the inner face of the big end than on the inner face of the small end, so that if it were flying free in space it would accelerate big end first?Is Iulian's Little-Endian negative reading consistent with other theory or experimental results?~KirkWatch Shawyers dynamic test. http://emdrive.com/dynamictests.htmlMoves toward small end. So both tests generate device movement in the same direction. Toward the small end.Also interesting, and just to remind folks, it appears that Iulian has the waveguide directly inside the fulstrum ending some what near the centre of the cavity. Sawyer, et al all seemed to use an antenna of some type that was close to one side of the cavity.In the 1st EM Drive device, Shawyer fed the microwaves in via a short waveguide to around the centre of the frustum. You can see the wave guide at the rear of the frustum.In the next Demonstration EM Drive, the magnetron generated microwaves were fed into the frustum at a point near the big end as can be seen in these 2 last images.The 3rd device, the Flight Thruster, used a coax Rf feed at near the big end.So it would seem near the big end feeding will work.