Quote from: Prunesquallor on 07/28/2015 06:53 pmNeither of the photos (the table top nor the vacuum chamber) seem to indicate an offset, although it is a bit difficult to tell.Hi all, been lurking for several months now, just dipping a toe in...If you look at the thermal imaging pictures on page four of the Tajmar paper, the right-most image shows a top-down view. Looks pretty clear that the waveguide is centered on the axis of the frustum...Party on...
Neither of the photos (the table top nor the vacuum chamber) seem to indicate an offset, although it is a bit difficult to tell.
Quote from: andygood on 07/28/2015 09:37 pmQuote from: Prunesquallor on 07/28/2015 06:53 pmNeither of the photos (the table top nor the vacuum chamber) seem to indicate an offset, although it is a bit difficult to tell.Hi all, been lurking for several months now, just dipping a toe in...If you look at the thermal imaging pictures on page four of the Tajmar paper, the right-most image shows a top-down view. Looks pretty clear that the waveguide is centered on the axis of the frustum...Party on...Thank you for the observation. Concerning the geometrical placement of the waveguide you are correct: Todd "WarpTech" was correct that the waveguide longitudinal central axis seems to be somewhat radially located from the center of the EM Drive.But this image actually makes the case for effective asymmetry much stronger: as what matters is the electromagnetic effect and the image shows that the temperature is not rotationally symmetric !The high temperature (shown by yellow-orange) extends to less than 40 degrees to the left, but it extends to over 90 degrees to the right.There is a very strong rotational asymmetry shown by the thermal image: whether the forces are due to thermal effects or due to electromagnetic fields that result in this asymmetry, that this asymmetry would produce a side force would not at all be surprising.....
Quote from: Rodal on 07/28/2015 11:40 pmQuote from: andygood on 07/28/2015 09:37 pmQuote from: Prunesquallor on 07/28/2015 06:53 pmNeither of the photos (the table top nor the vacuum chamber) seem to indicate an offset, although it is a bit difficult to tell.Hi all, been lurking for several months now, just dipping a toe in...If you look at the thermal imaging pictures on page four of the Tajmar paper, the right-most image shows a top-down view. Looks pretty clear that the waveguide is centered on the axis of the frustum...Party on...Thank you for the observation. Concerning the geometrical placement of the waveguide you are correct: Todd "WarpTech" was correct that the waveguide longitudinal central axis seems to be somewhat radially located from the center of the EM Drive.But this image actually makes the case for effective asymmetry much stronger: as what matters is the electromagnetic effect and the image shows that the temperature is not rotationally symmetric !The high temperature (shown by yellow-orange) extends to less than 40 degrees to the left, but it extends to over 90 degrees to the right.There is a very strong rotational asymmetry shown by the thermal image: whether the forces are due to thermal effects or due to electromagnetic fields that result in this asymmetry, that this asymmetry would produce a side force would not at all be surprising.....Another thing to note is that there are high temperatures around the seam between the fustrum and the end plate. Most likely hot gasses leaking from the fustrum. Leaks may play an important role in experimental artifacts give the low thrust levels expected from these experiments to date - especially those in a vacuum.
Wow, what an exciting day! Live blog from Dr BB, classical theorists slinging "crap" in their not-so-well thought out responses, noobs, posting and departing, and me finishing my assembly so I can celebrate the day with a video of my first thermal test of NSF-1701 under power!Time to savor an adult beverage...been in the shop all day and I have some reading to catch up on
Wow, what an exciting day! Live blog from Dr BB, classical theorists slinging "crap" in their not-so-well thought out responses, noobs, posting and departing, and me finishing my assembly so I can celebrate the day with a video of my first thermal test of NSF-1701 under power!Here it is guys and gals: Time to savor an adult beverage...been in the shop all day and I have some reading to catch up on
Quote from: WarpTech on 07/28/2015 06:03 amI just figured out that apparently, Dr. McCulloch's formula, Dr. @Notsosurofit's formula and my own formula using the cylindrical approximation, all amount to the same basic force. This force is proportional to, the change in energy from the small end to the big end, divided by the length, i.e, delta_E/delta_z.Where the 3 formula differ, is in how this force is multiplied by the group velocity v/c or (v/c)2, and what formula is used for the group velocity. Dr. McCulloch skips this concept entirely and simply inputs the energy as Power x time, using the speed of light and the length. No consideration of group velocity at all. He simply uses m*c^2 where m is derived by his theory.Dr. @Notsosureofit's formula, after completing the square and factoring the difference between two frequencies squared. The basic force above is multiplied by the average cut-off (I know @Rodal) over the input frequency:(ws + wb)/2*wThis was surprising and interesting. Note, that when the frequency is less than the average cut-off, (i.e, becoming evanescent) this factor is > 1.My formula, without Zeng and Fan results in a factor that also depends on the frequency, but has a much larger value near the cut-off: (w + wb)/(w - wb),ws and wb, are the resonant or cut-off frequencies at each end respectively.These factors are each multiplied by delta_E/delta_z, where delta_z is the length, and delta_E is the frequency shift from small end to big end. There is one more factor, and that is the impedance plots in Zeng and Fan for a cone. It is somewhere between the "infinite" value of my formula and the subtle value of @Notsosureofit's formula. It's late, I hope I didn't make any errors, I'll be back. ToddTodd you also need to consider Shawyers equation, that Tajam says correctly predicted his Force, which is based on Shawyer's Df that factors in the cutoff at each end of the tapered cavity to produce the guide wavelengths at each end. The delta of the end plate guide wavelengths represents the delta of the end plates forces.
I just figured out that apparently, Dr. McCulloch's formula, Dr. @Notsosurofit's formula and my own formula using the cylindrical approximation, all amount to the same basic force. This force is proportional to, the change in energy from the small end to the big end, divided by the length, i.e, delta_E/delta_z.Where the 3 formula differ, is in how this force is multiplied by the group velocity v/c or (v/c)2, and what formula is used for the group velocity. Dr. McCulloch skips this concept entirely and simply inputs the energy as Power x time, using the speed of light and the length. No consideration of group velocity at all. He simply uses m*c^2 where m is derived by his theory.Dr. @Notsosureofit's formula, after completing the square and factoring the difference between two frequencies squared. The basic force above is multiplied by the average cut-off (I know @Rodal) over the input frequency:(ws + wb)/2*wThis was surprising and interesting. Note, that when the frequency is less than the average cut-off, (i.e, becoming evanescent) this factor is > 1.My formula, without Zeng and Fan results in a factor that also depends on the frequency, but has a much larger value near the cut-off: (w + wb)/(w - wb),ws and wb, are the resonant or cut-off frequencies at each end respectively.These factors are each multiplied by delta_E/delta_z, where delta_z is the length, and delta_E is the frequency shift from small end to big end. There is one more factor, and that is the impedance plots in Zeng and Fan for a cone. It is somewhere between the "infinite" value of my formula and the subtle value of @Notsosureofit's formula. It's late, I hope I didn't make any errors, I'll be back. Todd
Re Post #5473 by PrunesquallorLaser pulses that are shot at the reflectors that the Apollo astronauts left on the moon start out at a diameter of 3.5 m. Atmospheric effects cause divergence to about 2 km at the moon.Edit: tei-po
(Reposting for posterity with the direct links to the paper removed:)I see three pieces of evidence from Tajmar's paper that high power RF effects such as corona breakdown, multipaction (or simple out-gassing) could be incinerating the materials in the cavity and generating particles thus creating the observed thrust:1) For the latest results, the fact that the thrust continues to exist after the removal of RF power and correlates well to temperature indicates to me that particle generation is due to thermal effects (such as burning an adhesive). From the paper:"The implementation of all isolation methods (thermal, magnetic, air circulation block) resulted in the cleanest measurement with an expected behavior such that the thrust appeared after turn-on, then steadily increased until power turn off. It then remained there and slowly decreased as the EMDrive cooled down. "2) The second piece of evidence comes from my suggestion here: http://forum.nasaspaceflight.com/index.php?topic=36313.msg1367663#msg1367663 that the cavity be disassembled and inspected for the damaging effects of the above phenomenon to verify if they are occurring or not. It seems this was done and the damage was found:" Indeed we measured that our Q factor was reduced to only 20.3 – probably due to the fact that our inner surfaces were now much more oxidized compared to the start of our test campaign after a visual inspection. "The visual evidence confirms that the effects I've mentioned are occurring. 3) Further, taking a look at the thermal imager pictures in Figure 3 it appears the seam of the cylindrical cavity is the hottest point which is where you would expect these effects to occur as was also suggested here http://forum.nasaspaceflight.com/index.php?topic=36313.msg1367663#msg1367663-JK
I got a kick out of two of the snippets Dr. BB posted as Tajmar was speaking. Funny to hear a scientist say stuff like this."So I went to the hardware store and bought a microwave."and later..."I had to go to the supermarket and pick up some vegetable oil."Was it a presentation, or was it a cooking show?
Quote from: jknuble on 07/29/2015 03:09 am(Reposting for posterity with the direct links to the paper removed:)Just some suggestions...TO THE MEEPERS; This is WHY we need an accurate simulation that shows what happens when the power is turned off. If the losses are small, the resonant wave will persist for some period of time until all the energy decays. We need to know what that decay time really is. In fact, I would predict that the thrust goes up the faster it decays. Thrust persisting after it is turned off is to be expected. The question is, how long does it take for the internal stored energy to dissipate? The higher the Q, the longer it will take to discharge. DIY's; It would be a good idea to monitor what is going on inside the frustum at all times. Such that you can determine the difference between a hot frustum, and one that is still charged with resonant energy. If there is persistent force after the power is turned off but there is nothing going on inside it, then you know you have an artifact. Just because it's hot doesn't mean it's an artifact. The temperature will follow the power dissipation. As it declines so will the temperature. The two are well correlated, so the fact that the thrust persists and decays with the temperature is to be expected. This alone is not an indication of an artifact. The artifact would be only IF there is no energy inside the frustum doing the work. Still, I would be hard pressed to believe that thermal radiation can exert forces at least an order of magnitude larger than a photon rocket. Regarding arcing, breakdown, etc. Today's video from @rfmwguy did not appear to me to show arching. Shell said she saw it, but I think that what looks like arcing is actually his laser on the thermal probe reflecting off the magnetron. I did not see arcing. Perhaps @rfmwguy will update us in the morning regarding this.Todd
(Reposting for posterity with the direct links to the paper removed:)
You know that I am about as pro real thrust as it can get, but if its an artefact I want to find it. So I had a thought re. Tajmar's measured lingering thrust dissipating in seemingly correlation to the hot magnetron cooling.Have the folks using magnetrons all used New magnetrons? That is, tubes that have not yet been burned in? If so then the inside of those things is likely be "dirty." Covered with a sheen of cleanser if not light oil. Not very dirty but what ever amount is burning off could travel down the waveguide to the cavity and leak out, or deposit on the inside of the relatively cool and polished copper. EW didn't use a magnetron, and still got a force signal. Is there a sneak path for hot stuff to come out of the waveguide into the cavity in that case?
It looks like the only person in this world other than Shawyer that likes Shawyer's theory is TheTravellerWe have McCulloch and now Tajmar strongly against Shawyer's theory