-
#820 by Rodal on 20 Sep, 2014 21:22
-
I saw the talk about Dark Hamsters, and shall be going back to my beer soon. Still:
In my model, one photon would be recycled by, say, 10^9 times.
Let's not, say, shall we? ...
We can't say "dicendo" anymore?
Does that mean we can only say "solo" from now on ? Or is it only forbidden to "say" in English? -
#821 by aero on 20 Sep, 2014 21:27
-
Ok, I used 17 watts in the following, so it will need to be reduced a little to allow for the dissipated heat power.
I calculated that the number of electrons, 2.54E+12 from before carry a charge of 4.07E-07 C so the current flow is 4.07E-07 amps. 17 watts of power, P = I*V gives Voltage across the cavity of 41.8 MV . This is a little higher (factor of 10 higher) than the striking voltage in air so glow discharge might be a problem.
Other than that, does physics allow this mechanism?
1) Thank you for performing calculations and providing numbers
2) I have to think about this.
3) I look forward to other readers providing comments, performing their own calculations and cross-checking.
You're welcome. I too hope that folks will comment based on knowledge of physics and also check the math.
As for the voltage being 10 times the striking voltage in air, I think that can be easily dealt with by modifying my assumptions. Reduce the electron velocity will decrease the electron acceleration, increase the current flow and reduce the cavity voltage. Alternatively, assume massive initial acceleration can maintain the thrust, which will trickle through the math to change the voltage required for the power. Anyhow, to reduce the voltage for the fixed power, the current must increase. About a factor of 10 in both cases since power is linear with voltage and current.
The key is the question, can the accelerated electrons be turned by the RF waves in the cavity without acting on the base? And this is a question where consideration of the Q factor comes into play. -
#822 by JohnFornaro on 20 Sep, 2014 22:25
-
Ok. Just popped open Becks #4, so still functioning.
From:The second case is a bit harder to prove, but still within reach of ordinary algebra:
Ek = ½mV² ... kinetic energy as a function of V, again.
V = at ... again, now substitute
Ek = ½ma²t² and remembering that [F = ma]...
Ek = F²t²/(2m)
If F = ma, then F^2 = (ma)^2 = m^2a^2, correct?
As to "solo dicendo"...
Yo no lo visto, ah. ah.
Yo no conozco, eh. eh.
Es peligroso... amore.
These guys in the song are brutalizing Spanish, are they not? Wait a sec. Beloved one tells me it's Italian:
"non l’ho visto, non lo cognosco, e pericoloso!!!!!!"
Yo no sabe nada. ¿Estoy embrutecedor álgebra?
-
#823 by Notsosureofit on 20 Sep, 2014 22:38
-
"The key is the question, can the accelerated electrons be turned by the RF waves in the cavity without acting on the base? And this is a question where consideration of the Q factor comes into play."
Not sure about relevance, but probably more dependent on which cavity mode(s) than Q. -
#824 by aero on 20 Sep, 2014 23:18
-
@Rodal -
This is my concept of what a cathode ray tube converted to a thruster would look like. It is a static picture. In this case it seems like the electrons would strike the anode with the full vertical velocity and so would generate no external thrust. That would be so because the positive anode voltage had caused increased velocity by adding horizontal velocity. This turns the beam direction but does not react the vertical velocity. But the wall does.
So - can the RF wave turn the electron beam without adding velocity?
-
#825 by Rodal on 20 Sep, 2014 23:32
-
@Rodal -
This is my concept of what a cathode ray tube converted to a thruster would look like. It is a static picture. In this case it seems like the electrons would strike the anode with the full vertical velocity and so would generate no external thrust. That would be so because the positive anode voltage had caused increased velocity by adding horizontal velocity. This turns the beam direction but does not react the vertical velocity. But the wall does.
So - can the RF wave turn the electron beam without adding velocity?
Did you notice that
A) the "Frustum" (or truncated cone) smaller end surface appears to be metallic (copper?) but the Frustum larger end appears to be NOT-metallic (circuit-board polymer?)
B) the dielectric resonator is located right against the smaller end (copper?) surface
see Fig. 15 and Fig. 17 in the report:
_____________________________________
Image from "Anomalous Thrust Production from an RF Test Device Measured on a Low-Thrust Torsion Pendulum" by David A. Brady*, Harold G. White†, Paul March‡, James T. Lawrence§, and Frank J. Davies**, July 28-30, 2014, Cleveland, OH, AIAA 2014-4029, Propulsion and Energy Forum, 50th AIAA/ASME/SAE/ASEE Joint Propulsion Conference
the article notes: "This material is declared a work of the U.S. Government and is not subject to copyright protection in the United States." (Also see: http://en.wikipedia.org/wiki/Copyright_status_of_work_by_the_U.S._government) as posted in http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdf
-
#826 by Notsosureofit on 20 Sep, 2014 23:43
-
Circuit board material 2 layers
-
#827 by Rodal on 20 Sep, 2014 23:47
-
Circuit board material 2 layers
Can you please elaborate or guess on purpose of Circuit board plastic "2 layers" instead of 1 layer ? -
#828 by RotoSequence on 21 Sep, 2014 00:02
-
Circuit board material 2 layers
So, glass reinforced plastic, with two intact copper layers inside? -
#829 by Rodal on 21 Sep, 2014 00:13
-
Circuit board material 2 layers
So, glass reinforced plastic, with two intact copper layers inside?
How do we know that there are "two intact copper layers inside?" -
#830 by Bob Shaw on 21 Sep, 2014 00:21
-
Guys, really! TANSTAAFL!
Let's stick with sensible stuff, like fully-reuseable return-to-launch-site boosters (with wings if possible)... -
#831 by Rodal on 21 Sep, 2014 00:24
-
That circuit-board material on the larger end of the frustum cavity is not there by accident...
-
#832 by RotoSequence on 21 Sep, 2014 00:25
-
Circuit board material 2 layers
So, glass reinforced plastic, with two intact copper layers inside?
How do we know that there are "two intact copper layers inside?"
I don't think we have a way of knowing for sure. Using PCB for the end of the thruster complicates things, since it's entirely possible for circuitry or antennas to be etched into the board's layers.
On the other hand, antennas would be odd, since microwave antennas use the quintessential parabolic dish. If nothing else, PCBs offer a uniform, flat, tough, and inexpensive copper surface. -
#833 by Rodal on 21 Sep, 2014 00:32
-
Guys, really! TANSTAAFL!
We ain't saying that there is any free lunch
On the contrary, we are trying to figure out what is paying the lunch ...
-
#834 by Rodal on 21 Sep, 2014 01:26
-
At breakdown, when the electric field exceeds the dielectric strength, electrons are indeed released. If the applied electric field is sufficiently high, free electrons may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown.
The dielectric strength of PTFE ("Teflon") is 20*10^6 Volt/meter, but it decreases with increased frequency and with defects.
The reported calculations show that the electric field (maximum value of the Electric Field shown in Fig. 14, p.10, as 4.7189*10^4 V/m) was 400 times below the 20*10^6 V/m dielectric strength of PTFE ("Teflon"). On the other hand, if the Teflon dielectric resonator contained an unusual amount and size of defects, its dielectric strength could have been a fraction of that value.
-
#835 by aero on 21 Sep, 2014 01:56
-
At breakdown, when the electric field exceeds the dielectric strength, electrons are indeed released. If the applied electric field is sufficiently high, free electrons may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown.
The dielectric strength of PTFE ("Teflon") is 20*10^6 Volt/meter, but it decreases with increased frequency and with defects.
The reported calculations show that the electric field (maximum value of the Electric Field shown in Fig. 14, p.10, as 4.7189*10^4 V/m) was 400 times below the 20*10^6 V/m dielectric strength of PTFE ("Teflon"). On the other hand, if the Teflon dielectric resonator contained an unusual amount and size of defects, its dielectric strength could have been a fraction of that value.
I'm reading the chart a little differently. Looks to me like the color bar on the left is for e-field values within the thruster while the color bar on the right is for e-field values within the RF drive pipe, hence the dielectric. The color bar chart on the right has an over the top label of 3.5922 x 10^4 and red color label of 3000. But I don't know what it means as that layout is unfamiliar.
If IIUC, there is a mechanism where the RF wave in the dielectric can cause avalanche breakdown which will liberate huge numbers of electrons. But your reading of the available data is that the dielectric/RF wave interaction was selected to avoid that condition. Further, avalanche breakdown in the as specified dielectric is greater than 3 x 10^6 V/m, the striking voltage in air.
-
#836 by Rodal on 21 Sep, 2014 02:27
-
At breakdown, when the electric field exceeds the dielectric strength, electrons are indeed released. If the applied electric field is sufficiently high, free electrons may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown.
The dielectric strength of PTFE ("Teflon") is 20*10^6 Volt/meter, but it decreases with increased frequency and with defects.
The reported calculations show that the electric field (maximum value of the Electric Field shown in Fig. 14, p.10, as 4.7189*10^4 V/m) was 400 times below the 20*10^6 V/m dielectric strength of PTFE ("Teflon"). On the other hand, if the Teflon dielectric resonator contained an unusual amount and size of defects, its dielectric strength could have been a fraction of that value.
I'm reading the chart a little differently. Looks to me like the color bar on the left is for e-field values within the thruster while the color bar on the right is for e-field values within the RF drive pipe, hence the dielectric. The color bar chart on the right has an over the top label of 3.5922 x 10^4 and red color label of 3000. But I don't know what it means as that layout is unfamiliar....
We know that the Fig.14 chart is an output from COMSOL's Multiphysics (http://www.comsol.com/) finite element analysis. These are familiar, standard COMSOL finite element display of values.
The whole Cannae test article was most likely analyzed by COMSOL all-together: there was no separate analysis for the RF drive pipe (and if it were, for whatever reason as for example if they divided the analysis in "chunks" because of computer time, there would need to be compatibility and boundary condition requirements enforced between the "chunks" as physically there is one physical problem to solve, as Maxwell's equations need to be solved over the whole domain). Fig. 14 (left) is a COMSOL display of values for a plane revolving around the rotational axis (around which the pillbox geometry is defined). Therefore Fig. 14 (left) does contain the Electric Field for the entire Cannae drive, including the "drive pipe", which is indeed being shown on that figure. The (COMSOL) finite element program always displays the maximum and minimum values that occur over the entire region being shown by COMSOL. The maximum of the Electric Field is shown to occur in the dielectric.
Fig. 14 (right) is a COMSOL display of particular circular cross-sectional areas perpendicular to the axis of revolution. The purpose of showing Fig. 14 (right) is to show the distribution of the Electric Field in the rotational direction. It shows that the Electric Field is practically rotationally symmetric. Because only 7 cross-sections are shown, it is not surprising that the maximum value of the Electric Field will not appear on Fig 14 (right), such a discrete display (showing only 7 cross sections) must display a smaller value than a continuous display (such as Fig 14 left) unless the Electric Field happened to be uniform (constant) along the axis of revolution (in which case, if the field would have been constant in the axial direction, Fig 14 right would have displayed the same maximum as in Fig 14 left).
Bottom line: I'm pretty sure that the maximum value of the COMSOL-calculated Electric Field in the dielectric is the one shown in Fig. 14 (left): 4.7189*10^4 V/m. -
#837 by Rodal on 21 Sep, 2014 02:54
-
In their report they express the fact it took them significant time to analyze the distribution of the Electric Field and that realizing its importance (including the field in the dielectric) was among their most important achievements.At breakdown, when the electric field exceeds the dielectric strength, electrons are indeed released. If the applied electric field is sufficiently high, free electrons may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown.
The dielectric strength of PTFE ("Teflon") is 20*10^6 Volt/meter, but it decreases with increased frequency and with defects.
The reported calculations show that the electric field (maximum value of the Electric Field shown in Fig. 14, p.10, as 4.7189*10^4 V/m) was 400 times below the 20*10^6 V/m dielectric strength of PTFE ("Teflon"). On the other hand, if the Teflon dielectric resonator contained an unusual amount and size of defects, its dielectric strength could have been a fraction of that value.
.....
If IIUC, there is a mechanism where the RF wave in the dielectric can cause avalanche breakdown which will liberate huge numbers of electrons. But your reading of the available data is that the dielectric/RF wave interaction was selected to avoid that condition.
Yes, at breakdown of the dielectric there could be an avalanche of electrons released.
If (and only if):
A) their PTFE (Teflon) dielectric resonator was free of defects
B) their COMSOL calculations are accurate
C) the applied Electric Field was well controlled
then the tests should have been well below the breakdown point.
However those are assumptions that should not be taken for granted. Even if the COMSOL calculations are accurate, we don't know how well was the electric field controlled in their set-up.
Also, the result I quoted is for the Cannae drive. They did not report the COMSOL numbers for the electric field in the frustum.
And also we don't know the quality of the PTFE dielectric resonators they used during their tests. It is known that there is a huge range (particularly for a polymer like PTFE) of quality of PTFE in the market (Teflon is just DuPont's tradename). -
#838 by Rodal on 21 Sep, 2014 03:43
-
At breakdown, when the electric field exceeds the dielectric strength, electrons are indeed released. If the applied electric field is sufficiently high, free electrons may become accelerated to velocities that can liberate additional electrons during collisions with neutral atoms or molecules in a process called avalanche breakdown.
The dielectric strength of PTFE ("Teflon") is 20*10^6 Volt/meter, but it decreases with increased frequency and with defects.
The reported calculations show that the electric field (maximum value of the Electric Field shown in Fig. 14, p.10, as 4.7189*10^4 V/m) was 400 times below the 20*10^6 V/m dielectric strength of PTFE ("Teflon"). On the other hand, if the Teflon dielectric resonator contained an unusual amount and size of defects, its dielectric strength could have been a fraction of that value.
Further, avalanche breakdown in the as specified dielectric is greater than 3 x 10^6 V/m, the striking voltage in air.
The breakdown of Air (3 x 10^6 V/m) doesn't appear to be as much of an issue as the COMSOL Electric Field calculations show the Electric Field in the cavity to be lower than 0.002 x 10^6 V/m, or 1500 times less than the Electric Field required for Air breakdown, so clearly the dielectric was about 4 times closer to breakdown than the Air. -
#839 by aero on 21 Sep, 2014 03:45
-
So, bottom line is, "Yes, there is a possibility of the release of a large number of real electrons within the dielectric end of the thruster, via an electron avalanche." Such an electron avalanche was not a design feature of the thruster. And further, we know nothing definitive about the tapered cavity thruster.
Is there another mechanism which may have released electrons numbering in the ball park of 10^13 electrons/second? Note that is not a large number of electrons as such things go. The electron lifetime would be on the order of 10^-8 to 10^-9 seconds so at any given instant there only a few 10's of thousands of electrons within the cavity. That is a very small number as plasma densities go. Isn't it reasonable to assume that some small number of air molecules ionized within the cavity to create that small number of electrons?
Of course ionized air would result in positively charged ions also but if the cause of the electron acceleration was the magnetic field of the RF wave, then it would not discriminate between ion and electron acceleration forces. Ions would have a larger gyro radius around the magnetic field lines, and gyrate in the opposite direction (?) from the electrons but ultimately they would end up in the same place I think.
I'm on a roll here so I'd better stop before I go stupid on you.
On the contrary, we are trying to figure out what is paying the lunch ...