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#3280 by zen-in on 21 Jun, 2016 23:42
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Was planning to have a lot done this week on my tests, but yesterday we got hit by golf ball (and larger) size hail. I lost the driveway (washed away), cars beat up and glass shattered, the roof is beat all up and leaking. So I'll be picking up the pieces for the next few days... sigh.
My Drive build is going to have to wait, I'll be back after a few days away.
Shell
Sorry to hear about the extreme weather and damage. I hope you are covered by insurance. Have you considered the possibility your experiments are modifying the weather in your area? -
#3281 by Monomorphic on 22 Jun, 2016 00:29
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Short video on the magnetic fields around the EMdrive and test stand. Magnetron is the obvious culprit. Should insulation between Emdrive and torsional pendulum eliminate field traveling down torsional pendulum?
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#3282 by SeeShells on 22 Jun, 2016 02:11
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Short video on the magnetic fields around the EMdrive and test stand. Magnetron is the obvious culprit. Should insulation between Emdrive and torsional pendulum eliminate field traveling down torsional pendulum?
Glad you posted this video, it shows a lot of the issues of mounting a magnetron onto a frustum.
I'd say get it off of the frustum and isolate the thermal and magnetic effects of the magnetron by driving the magnetron into a waveguide and tapping off using a coax to drive into your frustum, but if it's not doable then you might want to try to wrap it in Mu Metal to isolate the magnetic fields some.
http://www.ebay.com/itm/MuMetal-Ultraperm-Permalloy-Alloy-Shielding-Sheet-Mu-Metal-Audio-Shield-80-/152045122881?hash=item2366987541:m:mcKIFWQJVx3F91pr4jiH-qQ
Shell -
#3283 by Tellmeagain on 22 Jun, 2016 03:02
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for rfmwguy. Edited screenshot of his video. [UPDATE] I got it wrong. I took another look and it seemed I got it wrong. I assumed the wire in the transparent tube was the anode wire, while the black and white wires in the opaque tube were the two filament wires. It turns out rfmwguy put one of the filament wire in the transparent tube, and put the anode (ground) wire with the other filament wire in the opaque tube. I do not understand why he wired this way but my assumption caused my misunderstanding. Sorry, Rfmwguy. I removed the misleading picture.
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#3284 by Bob Woods on 22 Jun, 2016 03:45
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Was planning to have a lot done this week on my tests, but yesterday we got hit by golf ball (and larger) size hail. I lost the driveway (washed away), cars beat up and glass shattered, the roof is beat all up and leaking. So I'll be picking up the pieces for the next few days... sigh.
My Drive build is going to have to wait, I'll be back after a few days away.
Shell
Sorry to hear about the extreme weather and damage. I hope you are covered by insurance. Have you considered the possibility your experiments are modifying the weather in your area?
That comment is a definite 10 points!
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#3285 by FattyLumpkin on 22 Jun, 2016 03:49
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keithpickering , would you please explain more about surface to volume ratio and it's relationship to the Q of a cavity. Please use these figures: flattened frustum S/V is
4362 cm2 / 8835 cm3. Original geometry NASA frustum S/V ratio is 2438 cm2 / 8835 cm3.
Thank you, FL -
#3286 by X_RaY on 22 Jun, 2016 05:35
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keithpickering , would you please explain more about surface to volume ratio and it's relationship to the Q of a cavity. Please use these figures: flattened frustum S/V is
4362 cm2 / 8835 cm3. Original geometry NASA frustum S/V ratio is 2438 cm2 / 8835 cm3.
Thank you, FL
https://forum.nasaspaceflight.com/index.php?topic=39214.msg1474347#msg1474347Quote from: Rodal4) Quality of resonance (Q) scaling
The definition of quality of resonance factor (Q) can be stated as follows (https://en.wikipedia.org/wiki/Q_factor#Definition_of_the_quality_factor):
Q ≝ ω EnergyStored /PowerLoss
where
ω = angular frequency
EnergyStored =∫Electromagnetic Energy Density dV
PowerLoss = ((ω δ) /2) (∫ Electromagnetic Energy Density dA)
= Rs (∫ Electromagnetic Energy Density dA)/ μ
= ρ (∫ Electromagnetic Energy Density dA)/ (μ δ)
where
Rs = "surface resistance"
= ρ / δ
ρ = resistivity of the interior wall of the EM Drive resonant cavity
μ = magnetic permeability of the interior wall of the EM Drive resonant cavity
= μoμr
δ =skin depth (the penetration depth of the electromagnetic energy into the interior metal wall)
(https://en.wikipedia.org/wiki/Skin_effect)
in general, for arbitrary frequencies, the skin depth is:
where
ε = electric permittivity of the interior wall of the EM Drive resonant cavity
= εoεr
At angular frequencies ω much below 1/(ρε), for example, in the case of copper, for frequencies much below exahertz (10^9 GHz, the range of hard X-rays and Gamma rays), the skin depth can be expressed as follows:
Now, using the fact that
PowerLoss =((ω δ) /2) (∫ ElectromagneticEnergy dA)
one immediately obtains:
Q=(2/SkinDepth)( ∫Electromagnetic Energy Density dV/ ∫ Electromagnetic Energy Density dA)
Alternatively one can arrive at the same result, using the formula for power loss that depends on the "surface resistance" Rs:
PowerLoss = Rs (∫ Electromagnetic Energy Density dA)/ μ
PowerLoss = ρ (∫ Electromagnetic Energy Density dA)/ (μ δ)
one gets:
Q = ω μ (∫Electromagnetic Energy Density dV)/ (Rs ∫ Electromagnetic Energy Density dA)
Q = ω μ δ (∫Electromagnetic Energy Density dV)/ (ρ ∫ Electromagnetic Energy Density dA)
and using the fact (at angular frequencies ω much below 1/(ρε) ) that the angular frequency ω is a function of the square of the skin depth δ:
ω = 2 ρ / (μ δ2)
it is straightforward to show that the quality of resonance Q is:
Q=(2/SkinDepth)( ∫Electromagnetic Energy Density dV/ ∫ Electromagnetic Energy Density dA)
the electromagnetic energy density integrated over the cavity volume, divided by the electromagnetic energy density integrated over the cavity surface area, divided by the skin depth.
4a) Skin depth scaling
At frequencies much below 1/(ρε) the skin depth can be expressed as
SkinDepth = √(ρ/(μ π fmnp))
where
ρ = resistivity of the interior wall of the EM Drive resonant cavity
ε = electric permittivity of the interior wall of the EM Drive resonant cavity
= εoεr
μ = magnetic permeability of the interior wall of the EM Drive resonant cavity
= μoμr
fmnp = resonant frequency at mode shape m,n,p
= ωmnp/(2π)
Plugging in the expression for the frequency
fmnp=(c/R) amnp
into the skin depth expression, results in the following expression:
SkinDepth = √R√(ρ/(μ π c amnp))
or, using the previously derived expressions for amnp one concludes that the skin depth scales like the square root of any geometrical dimension, for constant resistivity and magnetic permeability of the interior wall of the cavity and for constant geometrical ratios, constant medium properties μr,εr, and for the same mode shape m,n,p.
In other words, for increasing dimensions of the cavity, preserving all geometrical ratios, and keeping material properties constant and for the same mode shape, the skin depth will increase with the square root of the dimension, while the frequency will decrease, as the inverse of the dimension.
4b) Quality of resonance (Q) scaling
Having revealed the scaling law for the skin depth, what now remains to be shown is the scaling for the energy integral ratio in the expression for Q:
Q=(2/SkinDepth)(∫Electromagnetic Energy Density dV/ ∫ Electromagnetic Energy Density dA)
The expressions under the integrals are dependent on each mode shape, as the electromagnetic energy distribution depends on mode shape, of course. However, we can notice that the lowest mode shapes (those with low values of m,n,p, for example TE012, TM212) have been of interest in the EM Drive experiments so far. So, for simplification purposes we can assume that the distribution of the electromagnetic field is of low order, and hence not that much variable throughout the cavity, for low mnp number mode shapes (for example m=0 means a constant distribution in the azimuthal circumferential direction of the cavity). Under this assumption one can (for approximation purposes) take the energy out of the integral:
(∫Electromagnetic Energy Density dV/ ∫ Electromagnetic Energy Density dA) ~
~ (Electromagnetic Energy Density /Electromagnetic Energy Density) (∫dV/ ∫ dA )
~ InteriorVolume/InteriorSurfaceArea
~ π R2L/(2 π R (R+L) )
~ R/(2(1+R/L))
and substituting this and the previously found scaling law for the skin depth, into the expression for the quality of resonance factor Q, leads to:
Q=(2/SkinDepth)(∫Electromagnetic Energy Density dV/ ∫ Electromagnetic Energy Density dA)
~(2/(√R√(ρ/(μ π c amnp)))) R/(2(1+R/L))
~ √R b
where the factor b is:
b = (1/((1+R/L)√(ρ/(μ π c amnp))))
or, using the previously derived expressions for amnp one concludes that the quality of resonance (Q) scales like the square root of any geometrical dimension, for constant resistivity and magnetic permeability of the interior wall of the cavity and for constant geometrical ratios, constant medium properties μr,εr, and for the same mode shape m,n,p.
In other words, for increasing dimensions of the cavity, preserving all geometrical ratios, and keeping material properties constant and for the same mode shape, the quality of resonance (Q) will increase with the square root of the dimension, also the skin depth will increase with the square root of the dimension, while the frequency will decrease, as the inverse of the dimension.
Furthermore, we previously proved that all three theories for the EM Drive (McCulloch, Shawyer and Notsosureofit) have expressions for the force/inputPower to be proportional to the quality of factor Q times a dimensionless factor g:
(F / Pin)EMDrive/ (F / Pin)photonRocket = Q g
(F / Pin)EMDrive = (1/c) Q g
and we previously proved that the dimensionless factor g (for all three theories: McCulloch, Shawyer and Notsosureofit) remains perfectly constant for constant geometrical ratios, constant medium properties μr,εr, and for the same mode shape m,n,p.
Therefore one concludes that the force per input Power (for all three theories: McCulloch, Shawyer and Notsosureofit) scales like the square root of any geometrical dimension, for constant resistivity and magnetic permeability of the interior wall of the cavity and for constant geometrical ratios, constant medium properties μr,εr, and for the same mode shape m,n,p.
In other words, to maximize the force per input power, according to all three theories: (McCulloch, Shawyer and Notsosureofit) the most efficient EM Drive would be as large as possible, this being due to the fact that the quality of factor of resonance Q (all else being equal) scales like the square root of the geometrical dimensions.
Small cavity EM Drive's (all else being equal) are predicted to have smaller quality of resonance Q and therefore smaller force/inputPower.
It is not clear whether this has been known to EM Drive experimenters, given the fact that the recent experiments by Prof. Tajmar at TU Dresden, Germany, (under advise from Roger Shawyer according to the report) were performed with a much smaller EM Drive, and the fact that there are several EM Drive researchers discussing really tiny EM Drives (as the group in Aachen, Germany) for use in CubeSats. Such EM Drives are predicted to be much more inefficient, having substantially lower force/inputPower.
« Last Edit: 02/07/2016 02:20 PM by Rodal » -
#3287 by Tellmeagain on 22 Jun, 2016 12:42
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Yes. The conversation is here,
What is being implied by this post?
Is it being implied that there is a potentially unsafe grounding problem in rfmwguy's video?
Posing danger to the equipment as well as potentially a danger to the operator?
If you intended to post a safety warning, it would be useful to make it more transparent, as the first post in this thread shows:
and viewers should be advised of the potential danger. Thanks
https://www.reddit.com/r/QThruster/comments/4p5nik/new_1701a_emdrive_test_stand_walkaround_62116/ -
#3288 by SeeShells on 22 Jun, 2016 13:43
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I would like to see a schematic of his wiring layout, but that being said at first look it seems that he is not connected via floating ground of 5kv.
Yes. The conversation is here,
What is being implied by this post?
Is it being implied that there is a potentially unsafe grounding problem in rfmwguy's video?
Posing danger to the equipment as well as potentially a danger to the operator?
If you intended to post a safety warning, it would be useful to make it more transparent, as the first post in this thread shows:
and viewers should be advised of the potential danger. Thanks
https://www.reddit.com/r/QThruster/comments/4p5nik/new_1701a_emdrive_test_stand_walkaround_62116/
Also these voltage doubler power supplies will put out just under 4kv not 5kv. The frame and body of the metal around the magnetron tube is at earth ground and as are the cooling fins contacting the body of the magnetron tube. If it wasn't you would cook your popcorn and yourself at the same time, bad press and you would loose your customer base.
It is VERY important to make sure you are not putting yourself in danger and you have a background in electronics and I wish other DYIers would be smart and know that they are dealing with killer voltages. It is critical to post this danger several, if not numerous times.
SAFETY FIRST -
#3289 by Monomorphic on 22 Jun, 2016 14:04
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I went back for a more detailed look at the strange magnetic field at the other end of the pendulum. Apparently, the long nut used for the counter-weight has become slightly magnetized at the tip where is was cut. This was causing the magnetic reversal when the compass got too close! There is no magnetic field "going down" the aluminum beam - as there shouldn't be.
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#3290 by JaimeZX on 22 Jun, 2016 16:40
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Was planning to have a lot done this week on my tests, but yesterday we got hit by golf ball (and larger) size hail. I lost the driveway (washed away), cars beat up and glass shattered, the roof is beat all up and leaking. So I'll be picking up the pieces for the next few days... sigh.
That's terrible! So sorry to hear it. Good luck with repairs and insurance companies! :/
My Drive build is going to have to wait, I'll be back after a few days away.
Shell -
#3291 by FattyLumpkin on 22 Jun, 2016 17:29
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X_Ray, a hypothetical for you: if I were to double the size of the NASA frustum which has a Q of +/- 22,000, what would the Q be of cavity with the same geometry, but x 2 larger?
Thank you, FL
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#3292 by X_RaY on 22 Jun, 2016 18:25
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X_Ray, a hypothetical for you: if I were to double the size of the NASA frustum which has a Q of +/- 22,000, what would the Q be of the 2 x larger cavity? Thank you, FL
Which size?
Twice the volume?
All the length and diameters times two?
Under which conditions? Same mode, means lower frequency. Same resonant frequency higher mode? Different mode?
With dielectric or without, also scaled or not?
I can not answer such questions without any additional info, but lets try an example:
Lets multiply the diameters and length by a factor of 2 also for the dielectric(while the εr remains constant ~2.2 for HDPE), using the same mode(TE012). Allow the reaonant frequency to be lower in this larger case. The Q would be scale approximately as:
22000*sqrt(2)=31112.7
or using the closer calculation of Dr.Rodal:
22000*sqrt(2)*0.990599=30820.20789
and the frequency:
1.8804GHz/2=0.9402GHz
https://forum.nasaspaceflight.com/index.php?topic=39214.msg1474351#msg1474351Quote from: Rodal10x larger
Input
bigDiameter = (110.1 inch)*(2.54 cm/inch)*(1 m/(100 cm));
smallDiameter = (62.5 inch)*(2.54 cm/inch)*(1 m/(100 cm));
axialLength = (90 inch)*(2.54 cm/inch)*(1 m/(100 cm));
Material: Copper alloy 101; resistivity = 1.71*10^(-8) ohm meter
Exact solution output
TE012 natural frequency = 0.216467 GHz
TE012 skin depth = 4.43121 micrometers
TE012 Q = 251,049.
frequency scaling: (2.1646723144342628`*^9/2.1646723144342667`*^8)/10 =1.
Q scaling: (78642.44767279371`/251049.34868706256`)*Sqrt[10] = 0.990599
Add:
Other cases are mostly more complicated.
Take a look to the following post, there I have calculated the higher order modes (greater p-value of TE01p) using the Brady cone dimensions.
(Means different frequency{wavelength} to volume and surface ratio).
https://forum.nasaspaceflight.com/index.php?topic=39214.msg1536095#msg1536095 -
#3293 by zen-in on 22 Jun, 2016 19:52
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Exciting New Data: I moved the high voltage leads to the opposte side of the torsional pendulum to see if thermal heating of the wires was causing the anomalous force - as doing so should reverse the direction of the force. To my surprise, the same anomalous force was seen with the wires moved to the opposite position!
One note: the only difference between this run and the previous, besides the reversed leads, is the addition of the oil dampening.
I'm quoting this earlier post by Monomorphic to show the video of the spectrum analyzer display for a magnetron. Some of the broadening of the spectrum may be from front-end overload. However I think there is a strong case here for showing the magnetron output to be broadband. This may be an inherent feature based on the construction of the device. It doesn't make any difference if you are popping corn but when a high Q cavity is used one has to wonder if Q is really a significant factor at all. Earlier there was discussions on the another conundrum concerning Q: If a high Q cavity along with a very low phase noise RF power source is necessary for an em-drive to produce thrust what happens if the Q is very high and very little energy is lost? Where does the power to drive the em-drive come from?
With the high phase noise RF generated from a magnetron it seems counter intuitive to use a high Q cavity. Yet another logical contradiction. -
#3294 by FattyLumpkin on 22 Jun, 2016 20:03
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X_Ray, thank you for your hard work. This is quite surprising: while a 40% increase in Q is significant, I would have expected at least a doubling. However, your work provides an excellent prediction of the expected Q where doubling the size of this particular cavity is concerned, and as noted in the "Experimental Results" calculated Q and actual Q were very close in the TE012 mode for this geometry. You and Dr. Rodal have also supplied me (the HS physics level mathematician) with a good rule of thumb for estimating the Q of a cavity that is either increased or decreased in it's original dimensions. Anyone up for building a frustum x 3 of the NASA original?
Once again, vielen dank für deine Arbeit! K -
#3295 by Monomorphic on 22 Jun, 2016 21:11
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With the high phase noise RF generated from a magnetron it seems counter intuitive to use a high Q cavity.
The current noisy RF is simply a product of the magnetron being the easiest source we have access to ($40 on amazon). Several of us DIYers are working on a new clean solid state 250 watt RF source. I just got the signal generator installed today and am working on calibration. -
#3296 by SeeShells on 22 Jun, 2016 21:55
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NOTHING wrong with a magnetron, it provides high power and a clean stable signal if you supply it with clean DC and thermally stabilize the magnetron by correctly cooling it.With the high phase noise RF generated from a magnetron it seems counter intuitive to use a high Q cavity.
The current noisy RF is simply a product of the magnetron being the easiest source we have access to ($40 on amazon). Several of us DIYers are working on a new clean solid state 250 watt RF source. I just got the signal generator installed today and am working on calibration.
Realize the semiconductor industry has been using magnetrons for years, stable clean high power RF. https://www.google.com/#q=semiconductor+industrial+magnetrons+ You would not have your cell phone if they didn't use them or many electronics.
While the semi industrial models are little higher power the basics remain for gaining a quality output.
The only thing is it's not easy to tune and you must make the cavity tunable and to negate any thermal expansion issues.
So we get a high power RF output >1000 watts for less.
I may go to a Solid State amp later but realize within that move there are another set of challenges to overcome.
Rain has stopped so back to fixing the leaks in the roof of my lab.
Shell -
#3297 by dustinthewind on 22 Jun, 2016 22:08
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I have been contemplating if it is thermal energy that is the source of propulsion. Our universe seems to be on the increase in entropy and along with that our universe is expanding. One wonders if the two are not related. The thought was that one end of the drive is more hot than the other? I thought it was but not sure how that compares to the total surface area. Thermal energy radiated forwards compared to that radiated backwards. If memory serves me correct NASA got more of a thrust when putting the dielectric top which made the top more cool? Or do I have it backwards and it is the top that was more hot?
The idea being that the light is losing energy to creation of thermal currents. This would not quite be a reflection and allows the radiation to travel through the cavity. The idea being that there is an unequal sharing of photon energy (thermally) in the cavity after many bounces which allows more efficient harnessing of photon energy. Where normally the energy lost after 1 reflection is equivalent to the Doppler shift of a reflected photon. The cavity having many reflections, and non-symmetric induction in the cavity (evidenced by unequal thermal distribution - if there is any) allows Doppler-shifting the light to thermal wavelengths in a non-symmetric manner. possibly similar to the downshifting of light in the recycled photon rocket?
Could this be what they were talking about when they were suggesting light is escaping the cavity. In a thermal sense? Such that technically, it's not a reflection?
http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807
On the exhaust of electromagnetic drive by Patrick Grahn1,a), Arto Annila2,3,b) and Erkki Kolehmainen4,c)
"Our claim that the EM drive expels paired photons in the same way as a heat engine exhausts thermal photons entails that the vacuum, as the ultimate dump, comprises of photons"
or see quote below:...
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#3298 by Rodal on 22 Jun, 2016 22:19
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I have been contemplating if it is thermal energy that is the source of propulsion. Our universe seems to be on the increase in entropy and along with that our universe is expanding. One wonders if the two are not related. The thought was that one end of the drive is more hot than the other? I thought it was but not sure how that compares to the total surface area. Thermal energy radiated forwards compared to that radiated backwards. If memory serves me correct NASA got more of a thrust when putting the dielectric top which made the top more cool? Or do I have it backwards and it is the top that was more hot?
1) Yes,, thermal energy in the form of thermal convection has been a source of experimental artifacts ever since Maxwell first conceived thermal radiation pressure. Thermal radiation pressure was not experimentally confirmed until the year 1900 in which Russian physicist Pyotr Lebedev came out with the ingenious idea of... carrying out an experiment in a vacuum chamber.
The idea being that the light is losing energy to creation of thermal currents. This would not quite be a reflection and allows the radiation to travel through the cavity. The idea being that there is an unequal sharing of photon energy (thermally) in the cavity after many bounces which allows more efficient harnessing of photon energy. Where normally the energy lost after 1 reflection is equivalent to the Doppler shift of a reflected photon. The cavity having many reflections, and non-symmetric induction in the cavity (evidenced by unequal thermal distribution - if there is any) allows Doppler-shifting the light to thermal wavelengths in a non-symmetric manner
Could this be what they were talking about when they were suggesting light is escaping the cavity. In a thermal sense? Such that technically, it's not a reflection?
2) Yes, it is notorious that the inventor of the EM Drive, Roger Shawyer has never reported a single experiment conducted in a vacuum chamber, given the well known history of thermal radiation pressure experiments and such experimental artifacts. Particularly since he has been working on his invention for decades.
3) It is also notorious that those that conducted EM Drive experiments in a vacuum chamber (NASA and TU Dresden) reported much smaller anomalous force/InputPower when performing the experiments in a vacuum chamber than under ambient conditions.
4) Asymmetric thermal (black body) radiation takes place by emitting photons, (as opposed to thermal convection that takes place by fluid motion of air molecules in the atmosphere) and hence it can never result in propulsion exceeding the force/PowerInput of a perfectly collimated photon rocket: 1/c.
Hence claims of anomalous force/InputPower in excess of 1/c cannot possibly be explained as being the result of asymmetric thermal radiation (asymmetric thermal radiation as for example being responsible for what used to be known as the "Pioneer anomaly" until it was satisfactorily explained by JPL's analysis).
But yes, they certainly can be explained as resulting from spurious experimental artifacts due to convection of air molecules by fluid/thermal transport from the hotter surface in experiments not performed in vacuum chambers.
Otherwise, how can thermal dissipation result in self-acceleration exceeding by orders of magnitude the one of a perfectly collimated photon rocket? -
#3299 by frobnicat on 22 Jun, 2016 22:41
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Exciting New Data: I moved the high voltage leads to the opposte side of the torsional pendulum to see if thermal heating of the wires was causing the anomalous force - as doing so should reverse the direction of the force. To my surprise, the same anomalous force was seen with the wires moved to the opposite position!
One note: the only difference between this run and the previous, besides the reversed leads, is the addition of the oil dampening.
Why running the power leads along the other side of pendulum's arm should reverse direction of a force due to heating of wires ? The wires link a fixed part to a moving part, there is little question that there are forces due to the wire at rest (before heating) on the moving part, downward obviously (weight of the wire) but also radial and tangential, probably lower in magnitude but not 0. The pendulum finds its initial rest position (before heating) with those forces present and the braided wires have certainly some amount of stress in metal leads and "plastic" insulator around. Heating will change the stiffness, in particular for the insulators, and for a given shape (that of rest position before heating) will change the forces, that is on top of thermal expansion that also changes forces from a given geometry (or change new rest geometry for a given force). Not sure what would be the dominant factor. Anyway, change in force (on moving part) due to heating will result from a tendency of the braided wires to bend more or less around the axis of the "J curve" (axis horizontal, almost orthogonal to pendulum arm) and also from a tendency to twist more or less around the axis of the wires (roughly vertical and parallel to axis of rotation of pendulum near the fixed part, roughly horizontal and parallel to pendulum arm near moving part).
The bending change of force would have reverse effects on measurements when the wires are moved to opposite side of pendulum's arm (or rather axis of rotation), but such force is almost radial and since the measurements are (ideally) about the resulting torque around the axis of rotation the impact of such force should be greatly reduced. On the other hand, the "twist" in the upper part of the wire (where it is running down from the fixed part) will directly translate to a measured torque and this is not reversed when running the power leads one side or the other of the axis.
So, I don't know if heating induced thermal expansion and/or stiffness change (in presence of stress before) of the power leads play a significant role in your measurements, but running the braided wires the other way around, while simple to do, does not conclusively address the issue I think. I know this is speaking from the comfort of my armchair, and a bit late as a critical comment also. Thanks for doing the experiments and sharing.
Side note : following remarks on stray magnetic fields around, specifically for those experiments with a magnetron flown with the frustum, the permanent magnet of magnetron will contribute defining rest position wrt. geomagnetic field and stiffness of pendulum. This would not be relevant to measurements (device on vs off) if it was just a constant offset of static force. But heating a permanent magnet changes its magnetic field (reversibly, in operating range) so this would change the new rest position (when device on) and mimic an anomalous force. I don't think this was discussed before. Again (from the comfort of my armchair) I think it is important for those experiments to get a grip on ambient magnetic field by using Helmholtz coil and doing parametric study (all other things being equal). By changing the ambient DC magnetic field one should observe a change in the off rest position, and a significant change in delta on/off when ambient crosses 0 if the anomalous measurement is dependent on such spurious coupling, or no significant change in delta on/off when ambient crosses 0 if anomalous measurement doesn't rely on coupling with local magnetic field.
Wish I had time for drawings, hope words are clear enough.
