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#600 by Bob012345 on 15 Jul, 2017 20:37
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I've stated from the start of my visits over 2 years ago, building several cavities and test stands that you can not have an enclosed system that self accelerates without having a path to the outside universe, otherwise stuff just bounces around according to dear old Maxwell. https://en.wikipedia.org/wiki/Maxwell%27s_equationsAll I'm really interested in is this:
I assume you have not taken any classes in electrodynamics, because if you had , it shouldn't be hard to find dozens of errors in Shawyer's papers.
https://www.newscientist.com/data/images/ns/av/shawyertheory.pdf
Now, tell me how that is wrong.
First error is the original statement of force difference between the end plates ignoring the force on the sidewalls.
Next, he points to the Lorentz force equation and puts in the "group velocity of the EM wave" rather than "the velocity of the charged particle" which is the actual definition of v in that equation.
After that he makes a claim that special relativity magically makes it an open system. There is nothing in special relativity that supports this.
And that is all just on one page. Need more?
Even Dr. white's theory of Quantum Vacuum particles makes it outside the cavity and as does the MACH effects, or any theory that maybe causes a gravitational link or disturbance to the outside.
Anyone who says that they have a theory of just particles of light or matter that stay inside of the enclosed frame environment of the cavity bouncing around giving thrusts is going to get flack and questions asked in do they understand completely Maxwell's equations. Maxwell's equations for electromagnetism have been called the "second great unification in physics" and I don't believe Shawyer's theories trump them.
Further reading.
https://en.wikipedia.org/wiki/Conservation_law
My Very Best,
Shell
Since the microwave energy gets converted to heat (phonons) in the cavity walls, and that heat must ultimately propagate to the outer surface and radiate away, it's not truly a closed system with respect to radiation and should be able to at least provide a reduced photon thrust in some direction by virtue of the Stephan-Boltzmann law of radiation. I don't think it's even possible to have a perfectly closed system with respect to electromagnetic radiation. Has anyone modeled that aspect of the cavity design? Thanks. -
#601 by X_RaY on 15 Jul, 2017 20:51
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This was discussed during the past threads from the beginning on. We talk about much more thrust than generated by a photon rocket.
I've stated from the start of my visits over 2 years ago, building several cavities and test stands that you can not have an enclosed system that self accelerates without having a path to the outside universe, otherwise stuff just bounces around according to dear old Maxwell. https://en.wikipedia.org/wiki/Maxwell%27s_equationsAll I'm really interested in is this:
I assume you have not taken any classes in electrodynamics, because if you had , it shouldn't be hard to find dozens of errors in Shawyer's papers.
https://www.newscientist.com/data/images/ns/av/shawyertheory.pdf
Now, tell me how that is wrong.
First error is the original statement of force difference between the end plates ignoring the force on the sidewalls.
Next, he points to the Lorentz force equation and puts in the "group velocity of the EM wave" rather than "the velocity of the charged particle" which is the actual definition of v in that equation.
After that he makes a claim that special relativity magically makes it an open system. There is nothing in special relativity that supports this.
And that is all just on one page. Need more?
Even Dr. white's theory of Quantum Vacuum particles makes it outside the cavity and as does the MACH effects, or any theory that maybe causes a gravitational link or disturbance to the outside.
Anyone who says that they have a theory of just particles of light or matter that stay inside of the enclosed frame environment of the cavity bouncing around giving thrusts is going to get flack and questions asked in do they understand completely Maxwell's equations. Maxwell's equations for electromagnetism have been called the "second great unification in physics" and I don't believe Shawyer's theories trump them.
Further reading.
https://en.wikipedia.org/wiki/Conservation_law
My Very Best,
Shell
Since the microwave energy gets converted to heat (phonons) in the cavity walls, and that heat must ultimately propagate to the outer surface and radiate away, it's not truly a closed system with respect to radiation and should be able to at least provide a reduced photon thrust in some direction by virtue of the Stephan-Boltzmann law of radiation. I don't think it's even possible to have a perfectly closed system with respect to electromagnetic radiation. Has anyone modeled that aspect of the cavity design? Thanks.QuoteThe observed thrust of experimental results has been argued to exceed the maximum efficiency of a perfectly collimated photon rocket, comprised between 3.33 and 6.67 µN/kW.
http://www.wikiwand.com/en/RF_resonant_cavity_thruster
However, you are right about there is no perfect isolated system within our universe. -
#602 by LowerAtmosphere on 15 Jul, 2017 22:20
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This was discussed during the past threads from the beginning on. We talk about much more thrust than generated by a photon rocket.
--All I'm really interested in is this:
I assume you have not taken any classes in electrodynamics, because if you had , it shouldn't be hard to find dozens of errors in Shawyer's papers.
https://www.newscientist.com/data/images/ns/av/shawyertheory.pdf
Now, tell me how that is wrong.
First error is the original statement of force difference between the end plates ignoring the force on the sidewalls.
Next, he points to the Lorentz force equation and puts in the "group velocity of the EM wave" rather than "the velocity of the charged particle" which is the actual definition of v in that equation.
After that he makes a claim that special relativity magically makes it an open system. There is nothing in special relativity that supports this.
And that is all just on one page. Need more?
Further reading.
https://en.wikipedia.org/wiki/Conservation_law
My Very Best,
Shell
Since the microwave energy gets converted to heat (phonons) in the cavity walls, and that heat must ultimately propagate to the outer surface and radiate away, it's not truly a closed system with respect to radiation and should be able to at least provide a reduced photon thrust in some direction by virtue of the Stephan-Boltzmann law of radiation. I don't think it's even possible to have a perfectly closed system with respect to electromagnetic radiation. Has anyone modeled that aspect of the cavity design? Thanks.QuoteThe observed thrust of experimental results has been argued to exceed the maximum efficiency of a perfectly collimated photon rocket, comprised between 3.33 and 6.67 µN/kW.
http://www.wikiwand.com/en/RF_resonant_cavity_thruster
However, you are right about there is no perfect isolated system within our universe.
One had only look at gravitational waves and high energy errant photons. Oh and the QV is of course always volatile with spontaneous aggregations of energy. There used to be a great deal of enthusiasm about the excitation of the quantum vacuum. Well https://en.wikipedia.org/wiki/Woodward_effect -
#603 by Req on 16 Jul, 2017 01:39
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I've stated from the start of my visits over 2 years ago, building several cavities and test stands that you can not have an enclosed system that self accelerates without having a path to the outside universe, otherwise stuff just bounces around according to dear old Maxwell. https://en.wikipedia.org/wiki/Maxwell%27s_equationsAll I'm really interested in is this:
I assume you have not taken any classes in electrodynamics, because if you had , it shouldn't be hard to find dozens of errors in Shawyer's papers.
https://www.newscientist.com/data/images/ns/av/shawyertheory.pdf
Now, tell me how that is wrong.
First error is the original statement of force difference between the end plates ignoring the force on the sidewalls.
Next, he points to the Lorentz force equation and puts in the "group velocity of the EM wave" rather than "the velocity of the charged particle" which is the actual definition of v in that equation.
After that he makes a claim that special relativity magically makes it an open system. There is nothing in special relativity that supports this.
And that is all just on one page. Need more?
Even Dr. white's theory of Quantum Vacuum particles makes it outside the cavity and as does the MACH effects, or any theory that maybe causes a gravitational link or disturbance to the outside.
Anyone who says that they have a theory of just particles of light or matter that stay inside of the enclosed frame environment of the cavity bouncing around giving thrusts is going to get flack and questions asked in do they understand completely Maxwell's equations. Maxwell's equations for electromagnetism have been called the "second great unification in physics" and I don't believe Shawyer's theories trump them.
Further reading.
https://en.wikipedia.org/wiki/Conservation_law
My Very Best,
Shell
Since the microwave energy gets converted to heat (phonons) in the cavity walls, and that heat must ultimately propagate to the outer surface and radiate away, it's not truly a closed system with respect to radiation and should be able to at least provide a reduced photon thrust in some direction by virtue of the Stephan-Boltzmann law of radiation. I don't think it's even possible to have a perfectly closed system with respect to electromagnetic radiation. Has anyone modeled that aspect of the cavity design? Thanks.
Bob, I think I speak for pretty much everybody when asking that you please do not make this into another Blacklight Power thread. Shawyer does not need or want your defense, and it's not even useful given that many people here are actually trying to advance the overall goal already. -
#604 by WarpTech on 16 Jul, 2017 14:23
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A micromechanical proof-of-principle experiment for measuring the gravitational force of milligram masses
QuoteThis paper addresses a simple question: how small can one make a
gravitational source mass and still detect its gravitational coupling to a nearby test
mass? We describe an experimental scheme based on micromechanical sensing to
observe gravity between milligram-scale source masses, thereby improving the current
smallest source mass values by three orders of magnitude and possibly even more.
We also discuss the implications of such measurements both for improved precision
measurements of Newton’s constant and for a new generation of experiments at the
interface between quantum physics and gravity.
....
For this reason, simply scaling down a Cavendish experiment is not sufficient
to measure the gravitational effects of small source masses.
Instead, we periodically modulate the gravitational potential created by a small
source mass in order to resonantly enhance the amplitude response of a cantilever test
mass.
‡ For simplicity we assume that the effective mass of the oscillator mode is identical to the gravitational
mass.
§ The frequency shifts in actual measurements of G are typically one order of magnitude higher, as
the geometry of a torsion balance pendulum is only vaguely approximated by our 1-dimensional, linear
model.
https://arxiv.org/pdf/1602.07539.pdf
I'm having a little difficulty understanding this in detail. Equation 3d seems to have different units of measure than 3a, b & c. Is x(t) a position, displacement or signal strength? Also, their terminologies like, "mechanical susceptibility" and "displacement power" are unfamiliar to me.
Anyone care to explain?
Thanks. -
#605 by Rodal on 16 Jul, 2017 19:05
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A micromechanical proof-of-principle experiment for measuring the gravitational force of milligram masses
QuoteThis paper addresses a simple question: how small can one make a
gravitational source mass and still detect its gravitational coupling to a nearby test
mass? We describe an experimental scheme based on micromechanical sensing to
observe gravity between milligram-scale source masses, thereby improving the current
smallest source mass values by three orders of magnitude and possibly even more.
We also discuss the implications of such measurements both for improved precision
measurements of Newton’s constant and for a new generation of experiments at the
interface between quantum physics and gravity.
....
For this reason, simply scaling down a Cavendish experiment is not sufficient
to measure the gravitational effects of small source masses.
Instead, we periodically modulate the gravitational potential created by a small
source mass in order to resonantly enhance the amplitude response of a cantilever test
mass.
‡ For simplicity we assume that the effective mass of the oscillator mode is identical to the gravitational
mass.
§ The frequency shifts in actual measurements of G are typically one order of magnitude higher, as
the geometry of a torsion balance pendulum is only vaguely approximated by our 1-dimensional, linear
model.
https://arxiv.org/pdf/1602.07539.pdf
I'm having a little difficulty understanding this in detail. Equation 3d seems to have different units of measure than 3a, b & c. Is x(t) a position, displacement or signal strength? Also, their terminologies like, "mechanical susceptibility" and "displacement power" are unfamiliar to me.
Anyone care to explain?
Thanks.
Equation 3d has the same dimensions as Eq. 3c
x is a position
(x0 − xsup) is a difference between positions, which gives you a displacement
"displacement power spectral density" Sxx has units of m2 /Hz = s* m2 (square meter times second)
Explanation: Power spectral density in general has units of whatever you measure, squared, and divided by frequency. So if you measure displacement in meters, it has units of m2/Hz.
If you measure acceleration in g's, "acceleration power spectral density" would have units of g2/Hz or if you measure in (m/s2) it would have units of (m/s2)2/Hz. If you measure velocity, "velocity power spectral density"will have units of (m/s)2/Hz.
"displacement power" Pxx has units of m2 (square meter)
Explanation: "displacement power" Pxx is defined in p.5 immediately prior to Eq. (3) as the integral of Sxx with respect to omega. So the integral of something measured in m2/Hz with respect to frequency, will have units of m2.
The transfer function TE is dimensionless (the way it is defined in the paper)
Q is dimensionless
"mechanical susceptibility" is defined in the text, p.5 after Eq. (2)
Think of these quantities as "analogs to..." Obviously m^2 is a unit of surface area, and not a unit of power, and so on. Ditto for the transfer function being dimensionless...
Or you can think of some implicit constants being defined as unity, like in General Relativity, we take the speed of light c=1, instead of 299 792 458 m / s, and we use in GR -t +x instead of -t*c +x , which looks wrong since time and position have different units, but it is right if you take c=1 .
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#606 by Mark7777777 on 16 Jul, 2017 20:34
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http://cannae.com/cannae-exhibit-at-the-henry-ford-museum/
JUNE 2, 2017
CANNAE EXHIBIT AT THE HENRY FORD MUSEUM
Our media partner House Industries has a display at the Henry Ford Museum, in Dearborn, MI. In the display are featured artifacts, images and design work for the Cannae Inc. thruster project. In this image, you can see a copper resonating cavity, some 3D printed satellite models and a variety of Cannae logos and images. House Industries also has 2 pages featured on Cannae in their new book, “House Industries: The Process is the Inspiration,” which is available on Amazon.
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#607 by WarpTech on 16 Jul, 2017 21:34
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No, it doesn't. 3c has units of m2 but 3d has units of m2/s. There's no way to multiply by c and get that. The Gamma at the end makes the units different.A micromechanical proof-of-principle experiment for measuring the gravitational force of milligram masses
...
https://arxiv.org/pdf/1602.07539.pdf
I'm having a little difficulty understanding this in detail. Equation 3d seems to have different units of measure than 3a, b & c. Is x(t) a position, displacement or signal strength? Also, their terminologies like, "mechanical susceptibility" and "displacement power" are unfamiliar to me.
Anyone care to explain?
Thanks.
Equation 3d has the same dimensions as Eq. 3cQuote from: Rodal
Okay, but equation [2] has units of m2, not m2/Hz, and the 2nd term on 2nd line of equation [2] has units of Watts/kg. It contradicts the definition.
x is a position
(x0 − xsup) is a difference between positions, which gives you a displacement
"displacement power spectral density" Sxx has units of m2 /Hz = s* m2 (square meter times second)
Explanation: Power spectral density in general has units of whatever you measure, squared, and divided by frequency. So if you measure displacement in meters, it has units of m2/Hz.
If you measure acceleration in g's, "acceleration power spectral density" would have units of g2/Hz or if you measure in (m/s2) it would have units of (m/s2)2/Hz. If you measure velocity, "velocity power spectral density"will have units of (m/s)2/Hz.Quote from: Rodal
"displacement power" Pxx has units of m2 (square meter)
Explanation: "displacement power" Pxx is defined in p.5 immediately prior to Eq. (3) as the integral of Sxx with respect to omega. So the integral of something measured in m2/Hz with respect to frequency, will have units of m2.
The transfer function TE is dimensionless (the way it is defined in the paper)
Q is dimensionless
"mechanical susceptibility" is defined in the text, p.5 after Eq. (2)
Think of these quantities as "analogs to..." Obviously m^2 is a unit of surface area, and not a unit of power, and so on. Ditto for the transfer function being dimensionless...
Or you can think of some implicit constants being defined as unity, like in General Relativity, we take the speed of light c=1, instead of 299 792 458 m / s, and we use in GR -t +x instead of -t*c +x , which looks wrong since time and position have different units, but it is right if you take c=1 .
Thanks for this. The right constant's to use to correct the units are not obvious. It makes it very difficult to follow and gain understanding when the equations fail dimensional analysis. It's fine to abbreviate in cases where it does not cause confusion. This is not one of those cases IMO.
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#608 by Rodal on 16 Jul, 2017 21:42
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Wrong. Eq. 3d has unit of m2 as I proved previously.
No, it doesn't. 3c has units of m2 but 3d has units of m2/s.A micromechanical proof-of-principle experiment for measuring the gravitational force of milligram masses
...
https://arxiv.org/pdf/1602.07539.pdf
I'm having a little difficulty understanding this in detail. Equation 3d seems to have different units of measure than 3a, b & c. Is x(t) a position, displacement or signal strength? Also, their terminologies like, "mechanical susceptibility" and "displacement power" are unfamiliar to me.
Anyone care to explain?
Thanks.
Equation 3d has the same dimensions as Eq. 3c
Pxx has units of m2
Q is dimensionless
TE is dimensionless
Sxx has units of m2/Hz=m2 / (1/sec)=m2 * s
Γ has units of 1/sec
If you multiply
Q xTE xSxx xΓ= 1 x 1 x (m2 /(1/s)) x (1/s) = m2
When you multiply (m2/Hz) * (1/s) you get units of m2, because the unit of time cancels out.
Please let's not make this into a never ending argument back and forth on whether you don't like their choice of units. You said that you had difficulty understanding the units. They are as I defined above. Equations 3 c an 3 d have the same units for displacement power: m2.
I am not going to discuss the units in the other equations as I have work to do and I thought I was just helping here, did not want to get into a never ending argument about the definition of units in somebody else's paper
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#609 by flux_capacitor on 16 Jul, 2017 22:57
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No much EmDrive news lately… Since this thread has somewhat calmed down a bit, for those interested in the hypothesis of negative energy states and negative mass in cosmology, and their profound implication for spacetime engineering and interstellar travel, I wrote a few posts starting page 4 and ongoing of the parallel NSF thread Theoretical FTL.
This is not directly connected to resonant cavities but as an aside on the (anti)gravitational hypotheses of the EmDrive as well as space flight applications, some readers here may find the subject interesting, in particular for those interested in the theoretical aspects of general relativity.
Thanks to dustinthewind for having triggered those posts with his former questions about the lack of primordial antimatter in the universe, dark matter, the direction of the arrow of time, and parallel dimensions, which perfectly fit the cosmological model presented.
There is also a theoretical discussion about how energy density and pressure are related in physics, and the sign (positive or negative) of those quantities in general relativity.
Comments are welcome of course: https://forum.nasaspaceflight.com/index.php?topic=13542.msg1701797#msg1701797
N.B.: A series of videos presenting the model are currently in the process of being subtitled in English. More on that later. -
#610 by chongma on 17 Jul, 2017 10:49
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recent summary video posted on youtube
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#611 by Rodal on 17 Jul, 2017 14:29
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I have a few more minutes now to show that Eq. 3b (the displacement power due to thermal noise) in p.5 of https://arxiv.org/pdf/1602.07539.pdf is also dimensionally correct and consistent with the units of Eq. 3c and 3d, all of these additive terms having units of square meter, the unit of displacement power Pxx:
Wrong. Eq. 3d has unit of m2 as I proved previously.
No, it doesn't. 3c has units of m2 but 3d has units of m2/s.A micromechanical proof-of-principle experiment for measuring the gravitational force of milligram masses
...
https://arxiv.org/pdf/1602.07539.pdf
I'm having a little difficulty understanding this in detail. Equation 3d seems to have different units of measure than 3a, b & c. Is x(t) a position, displacement or signal strength? Also, their terminologies like, "mechanical susceptibility" and "displacement power" are unfamiliar to me.
Anyone care to explain?
Thanks.
Equation 3d has the same dimensions as Eq. 3c
Pxx has units of m2
Q is dimensionless
TE is dimensionless
Sxx has units of m2/Hz=m2 / (1/sec)=m2 * s
Γ has units of 1/sec
If you multiply
Q xTE xSxx xΓ= 1 x 1 x (m2 /(1/s)) x (1/s) = m2
When you multiply (m2/Hz) * (1/s) you get units of m2, because the unit of time cancels out.
Please let's not make this into a never ending argument back and forth on whether you don't like their choice of units. You said that you had difficulty understanding the units. They are as I defined above. Equations 3 c an 3 d have the same units for displacement power: m2.
I am not going to discuss the units in the other equations as I have work to do and I thought I was just helping here, did not want to get into a never ending argument about the definition of units in somebody else's paper
Pxx (displacement power) has units of m2
Q (quality factor of resonance) is dimensionless
ωo (angular frequency) has units of 1/s
kB (https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_constant) has units of J/K=(kg*m2/s2)/K (where m=meter). Where I used the fact that energy has units of force times length, and force has units of mass times acceleration, therefore energy has units of mass times acceleration times length.
m (test mass) has units of kg
T (temperature) has units of K (Kelvin)
Γ has units of 1/sec
If you multiply
Q x(1/ωo 3)xkBxTx(1/m)xΓ= 1 x(s3)x[(kg m2/s2)/K]x(1/kg)xKx(1/s)=m2
you get square meters, the units of "displacement power" Pxx (*)
QED (quod erat demonstrandum)
Now, for Eq. 3a (the displacement power due to gravitational force) in p.5 of https://arxiv.org/pdf/1602.07539.pdf is also dimensionally correct and consistent with the units of Eq. 3b, 3c and 3d, all of these additive terms having units of square meter, the unit of displacement power Pxx:
Pxx (displacement power) has units of m2
Q (quality factor of resonance) is dimensionless
ωo (angular frequency) has units of 1/s
G (Newton's gravitational constant) has units of m3/(kg s2)
M (source mass) has units of kg
ds (displacement amplitude of sinusoidal excitation of the source mass) has units of length (m)
do (distance between source mass M to the test mass m) has units of length (m)
If you multiply
Q x(1/ωo 4)x(G M)2xds2/do6= 1 x(s4)x{[m3/(kg s2)] kg}2x m2/m6=m2
you get square meters, the units of "displacement power" Pxx
QED (quod erat demonstrandum)
-------------------
(*) The right hand side of the equation is expressed in terms of SI units. Therefore, while "m" in the left-hand side of the equation stands for mass -the symbol used by the authors-, notice that "m" in the right-hand side of the equation stands for the SI unit of length: meter.
"x" stands for "times": the multiplication symbol.
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#612 by WarpTech on 17 Jul, 2017 19:03
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...you get square meters, the units of "displacement power" Pxx
QED (quod erat demonstrandum)
-------------------
(*) The right hand side of the equation is expressed in terms of SI units. Therefore, while "m" in the left-hand side of the equation stands for mass -the symbol used by the authors-, notice that "m" in the right-hand side of the equation stands for the SI unit of length: meter.
"x" stands for "times": the multiplication symbol.
What you say is perfectly correct. I saw this yesterday "provided" Sxx has units of m2/Hz. However, if you look at the first line of their equation 2, I get m2, but line 2 of equation 2 gives m2/Hz. So there seems to be a factor of "time" missing in the first line. That is what caused the confusion. It must be that the Delta function has units of "s", because I don't see anywhere else it could be. -
#613 by Bob012345 on 17 Jul, 2017 19:06
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This was discussed during the past threads from the beginning on. We talk about much more thrust than generated by a photon rocket.
I've stated from the start of my visits over 2 years ago, building several cavities and test stands that you can not have an enclosed system that self accelerates without having a path to the outside universe, otherwise stuff just bounces around according to dear old Maxwell. https://en.wikipedia.org/wiki/Maxwell%27s_equationsAll I'm really interested in is this:
I assume you have not taken any classes in electrodynamics, because if you had , it shouldn't be hard to find dozens of errors in Shawyer's papers.
https://www.newscientist.com/data/images/ns/av/shawyertheory.pdf
Now, tell me how that is wrong.
First error is the original statement of force difference between the end plates ignoring the force on the sidewalls.
Next, he points to the Lorentz force equation and puts in the "group velocity of the EM wave" rather than "the velocity of the charged particle" which is the actual definition of v in that equation.
After that he makes a claim that special relativity magically makes it an open system. There is nothing in special relativity that supports this.
And that is all just on one page. Need more?
Even Dr. white's theory of Quantum Vacuum particles makes it outside the cavity and as does the MACH effects, or any theory that maybe causes a gravitational link or disturbance to the outside.
Anyone who says that they have a theory of just particles of light or matter that stay inside of the enclosed frame environment of the cavity bouncing around giving thrusts is going to get flack and questions asked in do they understand completely Maxwell's equations. Maxwell's equations for electromagnetism have been called the "second great unification in physics" and I don't believe Shawyer's theories trump them.
Further reading.
https://en.wikipedia.org/wiki/Conservation_law
My Very Best,
Shell
Since the microwave energy gets converted to heat (phonons) in the cavity walls, and that heat must ultimately propagate to the outer surface and radiate away, it's not truly a closed system with respect to radiation and should be able to at least provide a reduced photon thrust in some direction by virtue of the Stephan-Boltzmann law of radiation. I don't think it's even possible to have a perfectly closed system with respect to electromagnetic radiation. Has anyone modeled that aspect of the cavity design? Thanks.QuoteThe observed thrust of experimental results has been argued to exceed the maximum efficiency of a perfectly collimated photon rocket, comprised between 3.33 and 6.67 µN/kW.
http://www.wikiwand.com/en/RF_resonant_cavity_thruster
However, you are right about there is no perfect isolated system within our universe.
Yes, I know about the enhanced thrust over a pure photon rocket! I should have been more clear. I'm wondering if there couldn't be some mechanism of enhancement due to the configuration since photon thrust is power/c yet radiated power goes as temperature raised to the fourth power as per the Stephan-Boltzmann Law. Photons, phonons and power are not conserved quantities. Energy is. So for a given energy, can the photon power, and thus thrust, be enhanced by clever use of the SB law? That was implied in the modeling question. I suspect not but don't know. If so, it would certainly give new meaning to the term 'heat engine'
Thanks.
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#614 by Rodal on 17 Jul, 2017 19:28
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Their Eq. (2) is also correct dimensionally. The text calls the Appendix, where the origin of this equation is fully explained....you get square meters, the units of "displacement power" Pxx
QED (quod erat demonstrandum)
-------------------
(*) The right hand side of the equation is expressed in terms of SI units. Therefore, while "m" in the left-hand side of the equation stands for mass -the symbol used by the authors-, notice that "m" in the right-hand side of the equation stands for the SI unit of length: meter.
"x" stands for "times": the multiplication symbol.
What you say is perfectly correct. I saw this yesterday "provided" Sxx has units of m2/Hz. However, if you look at the first line of their equation 2, I get m2, but line 2 of equation 2 gives m2/Hz. So there seems to be a factor of "time" missing in the first line. That is what caused the confusion. It must be that the Delta function has units of "s", because I don't see anywhere else it could be.
1) the first factor in absolute value brackets in their Eq. (2), χ, has dimensions of inverse squared frequency (1/ω2), hence it has dimensions of time squared, in SI units: s2
2) the second factor in absolute value brackets in their Eq. (2), is defined as the amplitude of the system and it has dimensions of frequency squared ω2, which, when squared and multiplied by the first factor squared, χ2 , cancel out, so all you have left is dS2 which has dimensions of length square times the Dirac Delta function of frequency.
By standard definition, the Dirac delta function always has the same dimensions as the inverse of its argument,
(if you forgot this fact and you need an Internet reference, see for example: https://physics.stackexchange.com/questions/33760/what-are-the-units-or-dimensions-of-the-dirac-delta-function)
since the argument of the Dirac delta function in Eq.(2) is frequency, the Dirac delta functions has dimensions of inverse frequency in this case.
Therefore you are left with dimensions of length square divided by frequency, which is consistent with the definition of displacement power spectral density : m2/ Hz.
consistent with what I posted here: https://forum.nasaspaceflight.com/index.php?topic=42978.msg1702885#msg1702885
Explanation: Power spectral density in general has units of whatever you measure, squared, and divided by frequency. So if you measure displacement in meters, it has units of m2/Hz.
If you measure acceleration in g's, "acceleration power spectral density" would have units of g2/Hz or if you measure in (m/s2) it would have units of (m/s2)2/Hz. If you measure velocity, "velocity power spectral density"will have units of (m/s)2/Hz.
QED (quod erat demonstrandum)
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#615 by Monomorphic on 18 Jul, 2017 02:07
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3D printer arrived! I have it set up and fully calibrated. A test model was printed and the quality is very good!
I should be able to start printing the spherical end-plates as early as tomorrow. I want to double check the geometry of the 3D models as they were created several weeks ago.
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#616 by spupeng7 on 18 Jul, 2017 07:17
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3D printer arrived! I have it set up and fully calibrated. A test model was printed and the quality is very good!
I should be able to start printing the spherical end-plates as early as tomorrow. I want to double check the geometry of the 3D models as they were created several weeks ago.
Monomorphic,
I am concerned that this highly distracting device will delay your test campaign
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#617 by flux_capacitor on 18 Jul, 2017 10:12
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2nd paper about the EmDrive and McCulloch's MiHsC just published in EPL and freely available:
• McCulloch, M. E. (July 2017). "Testing quantised inertia on emdrives with dielectrics" (PDF). EPL. 118 (3). doi:10.1209/0295-5075/118/34003.
In this second EPL paper, the speed of light does not vary anymore within the cavity. Instead, more Unruh waves fit the wide end of the cavity than its narrow end, continuously shifting the center of inertial mass of the microwaves towards the wide end: the cavity then has to move towards the small end, for momentum to be conserved. As the speed of light does not change in the latter model, there is no more relativistic violation, which was the main criticism of the 1st paper. -
#618 by Monomorphic on 18 Jul, 2017 14:45
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There area a few ways of doing this, but to conserve filament and maintain geometry and strength, I will be using break-away supports. I'm also able to save on filament using a 20% infill. Each big end spherical quarter is ~136 cm3. For a total of 544 cm3. Which is something like 250 meters of filament.
Correction, Pronterface says each quarter is 63 meters of filament and will take 18 hours each to print!
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#619 by WarpTech on 18 Jul, 2017 16:53
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2nd paper about the EmDrive and McCulloch's MiHsC just published in EPL and freely available:
• McCulloch, M. E. (July 2017). "Testing quantised inertia on emdrives with dielectrics" (PDF). EPL. 118 (3). doi:10.1209/0295-5075/118/34003.
In this second EPL paper, the speed of light does not vary anymore within the cavity. Instead, more Unruh waves fit the wide end of the cavity than its narrow end, continuously shifting the center of inertial mass of the microwaves towards the wide end: the cavity then has to move towards the small end, for momentum to be conserved. As the speed of light does not change in the latter model, there is no more relativistic violation, which was the main criticism of the 1st paper.
I read the paper a couple of weeks ago. His equations apparently conserve momentum between the frustum and the EM momentum inside, but it still remains unclear how momentum escapes from inside the cavity to propel the frustum. He relies on his theory and the assumption that the difference in the Unruh spectrum between the two ends is enough to result in thrust. IMO, the real mechanism is the power being lost "inefficiently" to heating the copper and the air, where the exchange of momentum between the photons and the random moving atoms is non-conservative.
Todd
Thanks.
