Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2168078 times)

Offline ThinkerX

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Ok, time for me to humiliate myself here yet again (I have been doing so in these threads since last year at regular intervals).

Something I have thought about before and am thinking about once again is thermal effects.  Virtually every engineer or theorist to jump into these EM Drive threads has cited thermal artefacts as a possible solution for EM Drive thrust.  Doctor Rodal's first posts in the original thread dealt with this.   Back then, the reasoning was:

1) thermal effects could mimic significant thrust in ambient air conditions;

2) those same thermal effects would still produce thrust, albeit far smaller thrust, in a vacuum on earth;

3) but in space, the thermal effects, would, at best result in rotary motion (not propelling the device).

Thus far, we have multiple experiments producing significant thrust in ambient air. (in accord with point 1)  The one series of known vacuum tests produced far less thrust (in accord with point 2).  Now, nobody has yet shot an EM Drive device into space...but there is an interesting comparable case:

http://en.wikipedia.org/wiki/Pioneer_anomaly

Quote
Various explanations, both of spacecraft behavior and of gravitation itself, were proposed to explain the anomaly. Over the period 1998–2012, one particular explanation became accepted. The spacecraft, which are surrounded by an ultra-high vacuum and are each powered by a radioisotope thermoelectric generator (RTG), can shed heat only via thermal radiation. If, due to the design of the spacecraft, more heat is emitted in a particular direction—what is known as a radiative anisotropy—then the spacecraft would exhibit a small acceleration in the direction opposite that of the excess emitted radiation due to radiation pressure. Because this force is due to the recoil of thermal photons, it is also called the thermal recoil force. If the excess radiation and attendant radiation pressure were pointed in a general direction opposite the Sun, the spacecrafts’ velocity away from the Sun would be decelerating at a greater rate than could be explained by previously recognized forces, such as gravity and trace friction, due to the interplanetary medium (imperfect vacuum).

By 2012 several papers by different groups, all reanalyzing the thermal radiation pressure forces inherent in the spacecraft, showed that a careful accounting of this could explain the entire anomaly, and thus the cause was mundane and did not point to any new phenomena or need for a different physical paradigm.[2][3] The most detailed analysis to date, by some of the original investigators, explicitly looks at two methods of estimating thermal forces, then states "We find no statistically significant difference between the two estimates and conclude that once the thermal recoil force is properly accounted for, no anomalous acceleration remains

A known example of 'accidental thermal effects' 'propelling' spacecraft.

Wild speculation 1:

Consensus is there is a lot of energy being pumped into these frustums.  Witness the funky colored cones posted in this thread now and again, showing a concentration of energy/heat at the 'big end.'  So...could the EM Drive be some sort of focused or amplified thermal drive - a vastly 'improved' version of the forces behind the 'pioneer anomaly?'  This sort of leads to

Wild speculation 2:

If this is some sort of thermal drive, then perhaps...thrust remains constant only for so long as the temperature increases?  So, if the temperature plateau's, then thrust declines.  This might provide a solution of sorts to the Conservation of Energy issue.

If this is the case, then the EM drive might make for a useful orbital thruster, and perhaps power interplanetary probes...but would it be adequate for interstellar propulsion?



Offline deltaMass

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If the power supply, RF gear and cavity were in a box in space, you wouldn't need to speculate. You would simply need to observe whether the box went off-geodesic (after having removed solar origins of residual thrust)

Well, thermal origins also (like Pioneer)
« Last Edit: 05/10/2015 10:12 PM by deltaMass »

Online A_M_Swallow

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{snip}
Wild speculation 2:

If this is some sort of thermal drive, then perhaps...thrust remains constant only for so long as the temperature increases?  So, if the temperature plateau's, then thrust declines.  This might provide a solution of sorts to the Conservation of Energy issue.

If this is the case, then the EM drive might make for a useful orbital thruster, and perhaps power interplanetary probes...but would it be adequate for interstellar propulsion?



An interstellar probe would have to be powered by a nuclear reactor. Currently the cooling radiators for nuclear reactors are flat but curving metal, to produce the equivalent of a nozzle, is not hard.

Offline zellerium

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Something I have thought about before and am thinking about once again is thermal effects.  Virtually every engineer or theorist to jump into these EM Drive threads has cited thermal artefacts as a possible solution for EM Drive thrust.  Doctor Rodal's first posts in the original thread dealt with this.   Back then, the reasoning was:

1) thermal effects could mimic significant thrust in ambient air conditions;

2) those same thermal effects would still produce thrust, albeit far smaller thrust, in a vacuum on earth;


I did some simple calculations to determine how fast the gas particles in the vacuum chamber would need to be going to produce a 50 uN force.

Pressure in the chamber = 5 E-5 Torr which corresponds to ~10^17 molecules of air. Density of air in the chamber = 10^17/(6.022*E23 molecules/mol) * .02896 kg/mol = 4.8 E-9 kg/m^3

Assuming a steady force, constant velocity across the area of the frustum, and constant density, the Reynolds Transport theorem simplifies to F = rho*v^2*A

Net pressure on the large end of the frustum [Area .25*pi*(.2794m)^2 ] = 8.42 E-4 Pa

Therefore the velocity required to produce this pressure force is sqrt( 8.42 E-4/(4.8 E-9) ) = 419 m/s.
Using the expression for the thermal kinetic energy 1/2 mv^2 = 3/2 kBT, and that a molecule of air has a mass of 4.81 E-26 kg

I found that the temperature of the gas would need to be 203 K or -70 C.

So if large end of the cavity could create that magnitude of temperature differential could the entire thrust be due to the kinetic energy of the gas?
« Last Edit: 05/10/2015 11:31 PM by zellerium »

Offline frobnicat

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@ RODAL

Just got a minute but from your p expression;

If L1/c1 = L2/c2

del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))

might be a solution ??

Got to check the thinking later.

Night !

I find your previous expression

del f = ( f/(2*c^2)) * (c1^2-c2^2)

more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length,

while on the other hand

del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))

goes to zero for equal dielectric lengths, regardless of their dielectric constants.

The previous expression is only valid approximation for a "uniformly varying dielectric".  There is no L1 and L2 in that case.

What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )

I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.

After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster.

I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.

Todd D.

Looking for a mechanical analogy :
Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.

Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.

Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.

I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?
« Last Edit: 05/11/2015 12:06 AM by frobnicat »

Offline deltaMass

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@zellerium:
70oC temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.

Offline Rodal

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@zellerium:
70oC temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.
NASA's Eagleworks base is made of glass-fiber reinforced epoxy with an extremely thin layer (deposited) of copper

on the outside: 0.063 inch thick FR4 printed circuit board and ~35 microns thick of Cu on the inside

Thermal conductivity, through-plane   0.29 W/(m·K)

« Last Edit: 05/11/2015 01:07 AM by Rodal »

Continuing with the "Test In Space" theme.

We lack cheap space access for unmanned cargo. We have no railgun running from the coast of Ecuador up into the Andes to the east, and we have no Skylon/SABRE SSTO yet.  So we must pay many thousands of dollars per launched kilogram rather than what could be only tens of dollars.

What we do have is Cubesat and SpaceX. The problem is that the devices under consideration here won't fit even into the largest Cubesat. So let's talk miniaturisation.

What we have is photons in an asymmetric cavity. So let's use light instead of microwaves. I'll stop there for now.

I have been advocating a scaled down version in a Cubesat.  I'm sure the cavity can be 3D printed, out of copper or any other metal. 

Offline Rodal

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I know Shawyer and EW have tried a dielectric in the frustum. Are there any specifications for that dielectric? Material properties? Absorption properties at microwave frequencies?

I was looking at Pyramid Absorbers for microwaves, they can attenuate up to -55dB. A high power microwave source, pumped through a diode into such an absorber, seems to me should have a higher probability of thrust than the EM Drive and relatively simple to construct.


Todd D.

I found this document with the sourcing and the material properties assumed (based on the literature) for the High Density PolyEthylene used as a dielectric at NASA Eagleworks

http://forum.nasaspaceflight.com/index.php?action=dlattach;topic=36313.0;attach=635018
« Last Edit: 05/11/2015 12:49 AM by Rodal »

Offline deltaMass

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@zellerium:
70oC temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.
The base is made of glass-fiber reinforced epoxy with an extremely thin layer (deposited) of copper

on the outside: 0.063 inch thick FR4 printed circuit board and ~35 microns thick of Cu on the inside

Thermal conductivity, through-plane   0.29 W/(m·K)
A rough calc says that the Cu thickness has to be less than 44 microns. Looks like it's possible. But I calculated heat flow around the copper, not through-plane. Someone should recheck both.
P = 50 W, dT = 70oK

Offline Rodal

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@zellerium:
70oC temperature difference across the copper? But it's an excellent conductor of heat. I can't imagine that.
The base is made of glass-fiber reinforced epoxy with an extremely thin layer (deposited) of copper

on the outside: 0.063 inch thick FR4 printed circuit board and ~35 microns thick of Cu on the inside

Thermal conductivity, through-plane   0.29 W/(m·K)
A rough calc says that the Cu thickness has to be less than 44 microns. Looks like it's possible. But I calculated heat flow around the copper, not through-plane. Someone should recheck both.
P = 50 W, dT = 70oK



Prof. Juan Yang's reported temperature vs. time measurements with embedded thermocouples throughout their EM Drive cavity (without a polymer dielectric insert) under atmospheric conditions, that, curiously, show the highest temperature at the center of the small base (trace #1), followed, at a significantly lower temperature by the temperature at the periphery of the big base (trace #5).

Take a look at the temperatures measured by the thermocouple Trace #1




Offline deltaMass

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Yabut he' she's ackling at kilowatts. EW - not so much
« Last Edit: 05/11/2015 03:19 AM by deltaMass »

Offline WarpTech

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@ RODAL

Just got a minute but from your p expression;

If L1/c1 = L2/c2

del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))

might be a solution ??

Got to check the thinking later.

Night !

I find your previous expression

del f = ( f/(2*c^2)) * (c1^2-c2^2)

more physically appealing, since it goes to zero for equal dielectric constants, regardless or their dielectric length,

while on the other hand

del f = (1/2*f)*((c1*c2)/(L1*L2))*b^2*((1/dD1^2)-(1/dD2^2))

goes to zero for equal dielectric lengths, regardless of their dielectric constants.

The previous expression is only valid approximation for a "uniformly varying dielectric".  There is no L1 and L2 in that case.

What do you think might maximize the second expression ? (valid only for L1/c1 = L2/c2 )

I was discussing this last night and we made some interesting observations. In a variable dielectric, like in the frustum, when the waves are accelerating to a higher group velocity, they are losing momentum. This momentum is lost to the material "in the direction of the wave". It is similar to frame dragging. The wave is losing energy trying to drag the waveguide or the dielectric with it.

After the wave is reflected, it again tries to drag the dielectric or frustum with it, and this time it meets more resistance. It becomes an evanescent wave and decays faster.

I do not believe a small end cap is needed and the frustum should taper all the way down to the wave guide feeding it. The reflected waves cannot reach the small plate. That's what the thermal images show as well. Most of the energy I think should be trapped at the big end.

Todd D.

Looking for a mechanical analogy :
Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.

Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.

Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.

I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?

The ball (photon) doesn't fall back down the well. There is nothing to give it back enough energy to do so. It dissipates in multiple reflections between the walls and the big end. They are not getting more out than they put in, so it does not violate conservation of energy. They are simply getting more NET momentum on one direction than in the other direction because there is more dissipation and attenuation in one direction than there is in the other. Dissipative systems are typically "not" conservative, loses prevent a true equal measure from occuring in both directions.

Todd D.


Offline Rodal

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...

Looking for a mechanical analogy :
Let's play with a bended pipe and a ball rolling in it. The pipe can constrain the ball to various path, it can rise or fall, at various steepness. Height of the pipe at a given location defines gravitational potential energy of the ball there. The ball is launched with a given velocity, and then turns around the pipe if it is a closed circuit, or goes back and forth if the two ends of the pipe are high enough, should make no difference.

Assuming no friction, the ball goes-on forever. When rising the ball loses kinetic energy, slows, and imparts momentum to the pipe. When on the return path (different part of pipe if circuit path or same part of pipe if going back and forth), the same delta height will make ball regain same kinetic energy as lost when rising, accelerate, and imparts momentum again. When taking curves, ball also imparts momentum on pipe. Integrating all those momentum exchanges on a cycle yields 0 net momentum. Not depending on path details.

Assuming a closed circuit path and friction (dry, viscous, magnetic... whatever dissipative interaction), including parts with low friction (forth) and parts with high friction (back) and arbitrary height profile (potential well whatever). After a number of cycles the ball will come to rest. Integrating all the momentum exchanges of ball on pipe (changes of height, curves, friction) will yield a total momentum equal to the initial momentum of the ball when launched. Not depending on path details and what parts are more or less dissipative.

I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?

The ball (photon) doesn't fall back down the well. There is nothing to give it back enough energy to do so. It dissipates in multiple reflections between the walls and the big end. They are not getting more out than they put in, so it does not violate conservation of energy. They are simply getting more NET momentum on one direction than in the other direction because there is more dissipation and attenuation in one direction than there is in the other. Dissipative systems are typically "not" conservative, loses prevent a true equal measure from occuring in both directions.

Todd D.
Friction is a non-conservative force of course.  The energy that is not conserved turns into heat.

Follower forces are also non-conservatve (leading to very interesting instability problems).

Flutter instability --> follower force ---> energy harvesting



Water coming out of a flexible hose: is  a follower force



and here we have flutter induced by friction (something that the great Mechanician Koiter thought was impossible)



« Last Edit: 05/11/2015 02:01 AM by Rodal »

Offline deltaMass

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@arc:
Quote
Do higher em frequencies deliver increased net effect?
Excellent question. To which as yet we have no answers. This might be taken to imply that we should find out :)

But but but...there is no guarantee that these forces in any way combine to produce a larger force. And that's static I mean. I'm even more doubtful about the free space dynamics.

Statically we care about k, or N/W
Dynamically we don't know what we care about yet  8)
« Last Edit: 05/11/2015 04:55 AM by deltaMass »

Offline deltaMass

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It seems to me, having noodled some algebra, that all 4 combinations of
- energy is conserved/not conserved
- momentum is conserved/not conserved
are algebraically possible.

Ignoring Ms. Noether then, there is no reason why not.

The relevance here is that although we seem to have a violation of CofM, we can still maintain CofE
Or vice versa
« Last Edit: 05/11/2015 06:37 AM by deltaMass »

Offline CW

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Do we really understand the nature of CoE and CoM correctly? I'm curious why nobody else seems to wonder.. . or maybe it's just me.

;)
Reality is weirder than fiction

Offline deltaMass

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Our understanding of conservation is very deep, and Emmie Noether discovered it. She found that for every symmetry or invariance there exists a corresponding conservation law. That's a rather remarkable statement!

For momentum it's the symmetry of  space. For energy it's the symmetry of time. It's actually more mathematically complex than that, since it involves differentials of the Lagrangian.
http://en.wikipedia.org/wiki/Noether%27s_theorem

Susskind has some public lectures about all the gory details, on YouTube.
« Last Edit: 05/11/2015 10:31 AM by deltaMass »

Offline frobnicat

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...
I know a photon is not a ball but my question is, in "Newtonian layman's terms" how does the line of thinking you are developing making that analogy not valid, i.e. imply apparent deviation from conservation of momentum ?

The ball (photon) doesn't fall back down the well. There is nothing to give it back enough energy to do so. It dissipates in multiple reflections between the walls and the big end. They are not getting more out than they put in, so it does not violate conservation of energy. They are simply getting more NET momentum on one direction than in the other direction because there is more dissipation and attenuation in one direction than there is in the other. Dissipative systems are typically "not" conservative, loses prevent a true equal measure from occuring in both directions.

Todd D.

Classically, a dissipative system is conservative for both momentum and energy, it's just that for energy there is a (irreversible) conversion to a degraded form of energy, but there is no such thing as a mysterious part of total energy that would simply vanish. Even if not always convenient, an open system can be seen as part of a bigger closed system, and short of that the deltas total energy and total momentum of an open system can still be accounted, at least in principle, as integrated fluxes exchanged between open system and an outside.

My Newtonian ball of momentum pb can encounter an arbitrarily varying Force Fcb(t) of container on ball (vectors in bold). And dpb/dt=Fcb=-Fbc=-dpc/dt. That is instant conservation of momentum, and obviously integrating on successive instants just yields delta_pb=-delta_pc or  delta_pb+delta_pc=0, conservation of momentum on any time interval whatever the shape of varying Force Fcb(t). Where and how quantitatively your system is showing an apparent breaking of CoM at an "instantaneous scale" dt ? Short of that, details of trajectory is just, ahem, arm waving for propulsion purpose (aka Dean drive).
« Last Edit: 05/11/2015 11:53 AM by frobnicat »

Offline deltaMass

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Just an FYI that you have buttons to do superscripts and subscripts up there - sup and sub.
e.g. pb2

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