-
#380 by Rodal on 06 Sep, 2016 15:23
-
I hope there is no confusion re the various projects going on at Cannae...as folks know I've been conducting a study of them
So they are using miles instead of kilometers to specify an orbit?
1) for the cubesat, they say they are orbiting it at less than 150 miles...not kilometers. Mr. Feta told me there would be no super-cooling of this device. I'm left to conclude that the cooling of their thruster will be passive and that the thruster will be kept in shadow at all times
...
150*1.60934=241
So their orbit is really 241 km ?
That makes a difference ! Thanks
I calculated with 150 km. (I recall pointing this out weeks ago, when they first announced, that instead of using customary SI units, Cannae is using Miles, oh well
)
Just like the probe that crashed on Mars years ago (different units !)
Will recalculate with 150 miles tomorrow (Is it US Miles
)
US Survey mile = International mile =1.60934 km
Nautical mile = 1.852 km
Roman mile = 1.481 km
Chinese mile = 0.5 km
Let's work through the numbers for Cannae's proposed Cubesat mission, using 240 km instead of 150 km:
http://cannae.com/cubesat/
http://cannae.com/cannae-is-developing-a-cubesat-thruster/
http://www.popularmechanics.com/science/energy/a22678/em-drive-cannae-cubesat-reactionless/
The publicity picture appears to show a larger than 1x3U Cubesat, the second link talks about a 6U Cubesat
Orbit (assume circular orbit at published distance, interpreted as US Miles)
ro=150 USmile *1.60934 km/USmile ~ 240 km
Orbital velocity (Assuming circular orbit at 240 km)
G=6.67408 * 10^-11 m^3 kg^-1 s^-2
M=5.972 * 10^24 kg (mass of the Earth)
R=6.371*10^6 m (mean radius of the Earth)
r = R + ro
= 6.371*10^6 m+ 240*10^3 m
v=7765 m/sec
Drag Surface area: assume a minimum cross-sectional area, for a 1x3U Cubesat with cross-sectional drag surface of 0.10m x 0.30 m, perpendicular to the orbital plane(this assumes that the solar panels are always parallel to the orbital velocity vector)
Assume minimum configuration:
1x3U Cubesat (Notice that picture shows a larger Cubesat and link discusses a 6U Cubesat. The thrust necessary for larger Cubesats can be obtained by simple scaling of the appropriate cross-sectional area. For example, a 2x3U Cubesat will have twice the minimum cross-sectional area of a 1x3U Cubesat)
A=0.10m *0.30m
= 0.03 m^2
Drag coefficient
CD=2 (*)
Reynerson, "Aerodynamic Disturbance Force and Torque Estimation for Spacecraft and Simple Shapes Using Finite Plate Elements Part I: Drag Coefficient"
https://www.researchgate.net/publication/221910818_Aerodynamic_Disturbance_Force_
and_Torque_Estimation_for_Spacecraft_and_Simple_Shapes_Using_Finite_Plate_Elements_
Part_I_Drag_Coefficient/figures?lo=1
de Vries, "Cubesat Drag Calculations "https://e-reports-ext.llnl.gov/pdf/433600.pdf
Olttroge et.al.,"An evaluation of Cubesat Orbital Decay",
http://digitalcommons.usu.edu/cgi/viewcontent.cgi?article=1144&context=smallsat
Atmospheric Density
MSISE90 std atmosphere (for 240 km)
(References:
This link enables the computation and plotting of any subset of MSIS parameters:
http://omniweb.gsfc.nasa.gov/vitmo/msis_vitmo.html
http://ccmc.gsfc.nasa.gov/models/modelinfo.php?model=MSISE ) (*)
@Mean solar activity rhoMean= 9.91 x 10^-11 kg m^-3
@Maximum solar activity rhoMax= = 4.08 x 10^-10 kg m^-3
rhoMax/rhoMean=4.117
The solar cycle is very important,
since air density, and hence drag, is very much dependent on solar activity. It is what brought Skylab down ( https://en.wikipedia.org/wiki/Skylab#Solar_activity ):QuoteBritish mathematician Desmond King-Hele of the Royal Aircraft Establishment predicted in 1973 that Skylab would de-orbit and crash to earth in 1979, sooner than NASA's forecast, because of increased solar activity. Greater-than-expected solar activity heated the outer layers of Earth's atmosphere and increased drag on Skylab.
Observe the chart below for where we are now, and the predicted activity in the future:
Drag Force
DMax =(1/2) CD rhoMax A v^2
=(1/2) 2 (4.08 x 10^-10 kg m^-3) (0.03 m^2) (7765)^2
=7.38*10^(-4) N
DMean = DMax/4.117
=1.79*10^(-4) N
Solar radiation pressure is negligible: 4.5 (absorption) to 9 (reflection) μN /m2, so radiation force will be less than 0.27 μN. (For LEO, the radiation pressure from the Earth is hard to model as it depends on cloud albedo, but it is smaller than solar and thus also negligible).
Mass = 1.33 kg/U
3U =4 kg
6U =8 kg
Acceleration due to Atmospheric Drag
For 3U, mass=4 kg
aMax= DMax/Mass
=7.38*10^(-4) N/4 kg
=1.85*10^(-4) m/s^2
aMean=DMean/Mass
=1.79*10^(-4) N /4 kg
=4.40*10^(-5) m/s^2
Maximum Power available from sunlight = 10 watts
From Cannae's announcement: http://cannae.com/cubesat/QuoteOur thruster configuration for the cubesat mission with Theseus is anticipated to require less than 1.5 U volume and will use less than 10 watts of power to perform station keeping thrusting.
Effective power available , assuming a common-low-to-moderate-inclination circular orbit at 240 km altitude, as shown in this picture by Cannae:
and hence taking into account that solar panels will be experiencing eclipse ~ 50% of the time, and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
P=(1/2) 10 watts (*)
=5 watts
Assume no safety margin: SafetyMargin=1
Necessary thrust
TMax= SafetyMargin* DMax
= 7.38*10^(-4) N
TMean= TMax /(rhoMax/rhoMean)
= 1.79*10^(-4) N
Necessary Thrust/PowerInput
TMax /PowerInput= 7.38*10^(-4) /5 W
= 148 μN/W
TMean /PowerInput= 1.79*10^(-4) N /5 W
= 36 μN/W
Conclusion:
The orbit makes a big difference, concerning the requirements for such a mission. While a 150 km would require ~1 milliNewton/Watt to ensure no deorbiting, an orbit of 240 km requires substantially less thrust/PowerInput. Note that most Cubesat launches are at 300 - 400 km - The ISS maintains an orbit with an altitude of between 330 and 435 km by means of reboost manoeuvres using the engines of the Zvezda module or visiting spacecraft.
Cannae's mission for keeping in orbit for 6 months a Cubesat, assuming:
* minimum configuration 1x3U Cubesat, with cross sectional area of only 0.03 m2
* no safety margin
* mean Solar activity
* that the solar panels are kept always parallel to the orbital velocity vector (otherwise drag will be much greater)
requires a Thrust/PowerInput= 36 microNewton/Watt which is consistent with NASA's previously reported results for copper resonant cavities excited at ~2 GHz :
http://www.libertariannews.org/wp-content/uploads/2014/07/AnomalousThrustProductionFromanRFTestDevice-BradyEtAl.pdf
http://emdrive.wiki/Experimental_Results
However, maximum Solar activity would require about 150 microNewton/watt.
If the solar panels are not kept parallel to the orbital velocity vector at all times, drag will be much greater, and hence much greater thrust would be required.
Furthermore, this assumes no safety margin.
Also this is based on a minimum configuration 1x3U Cubesat, with minimum cross sectional area of only 0.03 m2
The Cannae publicity picture appears to show a larger than 1x3U Cubesat configuration instead, and if so, a larger cross-sectional area which would require a proportionally larger thrust force to overcome atmospheric drag.
The link http://cannae.com/cannae-is-developing-a-cubesat-thruster/ describes a 6U Cubesat. If it is a 2x3U Cubesat, then the minimum cross-sectional area is twice what is calculated above and therefore the atmospheric drag will require twice the thrust calculated above for a 1x3U Cubesat.
Also, worthy of note when planning a 6 month mission in Low Earth Orbit:I know I sound like a broken record, but I would really like to know how they plan to separate out thrust effects from the high variability of atmospheric density at those altitudes.
_________________________________________
(*) Thanks to Marshall Eubanks for providing these estimates. I am responsible for any errors in using them.
-
#381 by Star One on 06 Sep, 2016 15:45
-
Plan does not match aspiration then in your opinion?
-
#382 by Rodal on 06 Sep, 2016 16:48
-
Plan does not match aspiration then in your opinion?
This was just a numerical analysis on available information for their planned mission. The assumptions in the analysis were clearly stated.
Cannot provide further objective comments until there is more information, and hence clarity on exactly what is planned. To engage in further comments, need further engineering information on the planned mission.
As you see from the analysis, it makes a big difference whether the planned orbit is 150 km or 240 km.
Solar activity is crucial: it is what brought down Skylab ahead of NASA's early predictions.
If anybody knows of any public information that has not been taken into account in the above analysis let us know, and then we can elaborate further
-
#383 by Prunesquallor on 06 Sep, 2016 17:05
-
...
Cannae's mission for keeping in orbit for 6 months a 6U Cubesat, assuming:
* no safety margin
* mean Solar activity
* that the solar panels are kept always parallel to the orbital velocity vector (otherwise drag will be much greater)
requires a Thrust/PowerInput= 36 microNewton/Watt which is consistent with NASA's reported results for copper resonant cavities excited at ~2 GHz .
However, maximum Solar activity would require about 150 microNewton/watt.
If the solar panels are not kept parallel to the orbital velocity vector at all times, drag will be much greater, and hence much greater thrust would be required.
Furthermore, this assumes no safety margin.[/b]
Also, worthy of note when planning a 6 month mission in Low Earth Orbit:I know I sound like a broken record, but I would really like to know how they plan to separate out thrust effects from the high variability of atmospheric density at those altitudes.
_________________________________________
(*) Thanks to Marshall Eubanks for providing these estimates. I am responsible for any errors in using them.
Thanks Dr. Rodal. I just ran my atmosphere models and got slightly different absolute numbers, but similar variance. My point is always - why in the world would you try to demonstrate performance in an environment where external forces can vary (undetectably) by a factor of 4?
Best case, if you can show that thrust is exceeding drag, you still cannot accurately estimate the magnitude of that thrust unless it far exceeds the highest likely drag value. And any conclusion such as "well, the orbit is decaying more slowly that you would expect" would depend on assumption of actual atmospheric density in the exosphere (not to mention spacecraft attitude excursions which change Cd), both of which would be highly speculative.
Again, it seems to me that reducing error sources in the lab is highly preferable. -
#384 by Bob012345 on 06 Sep, 2016 18:48
-
A question I would like some thought to. Please don't get mad if this seems stupid.
These devices seem critically dependent on high Q to enhance and multiply the force differential which is extremely difficult to achieve since destructive interference has infinite possibilities to happen. I wonder if there is fundamentally a different way to get to the same end yet without needing to sustain resonance. What I'm thinking of is an asymmetrically designed device that acts more as a waveguide to recirculate the radiation as opposed to bounce it back and forth interacting with the ends. This mode of operation would be similar to photon recycling schemes recently validated by experiment.
BTW, concerning resonance, in recent photon recycling experiments by Y. Bae, an effective resonance was set up between mirrors so stable, the author could move the mirror around with his hand and maintain the resonance. He used a so-called gain medium in the loop. Do you builders have an analogy with microwaves? Thanks. -
#385 by RotoSequence on 06 Sep, 2016 19:02
-
Again, it seems to me that reducing error sources in the lab is highly preferable.
If they're pushing a millinewton per watt (baseless conjecture), they'd probably feel confident about their demonstration satellite. -
#386 by Bob Woods on 06 Sep, 2016 19:17
-
Again, it seems to me that reducing error sources in the lab is highly preferable.
If you're a commercial space venture and you need access to capital, a space demonstration is a lot sexier than more lab work.
-
#387 by JohnFornaro on 06 Sep, 2016 19:22
-
Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
-
#388 by krio on 06 Sep, 2016 19:46
-
If a copper Cannae drive would provide the same 1 milliNewton/Watt thrust/InputPower as what is claimed for the superconducting version, then what is the point of their superconducting version, with all its complexity
Fair question, it's just that I've never seen any specific claim from Cannae as to what their superconductor thrust/watt is.
The overall truth should come out after they launch their baby: Does it work at all, and if so, how efficient?
I know I sound like a broken record, but I would really like to know how they plan to separate out thrust effects from the high variability of atmospheric density at those altitudes.
Sent from my iPhone using Tapatalk
By turning it off and on again? -
#389 by SeeShells on 06 Sep, 2016 19:56
-
Someone correct me if I'm wrong but wouldn't a circular polar orbit place the cubesat in sunlight all the time and also place the extended solar panels in a minimum drag configuration?Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
Dr. Rodal, love your breakdown, you rock math.
Best to all,
Shell -
#390 by Monomorphic on 06 Sep, 2016 19:59
-
Someone correct me if I'm wrong but wouldn't a circular polar orbit place the cubesat in sunlight all the time and also place the extended solar panels in a minimum drag configuration?
Best to all,
Shell
I recommended a sun-synchronous orbit but then deleted the comment because Cannae's image of their orbit doesn't look sun-synchronous. -
#391 by punder on 06 Sep, 2016 20:06
-
Pardon me if this has been mentioned and shot down before, but what about a "control" satellite, in the same orbit? A duplicate of the experiment satellite in every way, except the EM drive isn't turned on.
I'm guessing collision issues and cost are the main arguments against, and perhaps drag models are so good it isn't needed. But it would make a statement that's dramatic and difficult to explain away (which is what I'm worried about with a single satellite, considering the small forces involved). -
#392 by Rodal on 06 Sep, 2016 20:13
-
Someone correct me if I'm wrong but wouldn't a circular polar orbit place the cubesat in sunlight all the time and also place the extended solar panels in a minimum drag configuration?Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
Dr. Rodal, love your breakdown, you rock math.
Best to all,
Shell
Thanks for the rocking
To be in constant Sunlight all the time,a Sun-synchronous orbit (SSO) is required that combines altitude and inclination in such a way that the satellite passes over any given point of the planet's surface at the same local solar time.
Typical sun-synchronous orbits are about 600–800 km in altitude, with periods in the 96–100 minute range, and inclinations of around 98° (i.e. slightly retrograde compared to the direction of Earth's rotation: 0° represents an equatorial orbit and 90° represents a polar orbit). Since they are not exactly 90° they are not purely polar orbits.
These orbits, (near 90 degrees inclination) are not minimum drag, as the atmosphere opposes the motion of the spacecraft even in a polar orbit. Rather, maximum drag is associated with low orbits, and to minimize atmospheric drag the best thing is to use higher orbits.
The market for such orbits is much smaller and they require higher thrust from the rocket to put them in orbit (*), so to me it sounds like a much more expensive proposition, than sending a Cubesat to a geocentric low-inclination circular orbit at 240 kmQuote from: WikipediaThe Sun-synchronous orbit is mostly selected for Earth observation satellites that should be operated at a relatively constant altitude suitable for its Earth observation instruments, this altitude typically being between 600 km and 1000 km over the Earth surface. Because of the deviations of the gravitational field of the Earth from that of a homogeneous sphere that are quite significant at such relatively low altitudes a strictly circular orbit is not possible for these satellites. Very often a frozen orbit is therefore selected that is slightly higher over the Southern hemisphere than over the Northern hemisphere. ERS-1, ERS-2 and Envisat of European Space Agency as well as the MetOp spacecraft of EUMETSAT are all operated in Sun-synchronous, "frozen" orbits.
(*) Compared to the advantage of launching near the Equator for a low inclination orbit to take advantage of the Earth's rotation. (Anything on the surface of the Earth at the equator is already moving at 460 meters per second (1670 kilometers per hour ) ). -
#393 by krio on 06 Sep, 2016 20:27
-
Pardon me if this has been mentioned and shot down before, but what about a "control" satellite, in the same orbit? A duplicate of the experiment satellite in every way, except the EM drive isn't turned on.
I'd imagine that depends on your budget strategy
I'm guessing collision issues and cost are the main arguments against, and perhaps drag models are so good it isn't needed. But it would make a statement that's dramatic and difficult to explain away (which is what I'm worried about with a single satellite, considering the small forces involved).
-
#394 by SeeShells on 06 Sep, 2016 20:36
-
Does the EM drive then distort gravity within the cavity and/or outside it as part of the 'exhaust'? And scaled up sufficiently would this in any way be visually apparent?
Just inside the cavity where the e.m. energy resides. I think that the effect of the exhaust is really tiny to be seen in some way. The smart thing to do is to shot a laser beam through the cavity in a interferometric experiment. EW just did this. It is ongoing work yet.
This hopefully will be covered in the AIAA paper then?
No, indeed. As far as I know, Dr. White was asked to remove any physical explanation in the paper to appear and publish it on a physics journal. I am not involved in EW activities.
Also, the interferometry studies are ongoing work yet and so, nothing will be said about.
It has intrigued me as well the first time I ran across the laser interferometer testing. They used a circular resonator for the testing and I had to ask how I could in some future testing do it on a Frustum. It was staring right at me in the quartz rod I have through the center of the frustum. Thinking you could use two equal lengths of fiber optics or even the quartz rod, one through the cavity and one on the outside to measure time differentials in beam travel.
Also if your using a TE012 or better yet a TE013 mode there is a center line through the frustum virtually free of fields, magnetic and electrical.
I'm not there yet although it is on the bucket list.
Back to testing.
My Best to all,
Shell -
#395 by SeeShells on 06 Sep, 2016 20:46
-
Thanks, that clears it up perfectly. I did ask.
Someone correct me if I'm wrong but wouldn't a circular polar orbit place the cubesat in sunlight all the time and also place the extended solar panels in a minimum drag configuration?Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
Dr. Rodal, love your breakdown, you rock math.
Best to all,
Shell
Thanks for the rocking
To be in constant Sunlight all the time,a Sun-synchronous orbit (SSO) is required that combines altitude and inclination in such a way that the satellite passes over any given point of the planet's surface at the same local solar time.
Typical sun-synchronous orbits are about 600–800 km in altitude, with periods in the 96–100 minute range, and inclinations of around 98° (i.e. slightly retrograde compared to the direction of Earth's rotation: 0° represents an equatorial orbit and 90° represents a polar orbit). Since they are not exactly 90° they are not purely polar orbits.
The market for such orbits is much smaller and they require higher thrust from the rocket to put them in orbit (*), so to me it sounds like a much more expensive proposition, than sending a Cubesat to a geocentric circular orbit at 240 kmQuote from: WikipediaThe Sun-synchronous orbit is mostly selected for Earth observation satellites that should be operated at a relatively constant altitude suitable for its Earth observation instruments, this altitude typically being between 600 km and 1000 km over the Earth surface. Because of the deviations of the gravitational field of the Earth from that of a homogeneous sphere that are quite significant at such relatively low altitudes a strictly circular orbit is not possible for these satellites. Very often a frozen orbit is therefore selected that is slightly higher over the Southern hemisphere than over the Northern hemisphere. ERS-1, ERS-2 and Envisat of European Space Agency as well as the MetOp spacecraft of EUMETSAT are all operated in Sun-synchronous, "frozen" orbits.
(*) Compared to the advantage of launching near the Equator for a low inclination orbit to take advantage of the Earth's rotation
Shell -
#396 by FattyLumpkin on 06 Sep, 2016 21:02
-
Shell, is an off-set antenna a loop or hoop that is suspended away from the wall or top or bottom caps?
-
#397 by Rodal on 06 Sep, 2016 21:09
-
As discussed in my post, if you don't keep the solar panels parallel to the orbital velocity vector at all times, the drag force is going to increase dramatically, and hence the needed thrust is going to increase concomitantly, particularly if you have the solar panels perpendicular to the orbital local plane, as the surface area of the solar panels is much greater than the considered minimum cross-sectional area.Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
To minimize drag forces one wants to present the smallest cross-sectional area to the flow, instead of presenting the largest cross-sectional area.
-
#398 by SeeShells on 06 Sep, 2016 22:53
-
Shell, is an off-set antenna a loop or hoop that is suspended away from the wall or top or bottom caps?
We'll see if it works and let you know.
Shell -
#399 by Prunesquallor on 06 Sep, 2016 23:11
-
As discussed in my post, if you don't keep the solar panels parallel to the orbital velocity vector at all times, the drag force is going to increase dramatically, and hence the needed thrust is going to increase concomitantly, particularly if you have the solar panels perpendicular to the orbital local plane, as the surface area of the solar panels is much greater than the considered minimum cross-sectional area.Effective power available taking into account that solar panels will be experiencing eclipse ~ 50% of the time
P=(1/2) 10 watts (*) and considering that solar panels must be kept always parallel to the orbital velocity vector, at all times)
=5 watts...
Does the illustration of the Cannae cubesat show the solar panels oriented properly?
I would have described the panel orientation as being perpendicular to the orbital velocity vector. For more efficiency, the panels would be aimed at the Sun when the cubesat was out of Earth's shadow.
To minimize drag forces one wants to present the smallest cross-sectional area to the flow, instead of presenting the largest cross-sectional area.
Well, that's a great point. One could:
1) maintain a sun-pointing attitude where you would get constant power (for half the orbit anyway) but have a varying drag coefficient (because your orientation with respect to the velocity vector is changing), or
2) maintain a constant attitude with respect to the velocity vector giving you a constant drag coefficient, but varying power input.
(Come to think of it, 1) won't work because the thruster orientation would be changing also)
One would have to assume there are enough batteries on board to maintain constant power to the thruster even during occultation and non-optimal array pointing.
Sent from my iPhone using Tapatalk
)
)