Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1797963 times)

Offline flux_capacitor

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Here is attached a high definition clean version of Yang's 2014 cavity drawing, so we can start guessing the correct dimensions and ratios from the length L of the coupling window (with the value of L1 or L2 provided by Flyby).

Online Rodal

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The difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wiki

Compared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !



Could you please superpose the geometry of the EM Drive Wiki  ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?

If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.

Blue outline is the Wiki version of Yang...
As you said... it's way off..

Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?

The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases.  It is 0.240 meters.

Look at page 811, Table 1  http://www.emdrive.com/NWPU2010paper.pdf

Hence it would be best if you superpose all the images so that they all have the same length: 0.240 meters

Then one has to use this image:



knowing that


1) L=24 cm

2) f = 2.45 GHz

3) D = (Dbig + Dsmall)/2

4) TE012  line should be used

« Last Edit: 08/01/2015 06:00 PM by Rodal »

Offline WarpTech

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@Notsosureofit,

EDIT: Never mind. I answered my own dumb question. I shouldn't post before I finish my coffee. :( The bandwidth used for Q is not the same, but f is. Therefore, it results in;

acceleration g = (c2/L)*(delta_f/f), where

delta_f = (2pi/f)*(fs2 - fb2)

Q = f/delta_b  (b for bandwidth)

N*T = (P/2pi*L*f)*(delta_f/delta_b)

This implies a lower frequency, large delta_f/L implies a short length, wide half-angle. Small delta_b implies narrow bandwidth. So a wide stubby frustum with both a smaller small end and a larger big end? Using a narrow band RF amplifier rather than a Magnetron. Just as @TT and Shawyer have said.

Thank you.
Todd
« Last Edit: 08/01/2015 09:33 PM by WarpTech »

Offline SeeShells

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The difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wiki

Compared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !



Could you please superpose the geometry of the EM Drive Wiki  ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?

If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.

Blue outline is the Wiki version of Yang...
As you said... it's way off..

Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a value that is way off?

No. I am the one and only one person that came up with the dimensions that are in the Wiki EM Drive wiki.  I know how I came up with those dimensions.  Either my interpretation of the tables or something else in her paper is off or she has a mistake in her paper.  Something is OFF by a huge amount.

There is NO basis on which to pretend that the dimensions in the EM Drive wiki represent Yang's geometry on the contrary. All the evidence is to the contrary

To start with, the all important cone half-angle is around 15 degrees instead of 6 degrees.

Yang's angle is close to NASA.

I look forward to Shell's test with a cone half-angle of only 6 degrees to prove this :)

Experiments have a way to settle things...

To make it simple:

Geometry close to a cylinder ===>  bad

Geometry close to Shawyer and NASA ===> good

I'm not looking to stir up the pot-ty here with this and the design will probably end up with something like RS's to quell everyone's fears. Currently I have vested time into the building of the yang design from the info I tried to glean from their publications (tough indeed) and came close to what Dr. Rodel posted on Wiki. I have more faith in his Wolfram calculations than my pen and paper. It would be a shame to just throw it into the trash without testing a couple ideas I have... beyond the Yang 6 degree model as we see in wiki.

One if them is I know I can get a higher Q from the design and second I would like to test out the longer frustum, looking at the small plate/large plate ratio 1:1.5 (or easier ~1/2) the large diameter by sliding another smaller plate up the cavity. (need the numbers I'll get them)

I would like to ask Dr. Rodal what does it look like when I slide the small plate forward to get close to the 50% size of the larger plate (keeping resonance of the cavity and for those here trying to understand too). If I set the large end on the floor and look down from the top of the cavity I'll see the same end relationships as in the current Yang and close to the current RS designs, right? But it will be longer. So I was interested in decreasing the diameter of the small endplate from the Wiki model and increasing the distance while remaining resonate. That's why I was jazzed to think I could see the meep model evaluation.

With this design I found out I could generate and test higher Q, I could test a longer cavity length, I could test for evanescent waves that maybe imparting thrust through the sidewalls of the perforated cavity and I could use different inserts for cavity length even insert dielectrics. I can even insert probes in to the top of the cavity past the small endplate to test for evanescent waves leaking through the endplates.

Interesting enough everyone should know me by now that seldom do I do something that I cannot kill two birds with one stone. This is a testbed and if it flops with little thrust and the data I get agrees with the calculations on frustum operational dimensionality we are doing, then I'll be happy, as there is no bad data.

I very much need to get out into the shop and saw and claw and glue and screw some fixtures! BBL

Shell

Offline SeeShells

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Great Internet collaboration!

We are getting there !

yayaya WHHOR WOOT WOOT.

I need to go get some work done... guys quit writing stuff I just have to read. ;)

Shell

Offline flux_capacitor

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knowing that

1) L=24 cm

Hemů taking the longer length L1 = 43.34 mm for the coupling window, according to latest Yang's drawing I get a cavity internal height of 14.87 cm :( The size of the coupling window may be wrong.

However if I follow my first idea of setting the waveguide height as the length of a WR340 waveguide used vertically as in Tajmar's experiment (86.36 mm) I get a cavity length =  22.35 cm, much closer to the 2010 data of 24 cm.
« Last Edit: 08/01/2015 06:21 PM by flux_capacitor »

Online Rodal

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knowing that

1) L=24 cm

Hemů taking the longer length L1 = 43.34 mm for the coupling window, according to latest Yang's drawing I get a cavity internal height of 14,87 cm :( The size of the coupling window may be wrong.

She has two lengths for TE012 in  page 811, Table 1  http://www.emdrive.com/NWPU2010paper.pdf

24 cm is the longer length, associated with a slightly higher Q and frequency

she also has a length of 17.5 cm .  This length of 17.5 cm agrees better with the 14,87 cm

However in her Table 2 this shorter length is associated with a much lower thrust level and much lower measured Q
« Last Edit: 08/01/2015 06:23 PM by Rodal »

Offline deltaMass

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Actually I got the impression from the vid that the heat capacity of the magnetron is quite low - it lost heat very rapidly after turn off. Oddly perhaps, the same could not be said of the frustum.

Or perhaps these observations are due to vid editing messing with the observed timeline?
It was about 1 minute clipped out as I took the cam off the tripod. Next time, I'll just let it run...you are correct, that is the best way.
Aha. OK so it lost heat in a "normal" way, good. But I understand your motives for editing. All you have to do is caption and/or voiceover how much time you spliced out. That keeps us all honest.

Offline Mulletron

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Only caught up to page 293. Eagleworks and Prof. Tajmar show the importance of atmo in/around the cavity. Thrust in vacuum is greatly diminished it has been shown (I have my ideas...see previous posts...and so do you as I've read enthusiastically), but to know what's up for sure, we need an airtight cavity. An airtight cavity has been suggested before and I second that.

Has anyone looked into non reciprocal interactions? Newton's third law is known to be N/A in certain circumstances (electrodynamics is one) but Noether still holds.

We need to look at the air inside the cavity.

What happens when action reaction symmetry is broken within an electromagnetic field to charged particle system? Does the universe step in to balance the books?

The way I see it is, even with different group velocities at each end, no configuration of internal stress will result in a net thrust. But we're seeing thrust.

Edit: 293

« Last Edit: 08/01/2015 07:47 PM by Mulletron »
Challenge your preconceptions, or they will challenge you. - Velik

Offline deltaMass

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rfmwguy.

I'm thinking replace the magnetron [see attachment below looks the same] with a RF transistor as used here

http://www.zdnet.com/article/freescales-radio-frequency-oven-the-end-of-the-microwave/

maybe it runs with less heat loss?
and runs cleaner r.f. waves
http://www.freescale.com/webapp/video_vault/videoSummary.sp?code=RF-SAGE-VIDEO

 see 2.37

They still require a heat sink. 3.7 deg. C/Watt is a lot of heat. These are also only good for about 200 to 400 mW per transistor, if you've got a good heat sink. If you want to get back up to 400W, you will need to parallel about a dozen of them on a well designed PC board, that is thermally conductive to the heatsink. At these frequencies, I have no idea how you would do that, or what proper design criteria are necessary for a microwave amplifier of sufficient power.
Todd
I take issue with your use of the arithmetic division operator.

Nick picking again...  oC/W.
Not that - that you say you need "about a dozen" 400 mW transistors to get 400 Watts.

Oops! 4W each, not 400mW. Off by a decimal. It was a long day.
Todd
Still a long day I think? 400/4 is not equal to 1 dozen  ???

Offline WarpTech

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knowing that

1) L=24 cm

Hemů taking the longer length L1 = 43.34 mm for the coupling window, according to latest Yang's drawing I get a cavity internal height of 14,87 cm :( The size of the coupling window may be wrong.

She has two lengths for TE012 in  page 811, Table 1  http://www.emdrive.com/NWPU2010paper.pdf

24 cm is the longer length, associated with a slightly higher Q and frequency

she also has a length of 17.5 cm .  This length of 17.5 cm agrees better with the 14,87 cm

However in her Table 2 this shorter length is associated with a much lower thrust level and much lower measured Q

Since she has the cylindrical sections at each end, there are 2 lengths. One is plate to plate, and the other is the length of the tapered section. It kind of messes with the equations in the theories because the relationship of k*r from the apex is no longer where the small end plate is located along the taper.
Todd

Offline X_RaY

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NSF-1701 random thoughts on thermal management. As someone else pointed out, and empty microwave oven will cause a magnetron to overheat. This will also apply with a frustum. Can't help but think that Yang adressed the issue correctly with a circulator on the waveguide, which redirects reflected energy into a load.

Another option seemed to be EW using a dielectric, and I propose it was a non-conductive "heat sink" as opposed to an electrical match (whether they realized it or not).

So, off for some more testing this weekend. I'll make a tear-down and inspection video next.
1. either, use an external tuner to get impedance match
2. or, use blots near your penetration point to get impedance match
3. or use a circulator and a load AND be sure you got impedance match (see 1.& 2.)
If you use only a circulator+load without impedance match between antenna and frustum, there will be a huge reflection at your antenna and the energy will flow direct into the load.
4. only the Source without matching impedance will also produce reflection, it travels back to the generator, this will cause much more heat.
it's not important if there is over coupling or uncoupling both leads to reflections.


The key is a useful way to measure the S11 and/or S21 while tuning process with the magnetron antenna feed.
You get almost all the energy into the cavity if there is a good impedance and not reflections.

BTW: "With knowledge of the coupling factor the unloaded quality Q_0 can be calculated from the measured loaded Q" see picture 1  (Q=Q_0/{1+K})

most impotrant: pictures 2 and 3
Last picture: high Q_eff vs low Q_eff for different coupling factors (1/Q_eff=1/Q_int+1/Q_ext)
« Last Edit: 08/01/2015 08:46 PM by X_RaY »

Offline WarpTech

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rfmwguy.

I'm thinking replace the magnetron [see attachment below looks the same] with a RF transistor as used here

http://www.zdnet.com/article/freescales-radio-frequency-oven-the-end-of-the-microwave/

maybe it runs with less heat loss?
and runs cleaner r.f. waves
http://www.freescale.com/webapp/video_vault/videoSummary.sp?code=RF-SAGE-VIDEO

 see 2.37

They still require a heat sink. 3.7 deg. C/Watt is a lot of heat. These are also only good for about 200 to 400 mW per transistor, if you've got a good heat sink. If you want to get back up to 400W, you will need to parallel about a dozen of them on a well designed PC board, that is thermally conductive to the heatsink. At these frequencies, I have no idea how you would do that, or what proper design criteria are necessary for a microwave amplifier of sufficient power.
Todd
I take issue with your use of the arithmetic division operator.

Nick picking again...  oC/W.
Not that - that you say you need "about a dozen" 400 mW transistors to get 400 Watts.

Oops! 4W each, not 400mW. Off by a decimal. It was a long day.
Todd
Still a long day I think? 400/4 is not equal to 1 dozen  ???

A long week, a concert and I really need a vacation! :(

Offline Flyby

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The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases.  It is 0.240 meters.

Look at page 811, Table 1  http://www.emdrive.com/NWPU2010paper.pdf

Hence it would be best if you superpose all the images so that they all have the same length: 0.240 meters

Then one has to use this image:



knowing that


1) L=24 cm

2) f = 2.45 GHz

3) D = (Dbig + Dsmall)/2

4) TE012  line should be used

I will update the drawing(s) tonight to match the total height of 240mm, but I got visitors for dinner... so it might take a bit... patience.. ;)

Offline Flyby

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Here is attached a high definition clean version of Yang's 2014 cavity drawing, so we can start guessing the correct dimensions and ratios from the length L of the coupling window (with the value of L1 or L2 provided by Flyby).

I tried that already 14 days ago, just before i went on holidays, when I started on building the top-view cross section of the Yang device...

There is a problem with the drawings not matching : the wall thickness of the longitudinal cut through the waveguide does not match the cross section, so I have to make a choice :

either have the exterior dimensions match up or have the interior match up.
Interior dimensions is suppose, no?

----back to the dinner table---ttyl
« Last Edit: 08/01/2015 07:36 PM by Flyby »

Online Rodal

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The difference between the two geometries is completely negligible compared to the difference between the geometry in the EM Drive wiki

Compared to the geometry in the EM Drive wiki, the two drawings you show line up excellent !



Could you please superpose the geometry of the EM Drive Wiki  ( http://emdrive.wiki/Experimental_Results ), which has a cone half angle of 6 degrees to make that clear ?

If you do that you will see how utterly different is the geometry in the EM Drive wiki for Yang, and that the estimate in the EM Drive Wiki is unreasonable.

Blue outline is the Wiki version of Yang...
As you said... it's way off..

Could it be that the composite shape of a cylinder and frustum, like we see in the more "technical" drawing, altered the data in such a way that reverse calculation gives a length value that is way off?

The only dimension given by Yang in her paper is the Length (also called height) measured perpendicular to the bases.  It is 0.240 meters.

Look at page 811, Table 1  http://www.emdrive.com/NWPU2010paper.pdf

Hence it would be best if you superpose all the images so that they all have the same length: 0.240 meters

Then one has to use this image:



knowing that


1) L=24 cm

2) f = 2.45 GHz

3) D = (Dbig + Dsmall)/2

4) TE012  line should be used

OK, using the above and

Cone half-angle = 15 degrees

I proceed as follows:

TE012 Equation (from Yang's graph):   y = 13.5 + 8.5 x

where

y = ((f D) ^2)*10^(-20)
x = (D/L)^2

replacing

L=24 cm
f=2.45*10^9 Hz

and solving the quadratic equation for D, I get

D=17.26915 cm

and since

(Db - Ds)/2 = (24 cm)* tan (15 degrees)

and

(Db + Ds)/2 =D = 17.26915 cm

and solving these coupled equations for Db and Ds, we finally get

Db = 23.69993 cm ~ 23.70 cm

Ds  = 10.83836 cm ~ 10.84 cm

L = 24 cm


Need to calculate whether these numbers give a natural frequency for TE012 of 2.45 GHz

if not, need to modify the diameters in order to get TE012 @ 2.45 GHz

The main imprecision comes from the coefficients of the equation   y = 13.5 + 8.5 x

Obviously, there is a whole family of solutions that satisfy the equation for TE012 in Yang's graph, for different values of the cone angle.  The larger the cone angle, the more different are the values of Db and Ds, all we know is their average D.
« Last Edit: 08/01/2015 08:50 PM by Rodal »

Offline WarpTech

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OK, using the above and

Cone half-angle = 15 degrees

I proceed as follows:

TE012 Equation (from Yang's graph):   y = 13.5 + 8.5 x

where

y = ((f D) ^2)*10^(-20)
x = (D/L)^2

replacing

L=24 cm
f=2.45*10^9 Hz

and solving the quadratic equation for D, I get

D=17.26915 cm

and since

(Db - Ds)/2 = (24 cm)* tan (15 degrees)

and

(Db + Ds)/2 =D = 17.26915 cm

and solving these coupled equations for Db and Ds, we finally get

Db = 23.69993 cm ~ 23.70 cm

Ds  = 10.83836 cm ~ 10.84 cm

L = 24 cm


Need to calculate whether these numbers give a natural frequency for TE012 of 2.45 GHz

if not, need to modify the diameters in order to get TE012 @ 2.45 GHz

The main imprecision comes from the coefficients of the equation   y = 13.5 + 8.5 x

Obviously, there is a whole family of solutions that satisfy the equation for TE012 in Yang's graph, for different values of the cone angle.  The larger the cone angle, the more different are the values of Db and Ds, all we know is their average D.

With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.
Todd
« Last Edit: 08/01/2015 09:22 PM by WarpTech »

Offline SeeShells

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NSF-1701 random thoughts on thermal management. As someone else pointed out, and empty microwave oven will cause a magnetron to overheat. This will also apply with a frustum. Can't help but think that Yang adressed the issue correctly with a circulator on the waveguide, which redirects reflected energy into a load.

Another option seemed to be EW using a dielectric, and I propose it was a non-conductive "heat sink" as opposed to an electrical match (whether they realized it or not).

So, off for some more testing this weekend. I'll make a tear-down and inspection video next.
Sharp observation!

It's like the potato in the microwave. You know the potato heats from evanescent waves... That's a lot of energy after the waveform collapses into it. Just saying.

Back to sawdust and glue! (psst a cold effervescent beverage too)

Shell

Offline SeeShells

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OK, using the above and

Cone half-angle = 15 degrees

I proceed as follows:

TE012 Equation (from Yang's graph):   y = 13.5 + 8.5 x

where

y = ((f D) ^2)*10^(-20)
x = (D/L)^2

replacing

L=24 cm
f=2.45*10^9 Hz

and solving the quadratic equation for D, I get

D=17.26915 cm

and since

(Db - Ds)/2 = (24 cm)* tan (15 degrees)

and

(Db + Ds)/2 =D = 17.26915 cm

and solving these coupled equations for Db and Ds, we finally get

Db = 23.69993 cm ~ 23.70 cm

Ds  = 10.83836 cm ~ 10.84 cm

L = 24 cm


Need to calculate whether these numbers give a natural frequency for TE012 of 2.45 GHz

if not, need to modify the diameters in order to get TE012 @ 2.45 GHz

The main imprecision comes from the coefficients of the equation   y = 13.5 + 8.5 x

Obviously, there is a whole family of solutions that satisfy the equation for TE012 in Yang's graph, for different values of the cone angle.  The larger the cone angle, the more different are the values of Db and Ds, all we know is their average D.

With these values, Yang went from the largest k*r values to the smallest, and is now utilizing a much larger differential position along the impedance curve.
Todd

Interesting. The small plate is getting close to 50% of the large. I need to calculate out for my Yang-Shell model where would I be in L with a 6 degree angle...

That's for later now out to sawdust and a effervescence cold something.

Shell 

Offline Flyby

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Here is the composite drawing all rescaled to 240mm.

magenta : cavity according the technical drawing frustum+waveguide
green : cavity according 2010 papers schematic drawing
blue : wiki cavity dimensions

I'll now try to match up the wave coupling window (in both directions) and see if that agrees with flux_capacitator.

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