Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1797096 times)

Online 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

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

Per Zeng & Fan's impedance graph, you would need the small end to have a k*r in the low 20's to have the nearly the same impedance differential.
Todd

Offline Flyby

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.....
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



Not sure where it goes wrong, but I'm getting way different numbers when I scale to 240mm height...
eighter the drawings or the calcs are wrong.... or both :-[

Db = 276.9mm
Ds = 185.5mm



Offline 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.
After a 5 mile run, I'm ready to put this to bed.

We proceed as follows:

We adopt the Cone half-angle value from Flyby 15.44 degrees, which is the Median of the three values obtained (15 degrees from SeeShells and the two values from Flyby 15.44 and 18.27 degrees)

Cone Half-Angle = 15.44

therefore the constraint is

Db - Ds = 2 (24 cm) Tan(15.44) = 13. 25746 cm

and vary Db and Ds subject to this constraint to get 2.45 GHz for TE012 from the exact solution

doing so, we get:

Db = 0.247 m

Ds  = 0.114425 m

L = 0.24 m


r1= 0.211022 m

r2= 0.455515 m

Cone half-angle = 15.44 degrees

these numbers represent only a small change from the previous numbers, which gives some level of encouragement, as the previous numbers were obtained from Yang's chart, just eyeballing it, without performing any frequency calculation

Notice that these diameters for Yang are close to the diameters of Shawyer's Flight Thruster, the only difference with Flight Thruster is that Yang has a greater length to enable resonating at TE012 for Yang at 2.45GHz instead of TE013 for Flight Thruster at 3.85GHz.   Recall that at the time that Yang embarked on her project, the Flight Thruster was the latest and highest effective force design by Shawyer.  The longer length of Yang being motivated perhaps by the need to resonate at 2.45 GHz instead of 3.85 GHz.

We adopt these numbers as the official Yang geometry and we correct the previous numbers I had entered in the EM Drive Wiki accordingly without further ado.

***A cone half-angle of 6 degrees does not make any sense to me for an EM Drive, it is like a cylinder ===> bad***
« Last Edit: 08/01/2015 10:47 PM by Rodal »

Offline Rodal

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...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

Which is very good my friend because k r rules the waves ! :) 

This agrees with my computations

This agrees with all the formulae:  McCulloch, Notsosureofit:  the larger the difference between the diameters the better
« Last Edit: 08/01/2015 10:41 PM by Rodal »

Offline Flyby

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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.

I can confirm that the drawings do not match in their wave coupling window size. I'm getting 147mm height instead of 240mm...

This is probably because that "wave coupling window" drawing is parametrically, so the chances of having it matching with the other drawing were slim to start with.

I'll try another with the total height, see if that gets anywhere..

Online WarpTech

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...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

Which is very good my friend because k r rules the waves ! :) 

This agrees with my computations

This agrees with all the formulae:  McCulloch, Notsosureofit:  the larger the difference between the diameters the better

Updated: Her's is still the smallest k*r but only slightly smaller now. I also updated the Tajmar size, and it is "closer" now too.
Todd


Offline Flyby

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.........

and vary Db and Ds subject to this constraint to get 2.45 GHz for TE012 from the exact solution

doing so, we get:

Db = 0.247 cm

Ds  = 0.114425 cm

L = 0.24 cm


r1= 0.211022 m

r2= 0.455515 m

Cone half-angle = 15.44 degrees

these numbers represent only a small change from the previous numbers, which gives some level of encouragement, as the previous numbers were obtained from Yang's chart, just eyeballing it, without performing any frequency calculation

Notice that these diameters for Yang are close to the diameters of Shawyer's Flight Thruster, the only difference with Flight Thruster is that Yang has a greater length to enable resonating at TE012 for Yang at 2.45GHz instead of TE013 for Flight Thruster at 3.85GHz.   Recall that at the time that Yang embarked on her project, the Flight Thruster was the latest and highest effective force design by Shawyer.  The longer length of Yang being motivated perhaps by the need to resonate at 2.45 GHz instead of 3.85 GHz.

We adopt these numbers as the official Yang geometry and we correct the previous numbers I had entered in the EM Drive Wiki accordingly without further ado.

Well, the problem i got now is that on 1 side you use the angle from the drawing as being correct, but when I scale the drawing to match the height, my Db and Ds are seriously different that what you have calculated....

So...You seem to have an agreement on your calculations and a disagreement with the drawings you based your initial calculus parameters on....doesn't that sound a bit ... odd?
I'm not comfortable with that.... either the drawing matches the calculated values or it doesn't...

Offline Rodal

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.........

and vary Db and Ds subject to this constraint to get 2.45 GHz for TE012 from the exact solution

doing so, we get:

Db = 0.247 cm

Ds  = 0.114425 cm

L = 0.24 cm


r1= 0.211022 m

r2= 0.455515 m

Cone half-angle = 15.44 degrees

these numbers represent only a small change from the previous numbers, which gives some level of encouragement, as the previous numbers were obtained from Yang's chart, just eyeballing it, without performing any frequency calculation

Notice that these diameters for Yang are close to the diameters of Shawyer's Flight Thruster, the only difference with Flight Thruster is that Yang has a greater length to enable resonating at TE012 for Yang at 2.45GHz instead of TE013 for Flight Thruster at 3.85GHz.   Recall that at the time that Yang embarked on her project, the Flight Thruster was the latest and highest effective force design by Shawyer.  The longer length of Yang being motivated perhaps by the need to resonate at 2.45 GHz instead of 3.85 GHz.

We adopt these numbers as the official Yang geometry and we correct the previous numbers I had entered in the EM Drive Wiki accordingly without further ado.

Well, the problem i got now is that on 1 side you use the angle from the drawing as being correct, but when I scale the drawing to match the height, my Db and Ds are seriously different that what you have calculated....

So...You seem to have an agreement on your calculations and a disagreement with the drawings you based your initial calculus parameters on....doesn't that sound a bit ... odd?
I'm not comfortable with that.... either the drawing matches the calculated values or it doesn't...

I am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.

There are many constraints to satisfy.  Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.

A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012

As in the movie: something's got to give :)

The most important geometrical constraint from the drawing is the cone half angle.  The others you mention are second order and not as important.

If it is any comfort to you remember that you thought that Yang's drawings were schematic.  There are more important things to base the dimensions on that this drawing.  The most important thing to get from her schematic drawings is the cone-half angle.
« Last Edit: 08/01/2015 11:10 PM by Rodal »

Offline Rodal

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...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

Which is very good my friend because k r rules the waves ! :) 

This agrees with my computations

This agrees with all the formulae:  McCulloch, Notsosureofit:  the larger the difference between the diameters the better

Updated: Her's is still the smallest k*r but only slightly smaller now. I also updated the Tajmar size, and it is "closer" now too.
Todd

Yang k * r is very, very close to Shawyer's Demo

As TheTraveller said that Yang was following Shawyer, this makes me much more comfortable.  It did not make any sense to me that Yang out of the blue would have a geometry resembling a cylinder instead of Shawyer's Demo and Flight Thruster
« Last Edit: 08/01/2015 11:08 PM by Rodal »

Offline Flyby

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I am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.

There are many constraints to satisfy.  Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.

A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012

As in the movie: something's got to give :)

The most important geometrical constraint from the drawing is the cone half angle.  The others you mention are second order and not as important.

If it is any comfort to you remember that you thought that Yang's drawings were schematic.  There are more important things to base the dimensions on that this drawing.  The most important thing to get from her schematic drawings is the cone-half angle.

I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.
The angle is tied to the dimensions of the big and small angle in relation to the height.
How can the angle be correct (for the given height) and not to the big and small bases?
After all, all i did is just rescale to the given height...

Let me add your new results into a new drawing, so we can see better....

brb

Offline Rodal

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I am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.

There are many constraints to satisfy.  Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.

A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012

As in the movie: something's got to give :)

The most important geometrical constraint from the drawing is the cone half angle.  The others you mention are second order and not as important.

If it is any comfort to you remember that you thought that Yang's drawings were schematic.  There are more important things to base the dimensions on that this drawing.  The most important thing to get from her schematic drawings is the cone-half angle.

I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.
The angle is tied to the dimensions of the big and small angle in relation to the height.
How can the angle be correct (for the given height) and not to the big and small bases?
After all, all i did is just rescale to the given height...

Let me add your new results into a new drawing, so we can see better....

brb

Look at it this way: plot Yang/Shell dimensions vs. these new dimensions and show us which one is closer to Yang's schematic drawing:  no contest !!!

and the proof of the pudding will be Yang/Shell test, let's see what thrust she gets from a cone angle of only 6 degrees

______

<<I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.>>

I strongly disagree:

The cone half-angle is an invariant under internal length-adjustment
The small and the big bases and the length will vary with reentrant ends, and they will vary with length adjustment
« Last Edit: 08/01/2015 11:15 PM by Rodal »

Offline RotoSequence

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...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

Which is very good my friend because k r rules the waves ! :) 

This agrees with my computations

This agrees with all the formulae:  McCulloch, Notsosureofit:  the larger the difference between the diameters the better

What stops one from wanting to take this statement to its maximum (probably wholly illogical) extreme?


Offline Flyby

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I think I found out what's going on here...

magenta : original technical drawing
red : newly calculated YANG values
yellow : abstracted frustum that does not take the top/bottom cylindrical shapes into count.
Note how, conveniently, that this yellow cone-half angle almost resembles the magenta cone-half angle.

I suspect that your calculations do not take those top/bottom edges into account...
I do not have enough insight in the matter to understand if those are important or not.


added:
I've changed the drawing to add dimensions to the red/yellow design
« Last Edit: 08/01/2015 11:31 PM by Flyby »

Offline Rodal

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...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

Which is very good my friend because k r rules the waves ! :) 

This agrees with my computations

This agrees with all the formulae:  McCulloch, Notsosureofit:  the larger the difference between the diameters the better

What stops one from wanting to take this statement to its maximum (probably wholly illogical) extreme?


I have no idea what that represents.  Please explain and elaborate for those of us on a different wavelength

Is that a cone ending at an apex?

I don't know what the slit represents
« Last Edit: 08/01/2015 11:30 PM by Rodal »

Offline Rodal

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I think I found out what's going on here...

magenta : original technical drawing
red : newly calculated YANG values
yellow : abstracted frustum that does not take the top/bottom cylindrical shapes into count.
Note how, conveniently, that this yellow cone-half angle almost resembles the magenta cone-half angle.

I suspect that your calculations do not take those top/bottom edges into account...
I do not have enough insight in the matter to understand if those are important or not.

No, of course that I don't take those cylindrical corners into account. What matters is the cone half-angle

On purpose I don't take those into account.

The cylindrical extensions make the geometry into an irregular geometry and they break the spherical waves.

Offline flux_capacitor

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I am extremely comfortable with these dimensions from many constraints, too numerous to explain at the moment.

There are many constraints to satisfy.  Satisfying natural frequency at 2.45 GHz is non-negotiable, so is TE012, so is to be near the plot she has for frequency vs D/L, and so is L=24 cm.

A cone with the dimensions you have from the Yang drawing cannot resonate at 2.45 GHz at TE012

As in the movie: something's got to give :)

The most important geometrical constraint from the drawing is the cone half angle.  The others you mention are second order and not as important.

If it is any comfort to you remember that you thought that Yang's drawings were schematic.  There are more important things to base the dimensions on that this drawing.  The most important thing to get from her schematic drawings is the cone-half angle.

I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.
The angle is tied to the dimensions of the big and small angle in relation to the height.
How can the angle be correct (for the given height) and not to the big and small bases?
After all, all i did is just rescale to the given height...

Let me add your new results into a new drawing, so we can see better....

brb

Look at it this way: plot Yang/Shell dimensions vs. these new dimensions and show us which one is closer to Yang's schematic drawing:  no contest !!!

and the proof of the pudding will be Yang/Shell test, let's see what thrust she gets from a cone angle of only 6 degrees

______

<<I'm not arguing your calculated results, but if you take the cone-half angle from the drawing to base your calculations on, then the the big and small bases should also match you calculations.>>

I strongly disagree:

The cone half-angle is an invariant under internal length-adjustment
The small and the big bases and the length will vary with reentrant ends, and they will vary with length adjustment

The issue is that newest Yang's drawings include two cylindrical parts at each end. If the internal length is considered to be 24 cm in all cases, then the half-cone angle is not the same with or without those cylindrical waveguides.

Below, on the left: Yang's frustum per Rodal's dimensions, with a half cone angle of 15.44░
One the right, the same frustum but with cylindrical extents. The half-cone angle is then above 21░ for the same maximum internal length of 24 cm.

Offline Flyby

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actually, look at my new drawing...
Dr Rodal's version of the new YANG cavity is in yellow and has a cone half angle is 15.45░ (where the initial magenta design had 15.44░). No idea if it is pure luck or if there is a correlation....

Offline Rodal

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actually, look at my new drawing...
Dr Rodal's version of the new YANG cavity is in yellow and has a cone half angle is 15.45░ (where the initial magenta design had 15.44░). No idea if it is pure luck or if there is a correlation....

Excellent

Thanks for the outstanding job :)
« Last Edit: 08/01/2015 11:39 PM by Rodal »

Offline Flyby

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No, of course that I don't take those cylindrical corners into account. What matters is the cone half-angle

On purpose I don't take those into account.

The cylindrical extensions make the geometry into an irregular geometry and they break the spherical waves.

Then, if I get this right, you do not agree with TT that these top/bottom edges are specifically designed to increase the Q of the cavity?

Or is it just a technical simplification of the required calculations to get a faster result ?

Excellent

Thanks for the outstanding job :)

euh.. come to think about it...
of course there is a correlation with the old angle... that's how you got the Db and Ds in the first place...
/Face palm... circular reasoning... :-[
« Last Edit: 08/01/2015 11:45 PM by Flyby »

Online 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

This should make what I'm saying a little clearer. Using @Notsosureofit's theory. From this, we can see how the design of the frustum could be maximized for thrust. What puzzles me is why we are working in microwaves when the equation clearly shows that lower frequency is better.  :)
Todd

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