### Author Topic: EM Drive Developments - related to space flight applications - Thread 2  (Read 2478258 times)

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3280 on: 05/14/2015 02:35 AM »
I'm only asking so as to help out with the maths. It doesn't mean I believe a word about its physicality.

For example: if the expression I'm trying to get verified is true, then you can get your blow-up condition. Other conditional relations produce different results (x0 < x1 < x2 or x1 < x2 < x0: x=lambda).

Then there's also the case that it blows up exactly when the numerator is also zero. As you know, 0/0 is an indeterminate quantity.

I find a nicer way to write Shawyer's Df is like this (again using x=lambda)

Df = x0*(x2 - x1) / (x1*x2 - x02)

You'll notice that Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple)
Someone said that Df varied between 0 and 1, and should =1 when x1 or x2 -> inf. It doesn't.

And obviously Df=0 when x1 = x2. So it can be zero for two separate reasons.
Doh!
« Last Edit: 05/14/2015 02:49 AM by deltaMass »

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3281 on: 05/14/2015 03:17 AM »
I won't bother reading all that. From equation (5) of the IAC paper it's clear that any of the wavelengths in the waveguide are greater than the free space value. Or lambda1 > lambda0 and lambda2 > lambda0
Therefore your blow-up condition is impossible.

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3282 on: 05/14/2015 03:18 AM »
....
As defined in the attachment dominant circular waveguide cutoff wavelength is 1.7 x diameter....

1.7 x diameter...

Yes of course, but what diameter? A cylinder has only one diameter. It has constant diameter.

Shawyer's EM Drive is not a cylinder with constant diameter.  Shawyer's EM Drive is a truncated cone.

A truncated cone has a variable diameter.  The diameters are different at each end.

Shawyer's reference (Cullen) does not deal with truncated cones.

What diameter ?

Look at what I derived in http://forum.nasaspaceflight.com/index.php?topic=36313.msg1373898#msg1373898:

...choosing the cutOffWavelength  to be TE110,

Look at where the factor of 1.7 comes from: it is Pi divided by X'11

cOW = gmD *(Pi/1.84118378134065)
= gmD *1.7062895542683174

1.7 times the Geometric Mean of the Diameters

cOW = cutOffWavelength

gmD =  Sqrt[bD * sD]

bD = big end diameter (m)
sD= small end diameter (m)

Shawyer only talks about the 2 ends of his cone and the wavelength of the Rf input.

Lambda0 is the wavelength of the input Rf signal in air or vacuum.

Lambda1 is the dominant cutoff wavelength for the bigger end of the cone.

Lambda2 is the dominant cutoff wavelength for the smaller end of the cone.

The focus on what is happening at each end of his cone cavity as the waves bounce off and transfer momentum, based on their individual group velocity, which varies as the dominant cutoff wavelength varies at each end. Higher group velocity at the big end and lower group velocity at the smaller end, delivering higher momentum xfer at the big end and lower momentum xfer at the smaller end.
It Is Time For The EmDrive To Come Out Of The Shadows

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3283 on: 05/14/2015 03:21 AM »
I'm only asking so as to help out with the maths. It doesn't mean I believe a word about its physicality.

For example: if the expression I'm trying to get verified is true, then you can get your blow-up condition. Other conditional relations produce different results (x0 < x1 < x2 or x1 < x2 < x0: x=lambda).

Then there's also the case that it blows up exactly when the numerator is also zero. As you know, 0/0 is an indeterminate quantity.

I find a nicer way to write Shawyer's Df is like this (again using x=lambda)

Df = x0*(x2 - x1) / (x1*x2 - x02)

You'll notice that Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple)
Someone said that Df varied between 0 and 1, and should =1 when x1 or x2 -> inf. It doesn't.

And obviously Df=0 when x1 = x2. So it can be zero for two separate reasons.
Doh!

Don't you have your Mathematica up and running ?

<<Df=0 when x1 or x2 -> infinity (proof available on request but it's dead simple) >> ??

Wrong.  (You need AND) See:

In[16]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity]

Out[16]= -(x0/x2)

In[17]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> Infinity]

Out[17]= x0/x1

In[18]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x0 -> Infinity]

Out[18]= 0

In[19]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0]

Out[19]= -(x2/x0)

In[20]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> 0]

Out[20]= x1/x0

In[21]:= Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x0 -> 0]

Out[21]= 0

In[22]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity],
x2 -> Infinity]

Out[22]= 0

In[23]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> Infinity],
x1 -> Infinity]

Out[23]= 0

In[24]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0],
x2 -> Infinity]

Out[24]= -\[Infinity]/x0

In[25]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> Infinity],
x2 -> 0]

Out[25]= x0 (-\[Infinity])

In[26]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x1 -> 0], x2 -> 0]

Out[26]= 0

In[27]:= Limit[Limit[x0*(x2 - x1)/(x1*x2 - x0^2), x2 -> 0], x1 -> 0]

Out[27]= 0
« Last Edit: 05/14/2015 03:55 AM by Rodal »

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3284 on: 05/14/2015 03:28 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
« Last Edit: 05/14/2015 03:31 AM by deltaMass »

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3285 on: 05/14/2015 03:37 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you that the lambda in free space should be smaller than in the waveguide in principle.

I will carefully double check tomorrow his definitions and my derivation
« Last Edit: 05/14/2015 03:53 AM by Rodal »

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3286 on: 05/14/2015 03:52 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you in principle.  I have to double check where he puts the cutoff wavelength...

All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.

The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.

Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.

If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

I'm actually amazed a flat ended conic cavity can generate thrust.
« Last Edit: 05/14/2015 04:06 AM by TheTraveller »
It Is Time For The EmDrive To Come Out Of The Shadows

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3287 on: 05/14/2015 04:05 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you in principle.  I have to double check where he puts the cutoff wavelength...

All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.

The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.

Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.

If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?

If so, the Design Factor should be trivial, much easier to calculate...
« Last Edit: 05/14/2015 04:09 AM by Rodal »

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3288 on: 05/14/2015 04:14 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you in principle.  I have to double check where he puts the cutoff wavelength...

All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.

The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.

Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.

If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?

He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation
It Is Time For The EmDrive To Come Out Of The Shadows

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3289 on: 05/14/2015 04:16 AM »
Df = 0 when lambda1 = lambda2
Df = 1 when min(lambda1, lambda2) = lambda0  (free space condition)

So it looks like the right Df

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3290 on: 05/14/2015 04:17 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible

Back to basics.  How does Shawyer define

lambda0 ? Does he define  lambda0 = loFreeSpaceWavelength = c/rfFrequency  ?

How does Shawyer define lambda 1 and lambda 2 ?

He does define lambda 1 not equal to lambda2, so they are different, we know that

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3291 on: 05/14/2015 04:21 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you in principle.  I have to double check where he puts the cutoff wavelength...

All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.

The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.

Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.

If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?

He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation

No, he doesn't use the standard definition because according to that formula there is only one waveguide wavelength and he is defining two of them.

Same question that deltaMass was asking

How does Shawyer define lambdag 1 and lambdag 2 ?

He does define lambdag1 not equal to lambdag2, so they are different, we know that

what is the difference?

In the formula you show the only things that appear are the free space wavelength and the cut off wavelength

But there is only one free space wavelength

There is only one cutoff wavelength

SO WHAT IS THE DIFFERENCE between lambdag1 and lambdag2
« Last Edit: 05/14/2015 04:31 AM by Rodal »

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3292 on: 05/14/2015 04:24 AM »
Df = 0 when lambda1 = lambda2
Df = 1 when min(lambda1, lambda2) = lambda0  (free space condition)

So it looks like the right Df

Shawyers definitions for Df = 0 and Df = 1 are attached.

Your Df = 0 assumption matches Shawyer but for Df = 1, small end plate lambda2 must be at cutoff and big end plate lambda1 close to lambda0.
It Is Time For The EmDrive To Come Out Of The Shadows

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3293 on: 05/14/2015 04:25 AM »
Rodal, you seem confused. Lambda0 is the free space wavelength = c/f. Equation (5) tells you the rest.

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3294 on: 05/14/2015 04:26 AM »
The Traveller: I know, and what I said doesn't contradict you, except that cut-off isn't necessary.

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3295 on: 05/14/2015 04:28 AM »
Rodal, you seem confused. Lambda0 is the free space wavelength = c/f. Equation (5) tells you the rest.

that's the same definition of free space wavelength I gave you

What is  lambdag 1 and lambdag 2 ?

Thanks

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3296 on: 05/14/2015 04:32 AM »
They are what TheTraveller described and what's in eqn (5). To whit:
vgN / c  =  lambda0 /  lambdagN
for N=1,2,3,4,5,6...whatever

...assuming the relative permeability and relative permittivity both =1
« Last Edit: 05/14/2015 04:34 AM by deltaMass »

#### TheTraveller

##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3297 on: 05/14/2015 04:34 AM »
You missed my post above. In order for the group velocity in the waveguide to never exceed c (and it never does!) it follows from eqn(5) that all lambda's in the waveguide are greater than the lambda in free space.
Therefore 1 - (lambda02/(lambda1*lambda2)) is always positive definite.

Therefore blow-up is impossible
I agree with you in principle.  I have to double check where he puts the cutoff wavelength...

All wavelengths inside the conic cavity are bigger than outside and the resultant group velocities are less than the velocity outside.

The guide wavelength and resultant group velocity constantly varies, driven by the constantly varying diameter of the conic section the wave is passing through.

Also the edges of the wave fronts are at right angles to the cone sides due to being spherical wave fronts as if they originated from and are returning to the vertex of the cone.

If however the conic cavity end plates are flat and not spherically matching the spherical wave fronts bouncing off them, well you may be pushing s##t up hill before things are working well inside the cavity.

So Shawyer does not have a cut-off wavelength appearing anywhere on his Design Factor ?

He uses what the industry uses. Guide wavelength as in the attached. It is related to cutoff wavelength as per the attached equation

No, he doesn't use the standard definition because according to that formula there is only one waveguide wavelength and he is defining two of them.

Same question that deltaMass was asking

How does Shawyer define lambdag 1 and lambdag 2 ?

He does define lambdag1 not equal to lambdag2, so they are different, we know that

what is the difference?

In the formula you show the only things that appear are the free space wavelength and the cut off wavelength

But there is only one free space wavelength

There is only one cutoff wavelength

SO WHAT IS THE DIFFERENCE between lambdag1 and lambdag2

The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.

Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.

Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.
It Is Time For The EmDrive To Come Out Of The Shadows

#### Rodal

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3298 on: 05/14/2015 04:37 AM »
....
The cutoff wavelength is defined for EACH end, based on it's diameter * 1.71. See attachment 2.

Lambda1 = guide wavelength equation from attachment 1 below using big end cutoff wavelength.

Lambda2 = guide wavelength equation from attachment 1 below using small end cutoff wavelength.

OK

That's NOT what deltaMass had written  (there is no N=1,2,3,4, in equation 5)

That's exactly the same factor of 1.7    I derived, based on TE110

Now I'm going to sleep.
« Last Edit: 05/14/2015 04:38 AM by Rodal »

#### deltaMass

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##### Re: EM Drive Developments - related to space flight applications - Thread 2
« Reply #3299 on: 05/14/2015 04:39 AM »
Notice that the game may change if sqrt(mur epsr) < 1. Usually it's ~=1, or a bit more , and that's the assumption I'm making with my inequalities games

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