Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1797165 times)

Online Rodal

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Started to do the layouts, have the stepper motor lead screw and carrier, drivers and controller, now getting ready to have the endplates water jet cut

Its great to see so many new projects getting underway.  I'm updating the Emdrive.wiki/Builders page with the details of your project and rmfwguy's.  As each builder gets testing underway, we'll create specific pages for your tests.

I did have a question for the builders:  Is anyone considering  replicating the Cannae drive? 

After reading back over the Brady et.al paper, they pretty much concluded the thrust was originating in the Rf feed tube, not in the pillbox cavity (COSMOL showed pretty much nothing going on in the cavity).  But Fetta's 2011 superconducting test had a different input design and does not mention using a dielectric.  ???  So to me there's still unresolved questions about the Cannae drive. I think it would be very interesting if someone planning to build an EmDrive also built a Cannae drive and tested them both on the same experimental setup.  Just a thought for the builders to file away as a possible future experiment.

Link to Fetta's superconducting design if you want to compare to what's in the EW paper: http://web.archive.org/web/20121104025749/http://www.cannae.com/proof-of-concept/design

For general info:  a number of problems with the superconducting Cannae drive test were pointed out by @zen-in in previous threads.   My recollection is that the consensus was that the results of this test (apparently only published in the Cannae webpage in 2012 but now no longer there according to Wikipedia) were questionable.
My recollection is that what was actually measured could very well have been an experimental artifact due to the way that the low superconducting  temperature was attempted to be reached. Apparently the only way to get this information now is to access the web archive, as you point out.

"So, use the info in the web-archive at one's own peril"
« Last Edit: 06/06/2015 02:43 AM by Rodal »

Offline kml

The first part has arrived, a surplus aluminum WR-650 waveguide section (6.5" x 3.25" x17").   Operating mode will be TE10 @1250MHz which can be driven by affordable 23cm Ham equipment (10w initially and later 160w).    Various dielectric materials will be tested at one end.   Used WR-650 waveguide parts do not seem to be very common so I am considering just drilling a hole for my coax transition.   At least one of the end shorts will likely need to be an adjustable insert.  Any net forces will be measured by an AND MC-10K 10.1kg/.001g mass comparator.    The list of dielectrics to be tested includes HDPE (n=1.5), Alumina (n=3), and Strontium Titanate (n=17).   Dielectrics will be drilled or notched at various depths for a low reflection transition.  I'm still looking for information on the ideal pattern for the dielectric transition and how long it needs to be.


Offline SeeShells

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Final or intermediate amp +39dBm (8 Watts) on its way!
Yes, good for you, but it kind of looks like a toaster. ;)

I was almost going to nickname my experiment Project Lampshade :)
You have to call it something I guess. ;) I was told mine looked like a kitchen microwave colander.

Not this bad an amp. but be sure your cavity is really RF-tight if you dont want to  got problems with the FCC or your healthiness. 39 Watt of @2.4GHz is a lot of RF-power... On the other hand people play with 400W or more :D that's quite dangerous even more. Seriously be sure you play safe with such equipment! Go a few meters away before put on the power. Only with spectrum analyzer and a little know how you are be sure about the emitted energy caused by self construction the cone and antenna.
There are people well-known by myself who burn a little hole into a RF sensitive foil with only 3W in a distance of a few cm!

Its like the case of the laser guys: Don't look with your left eye into the RF-beam! (in a distance of a few cm)  ;)
I know the dangers and yes I'll be careful. I'm even building an RF cage around them. I visibly cringed when I saw iulian turn his on.

Offline not_a_physicist

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What this shows is that constant Power is proportional to a constant Limiting Velocity, not a constant force. ...(clipped)... Given a constant input power, eventually the mass will reach a constant velocity < c.

Paradox solved!

I may be misunderstanding, but doesn't this privilege a reference frame? If you watch a spaceship using 1 watt moving to the right limited to 1 m/s, and I am driving left at 9 m/s, then in my frame of reference the spaceship is travelling at 10 m/s. That's the same 1 watt of input power leading to a different speed limit depending on who's watching.

Offline SeeShells

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So where did the extra "relative" momentum go? This physics is being totally neglected in the equation F = P/c. That ratio is a back of the envelope calculation of the momentum of light in free space. It has nothing to do with how well light can push on a material as it exits.

Todd
Well done. I'm in the middle of re-going through a refresher of all the hooks momentum has in physics, it's kind of fun. Such a simple equation p=mv.

Offline SeeShells

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Offline deltaMass

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...
a(t) = a0/(1 + (Pin*t)/(m0*c^2))

F = m(t)*a(t) = m0*a0
...
I'm afraid you've made a mistake and so your conclusion is wrong. You have to integrate a(t) to derive v(t), which yields Eout(t). This in turn readily exceeds Ein(t) when t > t0. I'll just cut to the chase (not showing the entire derivation). Let
Ein(t) := P t
a0 := k P / m0
T := m0c2/P
h := 2 m0 / (P k2)
then
m(t) = m0(1 + t/T)
v(t) = a0 T ln(1 + t/T)
so
Eout(t) / Ein(t) = (1/h)(ln(1 + t/T))2 / t , =1 when t = t0.

We need to solve this for t0:
(ln(1 + t0/T))2 = h t0

Mathematica v8 falls on its keester.
However we can plot the left and right side together and verify that a breakeven t0 indeed exists for certain values of h, T

So a brave and creative attempt, but it doesn't guarantee that overunity goes away.

Good catch thank you! That's what happens when I do Math on my lunch break.  :-[

I gotta hand it to you @deltaMass, you're sharp! Well, let's try a totally different approach then, shall we?

For instance: Let y be the relativistic gamma factor,

y = 1/sqrt(1 - (v/c)^2)

We "know" the following must be true to conserve energy.

m*c^2*(y - 1) = Pin*t

If there are no other losses or stored energy, then there is 100% conversion of input power to kinetic energy, no more, no less. Take the time derivative of both sides, keeping Pin constant.

m*c^2*dy/dt = Pin

Plug in  (EDIT: lost the a below)

dy/dt = (a*v/c^2)*y^3

and solve!

Pin = m*a*v*y^3  or

F = (Pin/v)*(1 - (v/c)^2)^3/2  (Edit 3)

What this shows is that constant Power is proportional to a constant Limiting Velocity, not a constant force. You can't have a system where you have constant N/kW and it just keeps accelerating. If N/kW is held constant, then it will reach a limiting velocity. Therefore, the photon rocket equation F = P/c is wrong! No such equation can exist when you are accelerating a mass. Nature won't allow it. Given a constant input power, eventually the mass will reach a constant velocity < c.

Paradox solved!

EDIT: I find it similar to hovering in a gravitational field of acceleration a". The gravitational potential has units of velocity squared, just like energy. But to stay at a constant potential (altitude) you must constantly run the engine to hover! (Funny, in my quantum PV warp drive theory, gravitational potentials are equivalent to different power spectrums of the ZPF.) Bingo!

Todd
Oops - you did it again.
Quote
m*c^2*dy/dt = Pin
isn't right. You ought to have the total differential because, recall, you have allowed a variable m(t). Thus
Pin = c2(y dm/dt +  m dy/dt)

and so forth...

Yes, there's a reason I call myself deltaMass  8)

But quite frankly, you only muddy the waters by introducing Special Relativity. Your equations ought to work at all speeds.


« Last Edit: 06/06/2015 04:38 AM by deltaMass »

Offline Dortex

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I may be misunderstanding, but doesn't this privilege a reference frame? If you watch a spaceship using 1 watt moving to the right limited to 1 m/s, and I am driving left at 9 m/s, then in my frame of reference the spaceship is travelling at 10 m/s. That's the same 1 watt of input power leading to a different speed limit depending on who's watching.

To make matters worse, if you're on the ship itself, the drive is stationary relative to you, so it never stops accelerating. It seems to me like we're making Einstein dig his own grave here by using relativity to refute relativity.  :-\

Offline birchoff

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...
a(t) = a0/(1 + (Pin*t)/(m0*c^2))

F = m(t)*a(t) = m0*a0
...
I'm afraid you've made a mistake and so your conclusion is wrong. You have to integrate a(t) to derive v(t), which yields Eout(t). This in turn readily exceeds Ein(t) when t > t0. I'll just cut to the chase (not showing the entire derivation). Let
Ein(t) := P t
a0 := k P / m0
T := m0c2/P
h := 2 m0 / (P k2)
then
m(t) = m0(1 + t/T)
v(t) = a0 T ln(1 + t/T)
so
Eout(t) / Ein(t) = (1/h)(ln(1 + t/T))2 / t , =1 when t = t0.

We need to solve this for t0:
(ln(1 + t0/T))2 = h t0

Mathematica v8 falls on its keester.
However we can plot the left and right side together and verify that a breakeven t0 indeed exists for certain values of h, T

So a brave and creative attempt, but it doesn't guarantee that overunity goes away.

Good catch thank you! That's what happens when I do Math on my lunch break.  :-[

I gotta hand it to you @deltaMass, you're sharp! Well, let's try a totally different approach then, shall we?

For instance: Let y be the relativistic gamma factor,

y = 1/sqrt(1 - (v/c)^2)

We "know" the following must be true to conserve energy.

m*c^2*(y - 1) = Pin*t

If there are no other losses or stored energy, then there is 100% conversion of input power to kinetic energy, no more, no less. Take the time derivative of both sides, keeping Pin constant.

m*c^2*dy/dt = Pin

Plug in  (EDIT: lost the a below)

dy/dt = (a*v/c^2)*y^3

and solve!

Pin = m*a*v*y^3  or

F = (Pin/v)*(1 - (v/c)^2)^3/2  (Edit 3)

What this shows is that constant Power is proportional to a constant Limiting Velocity, not a constant force. You can't have a system where you have constant N/kW and it just keeps accelerating. If N/kW is held constant, then it will reach a limiting velocity. Therefore, the photon rocket equation F = P/c is wrong! No such equation can exist when you are accelerating a mass. Nature won't allow it. Given a constant input power, eventually the mass will reach a constant velocity < c.

Paradox solved!

EDIT: I find it similar to hovering in a gravitational field of acceleration a". The gravitational potential has units of velocity squared, just like energy. But to stay at a constant potential (altitude) you must constantly run the engine to hover! (Funny, in my quantum PV warp drive theory, gravitational potentials are equivalent to different power spectrums of the ZPF.) Bingo!

Todd
Oops - you did it again.
Quote
m*c^2*dy/dt = Pin
isn't right. You ought to have the total differential because, recall, you have allowed a variable m(t). Thus
Pin = c2(y dm/dt +  m dy/dt)

and so forth...

I gotta ask, Is there any catastrophic side effect from EmDrive leading to a perpetual motion machine?

P.S. I do not consider needing to go back and fix up or re-write whole theories catastrophic.

Offline deltaMass

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Apart from the F = P/v (trash Einstein) and F = constant (trash Noether) dynamic models, I did propose a third model which is probably best named "frustrated momentum". In this case, the static F persists until a certain momentum value has been established, equal to an initial impulse given to the system. This momentum having been fully established, F returns to zero Newton.

Shawyer's video seems to fit the "frustrated momentum" bill. McCullough's theory also predicts it IIRC.

I think I found a solution to the Paradox:

The issue of a perpetual motion machine goes away when you accept that if the system has input power and no expulsion of mass, then mass is going to increase with time.

m(t) = m0 + Pin*t/c^2

Therefore, if F/Pin = N/kW is a constant, the acceleration will vary inversely to the mass as a function of time.

a(t) = a0/(1 + (Pin*t)/(m0*c^2))

F = m(t)*a(t) = m0*a0

v(t) = a0*t/(1 + (Pin*t)/(m0*c^2))

Now, the kinetic energy can never exceed the input energy.

Pin*t = m0*c^2 * (sqrt(m0*(a0*t)^2 / m(t)*v(t)^2)  -  1)

Its similar to special relativity, but for low velocity, taking relativistic mass into consideration. Even though the speed is far from c, mass is still increasing as energy is input to the system and is NOT expelled.

Comments?

Todd
I'm afraid you've made a mistake and so your conclusion is wrong. You have to integrate a(t) to derive v(t), which yields Eout(t). This in turn readily exceeds Ein(t) when t > t0. I'll just cut to the chase (not showing the entire derivation). Let
Ein(t) := P t
a0 := k P / m0
T := m0c2/P
h := 2 m0 / (P k2)
then
m(t) = m0(1 + t/T)
v(t) = a0 T ln(1 + t/T)
so
Eout(t) / Ein(t) = (1/h)(ln(1 + t/T))2 / t , =1 when t = t0.

We need to solve this for t0:
(ln(1 + t0/T))2 = h t0

Mathematica v8 falls on its keester.
However we can plot the left and right side together and verify that a breakeven t0 indeed exists for certain values of h, T

So a brave and creative attempt, but it doesn't guarantee that overunity goes away.

deltaMass, let's mind about units  ;)  [since you were saying last night that Rabbit, (in Notsosureofit's poem) had the wrong units  ;)]

a0 := k P / m0  here "k" has units of inverse velocity (time/length)

h := 2 m0 / (P k2) here "h" has units of time

therefore

(ln(1 + t0/T))2 = h t0


is not dimensionally correct
, because the natural logarithm (ln) is dimensionless. The square of a dimensionless quantity is also dimensionless. Notice that t0 must have dimensions of time because deltaMass defined T with dimensions of time, and because( t0/T)2 must be dimensionless to be able to compute (ln(1 + t0/T))2.

now

(ln(1 + t0/T))2 = h t0

since h has dimensions of time, and t0 has dimensions of time, this is stating

dimensionless^2 = time^2

which is obviously incorrect.

See http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385474#msg1385474 for a correct solution including the time at which the energy goes over unity
Sorry - you're right. That's because there was a typo; I dropped a constant. The definition of 'h' should have been

h := 2 m0 / (P (k T)2) here "h" has units of 1/time


Offline deltaMass

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Our posts crossed in the time warp. We are agreeing. You can relax now.

Except the problem is with Mathematica. Since 'h' is a constant of the motion, it doesn't change the fact that M'ca v8 can't solve it.

You say so yourself - "there is no closed form solution"

However, FindInstance[..] will find the solution numerically, using n=2 in the arglist (you need 2 because t0=0 is trivially a solution)

The bottom line is that Todd's attempt to guarantee avoiding overunity with variable mass fails.
« Last Edit: 06/06/2015 05:02 AM by deltaMass »

Offline dustinthewind

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I see no point in equalising volumes like that. All that's necessary is the venting orthogonal to the thrust (you have that) and two nozzles, diametrically opposite one another (you have that too). I don't think the height placement is critical.

(I checked your maths and it's correct btw)

My reasoning for equalizing the volumes was because I was considering the extreme condition that we place the valves at one end or the other.  As a volume of air is expelled that vacuum is then filled with new air.  In the extreme placement of the valves at one end then air must travel towards the valves.  The movement of the air towards the valves suggest a reaction on the cavity in the opposite direction.  My aim was to eliminate that. 

Offline deltaMass

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I can sort of buy that. Certainly the problem (almost) goes away if we decide not to weigh the cavity in order to deduce thrust, but measure horizontal thrust instead, as EW is doing
« Last Edit: 06/06/2015 05:00 AM by deltaMass »

Offline deltaMass

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Our posts crossed in the time warp. We are agreeing. You can relax now.

Except the problem is with Mathematica. Since 'h' is a constant of the motion, it doesn't change the fact that M'ca v8 can't solve it.

You say so yourself - "there is no closed form solution"

However, FindInstance[..] will find the solution numerically, using n=2 in the arglist (you need 2 because t=0 is trivially a solution)
I hope I'm not the only one that routinely uses perturbation analysis here?

Whenever you see a nonlinear equation, one shouldn't just give up. There is usually a perturbation analysis one can perform.

Most problems in nature, including General Relativity (which is a nonlinear theory) are nonlinear. The solution to first order is perfectly fine, because as I show, the overunity speed is way below the speed of light.

It would be incongruous for you to say to WarpTech on one hand "why are you using the nonlinear Special Relativity equations" and then when you see a nonlinearity you stop, instead of pursuing the perturbation to first order, which should take you all the way back to standard theory.
No idea why you continue to argue the toss with me and beat your breast about your expertise in perturbation theory. If I say "I'm very impressed, Rodal", will you quit?

We both had a go at deconstructing Todd's hypothesis. We both found (modulo my typo in a constant) the same result - namely,  that postulating variable mass doesn't fix what it was supposed to fix.

Offline deltaMass

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Quote from: birchoff
I gotta ask, Is there any catastrophic side effect from EmDrive leading to a perpetual motion machine?
P.S. I do not consider needing to go back and fix up or re-write whole theories catastrophic.
It depends what you mean by "catastrophic" of course. For example, Woodward does not think it catastrophic that his drive pulls apparently-unlimited energy and momentum from the distant stars. I would instead describe a device that provided unlimited local free energy as "miraculous".  ::)
« Last Edit: 06/06/2015 05:12 AM by deltaMass »

Offline dustinthewind

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Quote from: birchoff
I gotta ask, Is there any catastrophic side effect from EmDrive leading to a perpetual motion machine?
P.S. I do not consider needing to go back and fix up or re-write whole theories catastrophic.
It depends what you mean by "catastrophic" of course. For example, Woodward does not think it catastrophic that his drive pulls apparently-unlimited energy and momentum from the distant stars. I would instead describe a device that provided unlimited local free energy as "miraculous".  ::)

It is funny because they were suggesting at some speed the device reaches over-unity.  My brain immediately jumped to Doc Brown in Back to the Future and his flux capacitor.  His car needed to reach a certain velocity and then the flux capacitor would warp the car. 

I know it doesn't seem right because you could say the car is already going fast enough relative to say Jupiter or Pluto.  Same thing with respect to the cavity I would suspect. 

It is also funny to think the cavity could be thought as a flux capacitor but in the sense that it stores EM energy.  Then again a SMES (superconducting magnetic energy storage) unit or capacitor can also. 

Edit:  below @ deltamass Yeah, I saw the rotary illustration some time earlier in the threads.
« Last Edit: 06/06/2015 05:47 AM by dustinthewind »

Offline deltaMass

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One way to pin down this velocity dependent stuff is to imagine a rotary implementation. frobincat has already provided all the details of that, and I've discussed it too, going back to 1996 with my first chat with Woodward.

Online ThinkerX

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I wonder...would it be valid to test the thrust of this device on a near frictionless surface?

I was at a lake a few days ago.  Lots of kids splashing in the water with all sorts of boats and floating objects.  What caught my eye was a toddler - maybe three or four years old - pushing around a small boat.  On land, said toddler could not have budged the boat an inch.  But, in the water, he gave it a little push and away it went.  (said water wasn't more than knee deep for him, btw).

So, would a similar test be valid for the EM Drive?  Put it on a tiny boat in a tub or pool and see if it could move itself?  Or perhaps slick ice or something similar? 


Offline deltaMass

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Yes indeed. Or an air table. Or on dry ice mounts.

Offline dustinthewind

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One way to pin down this velocity dependent stuff is to imagine a rotary implementation. frobincat has already provided all the details of that, and I've discussed it too, going back to 1996 with my first chat with Woodward.

I wanted to suggest a possible parallel to propellant-less propulsion in a rotary sense.  Two examples come to mind that might qualify in this category.  One is the device in this video.  .  The device regardless of how fast it rotates can still have energy added to it by torquing against the force of the pull on the weights that pull towards the larger radius.  I think it qualifies as propellantless. 

Another is a swing but we are going to make it a hard long rod that can rotate 360 degrees and have an object at the end that can do work.  set it up like a swing so gravity pulls it down and now set it in motion.  When the device reaches max velocity it does work against the circular force.  When it reaches minimal velocity it contracts so it can do work again at max velocity. 

Would these devices be in the same category as a propellantless thruster rotating in circles? 
If so, what is to stop them from reaching overunity? 

whoops I didn't realize this but it looks like some one has already attempted such a thing.  I can't say it is valid or not.    It's possible energy could be being added externally.  Anyways, I think it might relate to the discussion of a propellantless em drive and conservation of momentum issues some times discussed. 
« Last Edit: 06/06/2015 07:23 AM by dustinthewind »

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