Author Topic: EM Drive Developments - related to space flight applications - Thread 3  (Read 1872638 times)

Offline smartcat

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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......

Itís 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

This would make it even more fun than it ever was.  A constipated frustum...

Offline X_RaY

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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......

Itís 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

This would make it even more fun than it ever was.  A constipated frustum...

splendiferous  ;D

Offline rfmwguy

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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 :)

Offline foob

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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)

Itís 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?

Still seems a little flaky. Assume an on-board power plant with 100% mass to RF conversion efficiency. The mass lost in the power plant would balance the mass gained in the thruster(aside from waste in the thruster). Thus all you have done is changed the center of mass of the spaceship while merrily jetting around the solar system.

Offline rfmwguy

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Final or intermediate amp +39dBm (8 Watts) on its way!

20 Watts: http://sunhans.com/wifi-signal-booster-20w-2-4ghz-40dbm/

Got a link for your 8 Watt unit?

Found one link: http://www.amazon.com/EDUP-EP-AB003-Repeater-Wireless-Broadband/dp/B00OOSRVPQ
Good price.

Review here which questions 8 Watt output:
http://w5vwp.com/wifidevices.shtml

8 watt amp:

Generic EDUP 8W WiFi Wireless LAN Broadband Router Signal Booster Amplifier 39dBm 2.4Ghz TDD
 by EDUP
Link: http://amzn.com/B00JVHDAPA

This increases my initial power plan about 5.8dB to about +38dBm. Still small, lightweight and battery operable. If I decide to ramp up final power even more, I can use this as an intermediate amp...probably be able to find or build a 100W final amp at reasonable costs.
« Last Edit: 06/05/2015 08:47 PM by rfmwguy »

Offline X_RaY

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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)

It’s 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

@ m(t) = m0 + Pin*t/c^2
Problem with this equation is, even you got a superconducting cavity, the energy inside increase to infinity if you pump energie into an infinite time. It will be more and more heavy and ends in a black hole.. On the other hand may be do you like to create a new universe inside the frustum? ;)
Don't be worry :) work on and may be one day we have a fitting result.
« Last Edit: 06/05/2015 09:04 PM by X_RaY »

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.

Offline deltaMass

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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. ;)
Now we have espresso and toast. I think we're set!  ;D

Offline X_RaY

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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)  ;)
« Last Edit: 06/05/2015 09:58 PM by X_RaY »

Online WarpTech

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Well, Energy reaches a maximum, Pin *Q as Po reaches Pin, so mass reaches a maximum, what, like
mo + Pin*Q/c^2 ?

No, the Q is not necessary in this calculation. It represents the accumulated "stored" energy, but the total input energy is still Pin*t. Now, I did not include heat losses, so all the energy input is either stored as EM standing waves, or stored as Kinetic energy of motion. You can't ignore the KE when calculating the total mass to be pushed.

Todd
« Last Edit: 06/05/2015 09:56 PM by WarpTech »

Offline X_RaY

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Well, Energy reaches a maximum, Pin *Q as Po reaches Pin, so mass reaches a maximum, what, like
mo + Pin*Q/c^2 ?

No, the Q is not necessary in this calculation. It represents the accumulated "stored" energy, but the total input energy is still Pin*t. Now, I did not include heat losses, so all the energy input is either stored as EM standing waves, or stored as Kinetic energy of motion. You can't ignore the KE when calculating the total mass to be pushed.

Todd

Congratulations you got build a super battery/capacitor. Sorry i don't believe in your equation  :-\

edit:
Therefore you got the solution for the energy problems in all the world. Storage energie of solar- and wind- systems and geothermal energy in a little resonator and get out of it if you need energy for your TV or microwave-truster, really nice ;)
« Last Edit: 06/05/2015 10:20 PM by X_RaY »

Online WarpTech

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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)

Itís 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?

Still seems a little flaky. Assume an on-board power plant with 100% mass to RF conversion efficiency. The mass lost in the power plant would balance the mass gained in the thruster(aside from waste in the thruster). Thus all you have done is changed the center of mass of the spaceship while merrily jetting around the solar system.

You are proposing a NEW problem, not the one that was presented by Dr. White and by deltaMass here on this forum, or in the paper by A. Higgins. If everyone wants me to do their Math homework for them, I'm going to start charging for my services!  ;)

Regardless, you are correct. The mass would remain constant if the on-board energy storage were 100% converted into stored RF energy and thrust. However, constant mass in this case still does not lead to free energy. The kinetic energy + stored RF energy can never exceed the initial stored energy of the on-board power plant.

Please, every engineer with half a brain knows there is no free-energy, not even from the QV. The thrust from a photon rocket is mistakenly taken to be F = P/c. This is wrong. The thrust of a photon rocket will depend on the change in frequency as the light leaves the material medium and enters vacuum. It's easy enough to show that inside a material, the momentum density D x B, depends on the properties of the medium.

D x B = eR*e0*E x uRxu0*H

where eR and uR are the relative permittivity and permeability in the material, and e0 and u0 are in vacuum.

eR >> u0, uR >> u0

When the light leaves the medium and enters the vacuum;

D x B = e0*E x u0*H

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




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)

Itís 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.

« Last Edit: 06/05/2015 10:57 PM by deltaMass »

Offline X_RaY

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 ???
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)

It’s 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?

Still seems a little flaky. Assume an on-board power plant with 100% mass to RF conversion efficiency. The mass lost in the power plant would balance the mass gained in the thruster(aside from waste in the thruster). Thus all you have done is changed the center of mass of the spaceship while merrily jetting around the solar system.

You are proposing a NEW problem, not the one that was presented by Dr. White and by deltaMass here on this forum, or in the paper by A. Higgins. If everyone wants me to do their Math homework for them, I'm going to start charging for my services!  ;)

Regardless, you are correct. The mass would remain constant if the on-board energy storage were 100% converted into stored RF energy and thrust. However, constant mass in this case still does not lead to free energy. The kinetic energy + stored RF energy can never exceed the initial stored energy of the on-board power plant.

Please, every engineer with half a brain knows there is no free-energy, not even from the QV. The thrust from a photon rocket is mistakenly taken to be F = P/c. This is wrong. The thrust of a photon rocket will depend on the change in frequency as the light leaves the material medium and enters vacuum. It's easy enough to show that inside a material, the momentum density D x B, depends on the properties of the medium.

D x B = eR*e0*E x uRxu0*H

where eR and uR are the relative permittivity and permeability in the material, and e0 and u0 are in vacuum.

eR >> u0, uR >> u0

When the light leaves the medium and enters the vacuum;

D x B = e0*E x u0*H

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

I know about and i work with this stuff namely Epsilon_R (Ķ_R and tan_Delta) in dielectric materials in presence of conical cavities (NDT specific applications, i am not aloud to tell on what exactly i am working on) .
So first i need to understood or to try understand you, please tell me is big "P" the momentum at the (+/-z direction foreword small ends) of photon times n(photons)
 or the total power in your equation? And k is the wavenumber? Right? OK, Let me think about it a short time please...
« Last Edit: 06/05/2015 11:14 PM by X_RaY »

Offline frobnicat

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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)

Itís 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?

Still seems a little flaky. Assume an on-board power plant with 100% mass to RF conversion efficiency. The mass lost in the power plant would balance the mass gained in the thruster(aside from waste in the thruster). Thus all you have done is changed the center of mass of the spaceship while merrily jetting around the solar system.

You are proposing a NEW problem, not the one that was presented by Dr. White and by deltaMass here on this forum, or in the paper by A. Higgins. If everyone wants me to do their Math homework for them, I'm going to start charging for my services!  ;)


The phenomenology of an em drive effect used for space flight leads to apparent generation of more energy than was put in, not with a beamed scheme but for a spacecraft accelerating on its own mass_energy resources . You propose a solution to this paradox, and when asked to take into account that power (mass_energy flow) poured into the em drive must come from another part of the same rigid system that constitute the spacecraft you say this is another problem. I'm sorry but this remark by foob (welcome !) is spot on, and nullifies your later musings until you take it into account from start.

There is no point in saying that throwing bricks from a heavy planetoid A into free floating pudding B will both impart momentum to B with averaged force/power higher than 1/c and yet respect CoE. Yes it will. But note that A+B is not a rigid system, ever expanding geometry of a space system can't be qualified as a single spacecraft. Also A[edited] will suffer a recoil. So this is not making any advancement to the resolution of the paradox at hand (if em drive effect is really useful for spaceflight of single spacecrafts, as its claimed phenomenology implies).
Quote

...
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

Please, every engineer with a whole brain knows there is no free-momentum. Photon rocket is action reaction. Counting momentum leaving forever the spacecraft gets you the gained momentum for what's left of the spacecraft energy_mass. Photons are leaving in a vacuum, momentum is accounted in the vacuum. Bouncing history of the photons and emitting medium characteristics are utterly irrelevant once the photon exhaust profile is known. The accountancy of leaving momentum flow should be made at a safe distance of the spacecraft, ideally at infinity, for instance to take into account the redshift of photons due to spacecraft gravity making them "falling back" to it a little bit (irrelevant, just kidding).

If you were to emit photons into some medium, then yes, medium characteristics would matter. But we are not discussing photon driven submarines...
« Last Edit: 06/05/2015 11:57 PM by frobnicat »

Offline rq3

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I suspect that your joke has been lost in translation over the years, continents, and languages. It actually goes: "Don't look at the laser with your remaining eye". The implication (joke here) is that the other eye is already useless due to carelessness or ignorance.

On another note, folks looking at 38 dBm amps should also look VERY carefully at the difference between gain and saturated power output. The advertised device seems to grossly conflate the two. It may have 38 dB of GAIN, but likely won't put out anything like 38 dBm of POWER.

« Last Edit: 06/06/2015 02:07 AM by Carl G »

Offline Rodal

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Defining scaled velocities and times:

vp =Pin/(ao mo) (here vp =1/k where "k" is deltaMass variable)

tp = c2/(vp ao) (this is the same variable as deltaMass "T")(reference "deltaMass" is :  http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385441#msg1385441 )


then the speed is:

v =(  c2 / vp) ln[(1+t/tp)/(1+to/tp)]

where to is the initial time

and the ratio of the kinetic energy to the input energy (assuming zero initial velocity) is

KE/(Pin t) = (1/2) (c/vp)2 (1+tp/t)(ln[(1+t/tp)/(1+to/tp)])2

It is obvious that this goes to Infinity with t going to Infinity, as the Log[t] goes to Infinity (very slowly)



Of course, time is arbitrary so we can take  to =0

which gives

v =(  c2 / vp) ln[1+t/tp] (this is the same result as deltaMass)

KE/(Pin t) = (1/2) (c/vp)2 (1+tp/t)(ln[1+t/tp])2 (this is different than deltaMass, who I think has an error)


Note that for t -> 0, this ratio goes to zero, because the square of the Log goes to zero much faster than the inverse of time.

Limit[ (1+tp/t)(ln[1+t/tp])2, t -> 0] = 0




So, a problem occurs when this ratio is greater than one ("overunity").  The time at which the problem occurs is the solution of this equation in terms of time "t" :

1= (1/2) (c/vp)2 (1+tp/t)(ln[1+t/tp])2 (this is different than deltaMass, who I think has an error)


This equation (solving for "t") has no closed-form solution in general.

However, a closed-form solution can readily be obtained for the case that overunity occurs at a time significantly smaller than tp,  t<< tp, such that

(1+tp/t) ~ tp/t

and

(ln[1+t/tp])2 ~ (t/ tp)2

Substituting those expressions, we obtain:

1= (1/2) (c/vp)2 (tp/t) (t/ tp)2

and substituting:


t =2 tp(vp/c)2
  =2 vp/ao
  =2 Pin /((ao)2 mo)


that's the time at which the Kinetic Energy becomes greater than the InputEnergy=InputPower*time,

this solution is only valid when this calculated time is significantly smaller than tp, which obviously occurs when vp is signficantly smaller than the speed of light.

Since in general  vp is signficantly smaller than the speed of light c, this solution


t =2 vp/ao


is of general applicabiltiy.




if we substitute this time in the expression for the velocity,

v =( c2 / vp) ln[1+t/tp]
   =( c2 / vp)(t/tp)        (for t/tp < <1 )
   =ao t

we get

v = 2 vp

as the velocity at which the kinetic energy becomes greater than the input energy

Notice that this is twice the amount previously calculated by @frobnicat for the case of constant mass.



NOTE added a posteriori: please see @frobnicat's message http://forum.nasaspaceflight.com/index.php?topic=37642.msg1385588#msg1385588 explaining the origin of the factor of 2 difference.
« Last Edit: 06/06/2015 12:10 PM by Rodal »

Online WarpTech

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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
« Last Edit: 06/06/2015 02:38 AM by WarpTech »

Offline Rodal

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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)

Itís 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

« Last Edit: 06/06/2015 01:53 AM by Rodal »

Offline sfrank

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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

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