Quote from: SeeShells on 06/05/2015 01:43 PMStarted to do the layouts, have the stepper motor lead screw and carrier, drivers and controller, now getting ready to have the endplates water jet cutIts 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

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

Quote from: SeeShells on 06/05/2015 09:10 PMQuote from: rfmwguy on 06/05/2015 08:33 PMQuote from: SeeShells on 06/05/2015 07:18 PMQuote from: rfmwguy on 06/05/2015 07:06 PMFinal 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 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)

Quote from: rfmwguy on 06/05/2015 08:33 PMQuote from: SeeShells on 06/05/2015 07:18 PMQuote from: rfmwguy on 06/05/2015 07:06 PMFinal 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.

Quote from: SeeShells on 06/05/2015 07:18 PMQuote from: rfmwguy on 06/05/2015 07:06 PMFinal 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

Quote from: rfmwguy on 06/05/2015 07:06 PMFinal or intermediate amp +39dBm (8 Watts) on its way!Yes, good for you, but it kind of looks like a toaster.

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

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!

Quote from: foob on 06/05/2015 08:34 PMQuote from: WarpTech on 06/05/2015 07:55 PMSo 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.ToddWell 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.

Quote from: WarpTech on 06/05/2015 07:55 PMSo 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.ToddWell 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.

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

Quote from: deltaMass on 06/05/2015 10:49 PMQuote from: WarpTech on 06/05/2015 07:55 PM...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 E_{out}(t). This in turn readily exceeds E_{in}(t) when t > t_{0}. I'll just cut to the chase (not showing the entire derivation). LetE_{in}(t) := P ta_{0} := k P / m_{0}T := m_{0}c^{2}/Ph := 2 m_{0} / (P k^{2})thenm(t) = m_{0}(1 + t/T)v(t) = a_{0} T ln(1 + t/T)soE_{out}(t) / E_{in}(t) = (1/h)(ln(1 + t/T))^{2} / t , =1 when t = t_{0}.We need to solve this for t_{0}:(ln(1 + t_{0}/T))^{2} = h t_{0}Mathematica v8 falls on its keester.However we can plot the left and right side together and verify that a breakeven t_{0} indeed exists for certain values of h, TSo 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*tIf 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 = PinPlug in… (EDIT: lost the “a” below)dy/dt = (a*v/c^2)*y^3and solve!Pin = m*a*v*y^3 orF = (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

Quote from: WarpTech on 06/05/2015 07:55 PM...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 E_{out}(t). This in turn readily exceeds E_{in}(t) when t > t_{0}. I'll just cut to the chase (not showing the entire derivation). LetE_{in}(t) := P ta_{0} := k P / m_{0}T := m_{0}c^{2}/Ph := 2 m_{0} / (P k^{2})thenm(t) = m_{0}(1 + t/T)v(t) = a_{0} T ln(1 + t/T)soE_{out}(t) / E_{in}(t) = (1/h)(ln(1 + t/T))^{2} / t , =1 when t = t_{0}.We need to solve this for t_{0}:(ln(1 + t_{0}/T))^{2} = h t_{0}Mathematica v8 falls on its keester.However we can plot the left and right side together and verify that a breakeven t_{0} indeed exists for certain values of h, TSo a brave and creative attempt, but it doesn't guarantee that overunity goes away.

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

m*c^2*dy/dt = Pin

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.

Quote from: WarpTech on 06/06/2015 12:56 AMQuote from: deltaMass on 06/05/2015 10:49 PMQuote from: WarpTech on 06/05/2015 07:55 PM...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 E_{out}(t). This in turn readily exceeds E_{in}(t) when t > t_{0}. I'll just cut to the chase (not showing the entire derivation). LetE_{in}(t) := P ta_{0} := k P / m_{0}T := m_{0}c^{2}/Ph := 2 m_{0} / (P k^{2})thenm(t) = m_{0}(1 + t/T)v(t) = a_{0} T ln(1 + t/T)soE_{out}(t) / E_{in}(t) = (1/h)(ln(1 + t/T))^{2} / t , =1 when t = t_{0}.We need to solve this for t_{0}:(ln(1 + t_{0}/T))^{2} = h t_{0}Mathematica v8 falls on its keester.However we can plot the left and right side together and verify that a breakeven t_{0} indeed exists for certain values of h, TSo 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*tIf 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 = PinPlug in… (EDIT: lost the “a” below)dy/dt = (a*v/c^2)*y^3and solve!Pin = m*a*v*y^3 orF = (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!ToddOops - you did it again.Quotem*c^2*dy/dt = Pinisn't right. You ought to have the total differential because, recall, you have allowed a variable m(t). ThusP_{in} = c^{2}(y dm/dt + m dy/dt)and so forth...

Quote from: deltaMass on 06/05/2015 10:49 PMQuote from: WarpTech on 06/05/2015 07:55 PMQuote from: deltaMass on 06/03/2015 06:16 PMApart 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^2Therefore, 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*a0v(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? ToddI'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 E_{out}(t). This in turn readily exceeds E_{in}(t) when t > t_{0}. I'll just cut to the chase (not showing the entire derivation). LetE_{in}(t) := P ta_{0} := k P / m_{0}T := m_{0}c^{2}/Ph := 2 m_{0} / (P k^{2})thenm(t) = m_{0}(1 + t/T)v(t) = a_{0} T ln(1 + t/T)soE_{out}(t) / E_{in}(t) = (1/h)(ln(1 + t/T))^{2} / t , =1 when t = t_{0}.We need to solve this for t_{0}:(ln(1 + t_{0}/T))^{2} = h t_{0}Mathematica v8 falls on its keester.However we can plot the left and right side together and verify that a breakeven t_{0} indeed exists for certain values of h, TSo 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 ]a_{0} := k P / m_{0} here "k" has units of inverse velocity (time/length)h := 2 m_{0} / (P k^{2}) here "h" has units of timetherefore (ln(1 + t_{0}/T))^{2} = h t_{0}is not dimensionally correct, because the natural logarithm (ln) is dimensionless. The square of a dimensionless quantity is also dimensionless. Notice that t_{0} must have dimensions of time because deltaMass defined T with dimensions of time, and because( t_{0}/T)^{2} must be dimensionless to be able to compute (ln(1 + t_{0}/T))^{2}. now(ln(1 + t_{0}/T))^{2} = h t_{0}since h has dimensions of time, and t_{0} has dimensions of time, this is stating dimensionless^2 = time^2which 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

Quote from: WarpTech on 06/05/2015 07:55 PMQuote from: deltaMass on 06/03/2015 06:16 PMApart 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^2Therefore, 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*a0v(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? ToddI'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 E_{out}(t). This in turn readily exceeds E_{in}(t) when t > t_{0}. I'll just cut to the chase (not showing the entire derivation). LetE_{in}(t) := P ta_{0} := k P / m_{0}T := m_{0}c^{2}/Ph := 2 m_{0} / (P k^{2})thenm(t) = m_{0}(1 + t/T)v(t) = a_{0} T ln(1 + t/T)soE_{out}(t) / E_{in}(t) = (1/h)(ln(1 + t/T))^{2} / t , =1 when t = t_{0}.We need to solve this for t_{0}:(ln(1 + t_{0}/T))^{2} = h t_{0}Mathematica v8 falls on its keester.However we can plot the left and right side together and verify that a breakeven t_{0} indeed exists for certain values of h, TSo a brave and creative attempt, but it doesn't guarantee that overunity goes away.

Quote from: deltaMass on 06/03/2015 06:16 PMApart 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^2Therefore, 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*a0v(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

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 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)

Quote from: deltaMass on 06/06/2015 04:49 AMOur 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.

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 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.

Quote from: birchoffI 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".

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.