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......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

Quote from: WarpTech on 06/05/2015 07:55 PMI 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......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? ToddThis would make it even more fun than it ever was. A constipated frustum...

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!

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?

Quote from: rfmwguy on 06/05/2015 07:06 PMFinal 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/B00OOSRVPQGood price.Review here which questions 8 Watt output:http://w5vwp.com/wifidevices.shtml

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.

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 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.

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

Quote from: aero on 06/05/2015 08:05 PMWell, Energy reaches a maximum, Pin *Q as Po reaches Pin, so mass reaches a maximum, what, likemo + 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

Quote from: WarpTech on 06/05/2015 07:55 PMNow, 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.

Quote from: foob on 06/05/2015 08:34 PMQuote from: WarpTech on 06/05/2015 07:55 PMNow, 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*Hwhere eR and uR are the relative permittivity and permeability in the material, and e0 and u0 are in vacuum.eR >> u0, uR >> u0When the light leaves the medium and enters the vacuum;D x B = e0*E x u0*HSo 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: foob on 06/05/2015 08:34 PMQuote from: WarpTech on 06/05/2015 07:55 PMNow, 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!

...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: 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...

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.

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