-
#1060 by Rodal on 27 Sep, 2016 17:38
-
1) The discussions about conservation of energy continue to take place without taking into account whether these are open systems. Any discussion about conservation of energy of an open system that does not take into account the external fields ,( for example one based on gravitation, like the Mach/Sciama/Woodward effect) is futile. Of course there will be an overunity problem if one does not take into account the external field (in the case of the Mach/Sciama effect: gravitation)! It would be like calculating conservation of energy for a gravity assist while ignoring the effect of gravity
.
Or like calculating conservation of energy for a solar sail while completely ignoring the solar pressure on it.
Or like calculating conservation of energy for a Laser propelled sail while completely ignoring the Laser (stationed on Earth or a supply vehicle) pressure on the sail.
Completely futile to discuss conservation of energy while ignoring the external field !
From elementary thermodynamics we know that conservation of energy applies to the whole closed system. One should not pick and choose a single component and discuss overunity while ignoring the interacting field effect on the other components.
2) If the material is piezoelectric and electrostrictive, all you need is to have a harmonic voltage excitation at a frequency omega (preferably the dominant natural frequency). The electrostriction is what automatically provides a 2*omega component. Few materials are piezoelectric, but all dielectrics are electrostrictive, so in general there is no need to separately excite at 2*omega a piezoelectric material. Due to damping (which is present in ALL materials) it is most effective to just excite at a single frequency and let electrostriction take care of the 2* omega term. Of course, the magnitude of the force depends on the electrostrictive and piezoelectric material constants for the material being tested.
3) Concerning whether or not there should be a terminal velocity.
Is there a terminal velocity for a gravity assist?
To have a terminal velocity you need to be moving in a fluid. Terminal velocity (as conventionally defined) is the highest velocity attainable by an object as it moves through a fluid. Terminal velocity occurs when the sum of the viscous drag force and the buoyancy is equal to the force of gravity acting on the object. When the forces are in balance then the total acceleration in steady state is zero, and hence one has a terminal velocity.
If you have a gravity assist maneuver around a gravitational body with no atmosphere (no fluid), there is no "terminal velocity" according to the usual definition of terminal velocity. There is a maximum velocity which is dictated by the equations of motion. If you are moving through a fluid, you have a terminal velocity, also dictated by the equations of motion, but now taking the fluid forces into account. -
#1061 by Bob012345 on 27 Sep, 2016 17:55
-
Consider a hypothetical rocket that can do 10000 burns at a delta v of 1000 m/s. Each burn requires a total energy of E, maybe only 1% goes to the ship. After the first burn, the kinetic energy of the ship wrt earth is 5E5/kg so assume the burn released 5E7 J/kg in its frame. What's the total energy released in the ships frame after 10000 burns? It should be 5E10J/kg. What's the kinetic energy wrt earth? It should be 5E13J/kg. No doubt some will claim nature conspires to prevent this by making the mass requirements great enough so it's never practical to make such a rocket.
It's not about "nature conspiring to prevent this", it's about math... assuming that CoE holds. One has to take into account not just the kinetic energy of the rocket, but also the kinetic energy of the propellant (which decreases since it's being ejected in the opposite direction if you compare it to an earlier moment when it was moving together with the rocket). The rocket has to be constantly losing mass (i.e. propellant), and if you calculate the total energy over time, it all adds up and no "over-unity" occurs. The rocket does not know when to stop accelerating, it slows down naturally because it's losing mass/propellant. If there is no propellant and the rocket is not losing mass, "over-unity" is an obvious consequence, unless there is a way for it to "steal" energy from some unknown field.
I've shown you that my hypothetical rocket does run into over unity. You have to show me that it's impossible for any rocket to do that many burns without running out of mass. -
#1062 by Gilbertdrive on 27 Sep, 2016 17:58
-
In his earlier work, Shawyer showed how acceleration can collapse Q and thus force which the collapses acceleration. Looking at the papers you referenced, it's very clear that Shawyer was discussing energy conservation in relation to how the cavity can support a acceleration as the acceleration effects Q and not energy conservation as we have been debating with regards to fundamental considerations of classical mechanics that apply to all situations with a constant force. Not also that the velocity in figure 3.1 is the average velocity over the whole trip, not the instantaneous velocity.
In his later work Shawyer developed and discusses compensation techniques that allow the cavity to maintain a constant acceleration over the whole trip (or at least nearly so allowing a terminal velocity of 0.67c at the target star). It is in this context that he designed the parameters of his Interstellar Probe. I do not believe he forget anything as you suggested.
I disagree. It is clearly the same debate that we had. He clearly calculate the Kinetic Energy. It is not about the loss of resonnance due to the doppler, that is another problem.
If I was wrong, the 333mN/kg that are indicated by Shawyer and that I calculated exactly would be a pure coincidence (The loss of resonnance due to the doppler shift would exactly loss so that CoE is verified in my calculus.)
Do you assume this coïncidence ?
Anyway, I just looked with more attention the equations. Shawyer write on page 7 :
Pk=Mva
Where Pk is the output power transfered to the ship during Delta T, M the mass of the ship, V the average speed during the Delta T, and a the acceleration. Since the Delta T is short, the use of the average speed during this short period of time is OK.
I can divide each side of the equation by Mv, assuming that speed is not zero.
I get a=Pk/(Mv)
The acceleration during Delta T is equal to the energy provided to the ship during Delta T divided by the mass of the ship, and the speed of the ship.
That is exactly what I am saying since some time. The acceleration of the ship is inversely proportional to the speed of the ship.
The formula Pk=mVa is written directly by Shawyer. Since V is not zero, I can divide each part of the formula, and get my a=Pk/(Mv) Since Pk is constant, or at least limited to the whole electrical power of the ship, I can tell that for this Shawyer paper, the acceleration of the ship is inversely proportional to the speed of the ship.
-
#1063 by Bob012345 on 27 Sep, 2016 17:58
-
What I am trying to say is that the folks that say there is an energy violation are comparing energy in different frames.
I still fail to see what is wrong with comparing energy in different inertial frames, as long as you're being careful. Of course kinetic energy is different in different frames, but the total energy must be conserved no matter what intertial frame you choose. It's easy to show that while the kinetic energy of a "constantly accelerating device" grows faster and faster in a "rest" frame as the device speeds up, this gets offset by other factors (kinetic energy of the propellant in the case of a conventional rocket, etc), so that the total energy spent over a unit of time is exactly the same, regardless of the reference frame.
It's not wrong to compare but to equate. Some are saying power expended x time in the ships frame must equal or exceed the total kinetic energy in the earth observer frame where the ship starts from rest. And if the velocity exceeds a critical number, the acceleration just falls off to conserve energy. I used to think that but now I don't. What magic mechanism alerts the ship that it's crossed that number wrt earth frame and decreases the acceleration?
Consider a hypothetical rocket that can do 10000 burns at a delta v of 1000 m/s. Each burn requires a total energy of E, maybe only 1% goes to the ship. After the first burn, the kinetic energy of the ship wrt earth is 5E5/kg so assume the burn released 5E7 J/kg in its frame. What's the total energy released in the ships frame after 10000 burns? It should be 5E10J/kg. What's the kinetic energy wrt earth? It should be 5E13J/kg. No doubt some will claim nature conspires to prevent this by making the mass requirements great enough so it's never practical to make such a rocket. That's saying an EMDrive like device should never work at all. But if it works at all, there is no feasible mechanism I am aware of to decrease acceleration for a fixed force at a fixed power.
Consider that rocket doing the first burn seen from two frames, earth and another frame moving at 0.5c in the opposite direction. From earth, the first burn gives the ship a kinetic energy of 5E5J/kg. From the other frame the kinetic energy gain of the ship is huge from that burn 1.5E11J/kg. But counting the exhaust, it's the same energy released by the ship. There are frames where the energy balance is negative and energy burn is more than kinetic energy in that frame and there are frames where energy burn is less than the ships kinetic energy in that frame. All frames agree how much energy the ship burned as actual fuel. In all frames one can use the work energy theorem and get a prefect balance between force, distance and energy. According to some, the acceleration form the burns deceases to preserve CoE in some frames, but not others. This is chaos.
Can you tell who on this forum claimed that the acceleration from the burns was decreasing in some frames but no others ? I am afraid there were misunderstanding.
It's implicit by saying that the EmDrive/Mach thruster cannot work because constant force/acceleration is not practically possible and would become an over unity device (in my opinion). -
#1064 by Bob012345 on 27 Sep, 2016 18:03
-
...
The power expended x time in the ship frame is not really jumping frames, but I get why you think it is.
It's not wrong to compare but to equate. Some are saying power expended x time in the ships frame must equal or exceed the total kinetic energy in the earth observer frame where the ship starts from rest. And if the velocity exceeds a critical number, the acceleration just falls off to conserve energy. I used to think that but now I don't. What magic mechanism alerts the ship that it's crossed that number wrt earth frame and decreases the acceleration?
...
What we are calculating is how much energy is in the big battery attached to the drive in the initial rest frame before it starts accelerating. When the battery is drained in the final moving ship frame (assume to 0), the battery also will have 0 potential energy as determined in the initial rest frame. (At relativistic velocities, there are some caveats, since the field of a moving point charge is not uniform in all directions, but for the example Gilbertdrive was looking at this would be a factor of 3*, when the failure of CoE is a factor on the order of 10000000.)
The next question about the magic mechanism is one you would have to ask one of the people who support a theory like Shawyer's. Theories like the Mach effect, degradable QV, etc. provide effective propellant interactions, or preferred frames that allow CoE to be conserved. Not being able to explain how the theory matches with CoE is one of the many problems with Shawyer's theory.
*the relativistic effect on potential energy is non-trivial, energy stored in the fields, radiation reaction, etc. has to be considered, and the field dependence with angle relative to motion means that 2 equal charges a fixed distance apart could have different stored energy depending on their angle relative to the direction of motion, but length contraction probably cancels this. I may work out the details for myself if I have time, but at most it should just be a factor of 1/sqrt(1-v^2 / c^2).
If the ship is under constant acceleration, it's constantly jumping inertial frames. Any each moment it is in a 'instantaneous rest frame'. -
#1065 by Bob012345 on 27 Sep, 2016 18:15
-
It is an authority argument. "If it is published, there is no big mistake"
At least, you should concede that the reviewer who accepted the Kinetic Energy to be calculated by 1/2*mv² for a speed of 0,67C made here a mistake, since the relativist formula was needed. The aim was to calculate a ratio that is around 0,31. so the relativist formula was making a real difference.
Also, the main point of my intervention was to show that in Shawyer theory paper, the thrust is supposed to decrease as acceleration from the time frame of the departure of the ship was increasing, so that CoE is verified. Are you convinced by the extracts that I have given, or do you think that, is this paper theory, the force is supposed to be constant ?
Here, the question is not what is true, it is what Shawyer is saying. Once we will agree on what Shawyer is saying in his theory paper, we can go to the other debate.
I have no idea how Shawyer's 2015 Acta Astronautica paper has passed review.
(To read the formulas below you'll have invoke your built-in latex interpreter...)
The efficiency calculation (Eqs. 9-13) is completely pointless and the efficiency as defined in Eq. 13 is always 1. Basically (for the sake of simplicity in a non-relativistic case) Eq. (9) is the well known formula P=Tv, where P is power, T is thrust (force) and v is the spacecraft velocity (and we'll use a reference frame where the spacecraft starts at rest, so v(0)=0). Then energy used for accelerating the spacecraft by time t_a is E_in(t_a)=\int_0^t_a P(t) dt = \int_0^t_a T(t)v(t) dt. Inserting T=ma (m is spacecraft mass, a is its acceleration) we get E_in(t_a)=\int_0^t_a ma(t)v(t) dt = m\int_0^t_a d/dt[v(t)] v(t) dt = m/2 \int_0^t_a d/dt[v(t)^2] dt = (m/2)v(t_a)^2=E_k. Shawyer's Eq. 10 for E_in expresses this using the 'average' velocity, which he never seems to define. In case of time-varying thrust T in Eq. 10 should also be some sort of average, but in the paper it seems that T is assumed constant, which means that v_{av} = v(t_a)/2 (Shawyer uses V_T for v(t_a). You can check that in that case Eq. 13 yields 1 after you use v(t_a)=a t_a=T/m t_a and plug in the expression for thrust from Eq. 8 ).
So, shockingly, classical mechanics seems to be consistent: if an object is accelerated with a given power, at the end its kinetic energy is the same as energy used for accelerating the object. However, in Shawyer's theory 'mechanical power' P_{mech} comes from electrical (microwave) power P_0 through mystically energy multiplying resonating cavity, as shown in Eq. 8. To really calculate the efficiency one should use E_{in}=t_a P_0 (possibly with some correction for efficiency in converting to microwave power). With high enough Q, it's easy to get over-unity efficiency.
The reason Shawyer gets e_{to}=0.363 instead of 1 for his 'spaceplane' seems to be that numbers in Table 4 are not consistent. If you plug numbers in the Table to his formulas, acceleration becomes about 16.5 m s^-2 and the final velocity after 1300 seconds about 21.5 km/s. His 7.8 km/s apparently comes from the text, where he discusses the mission profile with acceleration 6 m s^-2; however, that acceleration is not consistent with the parameters in the table.
There's of course the caveat that I may have misunderstood Shawyer's paper completely. That is possible because the paper is very confusingly written and full of errors (typos in equations, undefined symbols etc.).
You're correct about classical mechanics if I understand your argument. You used the mechanical power which is force x velocity. Integrate that over time for constant acceleration and you get the kinetic energy. Shawyer does not confuse electrical and mechanical power. Shawyer says the constant force/acceleration is caused by the electrical power. That force times the velocity is the mechanical power he integrates over time. Some here are saying electrical power x time equally total kinetic energy.
-
#1066 by Bob012345 on 27 Sep, 2016 18:58
-
I changed from 1000 to 10,000 and forgot to change that number. Sorry.Consider a hypothetical rocket that can do 10000 burns at a delta v of 1000 m/s. Each burn requires a total energy of E, maybe only 1% goes to the ship. After the first burn, the kinetic energy of the ship wrt earth is 5E5/kg so assume the burn released 5E7 J/kg in its frame. What's the total energy released in the ships frame after 10000 burns? It should be 5E10J/kg. What's the kinetic energy wrt earth? It should be 5E13J/kg. No doubt some will claim nature conspires to prevent this by making the mass requirements great enough so it's never practical to make such a rocket.
In fact, there is no need to use directly the CoE principle. The newtonian mechanics complies with CoE. So we can solve this problem by using Newtonian mechanics, and the result will always respect CoE. The fact that it is not pratical to do the rocket is not a theoretical problem. Since newtonian mechanics are consistent with CoE, we can not even find a theoretical counter example.
So, I propose to analyse your example, and detail why it does not work.
First, it is true that the same burn will give the same acceleration relatively to different inertial reference frames if we are in Newtonian Mechanic (not relativist effect). But that needs of course to have ships of the same mass. So, it was right to use this principle same burn/same acceleration for 2 identical ships going at different speeds from earth, but you can not use it for comparing 2 successive burns of the same ship. The second burn give more acceleration that the first one, since the ship is lighter. At the same time, when the ship is lighter, for the same speed, it's Kinetic energy is lower.
First, if each burn realeased 5E7 J/kg, the total energy released would be 5E7*E4=5E11, not 5E10.
I perfectly understand your analysis, and your assertion. But you have taken impossible data. The impossible data is… the fact that 1% of the Energy was given to the ship, while 10 000 burns are possible.
If 10 000 burns are possible, each burn uses less than the mass of the ship divided by 10 000.
I shall suppose that the first burn has a perfect efficiency. It means that all the energy spent goes in motion, not in heat.
I shall suppose that the mass ejected during the burn is the mass of the ship divided by 10 000, forgive me, only 9 999 burns are possible, and I shall suppose that the mass of the empty ship itself is 1/10 000. after the first burn, assuming it is a instantaneous burn, like if the ship shot a bullet, the speed of the ejected mass will be of 999,9m/s and the speed of the ship will be 0,1m/s in the earth referential.
I call mb the mass ejected during a burn
It means that the Kinetic Energy of the ship will be Ks=½*mb*9999*(0,1)^2 when the Kinetic Energy of the expulsed mass will be Ke=½*mb*(999,9)^2
I calculate the ratio Ks/Ke=(9999*(0,1)^2)/(999,9)^2=0.00010001
It is around 10^-4
That means that only a little more than 0,01% of the spent energy goes to the Kinetic Energy of the ship. Most of the Energy goes to the expulsed mass.
That is why expulsing few mass at high speed is not energy efficient. That is why even a perfect ionic motors would not be great for pushing a car. That is also why the amelioration of ionic needs more and more energy. If we send the ions 10 times faster, we get 10 times more acceleration for the ship, but it needs 100 times more energy. So, the same acceleration needs 10 times more energy !
The lighter the ship is, the more efficient will be the burn. The 9999th burn will expell as much mass that the remaining mass on the ship, and the half of the Energy spend goes to the Kinetic Energy of the ship, but the ship is loosing at the same time half of it's mass, so it divided by 2 the Kinetic Energy of the ship.
At the end, the Ship will have a very few Kinetic Energy. Most of Kinetic Energy will belong to the expulsed propellant.
I can detail more, and make the calculus of how much Kinetic energy the ship has at the end, but for that, you need to detail your ship.
-mass of the ship
-mass ejected for each burn
I knew there is not enough mass but the question is, is that nature's way of avoiding CoE issues as we are discussing or is that just because rockets are that way. If EmDrive works at any level, it should work at all levels. -
#1067 by Bob012345 on 27 Sep, 2016 19:05
-
In his earlier work, Shawyer showed how acceleration can collapse Q and thus force which the collapses acceleration. Looking at the papers you referenced, it's very clear that Shawyer was discussing energy conservation in relation to how the cavity can support a acceleration as the acceleration effects Q and not energy conservation as we have been debating with regards to fundamental considerations of classical mechanics that apply to all situations with a constant force. Not also that the velocity in figure 3.1 is the average velocity over the whole trip, not the instantaneous velocity.
In his later work Shawyer developed and discusses compensation techniques that allow the cavity to maintain a constant acceleration over the whole trip (or at least nearly so allowing a terminal velocity of 0.67c at the target star). It is in this context that he designed the parameters of his Interstellar Probe. I do not believe he forget anything as you suggested.
I disagree. It is clearly the same debate that we had. He clearly calculate the Kinetic Energy. It is not about the loss of resonnance due to the doppler, that is another problem.
If I was wrong, the 333mN/kg that are indicated by Shawyer and that I calculated exactly would be a pure coincidence (The loss of resonnance due to the doppler shift would exactly loss so that CoE is verified in my calculus.)
Do you assume this coïncidence ?
Anyway, I just looked with more attention the equations. Shawyer write on page 7 :
Pk=Mva
Where Pk is the output power transfered to the ship during Delta T, M the mass of the ship, V the average speed during the Delta T, and a the acceleration. Since the Delta T is short, the use of the average speed during this short period of time is OK.
I can divide each side of the equation by Mv, assuming that speed is not zero.
I get a=Pk/(Mv)
The acceleration during Delta T is equal to the energy provided to the ship during Delta T divided by the mass of the ship, and the speed of the ship.
That is exactly what I am saying since some time. The acceleration of the ship is inversely proportional to the speed of the ship.
The formula Pk=mVa is written directly by Shawyer. Since V is not zero, I can divide each part of the formula, and get my a=Pk/(Mv) Since Pk is constant, or at least limited to the whole electrical power of the ship, I can tell that for this Shawyer paper, the acceleration of the ship is inversely proportional to the speed of the ship.
With regards to the formula on page 8, the power is the mechanical power, not the electrical input and the formula is not an instantaneous dynamical equation. It concerns the entire trip. It assumes the acceleration was constant in the first place so it is not saying acceleration is inversely proportional to velocity at all.
Basically, he's deriving that the mechanical power is proportional to mass x acceleration x average velocity. This is always true for any system at constant acceleration. It's nothing new. -
#1068 by Josave on 27 Sep, 2016 20:53
-
I'd say the presentations are split about 50-50 between Mach Effect and EMdrive topic and results, and there is a concerted effort to integrate the theories for each approach.
Integrating the theories is an intriguing possibility, and I would like to hear more about that. Is anyone discussing McCulloch's theories?
https://www.technologyreview.com/s/601299/the-curious-link-between-the-fly-by-anomaly-and-the-impossible-emdrive-thruster/
http://arxiv.org/abs/1604.03449
By the way I see that Cramer is mentioned in this thread. Is he there?
New data in galaxies is confirming MiHsC theory of Dr. McCulloch. This seems a backup for using its modeling in increasing EmDrive size and exploring another resonant frequencies to scale up the thrust and facilitate the measuring.
http://physicsfromtheedge.blogspot.com.es/2016/09/mihscqi-vs-new-mcgaugh-lelli-data.html
-
#1069 by tchernik on 27 Sep, 2016 21:09
-
New data in galaxies is confirming MiHsC theory of Dr. McCulloch. This seems a backup for using its modeling in increasing EmDrive size and exploring another resonant frequencies to scale up the thrust and facilitate the measuring.
May the best theory win! -
#1070 by Rodal on 27 Sep, 2016 21:29
-
There may not be any such thing as a "best" theory to win. Rather, there maybe an overall unifying theory explaining all this. For example, before Witten (1995) unified string theory with M theory, it was thought that a large number of string theories were separate theories. Witten showed that they were all parts of the same theory.
New data in galaxies is confirming MiHsC theory of Dr. McCulloch. This seems a backup for using its modeling in increasing EmDrive size and exploring another resonant frequencies to scale up the thrust and facilitate the measuring.
May the best theory win!
Ditto for Schrodinger "wave mechanics" and Heissenberg "matrix mechanics" versions of Quantum Mechanics, initially posed as different theories, promptly shown to be the same thing.
Looking at aeronautical engineering: explaining what makes airplanes heavier than air fly, it was only later (after Prandtl) that the importance of vorticity and viscous fluid flow was understood to be the reason for it. And the reason why Lord Kelvin was wrong (thinking that heavier-than-air-flying was impossible: based on a perfect non-viscous flow model). (*)
This is usual in initial periods of uncertainty: only later are theories unified and better understood as a whole.
________________
(*) Present discussions of people pontificating that the EM Drive is impossible based on simplistic models of conservation of energy is amusing, as Lord Kelvin saying that heavier-than-air-flying was impossible: discussing conservation of energy for an open system that completely ignores the external field ! also making energy extrapolations while ignoring entropy !
-
#1071 by RERT on 27 Sep, 2016 21:47
-
#1072 by Kenjee on 27 Sep, 2016 21:55
-
Hope you don`t mind
-
#1073 by flux_capacitor on 27 Sep, 2016 22:07
-
By using the sand car example, did he mean that his MET rocket can fluctuate its mass, just like by releasing an astronaut, and then taking him back in again? This is beyond my pay grade too.
That is indeed the whole point of Mach Effect Thrusters, or METs. According to Woodward's transient mass equation:
A mass fluctuation arises in an object when it absorbs energy (like a capacitor) as it undergoes proper acceleration.
I agree that a capacitor can increase its mass by absorbing energy, because this is predicted by Einstein's mass-energy equation. However, the energy the capacitor absorbed is taken from the battery or other sorts of energy sources. By losing that same amount energy, assuming 100% efficiency, the battery's mass decreases with the same amount. If the battery is on board of the same rocket, how can that help?
The mass fluctuation in a MET does NOT come from the mass of electrons flowing from the power supply in and out of the vibrating capacitors. The answer has been given right back to EmDrive Thread 1 (!):Quote from: Ron StahlWoodward's thruster is not a transducer. It's a transistor. It doesn't convert electrical into kinetic. It controls the flow of gravinertial flux--the stuff that through the universe's gravitational field, gives matter its mass--in and out of the thruster. Each Mach Effect cycle, the active mass in the thruster goes through a full 2w cycle, so the mass gets heavier, than lighter then heavier then lighter--4 discrete changes in each Mach Effect event. Gravinertial flux is flowing from the rest of the universe, into the active mass when it gets heavier, and back out when it gets lighter, and that flux is linked to the entire universe. So Mach Effect Thrusters are not transducers that covert energy, they're transistors, like on a sailboat. On a large racing yacht, if you were to look at the electrical energy driving the winches for the sails, and look at the kinetic energy of the boat through the water, you would appear to have a conservation violation, since the vast majority of the power into the system is in the wind, not the winch. Same with MET's. The real power is in the gravinertial flux--the universal wind created by and controlled by the MET. So you have to look at the entire system--the universe-- to do any meaningful conservation calculations with MET's.
It is the same kind of explanation than what Dr. Rodal is explaining here, yet this debate about Mach effect vs CoE continues based upon false premises. -
#1074 by rq3 on 27 Sep, 2016 22:14
-
...
a) Lost wax 3dprinting
As the name indicates, it involves a secondary process of metal melting and needs a metal foundry set up.
Such a casted piece would need additional machining to obtain a high reflective surface.
...
This has been discussed before. Copper has to be cast in a vacuum. Fine Silver (999) is relatively easy to cast but the walls would have to be quite thick. Otherwise there would be voids. Another problem with the lost wax process is it is not practical to cast a hollow container where the interior plaster mold is not supported at all. I think the most practical method of building a Copper or Fine Silver fustrum is to spin-form it on a lathe. Second to that would be to roll out a cone pattern and join the two sides with a riveted strip, using Copper rivets. That will ensure electrical continuity at the seam. The end caps can also be screwed together, using Copper screws. A while back someone was in touch with a Copper cookware manufacturer who could do this type of work. It requires specialized tools to get a finished article with smooth surfaces.
I think quite a few Babylonians and Egyptians would have been surprised to know that copper has to be cast in a vacuum! -
#1075 by rq3 on 27 Sep, 2016 22:20
-
Has anyone considered CNC metal spinning?
That has been discussed before. My understanding with CNC metal spinning is that a die cast is used to form the metal onto. That is part of the reason I chose whole integers for the spherical radii of my endplates. 50cm and 25cm. I figure there are already die casts of spheres of those dimensions. I'm also looking into building or buying a spherometer.
The vast majority of metal spinning die are wood. Only if many (1000's) duplicate articles are to be spun on the same die is it economical to make a metal die. The die is usually called a "plug" or "form". And the vast majority of metal spinning is not CNC, it is done on a lathe with a wooden dowel or metal rod, manually. -
#1076 by Gilbertdrive on 27 Sep, 2016 23:28
-
With regards to the formula on page 8, the power is the mechanical power, not the electrical input and the formula is not an instantaneous dynamical equation. It concerns the entire trip. It assumes the acceleration was constant in the first place so it is not saying acceleration is inversely proportional to velocity at all.
Basically, he's deriving that the mechanical power is proportional to mass x acceleration x average velocity. This is always true for any system at constant acceleration. It's nothing new.
Pk is the part of the electrical power that is not lost.
There is written P0=Pe+Pk where P0 is the imput power of the emdrive system, and Pe the losses, so Pk is realy the part of the electrical power that is converted into Kinetic energy. And by definition, Pk can not be superior to P0, since the losses can not be negative (the best possible case, there are zero losses) Is it Ok with that precise point ? -
#1077 by Gilbertdrive on 27 Sep, 2016 23:40
-
Consider a hypothetical rocket that can do 10000 burns at a delta v of 1000 m/s. Each burn requires a total energy of E, maybe only 1% goes to the ship. After the first burn, the kinetic energy of the ship wrt earth is 5E5/kg so assume the burn released 5E7 J/kg in its frame. What's the total energy released in the ships frame after 10000 burns? It should be 5E10J/kg. What's the kinetic energy wrt earth? It should be 5E13J/kg. No doubt some will claim nature conspires to prevent this by making the mass requirements great enough so it's never practical to make such a rocket.
It's not about "nature conspiring to prevent this", it's about math... assuming that CoE holds. One has to take into account not just the kinetic energy of the rocket, but also the kinetic energy of the propellant (which decreases since it's being ejected in the opposite direction if you compare it to an earlier moment when it was moving together with the rocket). The rocket has to be constantly losing mass (i.e. propellant), and if you calculate the total energy over time, it all adds up and no "over-unity" occurs. The rocket does not know when to stop accelerating, it slows down naturally because it's losing mass/propellant. If there is no propellant and the rocket is not losing mass, "over-unity" is an obvious consequence, unless there is a way for it to "steal" energy from some unknown field.
I've shown you that my hypothetical rocket does run into over unity. You have to show me that it's impossible for any rocket to do that many burns without running out of mass.
I have shown you your rocket is not at all over unity.
I knew there is not enough mass but the question is, is that nature's way of avoiding CoE issues as we are discussing or is that just because rockets are that way. If EmDrive works at any level, it should work at all levels.
It is like if you try to cut a cake in many pieces, than to re-assemble it and get a bigger cake. Can we tell that it is the nature way of avoiding CoE issues ? It is just that the standard physical laws satisfy CoE. As the division of the cake in several parts, and it's re-assembly will no create more mass, the separation in many burns and several stages will no create more Kinetic Energy
-
#1078 by Gilbertdrive on 27 Sep, 2016 23:49
-
1) The discussions about conservation of energy continue to take place without taking into account whether these are open systems. Any discussion about conservation of energy of an open system that does not take into account the external fields ,( for example one based on gravitation, like the Mach/Sciama/Woodward effect) is futile. Of course there will be an overunity problem if one does not take into account the external field (in the case of the Mach/Sciama effect: gravitation)! It would be like calculating conservation of energy for a gravity assist while ignoring the effect of gravity
.
Or like calculating conservation of energy for a solar sail while completely ignoring the solar pressure on it.
Or like calculating conservation of energy for a Laser propelled sail while completely ignoring the Laser (stationed on Earth or a supply vehicle) pressure on the sail.
Completely futile to discuss conservation of energy while ignoring the external field !
You are right.
In fact, there are at this time two separates conversations about conservation of Energy. One about Mach effect and woodward papers, where I do not feel able to participate for now, another that I initiated about what Shawyer tells in this theory paper. It seems that I have a different understanding of what means this paper than some
Trying to solve this. When it is done I shall try to be more explicit about the context of Energy concern, with hypothesis indicated more clearly.
-
#1079 by zen-in on 28 Sep, 2016 05:15
-
The CoE argument is best understood mathematically. Considering the kinetic energy, Ke, of a moving object:
Ke = (MV2)/2
If this is increasing due to a constant force then
V = at where a= acceleration and t = time
or V = (Ft)/M where F = the force and M = the mass
so the change or Δ kinetic energy equation can be written as:
ΔKe = (F2Δt)/(2M) where Δt is the time increment
This equation shows that the kinetic energy of a reactionless drive increases as the square of the force.
However we have been told many times that an EM-Drive produces a force that is proportional to the power input, or
F = WKc where W = the input energy in Watts and Kc = a constant
re-arranging this equation one gets the following:
W = F/Kc
The total power (Watt-Hours, etc) used over a time interval to operate the EM-Drive is:
P = WΔt = (FΔt)/Kc
Clearly the input power over time varies linearly with F, the force produced while the increase in kinetic energy varies by the square. So plot a parabola and a straight line. The parabola will start off below the straight line and will always, eventually cross the straight line.
For energy to be conserved P ≡ Ke at every instant of time. (ignoring losses and stored energy considerations) But a parabola's Y value is equal to a straight line's Y value at only 0, 1, or 2 points.
. 
Trying to solve this. When it is done I shall try to be more explicit about the context of Energy concern, with hypothesis indicated more clearly.