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#900 by tleach on 24 Sep, 2016 00:53
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I can do a time domain analysis, but that requires quite a bit of computation. Due to the complex geometry of the cannae drive, it has quite a few triangles in the 3d mesh - which also adds dramatically to compute time.
Forgive me if someone else already raised this possibility, but would it be feasible to use some sort of cloud based computational resource to speed up render times? Something like Amazon's Elastic Compute Cloud (https://aws.amazon.com/ec2/) maybe? It looks like they have a "Free Teir" (https://aws.amazon.com/ec2/pricing/) that would give you 750 hours of per month for 1 year... -
#901 by tleach on 24 Sep, 2016 01:10
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Dr. Rodal, Todd, Paul March, Dr. Woodward, Dr. Hiedi Fern, Dr. Martin Tajmar along with many other other physicists, engineers and experts in propulsion.
Ahhhhh... To have been a fly on the wall during those discussion sessions! -
#902 by dustinthewind on 24 Sep, 2016 01:16
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2 : Emdrive works, and does not violate CoE, because it exists non classical reference frames to define appropriately Kinetic Energy, and the Emdrive does not give a constant acceleration, but an acceleration decreasing proportionnaly to the speed. New physics needed.
How can acceleration be decreasing proportional to the speed? This would mean that acceleration would be measured differently in different frames of reference (i.e. it would have to be higher in a frame co-moving with the rocket w/o acceleration). How is that possible?
It will take me a bit to describe this.
One can work out the percentage of energy exchanged in an instant upon impact using conservation of energy and momentum with two massive objects. If the photon has an effective mass that can possibly change, then the ratio of energy exchanged upon impact may change also. Photons, because of their very small effective mass, are very inefficient at exchanging energy upon impact. 100% efficient exchange of energy between two elastic objects happens when they are the same mass, but photons have a very small mass compared to ours.
Photon recycling by repeated collisions can make this energy exchange more efficient and can be observed in photon recycling laser propulsion but this is an open system where two objects are both accelerating in opposite directions. In such a system energy is drained more effectively from the light by Doppler-shifting.
It may be possible, inside a closed system or cavity, to modify the mass of light on one side of a cavity. This likely implies modifying the vacuum itself. If the mass of light in a closed cavity stays the same then the energy exchanged with a collision on one side of a cavity is transferred back to the photon upon the collision with the other side of the cavity. However, if the mass of light is heavier on one side of the cavity, then more energy is always transferred to one side of the cavity than the other. Also, the more times this light recycles and the greater the difference in mass of the light, then the more effectively energy is drained from the light in the cavity by Doppler-shifting.
Now this is a big if, because I don't know this is really happening, but I suspect it may be possible.
How momentum is being conserved in such a cavity is possibly that the higher index vacuum that is modifying the mass of light is experiencing a back reaction upon this light changing in mass. Experiments have been done in water and such a back reaction has been observed as well as greater impulses from light in water on a mirror so drawing such a parallel seems plausible to me.
This is where the answer to your question comes in:
Now lets look at the CMB which appears to give us an absolute reference frame for the maximum speed of light (hope I am right on this). At light speed this should appear for the observer to flatten into a pancake I would guess. Light approaching one from behind should appear severely red-shifted and I would assume reduced in energy or ability to provide impulse while light approaching from ahead should appear blue-shifted and provide more impulse. So a moving cavity would observe the light inside to provide less impulse forward by red-shifting and greater opposition by blue-shifting. I would suppose that this behavior of the light in the cavity would put a limit on the maximum velocity achievable when the two forces reach equilibrium.
This is what I suspect may possibly be going on but I am in no way implying this IS what is going on. However, it helped me understand why Q might be important and possible limitations of such a device. -
#903 by SeeShells on 24 Sep, 2016 01:38
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I was told they will be putting it up on their website for all to access.Dr. Rodal, Todd, Paul March, Dr. Woodward, Dr. Hiedi Fern, Dr. Martin Tajmar along with many other other physicists, engineers and experts in propulsion.
Ahhhhh... To have been a fly on the wall during those discussion sessions!
They videotaped the workshop presentations including the Q and A after each speaker. It will take some time to make clean edits for regular breaks and such. That totals to hours of presentations, you can now be a fly in your own home.
You have to love the internet and thank SSi.org taking the time to make this breakthrough propulsion workshop available to the world.
While I didn't present this time I was vocal.
My Best,
Shell
PS: A picture of Dr. Rodal presenting.
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#904 by HMXHMX on 24 Sep, 2016 02:11
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Is Professor Woodward on the photo ?
Regrettably no. He walks with two canes now and couldn't move quickly enough over the broken ground to get out to where the camera was located. We have a more complete group photo that will be posted next week, but without the double rainbow. -
#905 by meberbs on 24 Sep, 2016 03:46
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Yes, you do get massive free energy. Gilbertdrive's calculations were all done in one reference frame. The device starts out with a battery containing a certain amount of energy. At the end the battery is drained and the device has way more kinetic energy than it started with. Under a theory like Shawyer's where the device does not interact with anything else, this violates CoE.Shawyer's description of how the device works (which is what started this conversation) does not have it pushing on anything, so according to Shawyer's claims it would be a free energy device.
His claim is a small % of the momentum in the internal EmWave is transferred to the external frustum. Most of the energy in the frustum exits as heat as in the attached image from his peer reviewed paper.
Bottom line is the EmDrive generates a force and accelerates a mass while it consumes energy to do so.
What referential for Kinetic Energy ? in GR you do not have an absolute reference frame. What is your reference frame for Kinetic Energy ?
As I calculated a few pages ago, the 0.67C (relatively to the earth referential) mentionned in the Shawyer peer-rewieved paper means a Kinetic Energy of 3.1191599*10^20 joules
Since the 10 years of operation of the 200Kwe generator gives only 6.311*10^13 Joules
So, following Shawyer, at the expanse of 6.311*10^13 Joules a spaceship can get 3.1191599* 10^20 joules of Kinetic Energy. We get massive free energy !
No, you don't get massive 'free' energy. What you get is different energies in different reference frames. You can see that simply by the following example. Say a ship is moving wrt earth frame at just under 0.5c by 1000m/s and you observe it. The ship makes a burn to increase its velocity wrt you by 1000m/s to exactly 0.5c wrt you. For a 10,000kg ship, how much energy does the pilot need to add? It depends on which reference frame. In the ships frame the pilot only needs to do the same burn to gain 1000m/s as always which is 5E9 J. You would say its about 1.5E15 J. It does gain that but not by the ship expending that energy in its reference frame. It would be the same regardless of a rocket of an EmDrive engine except the EmDrive would be more efficient. No doubt some will claim the rocket fuel already has kinetic energy, which it does, but that's natures gift, not something the ship had to provide by internal energy release stored as fuel because it never had that energy content to begin with. If someone objects further then consider that the ship was launched from a planet already moving at just under 05c to start with.
The bottom line is that the actual energy to get to the stars is probably a lot less that convention dictates.
P.S. I understand Woodward and allies previously released an essay demolishing the free energy or over unity arguments.
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdf
Was that discussed at the workshop?
Also, thermal waste like TheTraveller mentioned just makes it worse, because that thermal energy that radiated away is more energy at the end, on the side of the equation that already has extra energy.
And before you mention relativity, you should be aware that might make it worse. If the 10 years of operation were in the earth frame (This is most likely, but I'd have to check the paper to be sure) then since the device experiences time dilation, the starting energy needed in the battery would actually be lower. (Remember the starting energy of the battery is with the device at rest in the earth-frame, the same frame the final kinetic energy is measured in.) -
#906 by TheTraveller on 24 Sep, 2016 04:34
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Shawyer's description of how the device works (which is what started this conversation) does not have it pushing on anything, so according to Shawyer's claims it would be a free energy device.
His claim is a small % of the momentum in the internal EmWave is transferred to the external frustum. Most of the energy in the frustum exits as heat as in the attached image from his peer reviewed paper.
Bottom line is the EmDrive generates a force and accelerates a mass while it consumes energy to do so.
What referential for Kinetic Energy ? in GR you do not have an absolute reference frame. What is your reference frame for Kinetic Energy ?
As I calculated a few pages ago, the 0.67C (relatively to the earth referential) mentionned in the Shawyer peer-rewieved paper means a Kinetic Energy of 3.1191599*10^20 joules
Since the 10 years of operation of the 200Kwe generator gives only 6.311*10^13 Joules
So, following Shawyer, at the expanse of 6.311*10^13 Joules a spaceship can get 3.1191599* 10^20 joules of Kinetic Energy. We get massive free energy !
No, you don't get massive 'free' energy. What you get is different energies in different reference frames. You can see that simply by the following example. Say a ship is moving wrt earth frame at just under 0.5c by 1000m/s and you observe it. The ship makes a burn to increase its velocity wrt you by 1000m/s to exactly 0.5c wrt you. For a 10,000kg ship, how much energy does the pilot need to add? It depends on which reference frame. In the ships frame the pilot only needs to do the same burn to gain 1000m/s as always which is 5E9 J. You would say its about 1.5E15 J. It does gain that but not by the ship expending that energy in its reference frame. It would be the same regardless of a rocket of an EmDrive engine except the EmDrive would be more efficient. No doubt some will claim the rocket fuel already has kinetic energy, which it does, but that's natures gift, not something the ship had to provide by internal energy release stored as fuel because it never had that energy content to begin with. If someone objects further then consider that the ship was launched from a planet already moving at just under 05c to start with.
The bottom line is that the actual energy to get to the stars is probably a lot less that convention dictates.
P.S. I understand Woodward and allies previously released an essay demolishing the free energy or over unity arguments.
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdf
Was that discussed at the workshop?
Correct.
In the ship's frame, the only one which matters, the energy needed to alter velocity in relation to some desired destination is always the same as the ship's mass does not alter. Well not so much as it matters.
BTW Jim Woodward makes a good comment:
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdfQuoteWe routinely hear a criticism of METs based upon an argument that claims: if a
MET is operated at constant power input for a sufficiently long time, it will acquire
enough kinetic energy to exceed the total input energy of operation. Assuming this
argument to be correct, critcs assert that METs violate energy conservation as the ratio of
the acquired kinetic energy to total input energy exceeds “unity.”
Contrary to this “over-unity” assumption, this argument is based on flawed
physics and, consequently, wrong. The fact that the argument applies to all simple
mechanical systems (in addition to METs) should have alerted critics to their mistake.
But it didn’t.
So, a dumb idea that should have been quickly buried is still with us. The
purpose of this essay is to carry out a long overdue burial. -
#907 by Tellmeagain on 24 Sep, 2016 04:53
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No, you don't get massive 'free' energy. What you get is different energies in different reference frames. You can see that simply by the following example. Say a ship is moving wrt earth frame at just under 0.5c by 1000m/s and you observe it. The ship makes a burn to increase its velocity wrt you by 1000m/s to exactly 0.5c wrt you. For a 10,000kg ship, how much energy does the pilot need to add? It depends on which reference frame. In the ships frame the pilot only needs to do the same burn to gain 1000m/s as always which is 5E9 J. You would say its about 1.5E15 J. It does gain that but not by the ship expending that energy in its reference frame. It would be the same regardless of a rocket of an EmDrive engine except the EmDrive would be more efficient. No doubt some will claim the rocket fuel already has kinetic energy, which it does, but that's natures gift, not something the ship had to provide by internal energy release stored as fuel because it never had that energy content to begin with. If someone objects further then consider that the ship was launched from a planet already moving at just under 05c to start with.
The bottom line is that the actual energy to get to the stars is probably a lot less that convention dictates.
P.S. I understand Woodward and allies previously released an essay demolishing the free energy or over unity arguments.
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdf
Was that discussed at the workshop?
Correct.
In the ship's frame, the only one which matters, the energy needed to alter velocity in relation to some desired destination is always the same as the ship's mass does not alter. Well not so much as it matters.
BTW Jim Woodward makes a good comment:
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdfQuoteWe routinely hear a criticism of METs based upon an argument that claims: if a
MET is operated at constant power input for a sufficiently long time, it will acquire
enough kinetic energy to exceed the total input energy of operation. Assuming this
argument to be correct, critcs assert that METs violate energy conservation as the ratio of
the acquired kinetic energy to total input energy exceeds “unity.”
Contrary to this “over-unity” assumption, this argument is based on flawed
physics and, consequently, wrong. The fact that the argument applies to all simple
mechanical systems (in addition to METs) should have alerted critics to their mistake.
But it didn’t.
So, a dumb idea that should have been quickly buried is still with us. The
purpose of this essay is to carry out a long overdue burial.
Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings. -
#908 by wicoe on 24 Sep, 2016 04:57
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Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings.
This thread might help you find out what's wrong with his arguments... https://forum.nasaspaceflight.com/index.php?topic=31037.0
There is no single reference frame "matters". All intertial reference frames matter. If an explanation/theory works in one intertial frame and breaks down in another, it is most likely wrong. -
#909 by TheTraveller on 24 Sep, 2016 05:02
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Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings.
This thread might help you find out what's wrong with his arguments... https://forum.nasaspaceflight.com/index.php?topic=31037.0
There is no single frame of reference "that matters". All frames of reference matter.
Only the ship's frame matters as it alters ship's velocity, relative to a desired destination, with a constant relationship between work done on the ship's mass and velocity.
To the ship, accelerating at a constant rate, there is no velocity, just acceleration. -
#910 by Tellmeagain on 24 Sep, 2016 05:10
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Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings.
This thread might help you find out what's wrong with his arguments... https://forum.nasaspaceflight.com/index.php?topic=31037.0
There is no single frame of reference "that matters". All frames of reference matter.
Only the ship's frame matters as it alters ship's velocity, relative to a desired destination, with a constant relationship between work done on the ship's mass and velocity.
To the ship, accelerating at a constant rate, there is no velocity, just acceleration.
Earth frame matters too. Because if a ship, after a dozen or so years of acceleration, hits the Earth，the Earth will break into a pile of asteroids. But I think there still can be math errors with the ship frame. Let me see whether I can be lucky. -
#911 by TheTraveller on 24 Sep, 2016 05:18
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Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings.
This thread might help you find out what's wrong with his arguments... https://forum.nasaspaceflight.com/index.php?topic=31037.0
There is no single frame of reference "that matters". All frames of reference matter.
Only the ship's frame matters as it alters ship's velocity, relative to a desired destination, with a constant relationship between work done on the ship's mass and velocity.
To the ship, accelerating at a constant rate, there is no velocity, just acceleration.
Earth frame matters too. Because if a ship, after a dozen or so years of acceleration, hits the Earth，the Earth will break into a pile of asteroids. But I think there still can be math errors with the ship frame. Let me see whether I can be lucky.
The Earth's frame only matters if an inbound ship needs to match velocity.
However the work needed to be done on the ship's mass to alter the relative velocity to 0, in the ship's frame, stays constant and that is the only effect that is important. -
#912 by meberbs on 24 Sep, 2016 05:29
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...
All inertial (constant velocity) frames matter, because physics is the same in all inertial frames. This is the fundamental principle of relativity. If you want to jump between frames, then you have to start being careful with transformations, and generally you can't compare energy between frames. Conservation of energy as normally discussed requires you to be talking about a single inertial reference frame. Being in the accelerating frame of the ship requires you to account for non-inertial effects, just like how the Coriolis effect appears if you are in a rotating frame.
In the ship's frame, the only one which matters, the energy needed to alter velocity in relation to some desired destination is always the same as the ship's mass does not alter. Well not so much as it matters.BTW Jim Woodward makes a good comment:
http://ssi.org/epi/Over-Unity_Argument_&_Mach_Effect_Thrusters.pdfQuoteWe routinely hear a criticism of METs based upon an argument that claims: if a
MET is operated at constant power input for a sufficiently long time, it will acquire
enough kinetic energy to exceed the total input energy of operation. Assuming this
argument to be correct, critcs assert that METs violate energy conservation as the ratio of
the acquired kinetic energy to total input energy exceeds “unity.”
Contrary to this “over-unity” assumption, this argument is based on flawed
physics and, consequently, wrong. The fact that the argument applies to all simple
mechanical systems (in addition to METs) should have alerted critics to their mistake.
But it didn’t.
So, a dumb idea that should have been quickly buried is still with us. The
purpose of this essay is to carry out a long overdue burial.
I hadn't researched Woodward's theories too thoroughly before, but I had the impression he was a general competent physicist, who was poking for (unlikely) loopholes in general relativity. After reading that paper I no longer have that impression.
After performing a step of just using the transitive property of equality (if a = b and b = c, then a = c), he says:QuoteNow we have done something stupid and wrong.
It is possible he is referring to the first step in that sequence, where the figure of merit is defined, but that figure of merit comes straight out of theories such as Shawyer's. If that is wrong then the theory that it comes from must be wrong.
he then remarks:QuoteBut this is the mathematics of those who make the “over unity” energy conservation violation argument about the operation of METs. The real question here is how could anyone, having done this
calculation or its equivalent, think that they had made a profound discovery about anything?
Simple, the step with defining the figure of merit only applies to METs and the like. Making a single assumption when every other step is valid, and ending up in a contradiction is a common way to disprove the assumption. That is what is done here, but Woodward tries to twist this inside out.
The last paragraph of the paper attempts to show the "correct" way of doing this. As part of that he says:QuoteWe know that, starting from t= 0, if we let the integration interval t get very large, the work equation integral will first equal and then exceed the energy calculated by the figure of merit equation. So we require that t be sufficiently small that this obvious violation of energy conservation does not happen.
He then tries to get around this limit by continuously jumping into the current rest frame of the device. This causes him to be comparing energy between reference frames which is not valid as Bob012345 has pointed out.
The only thing this paper has destroyed for me is the thought that Woodward has any credibility. Unless this paper is some kind of bad joke, or test to see who can find them flaws in it. -
#913 by OnlyMe on 24 Sep, 2016 06:33
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....
All inertial (constant velocity) frames matter, because physics is the same in all inertial frames. This is the fundamental principle of relativity. If you want to jump between frames, then you have to start being careful with transformations, and generally you can't compare energy between frames. Conservation of energy as normally discussed requires you to be talking about a single inertial reference frame. Being in the accelerating frame of the ship requires you to account for non-inertial effects, just like how the Coriolis effect appears if you are in a rotating frame.
....
Without comment on any of the various theories, the above quote describes the basis of the difficulty I have had with most of the CoM and to a lesser but not insignificant extent CoE debate.
As long as one is testing an EMDrive in a lab the arguments are entirely valid because it is tied to the inertial frame of reference of the lab. But, I have not seen any claim that an experiment has been allowed to run long enough to test the claim of constant uniform acceleration (which seems to me to be projecting a conclusion based a flawed theory of operation), so it is a mute point.
The real issue begins, if a functional EMDrive is put into space (the vacuum) and allowed to accelerate for an extended period of time. But then the only significant frame of reference is that of the EMDrive's or spaceship's frame which is accelerating, not inertial. Since kinetic energy is relative, to inertial frames, it should not be an issue, for an object in an accelerating frame... Unless it runs into an immovable object in its path.
I don't believe that once the drive or any spaceship which has moved sufficiently out of a gravity well that its inertial velocity is greater than escape velocity, the frame of reference of the gravity well's center of mass can be used as the drives frame of reference for purposes of any additional acceleration(s) either continuous or intermittent. The drive or spaceship becomes the only valid frame.
The only real issue I see is whether there is a limiting velocity for a massive object that is less than the speed of light... Special relativity says there is and it is the speed of light.., and everyone know where the energy requirements for acceleration in that case is... But it has only been tested and confirmed relative to an inertial lab frame of reference. At least with any degree of certainty.
If one of these drives is proven to work, I guess we will find out. -
#914 by Gilbertdrive on 24 Sep, 2016 13:53
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Interesting paper. I glanced at it and found the math was within my ability. So I printed it out to take a closer look. I think I will be able to find something wrong in his argument. But I may not be able to. Whatever the result is, I will write a swift report about my findings.
This thread might help you find out what's wrong with his arguments... https://forum.nasaspaceflight.com/index.php?topic=31037.0
There is no single frame of reference "that matters". All frames of reference matter.
Only the ship's frame matters as it alters ship's velocity, relative to a desired destination, with a constant relationship between work done on the ship's mass and velocity.
To the ship, accelerating at a constant rate, there is no velocity, just acceleration.
Earth frame matters too. Because if a ship, after a dozen or so years of acceleration, hits the Earth，the Earth will break into a pile of asteroids. But I think there still can be math errors with the ship frame. Let me see whether I can be lucky.
The Earth's frame only matters if an inbound ship needs to match velocity.
However the work needed to be done on the ship's mass to alter the relative velocity to 0, in the ship's frame, stays constant and that is the only effect that is important.
If the Em-ship take a turn at Alpha Centuri, than come back in the solar system with the speed of 0,67C, in less than a century, it comes back with far more energy that it's own generator has provided during this time. It is Overunity.
As meberbs has written :All inertial (constant velocity) frames matter, because physics is the same in all inertial frames. This is the fundamental principle of relativity.
The CoE needs to be verified in an heliocentric referential. At least if you do not take into account other stars
Anyway, Shawyer has mentionned clearly in "Emdrive Basis" that the acceleration was supposed to decrease as the Kinetic Energy was increasing. He just omitted to give a referential. But it can not be the ship itself, because the Kinetic Energy would always be zero. In Shayer theory, the Q factor is supposed to decrease with the speed (in a non given reference frame)
But in his paper, he forgot to take that into account when he calculated the speed of 0,67C.2 : Emdrive works, and does not violate CoE, because it exists non classical reference frames to define appropriately Kinetic Energy, and the Emdrive does not give a constant acceleration, but an acceleration decreasing proportionnaly to the speed. New physics needed.
How can acceleration be decreasing proportional to the speed? This would mean that acceleration would be measured differently in different frames of reference (i.e. it would have to be higher in a frame co-moving with the rocket w/o acceleration). How is that possible?
Maybe I should have been more explicit. I reformulate.
2 : Emdrive works, and does not violate CoE, because it exists non classical reference frames to define appropriately Kinetic Energy, so the Emdrive push against this reference frame as a car is pushing on the road, and the Emdrive does not give a constant acceleration, but an acceleration decreasing proportionnaly to the speed. New physics needed.
Also When a car accelerates on a road at constant power, it's acceleration decrease as the inverse of it's speed in the road reference frame.
This point is very important. It is a question of work.Quote from: WikipediaThe work done by a constant force of magnitude F on a point that moves a displacement (not distance) s in the direction of the force is the product
W = F s {\displaystyle W=Fs} {\displaystyle W=Fs}.
For example, if a force of 10 newtons (F = 10 N) acts along a point that travels 2 metres (s = 2 m), then it does the work W = (10 N)(2 m) = 20 N m = 20 J. This is approximately the work done lifting a 1 kg weight from ground to over a person's head against the force of gravity. Notice that the work is doubled either by lifting twice the weight the same distance or by lifting the same weight twice the distance.
Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work done on the body by the resultant force acting on that body. Conversely, a decrease in kinetic energy is caused by an equal amount of negative work done by the resultant force.
From Newton's second law, it can be shown that work on a free (no fields), rigid (no internal degrees of freedom) body, is equal to the change in kinetic energy of the velocity and rotation of that body.
Since the work is proportional to the displacement, it means that if the ship is moving 2 times faster, and the same force applied, the work is also *2. So, it needs two times more power to apply the same force on the car. This needs a force that apply on the referential of the road. This relation is not true in any other referential. If my car is driving on the roof of a moving train, the fact that the acceleration is divided by 2 when the speed in multiplied by 2. is true in the referential of the train. It is not true in the Earth referential. In the Earth referential, the car will get more Kinetic Energy that is spent internal Energy. Because it is true that Kinetic Energy depends on the referential. But CoE is not violated, because, in the earth referential, the train has been slow down by the car, so it has less Kinetic Energy, and the total Kinetic Energy of the system still satisfy CoE.
So, what is important here are the 2 points :
1 : When A push against B to get speed, using a constant power, it's acceleration is inversely proportional to the speed of A relatively to B.
2 : It is not true in other referential, but it the system A+B is taken into account, CoE still verified for any referential. -
#915 by Gilbertdrive on 24 Sep, 2016 14:24
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Each delta v requires the same burn, not more to account for the fact that future fuel is more effective. The reason we don't see more paradoxes with rockets is that they require so much mass that they run out before that point would be reached and they are so very inefficient so most kinetic energy is wasted.
Also, I don't consider the energy issue a paradox if you use the work-energy theorem. It's the mechanical power times the time that always equates to the kinetic energy change in every observer frame. So there is no paradox. There is just a mixing of energy from different frames that confuses people.
I think I have the right solution for this paradox that completes the answer that Meberbs did.
I shall first formulate the paradox a very clear way.
For classical rockets, unlike a car pushing against a road, the Delta V is the same for each burn.
So, in the earth reference frame, the faster the rocket is, the more Kinetic Energy gained, for the same amount of energy spent.
So, at a certain speed, the rocket gains more Kinetic Energy in the earth reference frame that the energy it has spent.
My solution of the paradox is.
Yes, it is true that in earth referential the rocket gains more Kinetic Energy that it has spent. But the rocket is not alone. The propellant that was expulsed at the opposite direction has now less Kinetic Energy in a earth reference frame. And the faster the rocket was, the more Kinetic Energy it has lost. It compensates the fact that the ship has gained more Kinetic Energy.
If I look the total Kinetic Energy of the system rocket + propellant, from an earth reference Frame, CoE will always be satisfied. In the earth point of view, the rocket has stolen Kinetic Energy to it's propellant, and there is no paradox.
In fact several messages already explicited that, but maybe another formulation can help. Tellmeagain and Meberbs have already made very good explanations.
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#916 by TheTraveller on 24 Sep, 2016 14:27
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...
On the ship A = F/M rules. The energy / work required to be done on the constant mass of the ship to effect velocity change never varies. The ship doesn't care what some other frame calculates as the necessary energy to cause velocity change. -
#917 by Gilbertdrive on 24 Sep, 2016 14:46
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How can acceleration be decreasing proportional to the speed? This would mean that acceleration would be measured differently in different frames of reference (i.e. it would have to be higher in a frame co-moving with the rocket w/o acceleration). How is that possible?
I already replied to this point a few minutes ago, but maybe I missed the essential, so I come back to this point.
If a car accelerates by pushing on a road, it's acceleration decrease as the inverse of the speed in the reference frame of the road.
It is not true in other referentials, not because the acceleration would be different, but because the speed is different. So the relation is not true anymore, but the acceleration is almost the same.
In say almost, because in fact, the planet does not have an infinite mass, so the road and the entire planet is accelerating a little in the opposite direction. But with a big planet like the earth, it is negligeable.
The acceleration will not be much different in different reference frames as long as speeds are non relativist.
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#918 by Gilbertdrive on 24 Sep, 2016 14:53
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On the ship A = F/M rules. The energy / work required to be done on the constant mass of the ship to effect velocity change never varies. The ship doesn't care what some other frame calculates as the necessary energy to cause velocity change.
If only the ship cares, why does Shawyer speaks about increase of Kinetic Energy. In the ship referential, Kinetic energy is zero... -
#919 by zen-in on 24 Sep, 2016 14:57
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The Earth's frame only matters if an inbound ship needs to match velocity.
However the work needed to be done on the ship's mass to alter the relative velocity to 0, in the ship's frame, stays constant and that is the only effect that is important.
You have to chose one reference frame. "alter the relative velocity to 0, in the ship's frame," if you chose the ship's frame the velocity is always zero and there is no relative velocity. If you use the ship's frame of reference you are expending energy and not seeing any increase in the kinetic energy. What you should do is to consider the frame when the ship is at rest; ie: before any acceleration occurs. Then calculate the final kinetic energy added after the acceleration has ended wrt to this rest frame. For any reactionless drive the kinetic energy will eventually become larger than the energy used to accelerate the spacecraft. This is the CoE conundrum all reactionless drives have. Using the wrong reference frame just introduces another error. The equations will never balance with this error so you still have the CoE conundrum and sometimes in the opposite sense.
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