Mike Smith Cosmos Posted July 9, 2013 Author Posted July 9, 2013 (edited) I previously suggested using small models, which were said to be unrealistic. It is probably futile, but why not do a computer simulation. The diagrams previously given are inconclusive, and the idea of doing math has not settled the issue. Ad infinitum discussion is not helping. There are only two possibilities left that may settle this issue. Either build one or simulate it. Building cost appears to be either prohibitive or promiscuous for everyone; thus, simulation appears to be the only method to solve this argument. I am not familiar with either MathCAD (purchase) or SMath Studio (free). I have not used either one, but I think they might be able to do this simulation. And, I suspect there is enough talent here to do it. http://smath.info/wiki/MainPage.ashx As I understand, the spring connecting the two masses is considered to have such a small mass that its mass can be ignored. Is that true? If so, then its purpose is to cause an oscillation of the two masses attached to either end, nothing more, is that true? . . Either build one or simulate it. Building cost appears to be either prohibitive or promiscuous for everyone; thus, simulation appears to be the only method to solve this argument. .I thoroughly agree with you, and have done precisely both of these in small measure. I went to University a second time in my late 50's and made what you say a Thesis Project . Partial arc swings measurement. with Mathcad simulation. Both were positive, but subject to interpretation . Yes you can see the content, The manuscript in Italy . Original in Plymouth University year 2000 . . the spring connecting the two masses is considered to... I have mentioned , there is no actual Spring . The Oscillating masses 2 . are fed by a transducer mechanism which can be viewed as elastic ( molecular bonding elastic ) , as it needs to drive the masses in some form of resonant vibration. In view of what is being commented by m656366 it is either necessary for the transducer to be moving along with everything else,narrow( or as you say negligible ) or following the arc.(or all). As the purpose of this device is effectively a 'Sky Hook' It is natural that something appears to be pulling things upwards. Arthur C Clark once proposed a space elevator with a counter weight circling way out in space . He was right on many other things he predicted like communication Satellites Edited July 9, 2013 by Mike Smith Cosmos
EdEarl Posted July 9, 2013 Posted July 9, 2013 There are a number of non-rocket based systems to move payloads into space. http://en.wikipedia.org/wiki/Space_elevator http://en.wikipedia.org/wiki/Space_fountain http://en.wikipedia.org/wiki/Launch_loop http://en.wikipedia.org/wiki/Space_gun and others However, I do not understand your answer to my questions. As I understand, the spring connecting the two masses is considered to have such a small mass that its mass can be ignored. Is that true? If so, then its purpose is to cause an oscillation of the two masses attached to either end, nothing more, is that true? Lets ignore semantics, whether it is elastic, spring or rubber band is not important. Does it have any mass? If it does have mass, how much in relation to the two oscillating masses. If it has no mass, it is not affected by gravity, inertia, centrifugal or centripetal force, which simplifies a model. The force of the elastic (whatever) will always be straight between the two oscillating masses. The elastic will not be curved. If it does have mass, the model is a bit more complicated. The elastic will curve due to some of the forces mentioned before, but will be tend to be stretched straight between the two oscillating masses. The smaller the mass of the elastic the less it will curve. Moreover, I suspect that an elastic with mass will also oscillate in several modes of up and down, side to side, around and around, and/or end to end. I have no idea what kind of affect it will have on the oscillating masses. The more massive the spring, the more that harmonics from the spring will affect the two masses.
Mike Smith Cosmos Posted July 10, 2013 Author Posted July 10, 2013 (edited) Here's what I mean... Don't think I am Ignoring you, I am following very closely your reasoning .I need to do a very careful balancing of Forces. I remember trying to pick myself up when I was a young boy ! However, I do not understand your answer to my questions. Lets ignore semantics, whether it is elastic, spring or rubber band is not important. Does it have any mass? If it does have mass, how much in relation to the two oscillating masses. If it has no mass, it is not affected by gravity, inertia, centrifugal or centripetal force, which simplifies a model. The force of the elastic (whatever) will always be straight between the two oscillating masses. The elastic will not be curved. If it does have mass, the model is a bit more complicated. The elastic will curve due to some of the forces mentioned before, but will be tend to be stretched straight between the two oscillating masses. The smaller the mass of the elastic the less it will curve. Moreover, I suspect that an elastic with mass will also oscillate in several modes of up and down, side to side, around and around, and/or end to end. I have no idea what kind of affect it will have on the oscillating masses. The more massive the spring, the more that harmonics from the spring will affect the two masses. Yes ,I do get your point and it has 'bugged ' me a bit , I must admit. But if the elasticity is bound up in the whole mechanism. EG like the giant slinky ( not real but illustrative ) then the "inertial out fling " which is some gastly non scientific term I know , but it's how I see it . Its the accumulation of all this mass on average moving faster or at 17,700 mph must mean that the inertia , momentum is overcoming the inward force of gravity, thus going toward the sky. The argument I would use would be. If it is true for a rocket put up by Nasa to 17,700 mph for it to be in orbit, it is true for a small part of that rocket that fell off. Similarly for an identical rocket traveling in the other direction. with its bit that fell off . If these two bits collided in a perfectly elastic collision , They would both bounce back each in the opposite direction. Both these bits would both still be in orbit , but in the opposite direction. If they now just miraculously got coupled together in an elastic coupling , then they would rattle back and forward in orbit. Now there are times during reversal that velocity would reduce momentarily. That is why there needs to be a peek velocity of 22,000 mph such that the RMS ( root mean squared ) is 17,700 mph. What I have just described is What the Device I am describing should be and do . Except that rather than all the rockets , bits falling off etc One sets up an oscillation device doing precisely the same . Edited July 10, 2013 by Mike Smith Cosmos
EdEarl Posted July 10, 2013 Posted July 10, 2013 (edited) I still have the same question, do you want to model an ideal system with no mass in the elastic, or a real system with all kinds of instabilities? BTW A real system will have the two oscillating masses be slightly different, equal within some tolerance. And, the connection of the elastic will be slightly off center within some tolerance. What tolerances do you wish? What affect will these inequalities have? For an ideal system, a model will show that the average velocity of the masses is important, not their instantaneous velocity. As a mass accelerates in the direction of orbit, the orbit becomes larger. As a mass decelerates the orbit becomes smaller. When the velocity is zero, the mass will fall straight toward the earth. As it accelerates in the other direction its orbit begins to increase again. As it slows down its orbit will become smaller. That sequence will continue until the effects of gravity when it is stopped, i.e., falling, add up to it falling into the earth. I believe it will actually fall even faster. The failure mode for a real system, because of imperfections, will be different, but ultimately the same. Just think about these, don't worry about answering. The I am merely predicting what the model will show. Perhaps, I am wrong. Edited July 10, 2013 by EdEarl
Mike Smith Cosmos Posted July 10, 2013 Author Posted July 10, 2013 (edited) Here's what I mean... This shows the forces the device employs when the spring is fully stretched, when trying to use orbital mechanics to counteract the effects of gravity: StretchedSpringForces.png Assuming that the spring system is set up to pull the masses back along the same orbital path, then the force exerted back on the spring has a net downward direction. The effects of the force of gravity is already accounted for here in the curve of the orbital path. The spring has to pull the masses relatively *upward*, and the masses are then pulling back relatively *downward*. Surprise!, the effects of gravity remain. Now to be fair, the opposite happens when the spring is fully compressed. It pushes relatively downward on the masses and the masses provide an upward reactive force. However, when the spring is more compressed, the masses are closer together, the force vectors are closer to horizontal on the diagram, and the upward force on the spring when compressed is less than the downward force on the spring when stretched. You can fiddle with the masses and the length of the arc and all that to trick yourself into making the effects of gravity seem to disappear relative to the other forces, but you can't do that in nature. Gravity still contributes the same despite other forces or velocities etc. This isn't the only way to explain why the device won't work, it's just one of many. My point is that if you provide some upward force to counteract gravity, yet disallow an equal and opposite reactive force pushing downward on something, then the device disobeys Newton's third law. Since you're not proposing a "reactionless drive" force here that could speculatively break that law, it won't work. Although the drawing puts integral parts far apart in reality all integral adjacent parts are tangential to the orbit which itself is at 90 degrees to the force of gravity , so should be neutral to gravity. Then the question is IF the parts of the device are trying to move in a straight line to a higher orbit does this ammount to a LIFT ? Why can I get a sweet smell of Tomatoes ? Edited July 10, 2013 by Mike Smith Cosmos
EdEarl Posted July 10, 2013 Posted July 10, 2013 (edited) Mike is it OK to replacmotor-elastic with a maglev system using superconductors and a super capacitor with everything shaded by the earth to keep it cool enough without refrigeration? I also recommend simplifying to a single mass oscillating vertically or horizontally, whichever you think is best. The curved system you have described has both H and V force components, but I do not know which you think is most important. If both, then the maglev can have both. The maglev is a linear motor-generator that can use the charge and recharge the super capacitor, which loses no energy from resistance. By separating vertical and horizontal force components the system is easier to analyze. A real system and ideal system are essentially the same. Edited July 10, 2013 by EdEarl
D H Posted July 10, 2013 Posted July 10, 2013 Although the drawing puts integral parts far apart in reality all integral adjacent parts are tangential to the orbit which itself is at 90 degrees to the force of gravity , so should be neutral to gravity. Then the question is IF the parts of the device are trying to move in a straight line to a higher orbit does this ammount to a LIFT ? The answer to your question is no. Forget about gravity for a bit. Imagine putting your device in a sealed container and in deep space, far from any gravitational source. Now power it up. What do you think will happen? Keep in mind the law of conservation of momentum. Now let's put it in orbit about the Earth. Power it up. Nothing happens, same as in deep space. Why can I get a sweet smell of Tomatoes ? I smell something a bit more foul. 1
dimreepr Posted July 10, 2013 Posted July 10, 2013 (edited) Although the drawing puts integral parts far apart in reality all integral adjacent parts are tangential to the orbit which itself is at 90 degrees to the force of gravity , so should be neutral to gravity. Then the question is IF the parts of the device are trying to move in a straight line to a higher orbit does this ammount to a LIFT ? Why can I get a sweet smell of Tomatoes ? This whole idea is essentially the idea put forward in the 70’s of opposing gyroscopes to create lift, it didn’t work then and won’t work now, simply because the invoked force is only an apparent/fictitious force. Edited July 10, 2013 by dimreepr
Mike Smith Cosmos Posted July 10, 2013 Author Posted July 10, 2013 (edited) The answer to your question is no. Forget about gravity for a bit. Imagine putting your device in a sealed container and in deep space, far from any gravitational source. Now power it up. What do you think will happen? Keep in mind the law of conservation of momentum. Now let's put it in orbit about the Earth. Power it up. Nothing happens, same as in deep space. I smell something a bit more foul. Forget about gravity for a bit. Imagine putting your device in a sealed container and in deep space, far from any gravitational source. Now power it up. What do you think will happen?I . 1. I imagine that nothing will happen ,other than the masses will oscillate back and forth in a straight line. Bring it back from outerspace where there is not much gravity to influence anything., into the neighborhood of earth, then I think this. 2. At an orbit radius , not moving around the earth, but oscillating back and forward each mass ( in anti phase ) by 4 " inches at 40khz at 22,000 mph peek velocity 17,700 mph RMS velocity THE DEVICE SHOULD HANG IN ORBIT HEIGHT , Not rotating around the Earth. 3. What I am not sure about is what happens at less that 17,700 RMS miles per hour , or greater that 17,700 mph. I would like to think it goes either Up or down away from 17,700 mph orbital. I think the mechanism is Mike Smith said As most parts on average of the device are moving at 17,700 mph . ( all be it a back and forth type of movement. the mass inertia and strait line momentum is exactly overcome by the force of gravity. This producing a balancing effect Inertia effect out , gravity effect in. This has the effect of changing the strait line projectory to that of a 'partial arc' orbital position. As such it has the capability of staying aloft , Geo stationary at 500 miles rather than 22,000 miles The principle here being that a mass must be moving at 17,700 mph in order for gravity to be equalised. or put another way for gravity to act on the mass in such a way as to take up the geometry of a part arc [as part of an orbital trace } Mike Smith July 10 2013 . Getting up and down May have to be by other means. Mike is it OK to replacmotor-elastic with a maglev system using superconductors and a super capacitor with everything shaded by the earth to keep it cool enough without refrigeration? I also recommend simplifying to a single mass oscillating vertically or horizontally, whichever you think is best. The curved system you have described has both H and V force components, but I do not know which you think is most important. If both, then the maglev can have both. The maglev is a linear motor-generator that can use the charge and recharge the super capacitor, which loses no energy from resistance. By separating vertical and horizontal force components the system is easier to analyze. A real system and ideal system are essentially the same. I am not quite clear what you are proposing and why you are proposing it. ? This whole idea is essentially the idea put forward in the 70’s of opposing gyroscopes to create lift, it didn’t work then and won’t work now, simply because the invoked force is only an apparent/fictitious force. There is something here in these rotating and oscillating Masses. Edited July 11, 2013 by Mike Smith Cosmos
EdEarl Posted July 10, 2013 Posted July 10, 2013 (edited) oscillating back and forward each mass ( in anti phase ) by 4 " inches at 40khz at 22,000 mph peek velocity 25 [math]\mu[/math]s per oscillation of a mass to 22,000 mph...impossible with current technology, and unlikely ever. The record for accelerating an object by a gun is held by the U.S. Navy, Project HARP. Using an old U.S. Navy 16 inch (406 mm) 50 caliber gun (20 m), later extended to 100 caliber (40 m), the team was able to fire a 180 kilogram slug at 3,600 metres per second (12,000 ft/s), reaching an altitude of 180 kilometers (591,000 ft). Only 8,050 mph. Edited July 10, 2013 by EdEarl
Mike Smith Cosmos Posted July 11, 2013 Author Posted July 11, 2013 (edited) 25 [math]\mu[/math]s per oscillation of a mass to 22,000 mph...impossible with current technology, and unlikely ever. The record for accelerating an object by a gun is held by the U.S. Navy, Project HARP. Only 8,050 mph. . I Like it Give me More ! But we attain these speeds by space flight , why not by oscillation (resonant oscillation ) .Even if it does have to be in a vacuum ? You notice in my early illustrations, I did put an explosion , I thought it might require a bit of initial Boost ( Hummf ? 3 times your US Naval Firing ! I like it a lot ! ) The passengers might have to go up later ! . . Mike PS I Knew I should have done the Maths I had the feeling it was a biggish project. ! PPS [ I am off to Barbados , looking for that discarded Gun } Edited July 11, 2013 by Mike Smith Cosmos
D H Posted July 11, 2013 Posted July 11, 2013 . 1. I imagine that nothing will happen ,other than the masses will oscillate back and forth in a straight line. Correct. Remember, though, that those masses are in a sealed container. You can't see them. All you can see is the container. If your oscillating masses are balanced, you won't see anything happen. If they aren't, you'll see the container move around a bit but it's mean (average) motion will still be zero. Bring it back from outerspace where there is not much gravity to influence anything., into the neighborhood of earth, then I think this. 2. At an orbit radius , not moving around the earth, but oscillating back and forward each mass ( in anti phase ) by 4 " inches at 40khz at 22,000 mph peek velocity 17,700 mph RMS velocity THE DEVICE SHOULD HANG IN ORBIT HEIGHT , Not rotating around the Earth. There is no difference between the empty space scenario and the freefall scenario. They are one and the same per the equivalence principle. There is something here in these rotating and oscillating Masses. No, there isn't. You can keep deluding yourself all you want. There's nothing here, and with that, I am done with this thread.
EdEarl Posted July 11, 2013 Posted July 11, 2013 (edited) 22,000 mph = 9 834.88 meter/second 25 s per oscillation = 25*10-6 second [math] a = \frac{\Delta v}{\Delta x} = \frac{9 834.88 m/s}{25 \times 10^{-6}s} = 3.933952 \times 10^8 m/s^2 [/math] 1 G = 9.80665 m/s2 [math] a = 3.933952 \times 10^8 m/s^2 [/math] is about [math] 4 \times 10^7 G [/math] assuming no arithmetic errors. A neutron star has gravity of about 1011G. And, I'm done with this thread. Edited July 11, 2013 by EdEarl
Mike Smith Cosmos Posted July 11, 2013 Author Posted July 11, 2013 (edited) There is no difference between the empty space scenario and the freefall scenario. They are one and the same per the equivalence principle. This bit is surely what its all about. ! When the device is placed say 500 miles up , the influence is similar to what it is here at ground level ( short of a change in radius from 4000 miles to 4500 miles. ) quite a strong force. Here the effect of this gravitational force will be enough to bend the path of each part of the device into a part arc identical to the orbit radius at 4500 miles. ( all this within its sealed container which must allow for this bending .) I fail to see why this " by equivalence " is not in orbit. The only difference its jiggling back and forward at 17,700 mph rather than in one direction. However I may be hopelessly WRONG by something I am completely overlooking DH. If that is the case please excuse me. Edited July 11, 2013 by Mike Smith Cosmos
md65536 Posted July 11, 2013 Posted July 11, 2013 This bit is surely what its all about. ! When the device is placed say 500 miles up , the influence is similar to what it is here at ground level ( short of a change in radius from 4000 miles to 4500 miles. ) quite a strong force. Here the effect of this gravitational force will be enough to bend the path of each part of the device into a part arc identical to the orbit radius at 4500 miles. ( all this within its sealed container which must allow for this bending .) I fail to see why this " by equivalence " is not in orbit. The only difference its jiggling back and forward at 17,700 mph rather than in one direction. However I may be hopelessly WRONG by something I am completely overlooking DH. If that is the case please excuse me. On the ground you have the ground holding you up, with an upward force that balances the downward force of gravity (otherwise you would keep accelerating downward). You are correct, you've overlooked something but it's been explained in different ways already. You've overlooked it in the sense of "To ignore deliberately or indulgently; disregard." You're saying that the mechanism that keeps this device up is based on the principle that keeps satellites up. But satellites are in free fall, and you don't like that idea, so you purposefully ignore that "model". If it's not in free fall then by definition it has forces other than gravity acting on it, and you're hand-waving past that by saying "There is something here in these rotating and oscillating Masses." But you're proposing a force with no reactive force, in violation of Newton's laws of motion. You have to show that the mechanism of the device is not confined by those laws, which you can't (cause it don't). You can't just ignore models because you don't like the gist of them. To quote Feynman, "You don't like it, go somewhere else! To another universe! Where the rules are simpler, philosophically more pleasing, more psychologically easy." Otherwise, you must abide by the rules of the universe, or show---not guess---that we've got them wrong. The thing you've overlooked is physics. Sadly, you must work with the models that say that the device doesn't work. You can choose different models but they can't both be correct and give different results. I've tried to explain this in terms of equal and opposite forces, EdEarl has shown that the masses would not accelerate in the direction you expect, and DH has shown that your device is in free fall (unless it disobeys Newton's first law). That's several things you've overlooked. You've overlooked why satellites stay in orbit, what forces actually act on gyroscopes, and how Newton's laws of motion apply. 2
Arnaud Antoine ANDRIEU Posted July 11, 2013 Posted July 11, 2013 You are correct, you've overlooked something but it's been explained in different ways already. You've overlooked it in the sense of "To ignore deliberately or indulgently; disregard." You're saying that the mechanism that keeps this device up is based on the principle that keeps satellites up. But satellites are in free fall, and you don't like that idea, so you purposefully ignore that "model". If it's not in free fall then by definition it has forces other than gravity acting on it, and you're hand-waving past that by saying "There is something here in these rotating and oscillating Masses." But you're proposing a force with no reactive force, in violation of Newton's laws of motion. You have to show that the mechanism of the device is not confined by those laws, which you can't (cause it don't). You can't just ignore models because you don't like the gist of them. To quote Feynman, "You don't like it, go somewhere else! To another universe! Where the rules are simpler, philosophically more pleasing, more psychologically easy." satellites ... earth But have you already heard of the sun?
Mike Smith Cosmos Posted July 11, 2013 Author Posted July 11, 2013 satellites ... earth But have you already heard of the sun? Arnaud What point are you making , I am not sure if you are supporting my proposal or against it ? Mike
D H Posted July 11, 2013 Posted July 11, 2013 I was going to stay out of this thread, but then I saw that someone downvoted md65536. Here's a +1 to counteract that -1, md65536!
Mike Smith Cosmos Posted July 11, 2013 Author Posted July 11, 2013 (edited) 22,000 mph = 9 834.88 meter/second 25 s per oscillation = 25*10-6 second [math] a = \frac{\Delta v}{\Delta x} = \frac{9 834.88 m/s}{25 \times 10^{-6}s} = 3.933952 \times 10^8 m/s^2 [/math] 1 G = 9.80665 m/s2 [math] a = 3.933952 \times 10^8 m/s^2 [/math] is about [math] 4 \times 10^7 G [/math] assuming no arithmetic errors. A neutron star has gravity of about 1011G. And, I'm done with this thread. Ed , I can understand your principle of speed change due to oscillation over 4 times 4 inches every cycle creating large acceleration , that was one of the reasons I looked into the gyroscope or circular rotating disc ( not requiring a change in direction ). Except of course circular motion is itself causing acceleration toward the center , and if I had my way large centrifugal forces outward ,away from the center . These forces are what tore my models to pieces. When I was a boy Centrifugal forces existed, they were in the text books. Atoms were explained with electrostatic forces (electron - to protons + ) pulling in with Centrifugal forces due to very large angular momentum causing a pushing out. They are still spoken about. But they (centrifugal forces) are being written out of text books. Now it is all about centripetal forces pushing inward , to steer straight line trajectories into a circle. I have tried to work with this on this thread showing Gravitational Force pushing the devise parts against their straight line inertia and momentum into a part of an orbital shaped arc. But now "free fall" is being pushed as model . Even by the current model of gravity pushing into a circle there is no "free" in it . Things want to go straight , especially fast things. It needs forces like gravity pulling mass away from their straight line trajectory. The faster they go,the more the momentum in a straight line, soon gravity is not strong enough, or structures are not strong enough and things move or blow apart. I am sure that with a bit of serious R & D in this area , many of these orders of magnitude can be overcome. Here we are at the threshold of needing to get out into space , cheaply and efficiently. All effort should be made to address any possibility of a 'Sky Hook'. What takes all the energy in conventional rockets is carrying all the Fuel. If you have a 'Sky Hook ' you can do it with a Washing Machine Motor . It might take the energy of 3 USA NAVY HARP Project Explosive devices ,to get a Sky Hook up there. But once in place its easy to raise things upwards using a Sky Hook ! On the ground you have the ground holding you up, with an upward force that balances the downward force of gravity (otherwise you would keep accelerating downward). I've tried to explain this in terms of equal and opposite forces, I have and am still working with your exhortation to do force diagrams, and appreciate your urging . The one that I am trying to reconcile now is :- The instant a moving mass travelling in a small section of space and time at right angles to the force of gravity ( indicated by a plumb line and weight ,) as pointing towards the center of the earth and indeed measurable by a simple Newton meter. Immediately after this instant moment, the force of gravity attempts to pull this mass away from its straight line trajectory.. Not a long time, but a very reduced yet finite time later. Because the tangent of the new position is not the same as the starting instant tangent , the effect of the force of gravity can be resolved into a vertical and horizantal components . I am trying to isolate the effect of being in a balanced orbit to being in a faster than balanced orbit. Edited July 11, 2013 by Mike Smith Cosmos
EdEarl Posted July 12, 2013 Posted July 12, 2013 (edited) 22,000 mph = 9 834.88 meter/second 25 s per oscillation = 25*10-6 second [math] a = \frac{\Delta v}{\Delta t} = \frac{9 834.88 m/s}{25 \times 10^{-6}s} = 3.933952 \times 10^8 m/s^2 [/math] 1 G = 9.80665 m/s2 [math] a = 3.933952 \times 10^8 m/s^2 [/math] is about [math] 4 \times 10^7 G [/math] assuming no arithmetic errors. A neutron star has gravity of about 1011G. And, I'm done with this thread. I made a mistake in the formula [math] \Delta x [/math] should have been [math]\Delta t [/math] as shown above. AFAIK the values are correct. Also, the energy required to move 1 kg to 9 834.88 meter/second is 96,724,865 joules. To oscillate 1 kg at 40KHz for one second requires about 3M tons of explosives, because it is unlikely you can find an elastic that can survive the forces of acceleration and deceleration you suggest. On the other hand, setting 90 pounds of explosives 40,000 times per second in exactly the right position is not possible either. Edited July 12, 2013 by EdEarl
Mike Smith Cosmos Posted July 12, 2013 Author Posted July 12, 2013 (edited) I made a mistake in the formula [math] \Delta x [/math] should have been [math]\Delta t [/math] as shown above. AFAIK the values are correct. Also, the energy required to move 1 kg to 9 834.88 meter/second is 96,724,865 joules. To oscillate 1 kg at 40KHz for one second requires about 3M tons of explosives, because it is unlikely you can find an elastic that can survive the forces of acceleration and deceleration you suggest. On the other hand, setting 90 pounds of explosives 40,000 times per second in exactly the right position is not possible either. Its only a one off. As once oscillation is set up, it is self perpetuating by the resonant oscillation. Alternatively, the oscillation can be cranked up over a period of time. A long time . This to some extent is what I was illustrating with the Charging vehicles . . . Edited July 12, 2013 by Mike Smith Cosmos
Arnaud Antoine ANDRIEU Posted July 12, 2013 Posted July 12, 2013 Arnaud What point are you making , I am not sure if you are supporting my proposal or against it ? Mike From my point of view you have a right to see. I just wanted to raise the idea of the tides exerted by the moon on the earth... and from sun? The balance would be it created between anti objects? anti-satellite <------- Earth -------> satellite Time dilation is acceleration with gravity, because the gravity create the expansion of the mass in the distance. This create time (mass in motion - quark star). From the center of the earth, the particle will Further more far, while more long it will wait for the high atmosphere with the same force of energy supplied. The separation of the dilation is observed at the tau particle. The big-bang is an giant neutron star plus. Then we can also speak of anti-bigbang. To be honest, anti-objects or anti-bigbang does not exist in the same form. This is simply expressed in EM waves.
md65536 Posted July 12, 2013 Posted July 12, 2013 But now "free fall" is being pushed as model . Even by the current model of gravity pushing into a circle there is no "free" in it .Look up what free fall actually means, and try to figure out if the device is in free fall or not. This is what I mean by overlooking physics. You're deciding that what you haven't considered, you don't like, and so it can be ignored. It's the same principle that keeps cartoon characters up when they don't yet realize they're supposed to fall, but it doesn't work in the real world. You can explain the behavior of a device in different ways, some easier than others, but it won't behave differently based on how you choose to explain it. Ignoring Newtonian laws doesn't free you from the effects they predict.
Mike Smith Cosmos Posted July 12, 2013 Author Posted July 12, 2013 (edited) Look up what free fall actually means, and try to figure out if the device is in free fall or not. This is what I mean by overlooking physics. You're deciding that what you haven't considered, you don't like, and so it can be ignored. It's the same principle that keeps cartoon characters up when they don't yet realize they're supposed to fall, but it doesn't work in the real world. You can explain the behavior of a device in different ways, some easier than others, but it won't behave differently based on how you choose to explain it. Ignoring Newtonian laws doesn't free you from the effects they predict. Well I have just looked up Free Fall on Google. Sort of what I expected. It refers to issues where only gravity is a force. It speaks of projectile, parabolic flights and people in aircraft etc. Before I looked it up I thought it refered to, and may still refer to a model that either Galileo or Isaac Newton came up with about shooting a Canon off a very high mountain sideways. With enough velocity the said person said it would go outwards and fall, but as it fell it got no closer to the earth , but kept on falling . All the way round the Earth . Describing an orbit. Not sure if I have found the correct link ( perhaps you could give me your link. If you do mean this (off a high mountain and keep falling all the way round the Earth.) Well I do not think it is applicable to the device I am talking about. a) I am not imagining firing it off the top of a mountain. b) It is not only force as there are others c) it is not going in a single forward flight but rather back and forward in oscillation. When I read about this version a few years back , I sort of , saw what being described , and could see how it worked out. I don't have a problem with it in the right circumstances, I just do not particularly like it as an easy one to use on these Oscillations or on Centripetal/centrifugal discussions. I am not dismissing it , [ as in I dont believe it ] . I just do not like it in this application as I want something to stand still in Gyosynchronos Orbit at 100 m and 10,000 meters and higher say 500 miles. I do understand the idea of fall caused by gravity. I am trying to make the fall of the device from one end of the device to the other as being the same fall you would propose by your " free fall" ( this very shallow part arc, however it is made up of all these bits ( this way very fast , slowing, speeding up to a peek, slowing again , different direction) all this is not easy to picture in a canon ball being fired off a mountain model. I have tried to view it as an electrical analogue. Where AC electric + and - mean 0 into full wave rectification + and + mainly. There may be a little -ve leakage but overall good + ve net This + to balance against gravity. You could be right , BUT if there is a chance that you are wrong and a net balance at least will give a floating Chamber in Gyosync Wherever . At this stage I am not discussing an action without reaction. I am saying inertia/momentum pushing out via attempted straight line . Gravity push in in radially in equal balance. This could be within our long reach :- . . Edited July 12, 2013 by Mike Smith Cosmos
swansont Posted July 12, 2013 Posted July 12, 2013 Time dilation is acceleration with gravity, because the gravity create the expansion of the mass in the distance. This create time (mass in motion - quark star). From the center of the earth, the particle will Further more far, while more long it will wait for the high atmosphere with the same force of energy supplied. The separation of the dilation is observed at the tau particle. The big-bang is an giant neutron star plus. Then we can also speak of anti-bigbang. To be honest, anti-objects or anti-bigbang does not exist in the same form. This is simply expressed in EM waves. ! Moderator Note OT. Keep it in a separate speculations thread. Nowhere else.
Recommended Posts