Cosmic Yoyo

12 ≠ 11? You sure bout that?

1. If you accelerate *to* 2. They are equivalent numerically. 3. This is a lay translation, trying to communicate the concept. It isn't fictitious. 4. Arbitrary duration that's defined by the units, which for time is seconds. If you're talking about f = mv, it's that force for one second and at any time within that second, the instantaneous force equals the force over the duration of one second. Force isn't time based itself, but time is used to define it as well as other units.

f = mv is what f = ma resolves to in one second. If you accelerate to 9.8 ms-1 in one second, you will be travelling at a velocity of 9.8 ms-1. Momentum and force are equivalents - if an object has a certain amount of momentum, it will require an equivalent amount of force to bring it to a stop. They don't have to have the same units to do this. By saying mv, you're saying full acceleration to the value of the unit time, which is just another way of saying the velocity. How long the force transfer takes is irrelevant here except for how the units relate to each other. When you start to talk about specifying the duration, you start to talk about impulse.

Oh dear. Swansont made the exact same mistake I did and now you're making it too Phi!

Ghideon Look at the page on momentum. You'll see on the right F = d/dt(mv). It's per unit time. So presumes the acceleration is one in one second, or whatever unit of time you're using for the acceleration. https://en.wikipedia.org/wiki/Momentum I can't describe two bodies pushing each other apart much more simply than that, but I've used Ironman and his double as the device because I don't need to explain how they do things. I also used the example of a massive barbell form with sliding weights. If you can think of simpler language, I'm open to suggestions. However, right now, Ironman and his double can push each other apart, are tethered, can lock their tether and can push against other objects. I really don't know how I can make it any simpler without having to have extraneous explanations to the point of bringing out finger puppets. It really is almost on that level. I can't help but think this is an effort to draw me into a rabbit hole. If people can answer my question in the negative - fine, answer it according to the rules of this site, i.e. bring evidence. If people can't answer, or want to hedge their bets, "I can't see why not" is a perfectly reasonable response. Anything else from now is just nit picking.

1. I don't understand why you're so annoyed at me describing things in a way that makes sense to me. I didn't say anything false. 2. f = mv which is the same as f = ma if complete velocity transfer is made, say if one goes from a stop to full speed. 3. I didn't think it was a requirement on this site to label things a particular way. I am sorry if I offend. I really thought it would be OK to provide physical context of a kick as a force. 4,5. Insipid examples are used because some people need these examples to follow a narrative. It's functional. Is it breaking the rules here? 6. If I said 'velocity that arises from applied force', you'd understand me. I say 'applied velocity' to truncate that. I really can't understand why you find this so irritating. How does which satellite acquire this velocity? See I can ask you are you referring to the Red Dwarf, or are you referring to Ironman and his double or just Ironman by himself. See how easy that is to distinguish elements? 7. The bounce is simply a force applied to another object. They key thing about a bounce if you like is that it's due to an ideally elastic collision. The term contrary motion I use is derived from the playing of a musical scale in opposing directions. It's basically a retelling of "for every action there's an equal and opposite reaction". The bounce is simply a way to transfer momentum. Edit - my presence here isn't to sell my idea, I'm asking the question does it work.

"If the masses are equal, they are experiencing the same exact force. This result is a fairly straightforward result of kinematics." No, gravity provides a different amount of force to objects at different altitudes. "Mass and velocity does nit give you a force. And you still have not explained what this alleged bounce is." Mass and change in velocity do. In my original post, I refer to the bounce force as planting a mawashi geri. "Where does this energy come from?" In the example, Iron man and his double's legs. "That makes no sense from a physics perspective " I was responding to your exact words. -"(you just add in a second velocity vector)" "You don’t have a “device”. You seem to have two particles in orbit. I think, because your description is far from clear." Ironman tethered to his mech buddy is not a device? "You don’t really “apply” velocities. You can impart a velocity by exerting a force." While obviously not technically correct, it's good to be able to distinguish initial velocity, applied velocity and the resultant total of the two. I'd rather you complain about that than having too much verbosity that will make people's eyes glaze over. "Contrary motion? of what? what is an ”internally activated” orbit? Again you mention a bounce without having explained what that means. " The bounce makes contrary motion if done right. Contrary motion means following the same shaped trajectory out of the manoeuvre that created the bounce or other interaction. When Ironman and his double kicked the Red Dwarf, their trajectories toward and away are mirrored over their shared orbital plane through a line running from the Red Dwarf to the centre of the Earth. They performed the bounce and made Red Dwarf elevate. In the mathematical solution, you can see it, it's the tear drop shape. At the tip of the tear drop, there's an abrupt change in velocity due to the bounce. Note the overall symmetry. The orbit is internally activated because that's the idea, internally generated propulsion. It's an orbit distinct from the original. "You haven’t described objects pushing each other apart. You have two objects in orbits. They are not, so far as I can tell, interacting with each other, hence there can be no push." Ironman and his mech buddy tethered together.

Swansont, let me do your second first. I'm talking about resultant forces from the planet. Two water molecules dropping from a tap will accelerate away from each other if they came out at different times. This is the tension I'm describing and it only exists because of the planet. For the bounce, what I have calculated is velocity. To get the force, then just add a value for mass. The separation movement requires energy, the more, the better. That's where the second velocity vector takes effect. All that needs to be done is send the two sides of the device on their way. There is nothing preventing this velocity being applied, all that needs to happen is for the two sides to travel away from each other away from the original orbital plane. Gravity will bring them back together twice per orbit, at apogee and back at perigee. Using contrary motion, it's possible to interrupt this internally activated orbit, apply the bounce to a target, then the resultant force of this bounce interaction if performed precisely will send the device back into its original orbit. It's good to have left mass out of these calculations up to this point, standard practice mostly still applies. To peel this whole idea away to a bare bones statement, it can be said that one way to defy gravity is to make two objects push each other apart in opposite directions that both have upward components.

Joigus I just mean terms you're more accustomed to. I know there's mistakes in the first post, I just had a bit of trouble finding anyone to check my work (I mean anyone). My DW did her best, it could have been worse! However, she has absolutely no capacity to check the details we're discussing. Your'e getting that raw.

Swansont The bounce is a way to transfer the 'generated' momentum. Nothing complicated. Honestly, I don't think tidal forces really hit home. Tension and compression are really good terms for inferring a relationship between two particles if there was a structure. If it's tensile, the particles want to accelerate away from each other. If its compression, they want to accelerate towards each other. If we try to use tidal forces to describe both, I think that's much worse.

Joigus Not sure if realised, but Swansont was on to something when he mentioned the inability to generate tangential propulsion. This drive has great difficulty with this. It's good at pushing in the direction opposite to the gravitational field. If you want to wait for someone to explain it in friendlier terms, no worries. The NASA material refers to tidal forces. I refer to it as gravitational tension. This is distinct from gravitational compression which is running orthogonal to the tensile force, or tidal force if that's what you want to call it. The moon heading off by itself - my understanding is that its the attraction to everything else that's causing this, not tidal forces per se. The tidal forces they refer to in the NASA material is the variation of gravitational effect due to different trajectories and locations. This isn't my expertise, but I hope that addresses what you brought up.

I'm here for those questions. I've generated two scenarios, one graphic and one mathematical. I calculate both, but the mathematical formulas show interactions a bit to small to easily see what's happening. That's why I added the graphic example. If you look closely, you can see what's going on behind the Earth and you can see the shifts in A by noting the points I've emphasised. The graphic example is obviously massive and I'm not proposing making anything that big - I made it to communicate a sense of proportion with respect to the physics being described. It's all to scale, the 3D views are axonometric, so translate it all by inverse of root two. Doing so allows you to measure with a ruler. The diagrams are solid, generated purely from the vis viva, all I'm doing is dumping in velocity and noting what happens to A (and by definition what happens to B) and drawing it out. The bounce is what one would use to keep a space elevator elevated, although simple satellite boosting is possible this way. If you had a whole bunch of Anti Gravity drives following each other around in orbit, they could provide a regular upwards bounce force, allowing a space elevator platform to operate at maybe 150 Km (geocentric) above the ground. I know this is far fetched and there's some obvious impracticalities, but the bounce is real and quantifiable. I'm happy to show you the design I am proposing and its sequence.

What am I doing wrong? Do you want me to post the pages individually?

Please read the pdf.

I attached it to the first post. This is what I've used to generate the eccentricity pumping diagram I showed you. Anti_Gravity_Mathematical_Solution.pdf

"Equations do a much better job of describing what’s going on. Accompanied by a diagram of the forces. " Thats what the mathematical solution I presented you does. That's what the vis viva formula does. That's what I've done to your logo. "The simplest explanation is that you are mistaken. It’s up to you to explain what your proposal is. Using proper physics terminology." That's like pulling rank. I'm not going to argue with you about nomenclature. The answer is here, I've done my best with the time I have to explain it.

I made a picture for you Swansont!

Just wanted to add, I'm really not trying to be rude by implying you don't understand this Swansont, I'm just explaining that this concept is new to you. In essence, reducing to the two particle simulation is the simplest system I can show you, and imagining a rigid body structure is the best way to understand the concept as a working body in space.

"If your idea depends on gravity then it's not going to work (or work well) in space, far from any sources, where gravity is small." Indeed, I have noted that the Anti-Gravity drive works poorly if there is little gravity to resist. Part of what this research has revealed is that places without much gravity are sort of like the doldrums for sailboats. "Gravity is radial, so I don't see how you get propulsion, which suggests a tangential force, from that" Eccentricity pumping? What I'm getting at is not a tangential force. The diagram shows a quantified propulsive effect. A sailboat uses sails, but you don't say it doesn't work at all because it can't sail into the wind. "This is way too verbose. Too many caveats." I can't help you there. What I've written there needs to be understood, my question can't be answered expertly without understanding most of that and why I've written it that way. What I'm expressing there is how two point like particles can be representative by reduction of many things. Bear in mind, this is all new to you, I've been working on this for a few years. "And, of course, it has nothing to do with anti-gravity. None of this does." Then describe what is happening in the eccentricity pumping diagram and how the increase in orbital energy is from something other than propulsion and how you don't think this force working in the diagram works in the opposite direction of gravity, i.e. anti-gravity. Eccentricity Pumping Diagram.pdf This is probably a good time to think a bit more broadly about what's happening, particularly in relation to energy. Rather than thinking about generating propulsion - this is the overall effect, but - think about it another way. A gravitational system (prime + orbiter) is a working system. I know we don't say gravity does work, but it's doing the equivalent of work and the result would be the same if you took away the prime and did the work yourself. The work gravity is doing is keeping the orbiter close. By performing the motions I'm describing, this work machine that is the gravitational system is interrupted. Imbibed kinetic energy prevents the prime from keeping the same strength grip (grip is lost with altitude and altitude gain can be achieved in an isolated system as I believe I've shown). As soon as you diminish the grip of gravity on something orbiting, its trajectory will be less deviated by gravity. The mirrored motion away from the orbital plane stymies the prime. It wants to chase and grab both, but doing so requires slightly divergent courses. The prime is played. It's played because it will take longer to achieve the same outcome, which is retaining the orbiter.

" a closed system of objects not affected by external forces is constant" Like I've been saying, it's not a closed system. The bit about gravity is that it's missing. An omission. It's not looking at what I presented correctly. "and therefore no variation in gravitational force due to the change of "r" or distance between the center of mass of two bodies is being achieved" I'm asserting apparent weight which incorporates mass separation is what should be used here. What is being used is not weight, but it is one component of weight, i.e. acceleration (edit - sorry, should be g here). Removing mass from this pretends it's not important irrespective of separation. Regarding the claim you find dubious - it's not the analysis that's changing per se, it's the presence of the opposing side that reveals a propulsive force, so it's a completely different analysis. You can't accelerate one of the particles without the other. No acceleration, no propulsion. I need to explain - Yes, this is the crux of the matter. The vis viva describes what happens. The simulation you could say is a peer reviewed expression of this principle (the software at least) and I provided secondary confirmation with my mathematical proof. Using two points is a reduction of the types of systems. It could represent a rigid structure which has a slightly larger width / separation than represented by the balls and a reflective mass distribution, i.e skinny middle, fat ends like a barbell. The rigid variant is useful for understanding these systems without introducing drift. By no means am I suggesting a rigid structure is what should be built. I would note however, eventually around the moon, I think a modified massive scissor lift variant could do a bit of work there. If you had a big enough barbell, and you could shift the weights like I've described, the effect could be observed. In the real world, gravity will fight the attitude and tends to bring objects major (mass) axis parallel with the gravitational field. A rigid structure would need heavy duty gyroscopic actions to counter these forces, if it's even possible. This is why the tether option is preferable aside from the whole city sized space ship issue. With a tether system, structure is mostly defined by velocity and inertia. Gravity's twisties can't do much to a tether. As for your last note, is it a propulsion system? I believe they both are and are antonymous with each other as well as being completely complementary. The NASA authors Landis and Hratch also did make the assertion of internal propulsion, although they described it as adding and removing energy to and from the orbit. It's right there in the abstract. I have my own reservations about what they've said and would attach caveats, but I think it's mostly sound. Edit - sorry, no they did assert propulsion Nasa Material (D3).pdf

Swansont I received this opinion at one point: "The two pair of mass being separated will have no effect to propulsion as the system is self-contained or closed. The mass is not pushed against something to achieve an opposite reaction nor there is a rate of change to the total mass of the system to generate thrust. The invention violates the conservation of momentum which states that the total momentum of a closed system of objects not affected by external forces is constant, thus the system will continue with the same velocity unless acted on by a force outside of the system. Furthermore, the specification teaches that for the invention to allegedly work, a pair of masses constrained to the apparatus are to be propelled, in the opposite direction and retracted, perpendicular to the gravitational field so that gravitational force varies due to the difference in “r” or distance between the center of mass of two bodies (first body is the apparatus and the second body is where the apparatus is orbiting, such as earth) resulting from the pair of masses of the apparatus being separated. However, as the pair of masses are still attached to the apparatus even when they are separated, only the distribution of force changes across the apparatus but there is no change in the location of the center of mass of the apparatus thus “r” or distance between the center of mass of two bodies remains unchanged and therefore no variation in gravitational force due to the change of "r" or distance between the center of mass of two bodies is being achieved." I think may have convinced the person who wrote this otherwise. Please bear in mind, the person who wrote this is a professional, so I expect others to draw the same what I believe is the wrong conclusion - and others have. This also is where the assertions of this being a reactionless drive come about. I'm not analysing 160km balls in space, I'm drawing the balls 160 Km in diameter so you can see them - illustrative purposes. They show what happens to regularly orbiting (point) particles - the blue balls - when two of them travelling together are made to oppose each other which adds kinetic energy. This is where energy is added to the system and a correct interpretation of apparent weight i.e. one that factors in mass separation will yield the correct cumulative/compound orbit. I'm happy to reinterpret this if I'm not hitting it on the head. It's often said objects in orbit are weightless. I don't look at it in that way, weight is countered by acceleration is how I interpret the state. The main takeaway here is that the location of the masses matters, the quote above implies that it doesn't, or specifically that a change in separation will produce no effect/produce no propulsive force. I'd also like to emphasise not getting confused by what's happening - it's only by viewing the pair of separated masses as a single body that is simply manipulating its form through mass separation that the propulsive force can be realised. When viewed individually, the particles behave exactly as they would be expected to with respect to the definition of weight because this definition is perfect for a point-like mass. Without each other, the pair are subject to the "no space propulsion but rocket propulsion" rules as usual. When opposed to each other, the prime loses its grip if the directions are orthogonal to the orbital plane (this is an ideal and presumes a tether of negligible mass, so not real world). Most other directions lead to divergence rather than the convergence - the particles accelerate away from each other. This I believe is how the Skyhook works most effectively - the tether system can create massive rotational moments and inertia from this divergence which withdraws this manifesting rotational energy from the orbital energy. This causes the Skyhook to descend into an orbit where the point of consolidation (full tether retraction) becomes the apogee. Edit - The point - this allows the skyhook to push another object downward, causing reversal of rotation and allowing an exit trajectory back to another apogee (not perfect contrary motion as I've asserted is possible with the Anti-Gravity drive, but pretty close - it can't be symmetric with the front or back acting as it must do to push downwards). The manoeuvre places the Skyhook back into the same shaped orbit as it had when consolidated, but is transformed around - i.e. the axis are modified.

Swansont I don't have an example of that specifically, but it's the inappropriate use of apparent weight and closed systems as criticism of this idea that imply separation distance is irrelevant. So I'm tackling this criticism I have received at the outset. A and B are half of the major and minor axis of ellipses respectively. If one is talking about the length and width of an ellipse, they're talking about double A and double B of an ellipse.

Swansont Yep. Sorry about the bad editing (edit - and mistake) on part 1, I'm spewing about that. Anyway, what this example provides is basically a static version of what the vis viva describes with combined orbits mirrored over the initial orbital plane. Calculating the forces operating on the combined centre of mass basically forces you to work through the diminishing of the overall downward force by both angular and elevatory modification. This is what leads to the fundamental driving force, which is the enlargement of the orbital ellipses of both the A and B components. Everything follows from there...unless I've missed something. I've attached a simulation I prepared before I worked through with the math solution (actually I totally botched this simulation up at first, but finally got it right). The numbers match pretty closely to the numerical example in the math solution. The particles are 160 Km in diameter, this is for clarity. It's pretty hard to get the model spot on, but the math solution indicates a lengthening of A of 62 Km which I hope you agree is pretty close to what can be seen in figure iv. These diagrams are interesting because they also provide the sense of drift that I'm talking about - the blue balls represent the initial orbit and comparing them with the white not only shows what's happening to A and B, they also provide an indication of the period change. Just to note, the balls are the same size in the simulation, even though they look a little different, I'm not sure how much of that is illusion and how much is render effect.

"So I had no idea what you were doing. Now I see 422,250,000 was a typo, and you meant 42,250,000 km^2" Sorry my bad. I'm a clumsy mathematician. It's not a balance scale, we're working out the load on the base. That's why we need two triangles.

Swansont "Going back to my example, let's say the angle is 45º, elevation at the beam centre is 6500Km" Ok, you don't like elevation here because it's above the ground so wrong. Fine. I'm referring to distance to centre. I'm not sure why you didn't pick that up from what I'm doing. Of course a 2Kg mass will weigh 1.92 Kg on scales with r of 6500. I don't know why you'd think that's strange. Elevation from the surface of the Earth is mostly a useless value here, though you're obviously technically correct, I'm using a little license here so it's not so wordy, people complain about wordiness, although it seems a strange thing to be complaining about when the ability to defy gravity is being explained and those reading realise I'm not a text book writer. The force triangle is mirrored because there's two weights, so you need one for each. 9.81ms-2 is what the scales are tuned for. If g is 9.41 at 6500Km, then divide it by 9.81 to get the ratio which will describe how the scales will behave.