swjr-swis 3,125 Posted February 21 Share Posted February 21 My forum and wider internet search-fu is failing me today, and I'm hoping some of the brighter and more educated minds inhabiting this forum can help me with this query: Quote For a set of four relays around the same celestial body, and disregarding N-body perturbational effects, does a set of valid orbits exist that will maintain the relays in a regular tetrahedron configuration relative to each other? It's easy to envision a rigid-body tetrahedron suspended in free space rotating around its own center of mass. Without resorting to any math it's easily conceivable that there are an infinite continuity of possible roll/yaw/pitch rates possible, up to the limits of the material shear strength anyway. However, this does not mean that the vertices of such a rotating tetrahedron are *individually* following a valid 'gravitational' orbit around that same center of mass. In fact one can easily envision inducing only a roll rotation around the normal of one of the faces, in which one of the vertices wouldn't even move at all, while the other three are following a circular trajectory centered well outside the body center of mass. Where it not a rigid body, such trajectories would be gravitationally impossible. This would be just one special case, there's probably an infinite number of similar, more complex rotations to be conceived. Relays in orbit are not rigidly attached to each other, so to maintain a tetrahedron configuration at all times, if it is at all possible, they'd have to individually travel a valid orbital path. For it to be a regular tetrahedron, they'd have to be circular orbits, and centered on the center of mass of the body they are all orbiting. Something tells me that it should be easy to reason whether this is physically possible or not, but my brain is not cooperating today. Nothing more vexing than an easy solution just out of reach. I've already found numerous references and papers on 'central configurations'. Almost all of them talk of either rigid structures/bodies (not applicable here), or stable configurations only in the sense that the individual objects maintain their relative configuration while collapsing towards a common point. I can't seem to find any reference that either proves (or disproves, equally acceptable) that the configuration I describe above is possible. I welcome any and all pointers/links to a relevant reference, or a direct explanation, if what I'm seeking is at all possible or not. If someone has already established a discrete set of orbits doing this in KSP, I would gladly accept that as proof too. Quote Link to post Share on other sites

paul_c 425 Posted February 21 Share Posted February 21 Maybe But it doesn't matter.......4 satellites can 100% cover a body (for comms) in a particular orbital configuration. I can't even visualise that configuration, so I have no idea if the shape remains stable or morphs etc. I do know that in the real world, if/when the 4 sat config is used, slightly elliptical orbits are actually used, because it maximises coverage in the face of inaccuracy or peturbations. Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 21 Author Share Posted February 21 4 minutes ago, paul_c said: But it doesn't matter.......4 satellites can 100% cover a body (for comms) in a particular orbital configuration. I can't even visualise that configuration, so I have no idea if the shape remains stable or morphs etc. It does matter for what I'm after, although rather than practical it's really more of a 'wait a minute... is this fundamentally even possible' type of questions. The one special case I am looking for, if possible and stable, would theoretically maintain full coverage without any interruption at all. Wether it would be practical is an entirely different matter. I know of full-coverage solutions relying on eccentric orbits, in which coverage is maintained except for the short moments when the individual relays race through their periapsis. This is in fact my usual way of creating a relay network, and they don't require a great degree of accuracy, which suits me fine. They are in pretty much all cases 'good enough'. That said, it doesn't answer my question, and I'm left with this nagging suspicion that the (im)possibility of such a configuration should be fairly easy to reason. Hence my seeking input from the community. Quote Link to post Share on other sites

paul_c 425 Posted February 21 Share Posted February 21 I don't think we're on the same page here. 4 satellites, can & do produce a stable, 100% coverage all the time everywhere, constellation. It is not an "absolutely precise" thing where they need to be in an exact position (and maintain it) otherwise it drops to 99.9%, there is a ton of slop in the positions. The eccentricity is used to maximise the slop, for pragmatic station-keeping or fuel-saving, or ease of putting into orbit concerns in the real world. That the "shape" formed by the 4 satellites is in itself rigidly stable and unchanging, is not a consideration. Obviously, because all 4 AREN'T on the equatorial plane (maybe 1 is though) they are not geostationary. ETA Don't confuse a 4-satellite constellation with Molniya or Tundra orbits (which use high eccentricity to provide high % coverage of a portion of the globe per satellite). Quote Link to post Share on other sites

OHara 796 Posted February 21 Share Posted February 21 10 minutes ago, paul_c said: That the "shape" formed by the 4 satellites is in itself rigidly stable and unchanging, is not a consideration. Well, the title of the thread indicates interest in a "stable tetrahedron" Of course independent satellites (that is, with no 6-Mm rods holding them in place relative to each other) will not stay in such a shape, unless they are all in a single plane. I think the question is "why not?" (maybe more Science and Spaceflight than Gameplay) The stable, rigid, shape would rotate about some axis, so each satellite needs a force to accelerate it toward that axis. Gravity pulls to the center of the orbited body, not along the closest direction to the desired rotation axis, so the tetrahedron collapses. Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 21 Author Share Posted February 21 6 minutes ago, paul_c said: I don't think we're on the same page here. We are definitely not on the same page, that was indirectly my point in the previous reply. You're offering me a practical answer to a theoretical/fundamental question. Not to dismiss the validity of your assertion in its own right, but it's simply irrelevant to my question here. Coverage, whether full or all-time or in any form, and practicality, are remotely secondary considerations of why I even started this train of thought in the first place. They were left behind completely once I questioned whether it was even at all possible. Right now, the only consideration (and reason for this thread, at all) is whether it is at all a mathematical/geometrical/physical possibility. Nothing less, nothing more. Quote Link to post Share on other sites

paul_c 425 Posted February 21 Share Posted February 21 Ok no worries, I can't really help much further. Good luck too! Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 21 Author Share Posted February 21 3 minutes ago, OHara said: maybe more Science and Spaceflight than Gameplay Absolutely and entirely, not maybe. 4 minutes ago, OHara said: Of course independent satellites (that is, with no 6-Mm rods holding them in place relative to each other) will not stay in such a shape, unless they are all in a single plane. I think the question is "why not?" You say 'of course'... but I'm missing the obviousness here. Why is it 100% certain that it only works if they are all in the same two-dimensional plane? 6 minutes ago, OHara said: The stable, rigid, shape would rotate about some axis, so each satellite needs a force to accelerate it toward that axis. Gravity pulls to the center of the orbited body, not along the closest direction to the desired rotation axis, so the tetrahedron collapses. And that is the case I illustrated in the OP. Now reason this for when it's not simply 'an axis', but its own center of mass. It will require rotations in multiple axes, possibly all three. Is there an obvious reason why it's impossible for all vertices to be describing a valid orbit due to the combination of rotations? Quote Link to post Share on other sites

HebaruSan 4,698 Posted February 21 Share Posted February 21 My mind keeps reaching for a proof along these lines... Suppose such a configuration exists instantaneously—there are four satellites in orbit that for however brief a moment form a perfect tetrahedron, so they're all equidistant and at the same altitude. Further assuming circular orbits, their velocity vectors must be of the same magnitude and tangent to the surface. Consider one of the satellites; we have 360° of possible directions it might be traveling, but the rotational and mirror symmetries of the tetrahedron mean we only need to consider 60° of that: It's either moving toward one of the other points, or toward the middle of an edge of the tetrahedron, or somewhere in between. If it's moving directly towards one of the other satellites, then that satellite must in turn be moving directly away along the same circular arc to keep them at the same distance; any other angle for the second satellite would result in them getting closer and the tetrahedron breaking. Locking two satellites into specific paths like this puts severe limits on where the other two can be; as far as I can tell from some quick mental visualization, those are not valid orbital paths (centers not at parent body's center). If our first satellite instead is moving toward the middle of an edge, then by symmetry it's moving directly away from a second satellite on the opposite side, which in turn must be moving towards the first satellite to maintain distances, yadda yadda yadda, also no valid orbits for the third and fourth. I leave the question of what happens in the 60° between heading-toward-a-vertex and heading-away-from-a-vertex to a greater mathemagician than myself. Maybe it can be reasoned by analogy with the points already argued, maybe not. Quote Link to post Share on other sites

kerbiloid 11,269 Posted February 21 Share Posted February 21 Molnya orbit for 4 sat groups, raising up to the vertices in shifts? https://en.wikipedia.org/wiki/Molniya_orbit Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 21 Author Share Posted February 21 14 minutes ago, HebaruSan said: the rotational and mirror symmetries of the tetrahedron mean we only need to consider 60° of that: It's either moving toward one of the other points, or toward the middle of an edge of the tetrahedron, or somewhere in between. Good point, and along the lines of the 'simple' reasoning I feel there must be to reach a conclusion. I'm sitting here trying to visualize for myself what spatial path the other vertices are forced to follow in the general case, but it's likely to be constrained much (or exactly) as you describe. 1 hour ago, OHara said: "why not?" And @OHara to be perfectly clear: I'm not dismissing that your 'of course' isn't exactly as obvious as you state. I'm just failing to envision it right now and wondering what part I am missing. Quote Link to post Share on other sites

OHara 796 Posted February 21 Share Posted February 21 53 minutes ago, HebaruSan said: Suppose such a configuration exists instantaneously [ . . . ] Further assuming circular orbits If we also suppose the configuration of satellites is stable, meaning the distance between any pair of satellites is constant, then each satellite moves in a circle around a common axis of rotation. . . . because motion of a rigid assembly of points always consists of a common translation, plus rotation about an axis. I'm sure there's some convincing formal proof of this in some kinematics textbook, but right now I can't think of a convincing argument -- it is too deep in my instinctive assumptions about how things move. That's why I thought immediately of the axis of rotation -- not necessarily any axis of symmetry of the arrangement of satellites, but some axis. Then for a tetrahedron it was obvious to me that not all those circles could be centered on the center of gravity. Quote Link to post Share on other sites

HebaruSan 4,698 Posted February 21 Share Posted February 21 (edited) OK, this might be a much easier proof... Start with zero satellites in orbit and treat it as a problem of constructing the tetrahedron from scratch. First satellite is unconstrained and can be put in any circular orbit we like; suppose equatorial just for familiarity and ease of visualization. Second satellite must be put into some circular orbit where it is always the same distance from the first satellite. If their inclinations are different, the distances will vary*, so the second satellite must also be in an equatorial orbit. Third satellite also has to be same distance from first and second, so by the same argument it also must be equatorial, but tetrahedrons are not flat, proof by contradiction. * This point ought to be demonstrated in more depth if this was a homework assignment, but it's intuitive enough to observe and difficult enough to prove to discourage me from the attempt. Edited February 21 by HebaruSan Quote Link to post Share on other sites

YNM 2,524 Posted February 21 Share Posted February 21 (edited) I found a US Patent from 1989 giving that the minimum required satellites to maintain coverage of one hemisphere (north or south) is 3, and they do make for a constant triangle. Spoiler Thick line is on the front of the globe. The 4-satellite whole globe coverage isn't quite as neat however - either that or the visualization is not adequate enough. (original paper, but behind a paywall.) Spoiler The latter is apparently known as Draim's tetrahedral... so QED ? Edited February 21 by YNM Quote Link to post Share on other sites

sevenperforce 7,470 Posted February 21 Share Posted February 21 1 hour ago, HebaruSan said: OK, this might be a much easier proof... Start with zero satellites in orbit and treat it as a problem of constructing the tetrahedron from scratch. First satellite is unconstrained and can be put in any circular orbit we like; suppose equatorial just for familiarity and ease of visualization. [snip] I think a mistake was made here. The first satellite is not necessarily unconstrained and cannot necessarily be put in any circular orbit desired. In particular, an equatorial orbit definitely will not work because it necessarily limits the rotational axis of the proposed tetrahedron to the rotational axis of the planet and you can prove trivially that such a configuration doesn’t work, as you did above. A better starting point would be to assume you have two satellites in inclined circular orbits and ask whether there is a set of two orbital planes such that the two satellites remain equidistant. I’m having trouble visualizing it perfectly but I am imagining two 45 degree inclined orbits with different arguments. Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 21 Author Share Posted February 21 2 hours ago, HebaruSan said: If their inclinations are different, the distances will vary*, so the second satellite must also be in an equatorial orbit. This is the essential -and exquisitely simple- element that was right there in front of me, but I was failing to visualize. There isn't any way to construct two orbits around the same body that differ in inclination but keep the distance between the orbitals the same. If it can't be done for just two, it certainly can't be done for four. And the inclination difference is required to construct any three-dimensional configuration, or they'd all be in the same plane, as @OHara already mentioned. Thank you all for the kick in the pants. I have taken my brain behind the shed and thoroughly reprimanded it for failing to see that obvious point. Bad, bad brain. 2 hours ago, OHara said: I'm sure there's some convincing formal proof of this in some kinematics textbook, but right now I can't think of a convincing argument I wasn't even looking for a formal proof in the full sense. Just the common logic reasoning of why it was (im)possible, as @HebaruSan worded. 14 minutes ago, sevenperforce said: I think a mistake was made here. Are you sure? That is the beauty of assuming perfect spheres and circles, after all - apply any rotation to make the starting orbit be any inclination you wish, and it would change nothing to the system as a whole. So I do think @HebaruSan is quite correct in simplifying by starting with an equatorial. Rotate that starting orbit at will, and it will essentially be the same explanation. Quote Link to post Share on other sites

HebaruSan 4,698 Posted February 21 Share Posted February 21 33 minutes ago, sevenperforce said: I’m having trouble visualizing it perfectly but I am imagining two 45 degree inclined orbits with different arguments. OK, then you can tilt the planet underneath them so that one of them becomes equatorial, without affecting the relationship of the orbits to each other. The coordinate system is arbitrary. Quote Link to post Share on other sites

sevenperforce 7,470 Posted February 22 Share Posted February 22 1 hour ago, HebaruSan said: OK, then you can tilt the planet underneath them so that one of them becomes equatorial, without affecting the relationship of the orbits to each other. The coordinate system is arbitrary. Okay, so let me think this through. I'm not quite willing to give it up just yet. The problem, if there is one, must be in "If their inclinations are different, the distances will vary." Yes, it seems intuitively accurate, but is it? Consider two circular Earth orbits A and B, each with an inclination of 45 degrees (just to allow us to use meaningful terminology). Imagine that Earth is frozen underneath and not spinning at all. The longitude of A's ascending node (the point at which the orbital line crosses the equator moving from south to north) is 0 degrees, so the satellite in orbit A will cross the prime meridian and the equator at the same time. The longitude of B's ascending node is 30 degrees east, which means the satellite in orbit B will cross the equator over Kikorongo, Uganda. Suppose both satellites cross the equator at the same time. Depending on orbital altitude, the two satellites will be approximately 3000 miles apart. I don't see any reason why they would ever NOT be 3000 miles apart. They are both following the exact same circle, merely rotated around Earth's axis by 30 degrees, and while their orbital planes certainly intersect, they would not get closer or farther away from each other at any point. At the highest point of its orbit, A will be at 45 degrees north and 90 degrees east, over Qitai County in the Changji Hui Autonomous Prefecture of China. At the exact same time, B will be at 45 degrees north and 120 degrees east, over Jarud Banner in Tongliao, Inner Mongolia. They will still be 3000 miles apart. Quote Link to post Share on other sites

HebaruSan 4,698 Posted February 22 Share Posted February 22 1 hour ago, sevenperforce said: The problem, if there is one, must be in "If their inclinations are different, the distances will vary." Yes, it seems intuitively accurate, but is it? Right, that's definitely the point to analyze, as I see it. 1 hour ago, sevenperforce said: <snip example> That's interesting; it should be straightfoward enough to set up equations for the coordinates of each of those satellites and solve for the distance between them to see whether the time-related terms drop out. Note to self to try that out... Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 22 Author Share Posted February 22 44 minutes ago, sevenperforce said: their orbital planes certainly intersect Because they intersect, it's always possible to rotate your position around the planet such that one of the orbits becomes equatorial, from your perspective. It really doesn't matter if it's equatorial from the planet's perspective, the system moves the same way regardless of frame of reference. Consider the special case of both sats exactly meeting each other at the points of intersection (AN/DN). Easy to see that the distance increases and decreases as they go through their orbit, yes? The distance is the hypotenuse of a rectangular triangle with the base on the projection of orbit B on the plane of orbit A, increasing and decreasing in a sinusoide function of the angular distance from the points of intersection. Now set one of them ahead, say, 90 degrees, with the other at a point of intersection. The distance between them is still the hypothenuse of a rectangular triangle, of which the standing side shows the same sinusoide increase and decrease, hence the hypotenuse (ie the distance) also increases and decreases as a sinusoide function. In other words: it continuously changes. We don't even need to calculate exactly by how much -even though we could- it is clear that it doesn't stay the same. Quote Link to post Share on other sites

K^2 2,746 Posted February 22 Share Posted February 22 8 hours ago, swjr-swis said: if it is at all possible No. It's definitely not possible for circular orbits, and I was thinking that maybe there's a way to cheat with elliptical orbits and some odd rotation for resulting tetrahedron, but it's actually impossible. Here is the proof. Masses shouldn't matter, so we can take them to be equal without loss of generality. There is precisely one arrangement for four satellites that are equidistant from each other, that of a regular tetrahedron. Now, we can't use all the properties of rigid bodies, but because we are requiring that satellites maintain this formation, we do know that total angular momentum of this tetrahedron will be same as if it was a rigid body. We also know that moment of inertia tensor of a regular tetrahedron is fully degenerate. That means that angular velocity vector of this tetrahedron is co-linear with the angular momentum. Finally, each satellite is in its own orbit, so angular momentum is conserved for each one. That means the sum of angular momenta, and consequently the angular momentum of tetrahedron, are conserved. What does it mean? It means that axis of rotation for the tetrahedron cannot change. That means trajectory of every satellite must be a circle around that axis. But we also know that each satellite must be traveling in a plane that contains center of mass of the planet. There is only one plane that contains circles that are perpendicular to given axis and contains a given point. That means that all satellites must travel in the same plane to satisfy our conditions. Since tetrahedron is not contained in a plane, this arrangement is impossible. QED. Keep in mind, above assumes an absolutely perfect, regular tetrahedron. It might be possible to have an arrangement where distances change, and might not all be quite the same, but still gives you a good coverage and is stable over time. This would probably involve one sat in equatorial circular orbit and three others in elliptical orbits with high inclination whose ascending nodes are all 120° apart and whose arguments of periapsis are all the same. These would form one of the faces of the tetrahedron, which will rotate 360° for every orbit. It's not pretty, but if your goal is global coverage, it might be good enough. Quote Link to post Share on other sites

sevenperforce 7,470 Posted February 22 Share Posted February 22 (edited) 10 minutes ago, swjr-swis said: Because they intersect, it's always possible to rotate your position around the planet such that one of the orbits becomes equatorial, from your perspective. It really doesn't matter if it's equatorial from the planet's perspective, the system moves the same way regardless of frame of reference. Yes, but that's beside the point. I'm sticking to "equal inclination but different longitude of ascending node" because that's easier to visualize Quote Consider the special case of both sats exactly meeting each other at the points of intersection (AN/DN). Easy to see that the distance increases and decreases as they go through their orbit, yes? The distance is the hypotenuse of a rectangular triangle with the base on the projection of orbit B on the plane of orbit A, increasing and decreasing in a sinusoide function of the angular distance from the points of intersection. We definitely don't want that specific special case. Quote Now set one of them ahead, say, 90 degrees, with the other at a point of intersection. The distance between them is still the hypothenuse of a rectangular triangle, of which the standing side shows the same sinusoide increase and decrease, hence the hypotenuse (ie the distance) also increases and decreases as a sinusoide function. In other words: it continuously changes. We don't even need to calculate exactly by how much -even though we could- it is clear that it doesn't stay the same. Set one of them ahead by the difference in the longitudes of their ascending nodes, then recalculate. You can do this for any two orbits, if you want to rotate one of them to equatorial. Place A at the intersection of the two orbital planes, and place B far enough back in its orbital path that its straight-line distance to the intersection is equal to the straight-line distance that B will be from the intersection when A reaches it. The straight-line distance between the two satellites do NOT lie in either orbital plane. That's the trick. And that, I think, is where the original assumption by @HebaruSan breaks. 5 minutes ago, K^2 said: Keep in mind, above assumes an absolutely perfect, regular tetrahedron. It might be possible to have an arrangement where distances change, and might not all be quite the same, but still gives you a good coverage and is stable over time. This would probably involve one sat in equatorial circular orbit and three others in elliptical orbits with high inclination whose ascending nodes are all 120° apart and whose arguments of periapsis are all the same. These would form one of the faces of the tetrahedron, which will rotate 360° for every orbit. It's not pretty, but if your goal is global coverage, it might be good enough. My guess is that with the right ellipticity, you can have a regular tetrahedron rotating around the planet's axis where each satellite is on the line from the origin to one of the tetrahedral vertices. Edited February 22 by sevenperforce Quote Link to post Share on other sites

K^2 2,746 Posted February 22 Share Posted February 22 7 minutes ago, sevenperforce said: My guess is that with the right ellipticity, you can have a regular tetrahedron rotating around the planet's axis where each satellite is on the line from the origin to one of the tetrahedral vertices. Well, no, I've proven it to be impossible for regular tetrahedron. Might be possible with something kind of close, and probably not with anything that keeps distances consistent for the entire period. Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 22 Author Share Posted February 22 (edited) 19 minutes ago, sevenperforce said: Place A at the intersection of the two orbital planes, and place B far enough back in its orbital path that its straight-line distance to the intersection is equal to the distance that B will be from the intersection when A reaches it. That straight line will still always be the hypotenuse of a rectangular triangle, which changes continuously as a sinusoide function. Regardless of the angle of intersection, the standing side of that triangle will change from zero to a non-zero value and back again, and thus the hypotenuse/distance will also change. 19 minutes ago, sevenperforce said: The straight-line distance between the two satellites do NOT lie in either orbital plane. it does though, at four moments (or two if they coincide - and somehow phase through each other): when they pass through the AN/DN of their respective orbits. Which is the moment of least distance between them. Edited February 22 by swjr-swis Quote Link to post Share on other sites

swjr-swis 3,125 Posted February 22 Author Share Posted February 22 26 minutes ago, K^2 said: angular momentum I have to admit, I'm not at ease with the properties of and rules governing momenta. Perhaps if I were this would be as clearly obvious as you make it sound. In particular, and to my shame, I have quite honestly no idea what the below sentence means, at all: 33 minutes ago, K^2 said: We also know that moment of inertia tensor of a regular tetrahedron is fully degenerate. In any case, thank you for offering another way of answering this question. Quote Link to post Share on other sites

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