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Stable tetrahedronal (central) configuration of four orbital relays - possible?


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26 minutes ago, K^2 said:

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.

I agree, you have proven it impossible for a regular tetrahedron. I am musing that it might be possible for a regular tetrahedral projection.

25 minutes ago, swjr-swis said:

That straight line will still always be the hypotenuse of a rectangular triangle, which changes continuously as a sinusoide function.

Wait, why is it the hypotenuse of a right (I assume that's what you mean by rectangular) triangle? For two equal-inclination orbits with only a slight difference in the longitude of the ascending node, it is the shortest leg of a very very acute triangle.

25 minutes ago, swjr-swis said:
38 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.

If the two satellites pass through their respective ascending nodes at the exact same time, then the straight line between them is definitely not in either of their orbital planes.

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2 minutes ago, sevenperforce said:

I am musing that it might be possible for a regular tetrahedral projection.

If so, we are deviating from what I asked. I know there are solutions that do not keep the regular tetrahedron configuration intact at all times.

 

4 minutes ago, sevenperforce said:

Wait, why is it the hypotenuse of a right (I assume that's what you mean by rectangular) triangle? For two equal-inclination orbits with only a slight difference in the longitude of the ascending node, it is the shortest leg of a very very acute triangle.

This is why I keep telling you to mentally reorientate the system (or your observational point of the system) to make one of the orbits equatorial. It's quite easy to see then that regardless your choice of inclinations or angular positions, the distance is a right (sorry, English is only my third language) triangle with the base in the equatorial orbit, the standing face an orthogonal projection onto the plane of that orbit, and the distance the hypotenuse of that triangle.

 

10 minutes ago, sevenperforce said:

If the two satellites pass through their respective ascending nodes at the exact same time, then the straight line between them is definitely not in either of their orbital planes.

I should've said the points of intersection of their orbits (which when reorientating the system is the AN/DN of one of the satellites).

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33 minutes ago, swjr-swis said:

If so, we are deviating from what I asked. I know there are solutions that do not keep the regular tetrahedron configuration intact at all times.

Yes, I am accepting @K^2's proof for the regular tetrahedron. But I'm not so sure that @HebaruSan's non-equidistance conjecture is accurate.

Quote

This is why I keep telling you to mentally reorientate the system (or your observational point of the system) to make one of the orbits equatorial. It's quite easy to see then that regardless your choice of inclinations or angular positions, the distance is a right (sorry, English is only my third language) triangle with the base in the equatorial orbit, the standing face an orthogonal projection onto the plane of that orbit, and the distance the hypotenuse of that triangle.

If English is your third language you are a LOT smarter than me...but what do you mean by standing face? I'm sorry; I must just not be following.

Quote

I should've said the points of intersection of their orbits (which when reorientating the system is the AN/DN of one of the satellites).

Okay, so let's look at the reoriented system.

Let's take satellite A as being the equatorially-orbiting satellite and B as being the non-equatorial one. Both are circular and have an orbital radius R from the center of Earth.

Define time t0 as the time when A is positioned at the intersection of the two orbits. Suppose that at time t0, B is some angular distance Φ away from the point of intersection. The shortest distance between the two satellites is the line segment 0 which lies in the orbital plane of B. Clearly, the length of  ℓ0 = 2*sine(Φ/2)*R.

Now, define time t1 as the time when B is positioned at the intersection of the two orbits. At this point, A has traveled some angular distance along the equator from the point of intersection. Since the two orbits have the same period, the time for B to traverse an angular distance of Φ is the same amount of time for A to traverse an angular distance of Φ, so the distance between A and B is the line segment ℓ1 and trivially the length of  ℓ1 = 2*sine(Φ/2)*R. So the two satellites are the same distance apart at t0 as they are at t1.

The question, then: is there a Φ such that ℓ = f(Φ) remains constant?

Edited by sevenperforce
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2 hours ago, sevenperforce said:

I agree, you have proven it impossible for a regular tetrahedron. I am musing that it might be possible for a regular tetrahedral projection.

Hm. Interesting thought. Given that masses of satellites can be adjusted, I can take any projection with constant dimensions and give it the same moment of inertia as regular tetrahedron, so clearly, you can't have it work if the distances never change, but I take it that's not your intention? You are suggesting satellites that are in elliptical orbits but stay over projections of tetrahedron vertices at all times?

I can't think of any way to prove that it's impossible off the top of my head, especially, if we don't require the center of tetrahedron to stay parked over the center of mass. But also, not coming up with any strategy of how to achieve anything like this configuration. Definitely worth further investigation.

I'm starting to think about numerical methods one might deploy to try and find solutions like this. With strict symmetries in place, it's relatively easy to go over possibilities by hand, but once we allow altitudes to change out of sync, feasibility space is quite large, so I think this would require a computer to solve.

2 hours ago, swjr-swis said:

In particular, and to my shame, I have quite honestly no idea what the below sentence means, at all

Moment of inertia tensor is said to be degenerate if all of its eigen values are the same. That also means that you can use any three orthogonal axes as principal axes for the body and they will all have the same moment of inertia about them. For any given moment of inertia tensor, the angular momentum and angular velocity are related thusly, L = Iω. Because in general, moment of inertia about different axes can be different, L and ω don't have to point in the same direction. However, if all axes have the same moment of inertia, then multiplying by I is the same as multiplying by a scalar - it only adjusts the magnitude, but not the direction, and so L and ω are always in the same direction.

This is crucial, because while L has to stay in the same direction due to conservation, ω in general doesn't. In the extreme case, this happens. This is called axis tumbling. If we were able to get our tetrahedron to tumble this way, there would be a lot more room to try and come up with motion that matches orbital motion of satellites for each vertex.

However, all platonic solids exhibit high degree of symmetry, and from perspective of angular momentum, can be replaced by a sphere without any loss of generality. Once a platonic solid, or anything with the same symmetries, is set to spin about an axis, it will spin around that specific axis without tumbling. So the argument I laid out would also apply to trying to arrange satellites in a cube or regular octahedron, dodecahedron, or icosahedron. There are also some other semi-regular shapes that are symmetric enough to have the same restriction, but that's getting needlessly specific.

Edited by K^2
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1 hour ago, K^2 said:

You are suggesting satellites that are in elliptical orbits but stay over projections of tetrahedron vertices at all times?

I can't think of any way to prove that it's impossible off the top of my head, especially, if we don't require the center of tetrahedron to stay parked over the center of mass. But also, not coming up with any strategy of how to achieve anything like this configuration. Definitely worth further investigation.

 

Perhaps relevant:

 

See also:

 

 

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Further search of Draim Tetrahedral / Tetrahedron lead me to these :

US Patent from 1989 :

Spoiler

US4854527-drawings-page-2.png
US4854527-drawings-page-3.png
US4854527-drawings-page-4.png
US4854527-drawings-page-5.png

Someone have made it in KSP :

Old thread of people calculating the orbital parameters on this very forum for various in-game celestial bodies :

It does rely on eccentric, inclined orbits, so I guess the "regular tetrahedron" part is definitely not possible, also I think it's impossible to do without having the distances between the satellites to be fixed in any way, shape or form due to the nature of eccentric orbits.

Edited by YNM
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4 hours ago, swjr-swis said:

Perhaps relevant

Hm, that actually made me think, "Why isn't it possible to just keep symmetry as each satellite passes near the center?" And I think this might actually be a consequence of a theorem with the second funniest name in all of the mathematics. Hairy Ball Theorem. Topology, you win again!

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5 hours ago, K^2 said:

Topology, you win again !

Amazing just how many problems can be solved with thinking of the problem in the form of that particular field...

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10 hours ago, K^2 said:

Hm. Interesting thought. Given that masses of satellites can be adjusted, I can take any projection with constant dimensions and give it the same moment of inertia as regular tetrahedron, so clearly, you can't have it work if the distances never change, but I take it that's not your intention? You are suggesting satellites that are in elliptical orbits but stay over projections of tetrahedron vertices at all times?

I can't think of any way to prove that it's impossible off the top of my head, especially, if we don't require the center of tetrahedron to stay parked over the center of mass. But also, not coming up with any strategy of how to achieve anything like this configuration. Definitely worth further investigation.

As discussed upthread (I believe), this is trivially approximable with a series of very eccentric Molniya-style orbits. Take a spherical non-rotating Earth and define two congruent tetrahedrons centered on Earth, one much larger than the other, and then rotate the smaller tetrahedron 180 degrees such that each of its vertices are opposite the center of earth from a corresponding vertex of the larger tetrahedron. Set each orbit with the perigee at the lesser vertex and the apogee at the corresponding greater vertex. Most of the time, the four satellites very nearly form a projected tetrahedron. But it is not a perfect tetrahedron and I believe the whole thing falls apart at perigee.

The reason I'm interested in pushing this a little farther is that while the OP specifically asked for the perfect regular tetrahedral solution, there's a slightly more relaxed solution possible. The real need expressed by the OP is to maintain perfect coverage over the surface of a sphere with only four observation points. A two-point solution leaves an unobserved ring (and leaves the satellites unable to communicate with each other) while a three-point solution leaves the poles unobserved. A four-point solution, on the other hand, works perfectly: each satellite can communicate with every other satellite and they have perfectly-overlapping symmetric coverage of the surface.

Spoiler

Esx-E83-ZWMAEnvb2.png

Let's suppose we imagine statites rather than satellites, so that the positions can be maintained. It can be noted that the symmetric coverage is independent of the altitude of the statites, assuming they are above a certain minimum altitude. If we imagined statites that could alter their altitude, they wouldn't even have to alter their altitude in any regular coordination. Their altitudes could be completely chaotic and the underlying coverage remains completely fixed.

10 hours ago, K^2 said:

I'm starting to think about numerical methods one might deploy to try and find solutions like this. With strict symmetries in place, it's relatively easy to go over possibilities by hand, but once we allow altitudes to change out of sync, feasibility space is quite large, so I think this would require a computer to solve.

We still may be able to find a solution by solving it piecewise. Let's take the limited set of solutions where the four satellites are in two pairs, with each pair sharing an orbital period. Each pair defines a line segment which follows some regular quasi-orbit around the planet. The problem then becomes whether two such line segments can maintain symmetry with each other such that their endpoints remain fixed on projections from the vertices of a free regular tetrahedron. The tetrahedron need not be rotating around a single axis at all; in fact, its rotation/tumble doesn't even need to be periodic.

One could start by considering the prior example of two circularly-orbiting satellites with the same inclination but different AN longitudes, positioned such that they both cross their ANs simultaneously. This defines a line segment which remains parallel to the equatorial plane at all times, rotates around 360 degrees during each quasi-orbit, and goes up and down in latitude. You can then change the length and rotation of the line segment by changing the orbital parameters of the two satellites. As long as you do not change their orbital period, the line segment's quasi-orbit remains regular and periodic.

I believe that the two line segments would need to also have the same periodicity which may further restrict the solution space.

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21 hours ago, sevenperforce said:

Define time t0 as the time when A is positioned at the intersection of the two orbits. Suppose that at time t0, B is some angular distance Φ away from the point of intersection. The shortest distance between the two satellites is the line segment 0 which lies in the orbital plane of B. Clearly, the length of  ℓ0 = 2*sine(Φ/2)*R.

[ . . . ]

The question, then: is there a Φ such that ℓ = f(Φ) remains constant?

The algebra for this case is not too bad.  The algebra didn't inspire any intuition for me, but maybe somebody else is inspired. (Extending it to figure the relative velocity of the satellites was almost inspiring to me.)

If the relative inclination of the orbits is I, and the angular coordinate of A along its orbit is υ, then in the coordinate system with the orbit of A in the x/y plane, then
position of A :  R (cos(υ) sin(υ), 0)
position of B:  R cos(υ+Φ(1, 0, 0)  + R sin(υ+Φ) (0, cos(I), sin(I))

and after some algebra the distance between them is 
    |A − B| =  R sqrt[ 2 − 2 cos(υ) cos(υ+Φ− 2 cos(I) sin(υ) sin(υ+Φ)]
after more algebra
    |A − B| =  R sqrt[ 2 − 2 cos(Φ) + (1 − cos(I)) (2 cos(Φ) sin²(υ) + sin(Φ) sin(2υ)) ]

The distance varies, as the angle υ increases with time, unless cos(I) = 1

On 2/22/2021 at 12:47 AM, sevenperforce said:

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.

This case is easier for my intuition. 
When the satellites cross the equator, they are  2R sin15°  apart. 
When they simultaneously reach their highest latitude of 45°, they are closer,  2R cos45° sin15°   apart.

In this case I can imagine the inclination varying from 45° up to 90°,  and at 90° inclination it is easy to see how they meet over the pole.

Edited by OHara
removed unnecessary spoiler
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45 minutes ago, OHara said:

This case is easier for my intuition. 
When the satellites cross the equator, they are  2R sin15°  apart. 
When they simultaneously reach their highest latitude of 45°, they are closer,  2R cos45° sin15°   apart.

In this case I can imagine the inclination varying from 45° up to 90°,  and at 90° inclination it is easy to see how they meet over the pole.

Fantastic work. That's exactly what I was missing.

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So I think I have a partial solution.

I hopped into KSP and played around with a couple of satellites for a bit to get an idea of how the line segment connecting two satellites (let's call them A and B) with equal period and inclination rotates in 3-space. I drew the lines in Paint for simplicity. Take a look:

Spoiler

tumble.png

As you can see, the line segment (let's call it AB) "tumbles" around the planet, returning to the exact same position with each circuit. Its midpoint essentially follows an orbit equidistant between the two other orbits but at a slightly lower inclination and it rotates 360 degrees with every orbit, lengthening as it approaches its AN/DN and contracting as it approaches its highest and lowest latitudes.

Trivially, we can place an equivalent pair of satellites (let's call them C and D) in an orbit such that their corresponding line segment CD is always at a right angle to AB. The two line segments are equal in length at two points in the orbit; otherwise one is always smaller while the other is always larger.  The four satellites ABCD thus continuously form a tumbling irregular tetrahedron with an insphere centered on the planet, which oscillates between two different tetrahedral isometries that I believe are a mirrored sphenoid and digonal disphenoid.

While this solution is not a rotating regular tetrahedron as requested by the OP, it is a tetrahedron which rotates and oscillates non-chaotically around the planet and can provide 100% coverage of the surface.

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2 hours ago, sevenperforce said:

As you can see, the line segment (let's call it AB) "tumbles" around the planet, returning to the exact same position with each circuit. Its midpoint essentially follows an orbit equidistant between the two other orbits but at a slightly lower inclination and it rotates 360 degrees with every orbit, lengthening as it approaches its AN/DN and contracting as it approaches its highest and lowest latitudes.

I have a feeling that what you're actually doing there is much more akin to LISA/eLISA.

LISA_motion.gif

If you put a satellite on the opposite of the fixed triangle I guess you can provide coverage for a whole planet if you put them far out enough, but I suppose it's not the most efficient way.

Edited by YNM
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22 hours ago, YNM said:

I have a feeling that what you're actually doing there is much more akin to LISA/eLISA.

Spoiler

LISA_motion.gif

 

If you put a satellite on the opposite of the fixed triangle I guess you can provide coverage for a whole planet if you put them far out enough, but I suppose it's not the most efficient way.

Ooooooh, fascinating!

If the center of the triangle maintains an essentially constant orbit, you could absolutely make that triangle bigger, pull it closer to the planet, and then have a corresponding satellite on the opposite end, creating a triangular pyramid. Not a regular tetrahedron but a stretched one. Total, nearly equidistant coverage.

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15 hours ago, sevenperforce said:

If the center of the triangle maintains an essentially constant orbit, you could absolutely make that triangle bigger, pull it closer to the planet, and then have a corresponding satellite on the opposite end, creating a triangular pyramid. Not a regular tetrahedron but a stretched one. Total, nearly equidistant coverage.

I think this is where you'd run into problems I suppose. There's a limit to how close can you pull them closer in and still have a valid 'orbit' for the center of the triangle. One of the Draim Constellation, specifically the one that consists of 3 satellites and provides coverage for one hemisphere, does make for perfect triangle; however the center of the triangle doesn't make a valid orbit that'd be possible to replicate with one that follows newtonian gravitation.

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4 hours ago, YNM said:

I think this is where you'd run into problems I suppose. There's a limit to how close can you pull them closer in and still have a valid 'orbit' for the center of the triangle. One of the Draim Constellation, specifically the one that consists of 3 satellites and provides coverage for one hemisphere, does make for perfect triangle; however the center of the triangle doesn't make a valid orbit that'd be possible to replicate with one that follows newtonian gravitation.

I spent entirely too long playing around with the Alt-12 menu, haha.

Here's a 1986 article by Draim but unfortunately I can only see the first page: https://arc.aiaa.org/doi/pdf/10.2514/3.20244

Of note:

"Throughout the 1970s, it was generally thought that the minimum number of satellites required to give continuous global coverage (including the polar regions) was five. A credible proof of this minimum number...is valid only for circular-orbit constellations.

"In 1984, the author presented the concept of a four-satellite elliptical-orbit constellation giving continuous global coverage. It was based on a three-satellite elliptic-orbit continuous hemispherical coverage constellation of 78 h or longer, with a fourth circuit orbit satellite having a period one-half that of the other three.

"The new four-satellite constellation described in this paper uses common period orbits and can maintain continuous global coverage at approximately one-half the altitude of the earlier four-satellite constellation. [T]he orbital periods, inclinations, and eccentricities of the satellites in the new constellation are identical."

And here's a 2012 article which DOES have full text: https://www.researchgate.net/publication/286375838_Common-period_four-satellite_continuous_global_coverage_constellations_revisited

"Draim’s analysis of the optimal, four-satellite, continuous single coverage satellite constellation is based on a tetrahedron formed by four planes–each plane contains three of the four satellites in the constellation (see Fig. 1). The basic requirement is for the planes of the tetrahedron to always encompass the Earth, without ever intersecting it, as the tetrahedron changes shape or warps during the constellation repeat groundtrack period."

Bingo. There it is.

Here's the design:

draim-tetrahedron.png

This is essentially the same thing I proposed upthread -- two pairs of satellites, each describing a line segment which orbits the planet while remaining at right angles to the corresponding line segment. He describes it: "Both pairs of satellites drift to the West. One pair of satellites moves in a clockwise fashion. The other pair moves in a counterclockwise fashion."

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8 minutes ago, sevenperforce said:

Draim’s analysis of the optimal, four-satellite, continuous single coverage satellite constellation is based on a tetrahedron formed by four planes–each plane contains three of the four satellites in the constellation (see Fig. 1). The basic requirement is for the planes of the tetrahedron to always encompass the Earth, without ever intersecting it, as the tetrahedron changes shape or warps during the constellation repeat groundtrack period.

It is not in any measure, shape or form regular, however, whether by projection from a hypothetical smaller tetrahedron or by pure geometry alone. I've posted on the whole thing much earlier, incl. a video of someone making it in KSP and a much older thread of the orbital parameters needed for the tetrahedron.

Edited by YNM
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9 minutes ago, YNM said:

It is not in any measure, shape or form regular, however, whether by projection from a hypothetical smaller tetrahedron or by pure geometry alone. 

Right, it's not a regular tetrahedron. It does, however, describe what I believe is a single mirrored sphenoid that grows and shrinks periodically.

This appears to have been beaten to death now, haha.

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2 hours ago, sevenperforce said:

Right, it's not a regular tetrahedron. It does, however, describe what I believe is a single mirrored sphenoid that grows and shrinks periodically.

Yeah... The angles between the lines are in no way regular (what would be needed for a solid body) nor the relative angles to each satellite from the center body origin (for the less strict rule of projection). I guess this is actually also true for the case of something like LISA/eLISA orbit plus an opposite satellite (since the distance to the three satellite from the opposite end will shrink and grow, and so does the relative angles from the opposite satellite). Although the Draim Tetrahedron is more effective (overall-angle wise from a point on the surface) than what OP had known for longer, it's still not a regular tetrahedron the way it's wanted in the first post nor the later supposition of a projection of a smaller tetrahedron. The logical mathematical proof itself hasn't quite been hashed out anywhere here of why it's not possible, though. We've only said that it's not possible from simulation or from limited examples.

Edited by YNM
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bHhEDUq.jpgFour satellites in Draim's irregular tetrahedron are in a 90-kB KSP save-file here (link)

This seems to be the closest one can have to a stable tetrahedron, changing shape somewhat as it tumbles, but staying reasonably non-flat. 

I sized the orbits to be Kerbo-synchronous --- which is low enough that from some places on Kerbin all satellites lie below the horizon sometimes --- because that case is interesting to watch.

{Edit: the minimum dimension for the four satellites' orbits that keeps all of Kerbin covered is  SMA = 4381634 ]

Edited by OHara
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Hmmm... Does it need to be orbiting a spherical planet with sane rotation rates, in hydrostatic equilibrium, with a sane escape velocity?

 

Admittedly the math for orbits around anything other than that gets a lot messier.

Edited by Pds314
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