ADVANCED Orbital alignment assistance needed

[Tallinu]

Hello! I'm not sure if this is really the right place for this advanced topic, and if anyone has a better idea for where to put this I'll happily post there or get it moved if possible.

I'm working on a mission plan to place four relay satellites in a tetrahedral configuration which will provide 100% uninterrupted coverage of a planet's surface, using Kerbin as my starting point for now (since I still need to figure out how to calculate the minimum semi-major axis required to provide such coverage around other planets and moons). This requires four eccentric (0.28), inclined (33 deg) orbits with the same orbital period and, for Kerbin, a SMA of 4,350,000 km or greater. I'm using 5,506,299 km because that results in an easy to deal with period of 12 hours. I'll include the exact orbital parameters needed farther down, which I have shamelessly lifted from a very old (2013) thread which I will also link.

Here are a few screenshots to give you a quick idea of what it looks like.

Four-satellite tetrahedral configuration for 100% comms coverage on Imgur, or open the following:

Spoiler

View from one angle near the equatorial plane, showing where the orbits with apoapsis in the southern hemisphere cross each other.

View from another angle, rotated around the planet's axis by about 90 degrees, giving a clear view of where the orbits with apoapsis in the northern hemisphere cross. Tetrahedral shape of connections between satellites can be seen more clearly here.

View from above north pole, with orbit inclinations flattened to zero (equatorial). These are the initial orbits the satellites will be deployed into prior to the plane change maneuvers which finalize their orbits, allowing them to provide uninterrupted polar coverage as well as equatorial.

I've figured out how to accomplish *almost* every step of the process! Plotting the inclination changes to be performed at the correct time/location is the only thing eluding me. It's not a simple matter of doing something at the current ascending/descending node since they all have to start out in equatorial orbits, and I couldn't just target another vessel and use the relative AN/DNs unless I launch two additional satellites and carefully position them in 90 degree polar orbits which are perpendicular to each other and have their Longitude of Ascending Node carefully tweaked to be positioned exactly along the semi-minor axis of the pair of satellites which need to change inclinations relative to that plane. But launching six satellites instead of four defeats the entire purpose of this plan, and if I knew how to figure out where the AN/DN of these polar satellites needed to be, I think I would already have what I needed to figure out when the inclination changes should be performed without requiring some external reference.

All I really need is a way to find out the number of seconds since periapsis (or AP, whatever) which will put the satellite in line with its orbit's semi-minor axis (not major, the one at a right angle to it), the latus rectum, which is to say, when the satellite is at +/-90 degrees True Anomaly. (For an elliptical orbit like this, 90 degrees is not simply 1/4 of the orbital period, like it is for a circular orbit.) Kerbal Engineer can show me what my vessel's current True Anomaly is, but it doesn't provide a way of finding out what the UT will be at that point in the orbit or how much time remains before it, so that doesn't help me plot a maneuver node in advance so that it occurs exactly at 90 degrees, which is necessary to allow the burn to be started at the right time.

For reference, here's where I'm at in terms of the mission plan.

The four satellites each need to reach their target apoapsis in sequence, 1/4 of their final orbital period apart. The period I'm using is 12 hours, so each satellite need to decouple and burn to raise its apoapsis at 3 hour intervals. Additionally, since the Argument of Periapsis for each satellite is rotated 90 degrees (drawing lines from their APs to Kerbin will form an X or + when viewed from far above a pole), the launch vehicle needs to complete 1/4 of an orbit more or less every 3 hours (180 minutes). A circular orbit with a period of 48 minutes will result in making 3 complete orbits in 144 minutes and another 3/4 of an orbit in the remaining 36 minutes.

So, first the launch vehicle gets into an equatorial orbit at an altitude of 305,314 meters, which gives an orbital period around Kerbin of 48 minutes. Just before every 3 hour mark the launch vehicle decouples a satellite, I take control of it and create a very precise maneuver node, first using MechJeb's maneuver planner to "raise apoapsis" after an arbitrary fixed time of several minutes and then tweaking the values using PreciseNode until the accuracy is down to the 10 meter range (just because I can), and then I set the UT of the node to a specific time. For the first satellite I just pick some convenient round number far enough in the future to allow time to get ready. I record this value, and then for the other satellites I simply add on the correct number of additional seconds (3 * 60 * 60 = 10800) to ensure that each node is executed with the correct timing.

The time taken for these satellites to reach apoapsis from the deployment orbit is around 3 hours 40 minutes, so the first will need to perform a burn shortly after the second is deployed, and so on. The AP burn brings the PE up to approximately 3364.5 km, and I adjust as needed to ensure that the orbital period is as close to the intended 12 hours as possible given the limits of precision in the values displayed by Engineer and MechJeb. Getting MechJeb's Synodic Period (with another of these deployed satellites as a target) to display a value of over 1000 years, for example, is not difficult and offers more than sufficient precision.

After all four satellites have performed these burns, the final step will be to change their inclinations so that they provide constant polar coverage as well as equatorial. And that's what I'm not sure how to do accurately, as described.

Here is a save file for a 1.6.1 sandbox game that shows the desired final configuration (it's really nifty to watch in the tracking station as the comm network lines maintain this rolling, stretching irregular tetrahedron, always completely surrounding the planet as the satellites sweep through their elliptical orbits). The state I need to get there from is simply the same thing but with all satellites having an inclination of zero. You'll find two quicksaves in there, one showing each configuration.

https://drive.google.com/open?id=1Y71svhDTCd6uBXDtN0mRpcN0l1dGn3z2

Additionally, I'd be grateful if anyone could provide suggestions regarding how to determine what the minimum SMA is for establishing 100% coverage with such a constellation around other planets or moons. Even having a simplified calculation that overestimates somewhat would be perfectly useful.

Finally, here are the exact orbital parameters and the thread I referenced for them (the real source, Draim's article, is sadly behind a paywall):

Spoiler

I originally went looking for information about the minimum number of satellites that could be used to provide 100% coverage of the surface at all times. I suspected it might be as low as four and that it would use inclined orbits. It apparently also requires eccentric orbits. As it turns out, I found a post on these very forums where someone had set up an example of this as a challenge. With the comm network, this satellite configuration has become useful even in stock KSP.

Here are the details of the orbital parameters required for this kind of constellation, in the terms used in KSP's save files. (LPE technically would be short for Longitude of Periapsis, which is the sum of LAN (Longitude of Ascending Node) and Argument of Periapsis. But I believe KSP actually stores Argument of Periapsis there despite the misleading field name.)

All satellites have the same eccentricity and inclination values:

ECC = 0.28
INC = 33.0

EPH (Epoch) and SMA (Semi-Major Axis) must also be the same (or nearly so) for all satellites in order to be properly positioned and maintain that positioning over time. That makes it tricky to cheat four satellites into proper orbital positions to see an example of how it should look, since you can't do them all at the same instant. So I did it by placing them in orbit, then editing a quicksave file and loading it.

Just using EPH = 0.0 for all four is fine. And for Kerbin, the minimum SMA for 100% surface coverage appears to be pretty close to the 4350000 km used in the thread.

The other three important parameters are different for each satellite (and MNA is Mean Anomaly, an angle expressed here in radians):

Satellite 1

LPE = 270.0
LAN = 0.0
MNA = 0.0

Satellite 2

LPE = 90.0
LAN = 90.0
MNA = -1.57078

Satellite 3

LPE = 270.0
LAN = 180.0
MNA = 3.14159

Satellite 4

LPE = 90.0
LAN = 270.0
MNA = 1.57078

 

 

Edited February 13, 2019 by Tallinu
Corrected erroneous use of "semi-minor axis"

[Xd the great]

Lazy way: 3 satellites at high eccentricity (high Ap, low pe), 1 polar low-altitude satellite.

Pretty much 4 satellites at any "widely separated" configuration will do.

If not, CUBESATS.

Or if you want, launch to a low orbit. Then use the manouvers to do 2 burns, 1 to reach desired ap at desired place, 1 to circularize at AP

[Curveball Anders]

Erh, Resonant Orbits?

https://meyerweb.com/eric/ksp/resonant-orbits/

 

[Tallinu]

2 hours ago, Xd the great said:

Lazy way: 3 satellites at high eccentricity (high Ap, low pe), 1 polar low-altitude satellite.

Pretty much 4 satellites at any "widely separated" configuration will do.

If not, CUBESATS.

Or if you want, launch to a low orbit. Then use the manouvers to do 2 burns, 1 to reach desired ap at desired place, 1 to circularize at AP

 

2 hours ago, Curveball Anders said:

Erh, Resonant Orbits?

https://meyerweb.com/eric/ksp/resonant-orbits/

 

Credit for trying to help, but yes, I am familiar with orbital maneuvers and standard relay satellite deployment patterns. And I am, in fact, making use of a particular variation of a resonant orbit (or maybe it's something more properly known by some other name), a kind of result which that calculator isn't designed to provide, to get all four satellites launched on their different trajectories at precise times.

I'm afraid I put "advanced" in the subject line for a reason. I'm working on this because I want to do something way cooler and more effective with fewer satellites. The kind of help I need is not "how do I shot relay satellite?" It's "how do I math Kepler to solve for X given Y" or "how do I avoid having to math Kepler and get this accomplished to the desired precision by some other means." I don't want to be just eyeballing the position of my maneuver node based on where the lines appear to cross, I want to do it right, or find a way to make the game or the mods do it right for me (that doesn't take launching extra equipment).

I've been trying to figure out how to get from point A to point B mathematically speaking, where A is the assortment of orbital parameters I have and B is the one piece of information I think I need to solve for. But, at least so far, I'm just too unfamiliar with the relevant equations to know which I need and in what order and form, compounded by my programming-trained brain always finding it harder to work with variables that don't have names which are intelligible words that remind me what the variable is and does just by reading it. About the only one I even know how to say is theta, because it was used so often back in trig class, lol!

Edited February 12, 2019 by Tallinu

[Zhetaan]

I can help you with some of the prediction.  If you want to know when you'll be at 90 degrees true anomaly, then you need Kepler's laws; that's non-negotiable.

I'll assume that you're starting from a true anomaly of zero; it's one element that is easy enough to see both in KSP's interface and in Kerbal Engineer.

True anomaly, θ, is given by the equation:

(1 - ε) tan² (θ / 2) = (1 + ε) tan² (E / 2)

Where E is the eccentric anomaly and ε = .28, the eccentricity.  In this case, we know that we want a ninety-degree (or π/2) true anomaly, so this equation is a matter of solving for E.

(1 - .28) tan² (π / 4) = (1 + .28) tan² (E / 2)  --- This is easy enough; tan of π/4 is 1
.72 = 1.28 tan² (E / 2)
.5625 = tan² (E / 2)
.75 = tan (E / 2)
.643501 = (E / 2)
1.287 rad = E

This is about 73.7 degrees.

Once you have E, you need to solve Kepler's Equation for mean anomaly, M:

M = E - ε sin E  --- This is the easy way to do it; if you were solving for position instead of time, you'd have to solve the transcendental

M = 1.287 - .28 sin 1.287
M = 1.287 - .28 (.96)
M = 1.287 - .2688
M = 1.0182 rad

This is about 58.3 degrees.

Mean anomaly is the mean motion times time, and 2π radians divided by the period is the mean motion, n; you mentioned earlier that the period is twelve hours, so I'll use that.  Twelve hours is 43,200 seconds, so:

2π / 43,200 = 1.45444 x 10^-4 radians per second.

1.0182 rad = 1.45444 x 10^-4 s^-1 * t
7000.63 seconds = t

Your time after periapsis passage at true anomaly of 90 degrees is 7000.63 seconds, or 1 hour, 56 minutes, and 40.63 seconds.

The next question is whether you are certain that you want the true anomaly to be ninety degrees.  θ of ninety degrees actually gives the location of the latus rectum, which is certainly useful for some purposes, but if you really want the location of the semi-minor axis, then you want E to be ninety degrees, not θ.  Of course, that's even easier to solve; you can dispense with the true anomaly completely and can begin with Kepler's Equation.

 

Edit:  Oh, why not--let's do it:

M = E - ε sin E --- sin of π/2 is 1
M = (π / 2) - .28
M = 1.2908 radians, approximately, or 74.0 degrees

1.2908 = 1.45444 x 10^-4 s^-1 * t
8874.87 seconds = t

This gives a time after periapsis passage of 8874.87 seconds, or 2 hours, 27 minutes, and 54.86 seconds.

Edited February 12, 2019 by Zhetaan

[Tallinu]

12 hours ago, Zhetaan said:

I can help you with some of the prediction.  If you want to know when you'll be at 90 degrees true anomaly, then you need Kepler's laws; that's non-negotiable.

I'll assume that you're starting from a true anomaly of zero; it's one element that is easy enough to see both in KSP's interface and in Kerbal Engineer.

I figured that would be the case. Just to make sure I'm not missing anything important, my understanding is that TA = 0 at periapsis, is that right?

12 hours ago, Zhetaan said:

True anomaly, θ, is given by the equation:

(1 - ε) tan² (θ / 2) = (1 + ε) tan² (E / 2)

Where E is the eccentric anomaly and ε = .28, the eccentricity.  In this case, we know that we want a ninety-degree (or π/2) true anomaly, so this equation is a matter of solving for E.

(1 - .28) tan² (π / 4) = (1 + .28) tan² (E / 2)  --- This is easy enough; tan of π/4 is 1
.72 = 1.28 tan² (E / 2)
.5625 = tan² (E / 2)
.75 = tan (E / 2)
.643501 = (E / 2)
1.287 rad = E

This is about 73.7 degrees.

Once you have E, you need to solve Kepler's Equation for mean anomaly, M:

M = E - ε sin E  --- This is the easy way to do it; if you were solving for position instead of time, you'd have to solve the transcendental

I need to figure out how to get characters like ε typed in without copying and pasting from something else. And how to remember what to call them (other than "curly E" or "curly W" LOL).

I've seen these equations everywhere, but always in the context of solving for position, and just couldn't sort out what I needed to do to get the answer I wanted. Looks like I was overthinking things, and except for the first one they were already in the required form with no "numerical analysis methods" (whatever that means) needed!

12 hours ago, Zhetaan said:

M = 1.287 - .28 sin 1.287
M = 1.287 - .28 (.96)
M = 1.287 - .2688
M = 1.0182 rad

This is about 58.3 degrees.

Mean anomaly is the mean motion times time, and 2π radians divided by the period is the mean motion, n; you mentioned earlier that the period is twelve hours, so I'll use that.  Twelve hours is 43,200 seconds, so:

2π / 43,200 = 1.45444 x 10^-4 radians per second.

1.0182 rad = 1.45444 x 10^-4 s^-1 * t
7000.63 seconds = t

Your time after periapsis passage at true anomaly of 90 degrees is 7000.63 seconds, or 1 hour, 56 minutes, and 40.63 seconds.

Awesome! Thank you so much for the walkthrough of this process. And the answer -- when I try it myself I can use that to check my work.

This should be able to get me the time before periapsis as well, right? Symmetry and all. If my satellite is at apoapsis (just after getting the 12h period matched), just subtracting 7000.63 from the UT at which periapsis will be reached should give me the position on the closer side, without having to wait and go past PE.

And I should be able to use this around other bodies as well, right? It doesn't appear to directly require any constants regarding gravity and such, just the parameters of the orbit and the period. Taken together, those effectively incorporate that kind of information already, I believe?

So what I need to do now is go back to make sure all of these are solved in the form I need, more or less the way you did but keeping the input of True Anomaly as a variable, then chain it all together with another variable for period, and I should have an equation where I plug in those two values and it spits out a time-since-periapsis as a result. I could then convert that into a bit of code to do the job quickly. Maybe even figure out how to write a KSP mod to do it. (If I decide to get really ambitious I could probably work up a pull request for MechJeb to let it automatically place maneuver nodes at specific points in the orbit...)

12 hours ago, Zhetaan said:

The next question is whether you are certain that you want the true anomaly to be ninety degrees.  θ of ninety degrees actually gives the location of the latus rectum, which is certainly useful for some purposes, but if you really want the location of the semi-minor axis, then you want E to be ninety degrees, not θ.  Of course, that's even easier to solve; you can dispense with the true anomaly completely and can begin with Kepler's Equation.

As it turns out, I was very much mistaken when I claimed that the position I wanted was along the semi-minor axis. In hindsight, that should have been pretty obvious, since the minor axis goes through the center of the ellipse, NOT either of its foci. I'm glad I mentioned the true anomaly, since that's the actual position I need in order to change the inclination correctly. The apoapsis and periapsis need to be the points most distant from the equatorial plane, with the orbit not tilted one way or the other around the major axis, to match the example configuration as closely as possible.

 

Edited February 13, 2019 by Tallinu

[Zhetaan]

11 hours ago, Tallinu said:

I figured that would be the case. Just to make sure I'm not missing anything important, my understanding is that TA = 0 at periapsis, is that right?

That's correct; all of the anomalies are measured under the condition that periapsis equals zero (and apoapsis is π radians (or 180 degrees); the anomalies all converge there, too).  And to answer the unspoken question:  if it's a circular orbit, the anomalies don't exist--though the hypothetical orbit described by the mean anomaly describes the actual orbit, so you can use that and simplify calculation immensely.  In Kepler's time, orbits were assumed to be circular, so any deviation from circularity was anomalous--hence the term anomaly.

11 hours ago, Tallinu said:

I need to figure out how to get characters like ε typed in without copying and pasting from something else. And how to remember what to call them (other than "curly E" or "curly W" LOL).

I use Character Map if I need it.  However, there's no special requirement to use difficult-to-type characters in these equations; I could have solved for the true anomaly as x just as easily as θ.  It's simply convention to use Greek letters because of either tradition or a need to use more letters than are available in the Latin alphabet.  That being said, the conventions are fairly common, so here's a cheat sheet:

• ε - Epsilon, used for eccentricity; e is often used instead, but since e is also a mathematical quantity and E is used for eccentric anomaly, it can get confusing, so any time that I calculate eccentric anomaly, I make certain that I use E for the anomaly and ε for the eccentricity.  You'll also find ε used for planetaery axial tilt; it's not an issue in KSP but you'll want to know that if you ever try your calculations on real planets.
• θ - Theta, often used instead of x when the unknown variable is an angle; it's also used for true anomaly since t is almost universally used for time and T is used in astronomy for orbital period.  You'll encounter it a lot in place of x in trigonometric functions, and it is used for angular measure in polar coordinates (which use the (r, θ) form instead of the (x, y) of Cartesian coordinates).
• ν - Nu, an alternative way to show true anomaly.  It looks like Latin v in Arial but its actual form is a bit more elaborate (it has serifs and everything!); needless to say, its similarity to Latin v means that it can be mistaken for velocity, so I prefer to use θ.  Incidentally, Latin f is yet another way to show true anomaly, but f can also mean frequency, and it is used to denote focal length for a lens--this isn't a concern in KSP, but in the kind of astronomy that requires telescopes, it can be confusing, so again, I prefer θ.
• π - Pi, and I'm sure you know this one.
• ω - Omega, used in physics for angular velocity; the difference between this and θ is that θ describes a static angle, whereas ω describes rotational speed (usually in terms of radians per second).  It's used in orbital mechanics as the argument of periapsis (the angle between the periapsis and the ascending node).  Note that capital omega (Ω) is used for the longitude of the ascending node, so there are two different omega parameters in orbital mechanics.  It turns out that there aren't enough letters even with Greek.
• μ - Mu (pronounced myoo) (or mew, as though it were a kitten), which is used in many different contexts.  It is the symbol for the SI prefix micro-, which in astronomy is likely only to arise if, for some reason, you need to describe micro-light-years.  Astronomy has to scale colossally, though, generally doesn't bother with prefices, and instead uses scientific notation almost exclusively.  You are more likely to encounter μ as the standard gravitational parameter, which is a measure of the gravitational influence of a large body and is actually a very important value to know when calculating orbital period and the like.  This parameter first arose as the proportionality constant in the relationship T² ∝ a³, which you may recognise as Kepler's Third Law (as an aside, the symbol ∝ means is proportional to).  When taking the proportionality constant into account in order to make an equation, it becomes T² = ka³, where T is the orbital period, a is the semi-major axis, and k is constant for all orbits about a given body (but it is different for every body).  The constant k relates to μ by this relationship: k is equal to 4π² / μ.  If that seems esoteric then know that the π factor appears because you're chasing around an ellipse; the angular velocity relates to the period because the period is the amount of time it takes to travel 2π radians (one full revolution) around the ellipse.  Since the period is squared in the Third Law, the period in terms of angular velocity needs must also be squared: hence, 4π².  A more intuitive way to describe the Third Law is this:  (T / 2π)² = a³ / μ, in which it is more obvious that the entire period factor refers to time per revolution.

11 hours ago, Tallinu said:

This should be able to get me the time before periapsis as well, right? Symmetry and all. If my satellite is at apoapsis (just after getting the 12h period matched), just subtracting 7000.63 from the UT at which periapsis will be reached should give me the position on the closer side, without having to wait and go past PE.

Correct.  Though if you are starting from apoapsis, then it may be simpler to subtract 7000.63 seconds from six hours to get time after apoapsis rather than time before periapsis.

11 hours ago, Tallinu said:

And I should be able to use this around other bodies as well, right? It doesn't appear to directly require any constants regarding gravity and such, just the parameters of the orbit and the period. Taken together, those effectively incorporate that kind of information already, I believe?

This will work around any body, and yes, the orbital period encodes the gravitational relationship between the primary and the orbit in its calculation; that's why an orbit of a given semi-major axis around a given body has the orbital period it does.  Furthermore, it's why we need further calculation to locate things on that orbit:  any orbit with that SMA about that body will have that orbital period, regardless of its shape, and the need to find the differences that relate to eccentricity is the reason that the concept of orbital anomaly was invented in the first place.  However, you won't need the orbital period until you are ready to include the mean orbital motion to get time from the mean anomaly; all ellipses of the same eccentricity are similar, so the angles will be the same in any case.  Bear in mind that you must know the eccentricity to get that far, but orbital eccentricity is trivial to calculate, especially if you have Kerbal Engineer to do it for you.  However, if not (or if you'd like to do it yourself):

ε = (r_a - r_p) / (r_a + r_p),

where r_a and r_p are the radii of the apoapsis and periapsis to the centre of mass (meaning the planet's centre if you're orbiting a planet), respectively.

Bear in mind that this requires the actual apsides, but KSP displays the altitudes above the surface.  You'll need to add the planetary radius to get the correct value, but that radius is available both on the wiki and in the KSP Knowledge Base.  For Kerbin, it's 600,000 metres.

 

I think that was all of your questions.  In any case, good luck.

Edited February 13, 2019 by Zhetaan

[Tallinu]

https://steamcommunity.com/sharedfiles/filedetails/?id=1654472000

I just got done setting up for the first of these maneuvers, and it looks great! Not exactly crossing the other satellite's orbit line but I doubt that's possible without performing the inclination change for the first two satellites before the other two (the ones opposite them) are even launched...

https://steamcommunity.com/sharedfiles/filedetails/?id=1654490026

Aaaand done! Periods correct down to a thousandth of a second thanks to thrust limiting down to 0.5% and slow-mo physics warp. Pretty much perfect.

1 hour ago, Zhetaan said:

so any deviation from circularity was anomalous--hence the term anomaly.

Yeah, reading about it did fill in a bit more historical detail for me, which was interesting. The whole epicycles thing, heh... And to think that today we have people "all around the globe" (lol) who supposedly believe the Earth is flat again.

1 hour ago, Zhetaan said:

Correct.  Though if you are starting from apoapsis, then it may be simpler to subtract 7000.63 seconds from six hours to get time after apoapsis rather than time before periapsis.

Right. I figured that one out.

1 hour ago, Zhetaan said:

This will work around any body, and yes, the orbital period encodes the gravitational relationship between the primary and the orbit in its calculation; that's why an orbit of a given semi-major axis around a given body has the orbital period it does.  Furthermore, it's why we need further calculation to locate things on that orbit:  any orbit with that SMA about that body will have that orbital period, regardless of its shape, and the need to find the differences that relate to eccentricity is the reason that the concept of orbital anomaly was invented in the first place.  However, you won't need the orbital period until you are ready to include the mean orbital motion to get time from the mean anomaly; all ellipses of the same eccentricity are similar, so the angles will be the same in any case.  Bear in mind that you must know the eccentricity to get that far, but orbital eccentricity is trivial to calculate, especially if you have Kerbal Engineer to do it for you.  However, if not (or if you'd like to do it yourself):

e = (r_a - r_p) / (r_a + r_p),

where r_a and r_p are the radii of the apoapsis and periapsis to the centre of mass (meaning the planet's centre if you're orbiting a planet), respectively.

Bear in mind that this requires the actual apsides, but KSP displays the altitudes above the surface.  You'll need to add the planetary radius to get the correct value, but that radius is available both on the wiki and in the KSP Knowledge Base.  For Kerbin, it's 600,000 metres.

Yeah, I had to figure out adding in and removing the planet's radius already. It's nice that they gave Kerbin such neat round numbers, heh.

The reason I was trying to figure these calculations out to begin with was that I couldn't find a way to do it (accurately) using Engineer or MechJeb, but hey, maybe I've actually learned something! I doubt I'll remember the details of the equations a week from now, and there's no way I could write any of them out from memory, but the relationships between the anomalies and what everything does and means is all pretty clear.

You've been a tremendous help with this! I don't suppose you have any thoughts on the additional issue I noted near the end of the OP, about figuring out what semi-major axis would be required around other planets? 

My best guess on that right now is to compare the radii of the planets and just scale up/down proportionally from the 4,350,000 that was used in the original details for Kerbin (in the thread I referenced), on the assumption that the geometry will end up the same. I'll still need to figure out what the period would be in the rescaled orbit and figure out if it can be done within the SOI of some of the smaller moons, or if the reduction in mass (and hence gravitation) causes even that (presumably) closest possible orbit to be outside the SOI of, say, Minmus the same way a synchronous orbit would. Although I have the advantage of not worrying about the planet or moon's rotation speed, only its radius, which is going to be smaller along with the gravity, so maybe it'll work out nicely...

 

Edited February 13, 2019 by Tallinu

[Zhetaan]

21 hours ago, Tallinu said:

You've been a tremendous help with this! I don't suppose you have any thoughts on the additional issue I noted near the end of the OP, about figuring out what semi-major axis would be required around other planets? 

My best guess on that right now is to compare the radii of the planets and just scale up/down proportionally from the 4,350,000 that was used in the original details for Kerbin (in the thread I referenced), on the assumption that the geometry will end up the same. I'll still need to figure out what the period would be in the rescaled orbit and figure out if it can be done within the SOI of some of the smaller moons, or if the reduction in mass (and hence gravitation) causes even that (presumably) closest possible orbit to be outside the SOI of, say, Minmus the same way a synchronous orbit would. Although I have the advantage of not worrying about the planet or moon's rotation speed, only its radius, which is going to be smaller along with the gravity, so maybe it'll work out nicely...

Without access to that particular paper, I am limited in what I can figure out.  The radius and the gravity tend to increase and decrease together, but there is no requirement that they do so--planetary density can vary quite a lot.  Saturn, for example, is less dense than water, because its volume belies the fact that there simply isn't that much mass to it.  In other words, Saturn is our fluffiest planet.  However, satellite coverage has more to do with the visible surface, which is of course related to radius, but it appears that the parameters for this tetrahedron are very flexible.  Nevertheless, I note that the table of parameters in the original challenge post seems to preserve the angular relationship from Draim's work, which is why the satellites in KSP have the semi-major axis that they do.

For a satellite orbiting Kerbin with a semi-major axis of 4,350,000 metres and an eccentricity of .28, the (true) apoapsis and periapsis are as follows:

Ap = 5,568,000 metres
Pe = 3,132,000 metres

These were derived from the relation Ap = (1 + ε) * SMA and Pe = (1 - ε) * SMA.

With the planetary radius of 600,000 metres, the angle between the satellite and a point on the circumference of the greatest-size cross section that faces the satellite (in other words, the farthest possible visible point) is given by:

Sight angle Ap:  arctan 600,000 / 5,568,000 = 6.15°
Sight angle Pe:  arctan 600,000 / 3,132,000 = 10.84°

Using a similar derivation for satellites orbiting Earth:

Earth's standard gravitational parameter is about 3.986004419×10¹⁴ m³/s².
Earth's average planetary radius is 6,371,000 metres.
The required minimum orbital period for Earth of 27 hours is equal to 97,200 seconds.

A 27-hour period gives a semi-major axis of = 45,691,651.57 metres, so:

Ap = 58,485,314.01 metres
Pe = 32,897,989.13 metres

Sight angle Ap:  arctan 6,371,000 / 58,485,314.01 = 6.22°
Sight angle Pe:  arctan 6,371,000 / 32,897,989.13 = 10.96°

I think we have our relationship.  So long as your sight angle is less than 6.22 degrees at apoapsis and 10.96 degrees at periapsis (smaller angles mean greater distance, which means better coverage), it can work.  Since we're using the same eccentricity for everything, we can also find the semi-major axis directly:

Hypothetical sight angle:  arctan 6,371,000 / 45,691,651.57 = 7.938°

To figure the minimum semi-major axis for a given body of known radius:

SMA = r_body / tan 7.938°

To check for Kerbin:

600,000 / tan 7.938° = 600,000 / .1394375 = 4,303,002 metres.  This is lower than the given value for Kerbin because, I think, the original poster decided to be safe, but again, without having access to the paper, it is possible that there is a practical reason for the given figure.  If you prefer to have that margin, then the hypothetical angle to use is actually 7.853°.

This is in line with your supposition that you can scale the semi-major axis by the radius times a constant; I simply took it a step farther to determine both that 1 / tan 7.853° is the value of that constant, and why it appears.

To check these figures, we can use the Mun (radius = 200,000 metres):

200,000 / tan 7.938° = 1,434,334.13 metres

Alternatively, with the safer angle:

200,000 / tan 7.853° = 1,450,058.58 metres

See whether you can get a good network there using that parameter, and if so, you'll have your solution.

 

Edit:

Alternatively, you can use this set of precalculated figures that I just found, of course, only after doing all of that:

https://forum.kerbalspaceprogram.com/index.php?/topic/155188-orbital-parameters-for-a-tetrahedral-satellite-constellation/

 

Edit edit:

Draim's original paper is behind a paywall, but he also got a patent for it, and because of the wonderful scope of U.S. patent law, the patent itself is available:  Tetrahedral Multi-Satellite Continuous-Coverage Constellation

 

Edit edit edit:

And here's an interesting animation of satellite coverage using this model:

 

 

Edited February 14, 2019 by Zhetaan

[Tallinu]

Wow, you've really gone above and beyond here!

19 hours ago, Zhetaan said:

Without access to that particular paper, I am limited in what I can figure out.  The radius and the gravity tend to increase and decrease together, but there is no requirement that they do so--planetary density can vary quite a lot.  Saturn, for example, is less dense than water, because its volume belies the fact that there simply isn't that much mass to it.  In other words, Saturn is our fluffiest planet.  However, satellite coverage has more to do with the visible surface, which is of course related to radius, but it appears that the parameters for this tetrahedron are very flexible.  Nevertheless, I note that the table of parameters in the original challenge post seems to preserve the angular relationship from Draim's work, which is why the satellites in KSP have the semi-major axis that they do.

For a satellite orbiting Kerbin with a semi-major axis of 4,350,000 metres, the (true) apoapsis and periapsis are as follows:

Ap = 5,568,000 metres
Pe = 3,132,000 metres

These were derived from the relation Ap = (1 + ε) * SMA and Pe = (1 - ε) * SMA.

With the planetary radius of 600,000 metres, the angle between the satellite and a point on the circumference of the greatest-size cross section that faces the satellite (in other words, the farthest possible visible point) is given by:

Sight angle Ap:  arctan 600,000 / 5,568,000 = 6.15°
Sight angle Pe:  arctan 600,000 / 3,132,000 = 10.84°

Using a similar derivation for satellites orbiting Earth:

Earth's standard gravitational parameter is about 3.986004419×10¹⁴ m³/s².
Earth's average planetary radius is 6,371,000 metres.
The required minimum orbital period for Earth of 27 hours is equal to 97,200 seconds.

A 27-hour period gives a semi-major axis of = 45,691,651.57 metres, so:

Ap = 58,485,314.01 metres
Pe = 32,897,989.13 metres

Sight angle Ap:  arctan 6,371,000 / 58,485,314.01 = 6.22°
Sight angle Pe:  arctan 6,371,000 / 32,897,989.13 = 10.96°

I think we have our relationship.  So long as your sight angle is less than 6.22 degrees at periapsis and 10.96 degrees at apoapsis (smaller angles mean greater distance, which means better coverage), it can work.

Did you get your Ap and Pe references swapped in the above line? Or is there some reason to use the Ap sight angle for Pe and vice versa?

19 hours ago, Zhetaan said:

Since we're using the same eccentricity for everything, we can also find the semi-major axis directly:

Hypothetical sight angle:  arctan 6,371,000 / 45,691,651.57 = 7.938°

To figure the minimum semi-major axis for a given body of known radius:

SMA = r_body / tan 7.938°

To check for Kerbin:

600,000 / tan 7.938° = 600,000 / .1394375 = 4,303,002 metres.  This is lower than the given value for Kerbin because, I think, the original poster decided to be safe, but again, without having access to the paper, it is possible that there is a practical reason for the given figure.  If you prefer to have that margin, then the hypothetical angle to use is actually 7.853°.

And considering that I actually used a SMA greater than 4,350,000m around Kerbin anyway (5,506,299m), in order to get easy to work with periods (12h), and plan to do the same around other bodies, there should be an even larger safety margin.

19 hours ago, Zhetaan said:

This is in line with your supposition that you can scale the semi-major axis by the radius times a constant; I simply took it a step farther to determine both that 1 / tan 7.853° is the value of that constant, and why it appears.

To check these figures, we can use the Mun (radius = 200,000 metres):

200,000 / tan 7.938° = 1,434,334.13 metres

Alternatively, with the safer angle:

200,000 / tan 7.853° = 1,450,058.58 metres

See whether you can get a good network there using that parameter, and if so, you'll have your solution.

What I had in mind was,

TargetBodySMA = KerbinSMA * TargetBodyRadius / KerbinRadius

which I'm pretty sure gives the result I'm looking for, linearly larger or smaller in proportion to the change in radius. If that does the job the way I believe it should, then I don't have to mess around with math that I can't do in my head.

It's great to have alternate verification though!

19 hours ago, Zhetaan said:

Edit:

Alternatively, you can use this set of precalculated figures that I just found, of course, only after doing all of that:

https://forum.kerbalspaceprogram.com/index.php?/topic/155188-orbital-parameters-for-a-tetrahedral-satellite-constellation/

 

Edit edit:

Draim's original paper is behind a paywall, but he also got a patent for it, and because of the wonderful scope of U.S. patent law, the patent itself is available:  Tetrahedral Multi-Satellite Continuous-Coverage Constellation

 

Edit edit edit:

And here's an interesting animation of satellite coverage using this model:

 

Dangit! Google never dug up Zyx's post for me.  And you had even posted there, too, at the time. "Indeed, you're looking at the prospect of four launches, at least for Kerbin." Well, I'm getting all the satellites there with one launch, but making that plan work is what I needed help with in the first place. XD Mainly I'm doing it that way because I wanted something that would work for moons and other planets, and getting a single vessel carrying multiple satellites to intercept a planet will be far easier than trying to manage multiple vessels, and means that the satellites themselves only need to carry enough delta-V for moving into the correct orbits once they detach from the carrier.

And it looks like he already did the work of figuring out the SMA for all the planets and moons! I'll probably check a few with that ratio method to see if they match, and if they do, I can just use his values as minimums and try to find nice periods above them. It seems the only one that he had an issue with was Duna because of Ike, and for some reason he went lower instead of higher. I'll have to have a look at Duna's SOI and see if the constellation could be set up in higher orbits instead, to avoid creating gaps in its surface coverage.

And that video and patent document are also interesting. I suspected there was probably a range of values that would work for things like eccentricity and inclination, and one of those illustrations confirms that, though it seems like 0.28 and 33 are pretty close to optimal there.

Again, thank you for all your help, you've done far more than I asked!

Edit:

I just had a thought that might save me a lot of trickier calculations if it's correct.

Assume that the eccentricity of the orbit remains the same, and only the period is altered (causing the SMA and both the periapsis and apoapsis heights to increase). For example, going from a 12 hour period to an 18 hour period. Am I correct in thinking that the time taken for a satellite to travel from periapsis to latus rectum (TA=+-90) scales linearly with the period? As in, 18/12=1.5 so t=7000*1.5?

In my notes from working out the right equation to do what you showed me, I have:

t=(2tan^-1(3/4)-(0.28*sin(2tan^-1(3/4))))*43200/(2pi)

It looks to me like the location of that 43200 value for period makes it a simple linear scaling factor for the result.

 

 

Edited February 14, 2019 by Tallinu

[Zhetaan]

1 hour ago, Tallinu said:

Did you get your Ap and Pe references swapped in the above line? Or is there some reason to use the Ap sight angle for Pe and vice versa?

I did swap them; that was my error.  The apoapsis sight angle will always be smaller because it is farther away.  I fixed it, in any case.

1 hour ago, Tallinu said:

What I had in mind was,

TargetBodySMA = KerbinSMA * TargetBodyRadius / KerbinRadius

which I'm pretty sure gives the result I'm looking for, linearly larger or smaller in proportion to the change in radius. If that does the job the way I believe it should, then I don't have to mess around with math that I can't do in my head.

That will work well enough for most situations.  It will give issues when you need to deal with either massive bodies orbiting your targets closely (because their sphere of influence will interfere with your desired orbit) or targets with low mass orbiting closely to their primaries (because the target's sphere of influence is too small to give you the correct orbit within the constraints).  I know that there is no solution for Laythe that fits the parameters (Laythe is fairly large for a moon but not compared to Jool at that orbital distance), but Duna should be possible provided that you orbit outside of Ike's influence.  The target value from the other thread is a semi-major axis of at least 7.25 times the planetary radius, (which is the value of 1 / tan 7.853°) so that's yet another verification method.  I saw nothing constraining the maximum value of the semi-major axis except insofar as the constellation needs to remain in orbit.

1 hour ago, Tallinu said:

Dangit! Google never dug up Zyx's post for me.  And you had even posted there, too, at the time. "Indeed, you're looking at the prospect of four launches, at least for Kerbin." Well, I'm getting all the satellites there with one launch, but making that plan work is what I needed help with in the first place. XD Mainly I'm doing it that way because I wanted something that would work for moons and other planets, and getting a single vessel carrying multiple satellites to intercept a planet will be far easier than trying to manage multiple vessels, and means that the satellites themselves only need to carry enough delta-V for moving into the correct orbits once they detach from the carrier.

Oh, I don't doubt that one launch would work better for anything interplanetary, at least in transfer costs.  That being said, you may still want to consider the flotilla alternative, or possibly a hybrid where you have one carrier to take the satellites into interplanetary space but then separate the satellites and set them to their own intercepts before the target encounter.  That may not be necessary but it's probably worth investigating; thirty-three degrees inclination is expensive no matter how you look at it.

1 hour ago, Tallinu said:

And it looks like he already did the work of figuring out the SMA for all the planets and moons! I'll probably check a few with that ratio method to see if they match, and if they do, I can just use his values as minimums and try to find nice periods above them. It seems the only one that he had an issue with was Duna because of Ike, and for some reason he went lower instead of higher. I'll have to have a look at Duna's SOI and see if the constellation could be set up in higher orbits instead, to avoid creating gaps in its surface coverage.

As I mentioned above, Laythe's sphere of influence is too small and Laythe itself too large for the tetrahedron to work with 33° inclination and .28 eccentricity.  Its sphere of influence is 7.45 times its radius, which fits one parameter, but that requires orbits to have eccentricity of no more than .027, which is too circular to be of help (the minimum for a Draim tetrahedron appears to be more than .1).  I have seen references to some other solutions that work with five satellites, but I am not familiar enough with the details to tell you whether they would work in Laythe's sphere.

Duna has a radius of 320,000 metres, so that would require a constellation semi-major axis of 2,320,093.7 metres.  Eccentricity of .28 gives an apoapsis for that orbit of 2,969,719.9 metres.  Ike's periapsis is 3,104,000 metres, and with its sphere of influence of 1,049,599 metres, that gives a potential closest approach of Ike's influence at 2,054,041 metres.  More likely, that conflict would occur farther out (33° inclination gives a lot of extra room in 3-space), but it would still occur.  You're right about the coverage gaps:  the table-given SMA of 2,080,000 metres reflects the three-dimensional geometry (it's too large for anything coplanar), but even so, 2,080,000 / 320,000 = 6.5, or alternatively a sight angle of 8.746°, which is too close to Duna.

On the other hand, if you go higher, then Ike's apoapsis is 3,296,000 metres, so it gives a maximal influence conflict at 4,345,599 metres.  Let that be the periapsis and that gives the satellites a semi-major axis of 6,035,554.2 metres.  You can get closer (again because of three dimensions) but that value guarantees no conflicts with Ike.  It's nearly nineteen times the ratio of the radius to the SMA, or a sight angle of only 3.035°, but I see no reason why that wouldn't work.  Satellites on that orbit would have an apoapsis of 7,725,509.4 metres, which is well within Duna's sphere of influence (nearly 48 million metres).  But there are still coverage gaps:  Ike gets in the way.  That's probably why the author chose the lower orbit.

On the gripping hand, the solution to that is simple:  put a constellation around Ike.  By definition, a body with one hundred percent coverage is transparent to the network--meaning that if you can see every point on a planet, then there's no reason why you can't see through to what's beyond that planet; it's only a question of whether you're looking.  So long as your Ike satellites have transmitters that can see Duna, then you'll get full coverage without needing to send anything extra beyond the more powerful antennas--which you need to send anyway if you want that constellation to talk to Kerbin.  If you put your Kerbin-capable relays around Duna and have Ike talk to them, then you still get full coverage without needing to do anything extra; the required minimum transmission distance to the relays is greater than the distance to Duna's surface.

3 hours ago, Tallinu said:

I just had a thought that might save me a lot of trickier calculations if it's correct.

Assume that the eccentricity of the orbit remains the same, and only the period is altered (causing the SMA and both the periapsis and apoapsis heights to increase). For example, going from a 12 hour period to an 18 hour period. Am I correct in thinking that the time taken for a satellite to travel from periapsis to latus rectum (TA=+-90) scales linearly with the period? As in, 18/12=1.5 so t=7000*1.5?

In my notes from working out the right equation to do what you showed me, I have:

t=(2tan^-1(3/4)-(0.28*sin(2tan^-1(3/4))))*43200/(2pi)

It looks to me like the location of that 43200 value for period makes it a simple linear scaling factor for the result.

Yes;  assuming that you hold eccentricity constant and travel from a specific true anomaly θ_a to a specific true anomaly θ_b on both orbits, the change in mean anomaly will be the same and the travel time will scale linearly with the period: t = MT / 2π.  It's not enough that the craft travel through the same angle; the caveat is that the trip must begin and end at the same true anomaly on both orbits.  Since you're going from 0° to 90° each time, it works.  However, understand that this is a special case since both endpoints are identifiable as other orbital parameters (Pe and AN/DN).

[Tallinu]

 

1 hour ago, Zhetaan said:

Oh, I don't doubt that one launch would work better for anything interplanetary, at least in transfer costs.  That being said, you may still want to consider the flotilla alternative, or possibly a hybrid where you have one carrier to take the satellites into interplanetary space but then separate the satellites and set them to their own intercepts before the target encounter.  That may not be necessary but it's probably worth investigating; thirty-three degrees inclination is expensive no matter how you look at it.

Actually, the biggest issue I'm addressing by having all four satellites be deployed by a single carrier vehicle (second to interplanetary encounter/capture) is to get the correct relative timing between the four satellites. Each must reach apoapsis at a different time, staggered by 1/4 of their period and in proper sequence. Deploying them from a lower circular orbit that makes 1/4 less than a whole number of full orbits in that amount of time allows each satellite to arrive at the target apoapsis from its deployment burn with the correct timing, raise its periapsis and tweak its period to be as accurate as possible, and from then on remain synchronized with the others, even as they all later perform their inclination changes -- the orbits are fairly high even at that point, so the delta-V cost of doing inclination last was acceptable.

A small lightweight relay satellite can easily have 2000 to 3000 delta-V with just a few oscar tanks and a spark engine, though relays meant to reach Kerbin from the outer planets would by nature require larger, heavier antennas (or arrays of them) and therefore more fuel and thrust -- although a better way to go about it, especially around Jool, would likely be to have a pair of large, powerful relays as the main link back toward Kerbin, while small satellites with RA-2 antennas like the ones I've been experimenting with provide local coverage of each moon, bouncing signals around to reach the larger relays as needed.

1 hour ago, Zhetaan said:

That will work well enough for most situations.  It will give issues when you need to deal with either massive bodies orbiting your targets closely (because their sphere of influence will interfere with your desired orbit) or targets with low mass orbiting closely to their primaries (because the target's sphere of influence is too small to give you the correct orbit within the constraints).  I know that there is no solution for Laythe that fits the parameters (Laythe is fairly large for a moon but not compared to Jool at that orbital distance), but Duna should be possible provided that you orbit outside of Ike's influence.  The target value from the other thread is a semi-major axis of at least 7.25 times the planetary radius, (which is the value of 1 / tan 7.853°) so that's yet another verification method.  I saw nothing constraining the maximum value of the semi-major axis except insofar as the constellation needs to remain in orbit.

[...]

As I mentioned above, Laythe's sphere of influence is too small and Laythe itself too large for the tetrahedron to work with 33° inclination and .28 eccentricity.  Its sphere of influence is 7.45 times its radius, which fits one parameter, but that requires orbits to have eccentricity of no more than .027, which is too circular to be of help (the minimum for a Draim tetrahedron appears to be more than .1).  I have seen references to some other solutions that work with five satellites, but I am not familiar enough with the details to tell you whether they would work in Laythe's sphere.

Really? Let's see, the SMA given in his thread is 3,625,000m and according to the wiki, Laythe's SOI is 3,723,645.8m and its radius 500,000m. So that's 5/6 of Kerbin's radius. Scaling 4,350,000 by that factor does indeed give 3,625,000. Despite the fact that Laythe's SOI is vastly smaller than Kerbin's, that result is still lower, but the orbit isn't circular, so...

Okay, I think I finally found what I need to do to figure out the Ap and Pe (to center of mass) from eccentricity and semi-major axis. Google was all like "Here's how you calculate the things you already know!" Eventually found https://jtauber.github.io/orbits/017.html and managed to piece it together from various pages there...

Focal distance c = a * e
3625000*0.28=1015000
It looks like Pe = a - c/2
3625000 - 1015000 / 2 = 3625000 - 507500 = 3117500
And then I think Ap = 2a-Pe
2 * 3625000 - 3117500 = 7250000 - 3117500 = 4132500

Dang, it's less than 10% outside the SOI! And for the most interesting and science-rich of Jool's moons, too... Well, that one will just have to make do with three equatorial satellites. Anything landing on the poles will probably be able to see the strong primary relays north or south of Jool. A set of three of those, equidistant in a polar orbit, should ensure that one or more are always visible.

1 hour ago, Zhetaan said:

Duna has a radius of 320,000 metres, so that would require a constellation semi-major axis of 2,320,093.7 metres.  Eccentricity of .28 gives an apoapsis for that orbit of 2,969,719.9 metres.  Ike's periapsis is 3,104,000 metres, and with its sphere of influence of 1,049,599 metres, that gives a potential closest approach of Ike's influence at 2,054,041 metres.  More likely, that conflict would occur farther out (33° inclination gives a lot of extra room in 3-space), but it would still occur.  You're right about the coverage gaps:  the table-given SMA of 2,080,000 metres reflects the three-dimensional geometry (it's too large for anything coplanar), but even so, 2,080,000 / 320,000 = 6.5, or alternatively a sight angle of 8.746°, which is too close to Duna.

On the other hand, if you go higher, then Ike's apoapsis is 3,296,000 metres, so it gives a maximal influence conflict at 4,345,599 metres.  Let that be the periapsis and that gives the satellites a semi-major axis of 6,035,554.2 metres.  You can get closer (again because of three dimensions) but that value guarantees no conflicts with Ike.  It's nearly nineteen times the ratio of the radius to the SMA, or a sight angle of only 3.035°, but I see no reason why that wouldn't work.  Satellites on that orbit would have an apoapsis of 7,725,509.4 metres, which is well within Duna's sphere of influence (nearly 48 million metres).  But there are still coverage gaps:  Ike gets in the way.  That's probably why the author chose the lower orbit.

On the gripping hand, the solution to that is simple:  put a constellation around Ike.

Indeed, I was just thinking that as I read the previous paragraph. (Also: I get that reference! High-five! )

1 hour ago, Zhetaan said:

Yes;  assuming that you hold eccentricity constant and travel from a specific true anomaly θ_a to a specific true anomaly θ_b on both orbits, the change in mean anomaly will be the same and the travel time will scale linearly with the period: t = MT / 2π.  It's not enough that the craft travel through the same angle; the caveat is that the trip must begin and end at the same true anomaly on both orbits.

Good, I had a feeling it was something like that.

1 hour ago, Zhetaan said:

However, understand that this is a special case since both endpoints are identifiable as other orbital parameters (Pe and AN/DN).

And now I'm confused. I'm not sure what the difference is, compared to what you said earlier in that paragraph. How does "being Pe" or "being AN/DN" change anything, compared to going from, for instance, 30 degrees to 45 degrees TA on two orbits with the same eccentricity but different periods?

 

[Zhetaan]

4 hours ago, Tallinu said:

And now I'm confused. I'm not sure what the difference is, compared to what you said earlier in that paragraph. How does "being Pe" or "being AN/DN" change anything, compared to going from, for instance, 30 degrees to 45 degrees TA on two orbits with the same eccentricity but different periods?

You have my apologies; I oversimplified.  You're correct in that the period of the orbit cares not a whit about the location of the nodes--though the comparison between two scaled orbits does care about the location of periapsis insofar as that defines zero true anomaly.  My intent in saying that was only to recall that for this specific problem, there is a special reason why the ascending node must occur at ±90° true anomaly:  the Draim constellation constrains not only the inclination and eccentricity, but the argument of periapsis as well.  This is the angle between the ascending node and the periapsis as measured on the orbital plane (as opposed to the longitude of ascending node and longitude of periapsis, which are measured on the equatorial reference plane).  Specifically, all of the orbits in the tetrahedron have an argument of periapsis of either 90° or 270°, which puts the periapsis at right angles to the line of nodes with respect to the primary as the vertex.  This is why, in this case, the line of nodes coincides with the latus rectum of the ellipse, but for any other argument of periapsis, this would not be so.

Since your original question was not how to find the latus rectum, but how to find the correct point at which to change inclination, I was pointing out that although you have a general solution to scale orbits of constant eccentricity by their orbital period regardless of their orientation and can also find the latus rectum of any orbit with relative ease, the equation will only find the correct burn point if you begin with the correct argument of periapsis:  90°, 270°, or undefined (which means starting from equatorial orbit).

Granted, I'm also aware that your entire purpose was to simulate the Draim constellation, so alternative arguments of periapsis were never in your plans.  Perhaps I was being over-thorough ... you may have noticed that I have that habit.

[Tallinu]

46 minutes ago, Zhetaan said:

My intent in saying that was only to recall that for this specific problem, there is a special reason why the ascending node must occur at ±90° true anomaly:  the Draim constellation constrains not only the inclination and eccentricity, but the argument of periapsis as well.

Right. I just didn't understand that when you said "this is a special case" you were referring back to this constellation setup again, rather than orbit segment time scaling in more general terms, which is where my mind was leaning (as in that last paragraph) as I'm learning how these things work.  I always understood intuitively why the TA of 90 was needed in this case, even when I was originally describing it wrong, lol! It needed to be symmetric, with the AN/DN equidistant from both Pe and Ap. Anywhere else would not put the nodes in the right place, the orbital plane would be tilted around the wrong axis, and sufficient error would prevent 100% coverage.

Edited February 15, 2019 by Tallinu