Help with inclination of a Molniya-style orbit

[eatU4myT]

I'm planning to put up a satellite into a Molniya-style orbit.  I know that as Kerbin is a perfect sphere, the real life restriction on the inclination is removed.  What I'm trying to work out is what the realtionship is between inclination and the centre of area that the apoapsis hangs over.  I understand that as the inclination increases, so the maximum latitude that the ground track will reach increases, but is there a (realistically acheivable by me) way to model what average latitude it will hang over?

Thanks

[Zhetaan]

@eatU4myT:

You've got the right idea, but the wrong parameter.  Inclination and apoapsis don't have a mathematical relationship.  What you need to use is the argument of periapsis.  There are other ways to derive a more exact position from Keplerian elements, but I'm going to assume that you want a quick-and-dirty approximation, so ArgPe it is!

I'm also going to assume that you have the means to obtain the argument of periapsis, because you need the means to obtain the ground latitude and most things that give you one can also give you the other.

Anyway, the argument of periapsis is the angle measured from the ascending node to the periapsis in the plane of the orbit and along the direction of motion.  Since the ascending node is always at zero latitude, that gives a convenient point of reference.  Also, the periapsis and the apoapsis will have the same absolute latitude, so there's no need to add extra steps for conversion--if it's important, then just remember to change the sign.

Unless I am completely missing the mark (I'm making up maths as I go here), for a given inclination, the periapsis varies in its latitude according to the inclination multiplied by the sine of the argument of periapsis.

What this means is that, for example, if the inclination is zero, then the latitude is zero--which we expect.  If the inclination is ninety but the argument of periapsis is zero, then it means that the periapsis is co-located with the ascending node, i.e., at the equator, so the latitude is also zero.  If the inclination is ten degrees and the argument of periapsis is ninety degrees, then the sine is equal to one and the latitude will equal the inclination, i.e. be at ten degrees.  If you have a Molniya orbit with an inclination if 63.4 degrees and an argument of periapsis of 270 degrees, then the latitude of periapsis will be -63.4 degrees, and that gives a latitude of apoapsis of +63.4 degrees.

 

Edited to Add:  And if I am completely missing the mark, then someone will be along shortly to correct my egregious errors.  Either way,  you'll have a good answer soon.

Edited September 5, 2018 by Zhetaan

[eatU4myT]

Thanks for taking time to respond, certainly useful to understand more about the maths behind the maximum latitude, however what I'm really looking for is a way to work out the average latitude it will dwell above in terms of time.

Say, for example, I have found an interesting site I want to land at and rove around and it is at latitude 50 degrees. What inclination or maximum latitude do I need to put the orbit in so that on average it will spend most of the time somewhere near over my 50 degree latitude site? Obviously it will spend some time at a slightly higher latitude, some time slightly lower, and then a very short amount of time down in the other hemisphere, but the hang time will be on average at 50 degrees. Does that make sense? I don't think I'm explaining it very well!

I'm probably being optimistic that I'll be able to get an easy solution to this really...

Thanks

[Sharpy]

Your inclination corresponds to the maximum latitude of the area you're lingering above - and so, your inclination should be such that it's the outermost (northernmost for northern hemisphere) edge of that area. The satellite will linger there the longest, but the latitudes south of it will be covered in a time that corresponds... I think according to cosine proportion of the latitude.

Then, "at epoch time" have your argument of periapsis, equal to the longitude of the point you want to hover above, Right ascension of ascending node that plus 90 degrees, and and true anomaly  another 90 (for total +180) degrees away.

And then make sure your orbital period is 4601772.5 seconds long, sharp (half of sidereal day). Earth Molinya orbits use a trick connected with oblateness of Earth, to give them a precession, so the period is half of synodic day,  and so they can linger over given area at the same time of day/night, but KSP doesn't model precession so you're stuck with sidereal and Kerbin lighting changing beneath over the course of the year. Periapsis to "something small", whatever; apoapsis - whatever it happens to be that grants that precise period.

The Molinya orbit monitors 2 areas of the globe, 180 degrees of longitude apart, so come next orbit your satellite will hang right above your designated area.

The typical constellation of Molinya is 3 satellites, differing in argument of periapsis (0, +120, -120 degrees apart) and mean anomaly same 0, +120, -120 degrees apart. Can't say exactly what true anomaly should be, but you could say the three are "the same" with epoch differing by 1/3 of sidereal Kerbin day. The orbits together should look kinda like a three-petal tulip flower.

 

Now, if you want "most time above", you have to define your observation cone. Otherwise you're getting a very unhelpful answer of "exaclty equal, and you're spending infinitesimally short time above it, but the longest of all other infinitesimal times". Define the cone.

In short, you want to reach the observation point with the southern-most edge of the cone when at periapsis (for northern hemisphere). Which shouldn't be difficult if you know the radius of the observation zone; just find the latitude of the center if the southern edge touches your observation zone. If instead of radius/diameter, you have the angle, it's a bit trickier as it will require the apoapsis altitude to find the radius, and instead of apoapsis altitude we're using orbital period. I could help you calculate the values if you need that and don't have the cone diameter; in general, semi-major axis ^3 is proportional to period ^2, and Ap+Pe (+ Kerbin diameter as they are altitudes, not distances from center) = 2x semi-major.

[eatU4myT]

2 hours ago, Sharpy said:

"corresponds... I think according to cosine proportion of the latitude"

 

"Define the cone"

This first bit is the sort of thing I was originally thinking of, as a means of finding the average latitude with time.

However, reading this second bit makes me realise that the whole issue is probably pretty minimal, as the cone in KSP is pretty much an entire hemisphere (unless I'm mistaken, I don't think signal strength decreases towards the edge of the satellite's field of vision, only with distance?)

Thanks

[Zhetaan]

This reply ran a bit long, so I decided to put my answers in spoiler tags.

21 hours ago, eatU4myT said:

Say, for example, I have found an interesting site I want to land at and rove around and it is at latitude 50 degrees. What inclination or maximum latitude do I need to put the orbit in so that on average it will spend most of the time somewhere near over my 50 degree latitude site?

Spoiler

'Average hang time' is determined by the eccentricity of the orbit and is another completely independent parameter.  Kepler's laws state that a line drawn from the central body (technically the focal body if the orbit is not circular) to the satellite orbiting it will sweep through equal areas in equal times.  Practically, this means that a satellite will sweep through large areas slowly and small areas quickly, and since the apoapsis is the farthest point on the orbit, it has the largest sweep area and thus the slowest speed.

Therefore, the latitude with the longest hang time is the one over which you place the apoapsis.  For a permanent installation, there really isn't much reason to give your orbit more inclination than it needs to reach a given latitude, so often the apoapsis is also placed such that the argument of periapsis is either ninety or two hundred seventy degrees.  (As a side note, I was wrong earlier:  my current working solution for maximum latitude is the arcsine of the product of the sines of both the inclination and the argument of periapsis, i.e.:  Lat_Max = arcsin (sin θ * sin ω), where θ is the inclination and ω is the argument of periapsis.)

If the radios work with a cone rather than a single straight line, you can get away with more or less inclination if you want to be miserly with the amount of time that you get, but that problem is one of just changing the 'effective' inclination to account for the extra coverage.  To do that, take the tangent of half the cone angle, multiply it by the satellite's altitude, divide by Kerbin's radius and take the arctangent of that:  the resulting angle is how many degrees you can add to the inclination and still get a downlink.  However, I cannot guarantee that you will be able to keep it for quite so long--rotational velocity of the surface can play games with you.

21 hours ago, eatU4myT said:

Obviously it will spend some time at a slightly higher latitude, some time slightly lower, and then a very short amount of time down in the other hemisphere, but the hang time will be on average at 50 degrees.

Spoiler

That is not so obvious.  For the most part, if your apoapsis is at the maximum of inclination, you don't need to overshoot--meaning that on the average in your example, it will spend some time approaching 50 degrees and some time retreating from 50 degrees, but it doesn't need to exceed 50 degrees because it's not just the location, but the direction that matters here.  For a more real-world example of that kind of principle, think of it as though you were rolling a ball over a small hill.  The ball slows as it crests the hill before it accelerates down the other side, so it ends up spending the most time at the top of the hill.  However, the fact that it spends the most time at the top on average does not mean that it must spend some time slightly below the top of the hill and then some time slightly above the top of the hill, if that makes sense.

21 hours ago, eatU4myT said:

I'm probably being optimistic that I'll be able to get an easy solution to this really...

Spoiler

There is a trivial solution:  if you can get either the inclination or the argument to equal ninety or two hundred seventy degrees, then the other parameter must equal the latitude of interest.

To use your example, if you want to hover over 50 degrees (I'll assume north latitude) then you need to set the apoapsis latitude to 50 degrees.  You can do this with an  inclination of 50 degrees and an argument of periapsis of 270 degrees, or with an inclination of 90 degrees and an argument of periapsis of 230 degrees, but you are correct that there are other solutions.  If you're flying at an inclination of 60 degrees, for example, then you need an argument of periapsis of 242.2 degrees.  That corresponds to, to coin a term, an 'argument of apoapsis' of 62.2 degrees.  The general idea is that the inclination must be between the target latitude and ninety degrees (or it can be retrograde by the same amount, though that's a bad idea) and the 'argument of apoapsis' varies inversely with the same constraint.

21 hours ago, eatU4myT said:

[W]hat I'm really looking for is a way to work out the average latitude it will dwell above in terms of time.

Spoiler

I'll assume that you know about synchronous orbits and that what you really want to know is how to set up the orbit so that the apoapsis hovers over your chosen spot rather than the correct latitude on some other part of the hemisphere.

For that, @Sharpy has the answer, though I'll caution you that KSP doesn't go out of its way to tell you what those parameters are.  Alternatively, you can work out the mean anomaly and use that, but to be honest, you can get good results by treating it the same as an orbital rendezvous:  get your inclination the way that you want it but leave the orbital period a little short, and when your orbit puts your apoapsis directly above your target, burn to the correct semi-major axis to get the right orbital period.  Remember not to lower your periapsis into the atmosphere.

 

Edited September 6, 2018 by Zhetaan