Calculating ideal polar orbits (stock)

[ajburges]

Discovered that the old thread of ideal scansat orbits got consumed by a forum change at some point so I'm working on a Google Sheet to help derive polar orbits with useful resonances with revolution period. I have calculators that can solve for altitude and resonance while calculating surveyed arc for given FoV, but the utility is limited beyond checking solutions.

A few questions come up:

• What is the proximity toggle for orbital survey waypoints? In stock I think the most useful resonance orbit would minimize the time to visit all possible waypoints for contracts.
• Is there a better way to find an orbit of particular approximate resonance than brute force? The presence of non-discrete modulo arithmetic is a barrier I don't know a way to reverse.

[OHara]

2 hours ago, ajburges said:

Is there a better way to find an orbit of particular approximate resonance than brute force?

If I understand what you mean, then yes: Kepler's third law.    a³ / T² = GM / (2π)²

GM is the 'gravitational parameter' on the wiki page for Kerbin.   a is the semi-major axis visible on map screen.  The period T is also visible in the 'maneuver' tab

Your orbital period T is, say, 2/9 times the rotational period of Kerbin, 21 549 seconds.  (The rotational period is not quite a 6-hour Kerbin day because the direction from Kerbin to to the Sun changes a bit during one day.)  The fraction 2/9 would spread 9 orbits over Kerbin's surface, finishing the pattern in 2 days.

If I understand, you can figure out the number N of orbits needed to cover the surface given the spacing you need.  Then among the orbital periods 1/N, 2/N, 3/N, etc. excluding those where the fraction reduces, you'll want the first one with a > 6000km + 70km to clear the atmosphere

2 hours ago, ajburges said:

What is the proximity toggle for orbital survey waypoints?

The folks who play career mode will know, but I'll guess 15km, because Contracts.cfg says "15000 // The minimum range to trigger waypoints on high altitude surveys"

[ajburges]

To clarify on finding an approximate resonance:

I would like a better way to find orbital periods that follow a ground track that advances by x degrees every planetary revolution and/or by y degrees when a*x (mod 360) < x. I already have Kepler's law incorporated into my solver, but I can come up with a good way to solve given x and/or y.

[Guest]

If this is for scansat purposes, don't forget you're limited by the equipment provided as well; the M700 only works between r/10 or 25km and 5r or 1.5Mm (presumably being as close to that upper limit as possible would be the best).

14 hours ago, ajburges said:

non-discrete modulo arithmetic

I'm sure that's the name of a mod?

[ajburges]

Didn't mention it in post but the title did specify stock. The useful instruments are the Kerb Net/Anomaly detector (FoV limited), experiments for survey contracts (are they lat/long limited or horizontal distance limited), and the resource scanner (practically not limited).

Rubber ducking it out, I suppose the best sock polar orbit is the one (per scientific region) with the lowest time to come within survey distance of all equatorial points. Unfortunately, this varies greatly depending on the frequency of your orbital harmonics.

Edited March 21, 2021 by ajburges

[OHara]

People use "modulo arithmetic", @FruitGoose, to mean arithmetic where you only care about the remainder.   

After n orbits of a satellite, the planet beneath will have rotated m full times, plus a little bit more.  That little bit more would be the remainder of nT ÷ mD, if the orbital period is T and the rotational period is D.  If you want the satellite to eventually cover every 5° of surface you want the remainder to be 5°/360°.   But that leaves a tricky problem to pick the orbital period T when you have so many possibilities for n and m.  

I put a ScanSat image at right because that is the easiest way to draw the orbital tracks across the ground.  The idea is to put the ground tracks so that painting ± a survey distance from the ground track eventually covers the whole planet (which ScanSat happens to draw for us)  and to set up the orbit so that the 'eventually' is as soon as possible.  People solve this problem using the ScanSat mod by adjusting the orbit until the equator-crossing markers are spaced as they want.

What I was assuming in my earlier post with 1/N, 2/N, etc. , was that I had might as well divide the spacing between the first and final ground track evenly among all orbits.  Then the scanning pattern will repeat exactly after N orbits and the planet will have rotated exactly M times.  NT = MD 

After k orbits the planet has made kM/N rotations, which is some whole number plus a fraction (kM mod N ) / N.   If M and N are mutually prime (i.e. the fraction N/M cannot be reduced like 12/9 -> 4/3) then as k runs 1,2,3,etc., (kM mod N) visits all positive integers less than N.   So we pick 1/N ≤ x/360° to space the eventual set of ground tracks by x degrees longitude, or closer.

The pattern finishes in M rotations of the planet, so we want the smallest possible M.  Atmosphere or mountains puts a lower limit on the orbital radius a = (T² GM / (2π)²)^(1/3), so atmosphere also puts a lower limit on the orbital period T.   Since T = (M/N) D there is a lower limit on M.     For Kerbin, for example,  M ≥ 0.086×N to stay out of the atmosphere.

So then to cover Kerbin completely with a scan that covers ±5° we need orbits spaced by 10° : 
N ≥ 360°/10° = 36
M ≥ 0.086× N =3, but 3/36=1/12 and 4/36=1/9, so 5/36 rotations is the first orbital period that will cover without overlapping ground tracks, taking 5 days to complete.

But sometimes slightly tighter spacing, slightly bigger N, allows a lower orbit for a net win:
N = 37 spaces ground paths a bit closer than 10°
M ≥ 0.086× N =4, and that is mutually prime to 37, so an orbit with period 4/37 rotations (2330 s) gets the job done in 4 days.

Those orbits cover the whole planet on just the north-going orbits, or while the planet is in daylight, which might be what you want. 

If you can scan on the dark side of the planet as well, then in addition to the (kM mod N) north-going passes over the equator there are south going passes at (kM + (M+N)/2 mod N).  These extra passes interleave helpfully if (M+N)/2 mod N is an integer, so the equator is covered in half the orbits it would take to finish the planet.  The smallest M where M and N follow all these rules is (N = 39;  M = 5) and an orbital period 5/39 rotations will completely cover Kerbin's equator in about 2.5 days, leaving only small uncovered slivers to be completed in the remaining 2.5 days.

Edited March 22, 2021 by OHara
typo

[Guest]

14 hours ago, OHara said:

People use "modulo arithmetic", @FruitGoose, to mean arithmetic where you only care about the remainder. 

Yes, I was being facetious (a common theme to many of my posts  ).

Yes you are quite right. A non-synchronous near polar orbit as low as possible should cover the most area in the quickest time, not sure what I was thinking about previously. 

Edited March 22, 2021 by Guest
Missed key part

[ajburges]

OHara, thanks for that simple mechanism I got lost in the weeds and overlooked that simple division would give me the orbits then it was simply a matter of choosing a coprime number of revolutions to pair with it.

I still have some questions about waypoint activation, but I'll ask that in seperate thread.

Edited March 28, 2021 by ajburges