Can't time the launch properly - some advice pls?

[BlackBicycle]

I am trying to launch a comms constellation of 3 sats, 120 deg from each other. Heading 90, altitude 450km. The orbital period is exactly 1 hour at that altitude. My vehicles include a capsule and a powersat.

So, I launch the first one, get it to orbit and deploy the sat. I then return to Kerbin. After a while I check the mission timer on that sat, and when it reads 1 hour 40 mins into the mission, I launch the next sat. It should end up 120 deg. from the first one, right? Only in my case, it ended up 60 degrees (in front) of the first one.

Am i missing something here?

[RangerDanger75]

I had this problem while I was trying to set up some satellite constellations of my own. The reason you\'re numbers are off is because you aren\'t taking the rotation of Kerbin into consideration.

I\'m guessing it takes about 20 min for the satellite to go from the launch pad to the desired orbit? Here\'s what I would do if you\'re not against using MechJeb or another plugin that gives you latitude and longitude.

Assuming it takes 20 minutes to get into orbit, find the change in longitude a satellite at 450km travels in 20 minutes. Next, launch a satellite into the 450km orbit, noting the longitude of the 450km transfer orbit apoapsis once you reach it. Finally, note the the longitude of the KSC and you\'ll have all the numbers you need to figure it out.

In order to figure out when to launch here\'s the formula you\'ll need:

SAT_{orbit,initial} = KSC_long + delta-SAT_you - (delta-SAT_orbit + 120^o)

Where,

KSC = Longitude of KSC

delta-SAT_you = change in longitude between launch pad and 450km apoapsis

delta-SAT_orbit = change in longitude of the satellite already in orbit

SAT_{orbit,initial} = longitude the satellite in orbit should be at in order to achieve a 120^o lead on it

Boring Explanation/Derivation:

In order to solve this, you want your final position to equal the orbiting satellite\'s final position + 120^o

SAT_you,final = SAT_orbit,final + 120

All your final position is, is KSC\'s longitude + your change in longitude

SAT_you,final = KSC_long + delta-SAT_you

And the orbiting satellite\'s final position equals it\'s initial position + it\'s change in longitude

SAT_orbit,final = SAT_{orbit,initial} + delta-SAT_orbit

Calling back on the first equation and substituting some equations...

SAT_you,final = SAT_orbit,final + 120^o

KSC_long + delta-SAT_you = SAT_{orbit,initial} + delta-SAT_orbit + 120^o

Solving for the initial position of the orbiting satellite

SAT_{orbit,initial} = KSC_long + delta-SAT_you - (delta-SAT_orbit + 120^o)

Easy as that!

P.S. Make sure you obtain all these values using the exact same craft, otherwise you\'ll get up to orbit at different times which will throw the whole thing off!

And please let me know if anything isn\'t clear. It\'s easy to forget how difficult this can be to grasp.

[maltesh]

Alternatively, launch all three satellites in one trip.

Get into a circular orbit at 450 km. Drop the first satellite.

Burn to push your apoapsis out to 896 km. Period of orbit is now 1 hour and 20 minutes. When you come back to 450 km, you\'ll be 120 degrees behind the first sat. Circularize and drop the second satellite.

Repeat for the third satellite, then head home.

[RangerDanger75]

My god, that is brilliant! I don\'t know why I\'ve never thought of that. The only problem I see is not being able to perform any corrections once the satellites have been deployed for a while.

[BlackBicycle]

Thank you both very much : D

Yep, for some reason I considered the spaceport stationary for my needs. Apparently, it isn\'t.

The multi-payload solution is also very interesting, I\'ll definitely give it a try at some point.

[BlackBicycle]

I am working on it : ) 25 mins sharp till circularisation burn end. 125 deg in 25 mins when in orbit (hmm, apparently that calc i am using does not calculate the orbital period correctly, it\'s 72 minutes rather than 60 : l ah well.

The longitude delta between launch and apoapsis, hmm... shouldn\'t it rather be 'between launch and end of circularisation burn'? I am using a weak engine on the last stage, so the burn takes around a minute.

Also: is there a mod, apart from MechJeb (which i use and am very happy with) that gives you info on objects that don\'t have mjeb in them? those powersats don\'t have anything on board, they are a self-contained unit. I just need something to show me their longitude. So far I have to keep the crew up in space to get telemetry, and i think they are going green with boredom...

Update: yay! I achieved a perfect triangle of sats! And - in true Kerbal style only then did I realize that 450km in not enough for the sats to see each other above the horizon : D

Thanks again guys : )

[Mr. Me]

The only problem I see is not being able to perform any corrections once the satellites have been deployed for a while.

I suggest that once you get a satellites into orbit, send up a ship with RCS tanks that docks to it using landing legs, if you means orbital corrections.

[BlackBicycle]

Another snag : )

After calculating that i will need at least a 2710km orbit to cover the planet from 3 sides, I decided I might as well go kerbostationary, which is 2868km. Now I need some way to calculate the orbital change I have to do to lag 120 degrees after arrival on the stationary orbit. Maltesh, how did you calculate the orbit change in the previous example (about multipayload launch)?

[Endeavour]

I don\'t think that Maltesh\'s method is possible for an apoapsis of 2868.4km.

[maltesh]

I calculated an orbit whose semimajor axis resulted in a period 4/3 that of the value you gave (one hour), and found the altitude of the apoapsis for an orbit with that semimajor axis and a periapsis altitude of 450 km.

For 2868.4 km (6-hour orbit), you need an 8-hour orbit to pop into.

Circulize at 2868.4, drop first satellite.

Burn for apoapsis of 4333 km, Period is 8 hours. When you get back to 2868.4, you\'re 120 degrees behind. Circularize and drop the second satellite.

(edit: Forgot to include the circularize step here)

Repeat for the third.

If you want to do it quicker, you can keep 2828.4 as your apoapsis, and burn to pull your periapsis down to 1,228 km. You\'ll go around once in four hours, and get back to 2868.4, but you\'ll be 120 degrees ahead of the sat you just dropped.

[BlackBicycle]

Thank you very much, got it all. Doing it atm. Man it\'s hard to think on saturday morning : )

[maltesh]

The larger orbit you burn to has a period 4/3 that of the circular orbit. In the time it takes you to go around once in the larger orbit, an object in the circular orbit has gone around once, and then moved ahead another 1/3 of its orbit. So when you get back to your original altitude, and circularize, you\'re 120 degrees behind the first satellite you dropped.

I imagine there aren\'t many articles on putting satellites evenly-spaced in orbits in this fashion, because in the real world, rocket science is harder, you have much better instrumentation for launch timing, and the method risks losing all three satellites at once in a launch mishap.

[RangerDanger75]

I calculated an orbit whose semimajor axis resulted in a period 4/3 that of the value you gave (one hour), and found the altitude of the apoapsis for an orbit with that semimajor axis and a periapsis altitude of 450 km.

How did you take into account the fact that the velocity wouldn\'t be constant in an elliptical orbit as opposed to the constant velocity of a circular orbit?

[BlackBicycle]

How did you take into account the fact that the velocity wouldn\'t be constant in an elliptical orbit as opposed to the constant velocity of a circular orbit?

There is no need for that. You change the period of rotation so that it takes 1/3 longer or shorter, do it for ONE orbit, then circularize back.

P.S. It would\'ve been nice to have a calculator, where you enter your circular orbit altitude, above which body, the lag or lead in degrees you need to achieve, and it gives you the new apsis altitude needed.

[Menelmacar]

Thanks to all in this thread, I had a similar problem to BlackBicycle\'s, I ended up using maltesh\'s method because the maths are very simple.

I put 4 sats in half-kerbo-sync orbits in one trip, 90 degrees apart and the method worked so much better than my previous attempts to use mechjeb\'s Lattitude output (which is on a scale of 0>180>-180>0). Later I managed 6 sats in one trip all kerbo-synced 60 degrees apart.

Honestly I\'m terrible at maths but this method worked for me because I can visualise it - I got the period of the orbit I wanted the sats at from KSP Orbit Mechanic which I then used to find an elliptical orbit with a Pe at my sat orbit altitude and a period of (numberofsats+1)/numberofsats x period of ultimate satellite orbit (I do this by putting numbers into Orbit Mechanic until I get the ellipse with the period I want - I know there\'s better ways but I\'m really not a fan of maths). With this I could then circularise, deploy, push my apopasis to the one of the elliptical orbit I figured out and wait until I return to Pe, circularize, deploy and repeat until I\'m out of satellites

So thanks again guys, just finished getting a few more comsats up around the Mun for a few unmanned rover missions (that speed of light delay from the relay plugin basically disables your ASAS at around 0.6s lag... unlike some on this site I think the Kerbals will agree with me when I say I think Kerbals should always live to see another explosion)... and I made my first trip to Minmus, I\'m sitting in really inclined but circular orbit there and can\'t wait to come back for comms link so I can go down and explore - I\'ve never been there and so far I\'ve managed to not hear anything much about it so it is exciting. ???

[RangerDanger75]

Thanks to Maltesh, I just learned all about Kepler\'s third law!

For those interested

http://en.wikipedia.org/wiki/Semi-major_axis#Orbital_period