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Orbital rendezvous issues (Kerbin orbit)


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I am new to the game. In Career mode, I am now trying to orbital rescue missions were I must get close enough to a craft/Kerbal to transfer between the vehicules, and re-enter the atmosphere.

I know the orbital basics and how to create maneuver nodes. So far I have managed to capture one Kerbin, but luck played a bit part.

Once I have a stable orbit at approximately the same altitude as the target craft, I don't understand how I can catch up (or get caught up) with it with a range small enough that I can RCS my relative velocity out.

I can set basic intercepts but by playing around I can mainly only get my closest intercept to ~20km, or waste so much fuel that I am left with none, or I can cross the path but only extremely briefly.

The orbits are also very close to Kerbin which makes it difficult to distinguish the paths clearly in the Orbital Map view (by zooming to the maximum, I can see paths are made up of straight sections for display).

I have found this thread about orbital maneuvers: 

 

I understand the gist of it, but am mostly still stuck in the very same orbits as my targets, but with a phase angle.

Additionally I have found this thread, where user Iodestar provided an example of a calculation for adjusting and out of phase orbit, but I am missing important intermediate steps and cannot reproduce the calculations.

Any help would be appreciated. Stock game, Career mode, some elements of the 90 level tech tree unlocked.

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 Welcome to the game, @wobartoz I can certainly help you with reproducing the calculations, so that is what I will do.  This is the original post:

On 7/22/2014 at 10:26 PM, lodestar said:

Reaching a nearly perfect orbit overlap when you are far apart isn't very useful, but if your orbits are already nearly exactly equal, you can figure out your distance and phase angle. Once you know that, you can change your orbital period such that after n orbital revolutions you'll be very close. If you are in front of the lander you want to rendezvous with, you have to increase your orbital period so the lander will catch up with you. If you are behind, you have to decrease it so you'll move faster and catch up with it. You can do it in small steps and wait, but what's the point of playing Kerbal Space Program if you're not putting your knowledge of orbital mechanics to good use? All you need to know is how to calculate your orbital period from your orbit, and the other way around, how to calculate your orbit from your orbital period, and you can do it in a single pass.

As an example, let's say you you are in a 40km circular orbit around the Mun, you're in front of the lander, and your angle of separation is 77°.

1. With that orbit, you have an orbital period of 2894 seconds. Divide that by 360 and you have around 8 seconds/degree.

2. If your angle of separation is 77°, you need to increase your orbital period by 77*8=616 seconds, or to 3510 seconds.

3. To have an orbital period of 3510 seconds around the Mun you need a Semi Major of 272.92km. With a 40km periapsis that means you need to raise your apoapsis to 105.84.

4. So, raise your apoapsis to 105.84, and after one revolution, you should be very close to the lander.

5. When you are closest to your lander, target it, point to the target velocity prograde marker and match velocities by burning until the relative velocity is near 0m/s.

6. To rendezvous, point to the target prograde marker and burn until you get a few m/s relative velocity.

7. When you are at the closest approach, point to the velocity retrograde marker and match velocity again.

8. If you are within 100m or less, you can fire up RCS and dock. If not, go back to 6 and repeat until you are close enough.

Let's take this in steps.  Also note that I will use American notation conventions (commas for thousands separators and periods for decimals).  First, we have a lander in a 40 km circular orbit of the Mun and another vessel ahead of it in the same orbit, with a separation angle of 77°.

1.  Calculate the orbital period.

The equation for this is T = 2π √(a3 / μ), where:

T = orbital period,
a = the semi-major axis of the orbit, and
μ = the gravitational parameter of the body that you are orbiting

KSP displays your orbital altitude as an altitude above datum, which is the zero point or 'sea level' of surface elevation.  The problem is that semi-major axis for an orbit is calculated with respect to the centre of mass, not the surface.  You need to add the radius of the Mun, which is 200,000 metres.  You can find that in the wiki, but it is also displayed in the information pane for the Mun that you can find in the Tracking Station.  The formula for semi-major axis in KSP is (Apoapsis + Periapsis + celestial body diameter) / 2, so for a 40 km circular orbit of the Mun, that is (40,000 + 40,000 + 400,000) / 2 = 480,000 / 2 = 240,000 metres.

The gravitational parameter is also unique to a given celestial body, and is equivalent to the universal gravitational constant times the mass of the celestial body.  This is also something that you can find either in the wiki or in the Tracking Station, and for the Mun is 6.5138398×1010 m3 / s2.

Putting that together:

T = 2π √(a3 / μ)
T = 2π √([240,000]3 / [6.5138398×1010])
T = 2π √([1.3824×1016] / [6.5138398×1010])
T = 2π √(212,225.05)
T = 2π * 460.6789
T = 2,894.5 seconds

2.  Calculate the separation in terms of the orbital period.

Since we're working in degrees, I will continue to do so, but know that many orbital calculations require radians.  2,894.5 seconds per 360° orbit gives 8.04 seconds per degree--we can round this down to 8 seconds exactly for our purposes.  77° at 8 seconds per degree means that, assuming that you are ahead in the orbit, you are 616 seconds ahead of the target in your orbit.

To allow your target to catch up to you in minimum time, you will want to increase your orbital period by that much.  Then, when you return to the same point, the target will have advanced by that exact amount and you will complete the rendezvous by matching velocities (which is the same thing as returning to your previous circular orbit).  However, it is not necessary to do it that way.  You can choose, for example, to increase your orbital period by 308 seconds and allow the target to catch up over two orbits.  That will save fuel, but at the cost of time.  Sometimes, it may be necessary to do it that way, such as when your phasing orbit will cause you to strike a celestial body, eject you from the sphere of influence, cost too much of a limited fuel budget, or otherwise cause issues.

Anyway, in this case, since we want to slow down and allow the trailing object to catch up, we need to increase the orbital period by 616 seconds.  This is as simple as 2,894.5 + 616 = 3,510.5 seconds.

3.  Calculate an orbit that has the required orbital period.

In this case, we want to use the same equation for orbital period, but we need to work through it the other way:  we have the period and we want the semi-major axis.  With algebra, we can derive a new equation to be:

a =  3√(μT2 / 4π2)

The variables are all the same as before, but please note that the outermost operation is cube root, not square root.

a = 3√([6.5138398×1010] * [3,510.5]2 / 4π2)
a = 3√([6.5138398×1010] * [12,323,610.25] / [39.4784176])
a = 3√([8.02740229×1017] / [39.4784176])
a = 3√([2.03336476×1016])
a = 272,942.9 metres

Remember that semi-major axis is equivalent to (Apoapsis + Periapsis + celestial body diameter) / 2.  In this case, we want to keep the periapsis for ease of rendezvous later, and we are naturally going to remain in orbit of the Mun, so the only variable to change is the apoapsis.  This problem simplifies to (2 * a) - Periapsis - celestial body diameter = Apoapsis, or (2 * 272942.9) - 40,000 - 400,000 = 545885.8 - 40,000 - 400,000 = 105,885.8 metres (or 105.8858 km) for your new apoapsis altitude.

4 - 8. Complete these steps as given in the quoted post.

One item of extreme interest here is that, although the calculation is valid, it depends on foreknowledge of the phase angle or angle of separation between your vessel and its rendezvous target.  In the example, this was given as 77°, but KSP does not give you this information.  You will need to either install a mod that will give the information (Kerbal Engineer will do this), estimate it by eye, or else use a protractor on the computer screen.  You can calculate it based on the separation changes resulting from tiny orbital manoeuvres, as well, but that is often more trouble than it is worth.  The typical approach for this is instead to raise your apoapsis by some arbitrary amount, wait for enough orbits to get a 'close' encounter, lower the apoapsis to get a near-perfect encounter according to the close approach markers in the map view, and rendezvous.  The idea is that you get some fuel savings by choosing a lower phasing orbit to offset the cost of the minor adjustments.  The net effect is equivalent to estimating the required phasing orbit and then refining that estimate by iteration until you get a rendezvous.

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My very short guide:

 

1) Get into orbit, with same inclination as target vessel. Remember to target it, left click on vessel!

2) Put your AP high enough to equal the target's orbit. Time warp to it.

3) On your AP, start burning prograde, your orbit should start getting elliptical and higher. Keep an eye on when Closest position is very close to target future position, marker distance should be near 3km or less.

4) Time warp before closest position, change orbit mode to TARGET (the green text on top of the navball that says your velocity). Point to the target's retrograde marker and burn SLOWLY to match the velocity. If done correctly you could even reach closer (Pro gamer move, kinda(?))!

5) Now that velocities match and you are 2 or 3 km from the target, burn slowly towards it .

6) Profit!

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Everyone will have their own take on this, but some random things I find useful to keep in mind:

You greatly enhance your chances of a close encounter by being on the same (or close) inclination. Change of inclination (Normal/Anti-normal) is most efficient at apoapsis, and moreso if you can combine it with a prograde/retrograde burn if the result combines.  Competing with this is the fact that you can get a cheaper inclination burn if you've already raised your apoapsis.

To catch up with a vessel on the same orbit as you, it's necessary to drop to a lower orbit or at least lower for part of the orbit.  Burning retrograde lowers the opposite side of your orbit, so although it drops your orbital speed, will increase your angular velocity.

If you are actually just a bit ahead of your target, or perhaps need to do an inclination burn which would raise your orbit anyway, you can burn prograde to increase your orbital altitude.  This results in a lower angular velocity (for at least part of your orbit) which will mean the target will catch up with you.

I would always do these manoeuvres with your most efficient main engines.  RCS efficiency (Isp) sucks for bigger movements, it's better to use less engines or reduce thrust limit on engines than use RCS for anything but the tinyest adjustments.  RCS comes into it's own for tweaks to planetary transfers or last moment tweaks on  final docking approach.  I generally get myself to within 100m and < 2m/s of my target with main engines, if not go all the way.

If you are using RCS, a handy tip I've picked up here is to disable attitude control (yaw, pitch, roll) on your RCS so it won't be trying to do the job of your reaction wheels while you want to use it for translation.  Attitude adjustments by RCS invariably do a certain amount of translation and hence make a right mess of your carefully planned rendezvous.

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1 hour ago, Linkageless said:

You greatly enhance your chances of a close encounter by being on the same (or close) inclination. Change of inclination (Normal/Anti-normal) is most efficient at apoapsis, and more so if you can combine it with a prograde/retrograde burn if the result combines.  Competing with this is the fact that you can get a cheaper inclination burn if you've already raised your apoapsis.

If you set something as your target, you can see two green markers on your orbit, with a dotted line reaching from your orbit to the orbit of the targeted body through the two markers. that is the most efficient point to burn according to an ingame tutorial, a tutorial by scott manley and many docking attempts by me.

Spoiler

but you might be right, I'll have to try to be sure

1 hour ago, Linkageless said:

If you are using RCS, a handy tip I've picked up here is to disable attitude control (yaw, pitch, roll) on your RCS so it won't be trying to do the job of your reaction wheels while you want to use it for translation.  Attitude adjustments by RCS invariably do a certain amount of translation and hence make a right mess of your carefully planned rendezvous.

 wait you can do that?

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2 hours ago, Linkageless said:

If you are using RCS, a handy tip I've picked up here is to disable attitude control

 

31 minutes ago, Dirkidirk said:

wait you can do that?

Most definitely. I believe you need advanced tweakables on.

There's also a module manager patch that will set it as the default. I'd paste it here but I'm on my phone.

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16 hours ago, Dirkidirk said:

you can see two green markers on your orbit,

Ah yes, sorry, green 'AN' and 'DN' markers for Ascending Node and Descending Node.   It's easy to take this knowledge for granted.

 

16 hours ago, 5thHorseman said:

I believe you need advanced tweakables on.

I'm pretty sure you're right, something I also forget I take for granted. :)  With that setting turned on, you'll see a 'Show Actuation Toggles' option in the part action menu (ie right click on part on non-console versions).

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Thank you all for your answers. @5thHorseman's tutorial proved to be very tangible.

From what I got, I was trying to hard to catch up with a vessel when already being on the exact same orbit. What I get from this is, I should have an arbitrary orbit, elliptical or not but different from my target, and then aim for a rendezvous maneuver. For that, I could math my way out of it but KSP provides very convenient tools to guesstimate it and visually plan the encounter.

@Zhetaan: Thanks for the math breakdown. I will still look into it for the sake of learning orbital mechanics. It's been a while I had to whip out the TI-83+ for a game.

 

 

 

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On 1/19/2020 at 9:17 PM, wobartoz said:

What I get from this is, I should have an arbitrary orbit, elliptical or not but different from my target, and then aim for a rendezvous maneuver.

Yes, that is it precisely.  There is a concept in celestial mechanics called the synodic period which relates to the frequency of conjunctions or oppositions between two planets.  Specifically, it describes the time it takes for a second planet to return to a particular orientation about the primary with respect to the perspective of the first.  This is relevant to orbital mechanics because the same period describes the frequency (though not the location) of transfer windows. Since interplanetary transfer is advanced rendezvous, it also relates to the similarity of orbits for purposes of rendezvous--and KSP does not provide the most convenient tools for interplanetary transfer.  Details are in the spoiler.

Spoiler

 

The formula for it is this:

1 / Tsyn = (1 / T1) - (1 / T2)

Tsyn is the synodic period, and T1 and T2 are the periods of the two planets.  One caveat is that T1 must be the smaller period (or the inner orbit if you prefer to think of it that way).

I raise this point to illustrate exactly why it is important to have dissimilar orbits for purposes of rendezvous.  If we hold the inner orbit to be some fixed value (because, for example, it's our target), then T1 is a constant and the synodic period relates to the difference between that orbit and whatever our vessel's orbit may be.  Without using actual figures, we can still learn a lot about this relationship.

For example, whatever the value of T1 is in seconds or other time units, let's just call it 1 period.  T2 must therefore be larger than 1, but we're not yet interested in how much larger it is.  Instead, what we can discern is that if T2 is the larger period, then the reciprocal 1 / T2 must be the smaller fraction because increasing the denominator of a fraction decreases the fraction's overall value.  What can we learn at the extreme limits of this?  Let's assume that T2's period is infinite--this cannot happen in reality, but mathematically, it describes a situation where the object that T2 represents stands still in the sky.  In the limit, as the denominator increases to infinity, the overall fraction reduces to zero, so the synodic period is the same as T1.  Let's also see what happens when T2 equals T1 (remember that the rule is that T1 is smaller, so that sets a minimum):  in that case, the difference is zero.  Again, in the limit, if the overall fraction reduces to zero (and we cannot modify the numerator for some reason), then the denominator must increase to infinity, which means that the synodic period is infinity:  there is no conjunction, opposition, or transfer window.

To put this into more viscerally understandable terms, imagine that you are a runner making laps about a track.  If you run around the track in, for example, two minutes, then the synodic period describes how frequently you pass (or are passed by) other runners, from your perspective.  A planet with infinite period would be like a spectator at the edge of the track:  you pass that spectator every two minutes.  As you choose destination planets that are farther away from you, the synodic period reduces to your period.  For planets that are extremely far away (such as Plock in Outer Planets Mod or a nearby star in mods that have them), the synodic period is effectively once per local year because the relative motion of these bodies from year to year is so slight.

In the other extreme, a runner with your exact speed of one lap per two minutes never passes you.  Thus, you can never rendezvous.

In other cases, we can see that the closer the other runner gets to your speed, the less frequently you pass:

You pass a runner with one tenth of your speed (one lap every twenty minutes) every (1/2 - 1/20)-1 = (10/20 - 1/20)-1 = (9/20)-1 = 20/9 = 2.222 minutes.
You pass a runner with one fifth of your speed (one lap every ten minutes) every (1/2 - 1/10)-1 = 10/4 = 2.5 minutes.
You pass a runner with one quarter of your speed (one lap every eight minutes) every (1/2 - 1/8)-1 = 8/3 = 2.667 minutes.
You pass a runner with one third of your speed (one lap every six minutes) every (1/2 - 1/6)-1 = 6/2 = 3 minutes.
You pass a runner with one half of your speed (one lap every four minutes) every (1/2 - 1/4)-1 = 4/1 = 4 minutes.
You pass a runner with two thirds of your speed (one lap every three minutes) every (1/2 - 1/3)-1 = 6/1 = 6 minutes.

To return this to orbital mechanics and rendezvous, it should be plain to see that orbits need to be dissimilar in order to have frequent encounters or transfer windows.  Unfortunately, dissimilarity (and correcting it) costs fuel.

 

 

On 1/19/2020 at 9:17 PM, wobartoz said:

Thanks for the math breakdown. I will still look into it for the sake of learning orbital mechanics. It's been a while I had to whip out the TI-83+ for a game.

You are most welcome.  We're all here to help.

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