How do I match orbit inclination with a contract orbit

[claudemeister]

I am not sure of how to match my own orbit's inclination with the orbit that is shown on the map by a contract.

I had to send a satellite into a specific orbit which was shown on the map. I had no problems with the apoapsis and periapsis (except for matching the exact spot for the ap and pe, I just eyeballed it, would appreciate a hint here as well) but when it came to matching up the inclination I was really struggling.

The contract said the desired orbit had an inclination of 170 or so degrees but relative to what? I also had the argument of the periapsis and longitude of ascending node, just didn't know what they meant (still don't know). If I aligned the map/ planet with the orbit it looked like about 290 degrees. If I launched and put the navball on 170 I got nowhere. Can someone tell me how I can launch at the correct angle or sort of figure it out? And also how to adjust my orbit too once in the air, like you would do for a moon (using maneuver nodes doesn't work because I can't set an orbit as a target so I'm not sure how much normal of delta-v to add/ subtract)?

In the end I just had to eyeball things by aligning the planet with the orbit and stuff like that which was fine but still way to much hassle in comparison to other things in this game. Also if you add normal or anti-normal the contract orbit tells you your ascending and descending nodes but maneuver nodes would be useful in this situation. I have Kerbal Engineer installed so I think there has to be a better way of doing these sorts of missions just not sure how. Would MechJeb help in this situation?

 

Thanks in advance for any answers although I'll probably thank you again later

[Snark]

Hello, and welcome to the forums! 

2 hours ago, claudemeister said:

The contract said the desired orbit had an inclination of 170 or so degrees but relative to what?

Relative to a prograde equatorial orbit.   A satellite with 0 degree inclination is orbiting in a perfect equatorial orbit, from west to east.  (The Mun's orbit has 0 degrees inclination, for example.)  90 degrees would be polar.  180 degrees would be equatorial, but retrograde, i.e. east to west, the opposite direction that the Mun orbits.

So if you see "170 degree inclination", it means an orbit that's 10 degrees off equatorial, backwards.

Normally, if a contract wants you to launch into an orbit with significant inclination, the best way to do that is to wait until Kerbin revolves to the point where KSC is directly beneath the target orbit, and then launch directly into the desired orbital plane.

If you're already orbiting, though, you can change your orbital inclination to match the desired one by waiting until your ship is located at the ascending node (AN in map view) and thrusting , or else is located at the descending node (DN in map view) and thrusting .

Thrusting or is only good if your inclination adjustment's not too huge, though.  For example, if you've already launched your satellite in a prograde orbit, you need to practically reverse direction completely in order to get to 170 degrees inclination, in which case thrusting mostly until you kill your velocity and then start orbiting the opposite direction.  That's going to take one heck of a lot of dV, though-- several kilometers per second.  It may be easier simply to give up on the satellite and launch a new one in the correct direction.

[4x4cheesecake]

2 hours ago, claudemeister said:

I have Kerbal Engineer installed

If you know how to use this wonderful mod, it is all you need to match orbits. I'll try to explain it to you:

First of all, something about inclination in general: The inclination is always relative to the equator of the body OR relative to your target. If a contract wants you to place something in a orbit will always refere to the equator. If you want to rendevouz with another ship and set the other ship as your target, the important inclination is the one of the ship you want to rendevouz. Since the latter one is about another topic, I'll ignore it here
0° inclination means, that your orbit ist perfectly aligned with the equator AND your orbit got the same direction of the rotation of the body (usually heading east). If the orbit of your ship rotates in the opposite direction (heading west), your inclination is 180° and a polar orbit got a 90° inclination (or 270° if you orbit in the other direction).

Imagine the equator and your orbit as a 2D plane while the equatorial plane is in a constant position, you can change the orbit/plane of your ship but there are always 2 spots, where these plane will cross (and an infinite amount if you got perfectly 0° inclination). These are your ascending and descending nodes. The ascending node is the one where you cross the equatorial plane and move 'up' afterwards (relative to the equator), the descendinge node is the opppsite.
If you aligne the ascending node with the core of the planet, you can imagine a line to connect these points and as soon as this line cross the surface of the planet, you find the 'Longitude of the ascending node'.

Why is the 'Longitude of the ascending node' interesting? Because it defines the axis you want to use to incline your orbit and the position of your ascending node. KER (kerbal engineer) can be used to read out the position of your ship (latitude and longitude) so you will always know where you need to burn normal / antinormal to incline your orbit to match the one from the contract.
BUT: I think it is easier to eyeball it Fokus the body you are orbiting so your camera will not move automatically, position the camera so that your own orbit and the required orbit are no longer elipical but just a straight line. These line will cross at some point, these are your ascending/descending nodes (depending on your position). You can place a maneuvernode there or just burn normal /antinormal as soon as your ship crosses this point until both orbits got the same inclination.

Argument of periapsis:
Well, I have no easy explanaition for this one...it is the angle between your ascending node and your periapsis. Is that a useful information for you? I don't know but mathematically you need this angle to define the position of your orbit...

Lets take a closer look at your contract:

You need an inclination of 170° which is just 10° off of a perfect retrograde orbit, so you want to launch your ship heading west instead of east. I would suggest not to try to launch into a 170° inclined orbit, it is not worth the hassle. If you want to try it anyway, you have to wait until the launchpad is close the ascending/descending node (this is very important!!), than launch your ship heading west. As soon as you reach an altitude of 20 - 30km, you can start to do perform some minor adjustments to your trajectory by pointing the ship a bit more to the north. This should bring you close to the required orbit.

How can KER help you?
In my opinion, you will just need the inclination read out of KER, you can eyeball everything else. Use the method I have mentioned to find your ascending/descending node and burn normal/antinormal until the inclination read out reaches 170° (if you launch your ship heading west, you will start at an inclination around 180°). As you said before, it is pretty easy to get the correct Ap/Pe and thats all you will need. Don't worry too much about the longitude of the ascending node and the argument of periapsis, these values are 'just' required to describe the required orbit mathematically but are less useful to actually reach the orbit (pretty sure this is different in reallity^^)
Of course, you can also use the KER read outs of these values and burn prograde/retrograde in some distance to your Ap to move the Pe around until the read out matches the requirement...but like I said, it is easier to eyeball this stuff

I hope some of these explanations are understandable for you

edit:

Edited August 16, 2018 by 4x4cheesecake
Added some stuff to the KER part

[StrandedonEarth]

When it come to these "satellite in specific orbit" contracts, the An/Dn on the target orbit will show your relative inclination,  but is not affected by maneuver nodes (until executed). While burning normal/anti-normal will change your inclination, it will also tend to raise your orbit, requiring some retrograde to compensate. Another handy mod for plotting this sort of maneuver is Precise Maneuver, or the similar mod (fork? clone? takeover? not sure which is which anymore) Precise Node, which has a button for changing inclination without changing the Ap/Pe.

[Blaarkies]

6 hours ago, Snark said:

That's going to take one heck of a lot of dV, though-- several kilometers per second.  It may be easier simply to give up on the satellite and launch a new one in the correct direction.

Something like 2 x orbital speed, (i believe 4400m/s ?), not even counting gravity drag. @claudemeister But you could always do it for much cheaper, by first thrusting into an elliptical orbit, then at Ap thrusting Retrograde to change the orbital direction, and at Pe again thrust to slow down into the intended circular orbit.

This at least costs less than 2000m/s dv without any aero-braking. Im not sure what the optimum solution would be though, has anyone seen the math for this somewhere?

[Xd the great]

16 minutes ago, Blaarkies said:

Something like 2 x orbital speed, (i believe 4400m/s ?), not even counting gravity drag. @claudemeister But you could always do it for much cheaper, by first thrusting into an elliptical orbit, then at Ap thrusting Retrograde to change the orbital direction, and at Pe again thrust to slow down into the intended circular orbit.

This at least costs less than 2000m/s dv without any aero-braking. Im not sure what the optimum solution would be though, has anyone seen the math for this somewhere?

The math? Vector sum.

[Blaarkies]

24 minutes ago, Xd the great said:

The math? Vector sum.

A - Burn into an elliptical orbit.
B - Burn retrograde at Ap, for 2x orbital speed

Mission plan: A + B + A = total dv required
...but the larger A becomes, the smaller B becomes, so the question is "How much should A be (for the elliptical orbit) to minimize the total required dv?"
so, rather Vis-viva equation maybe?

Edited August 16, 2018 by Blaarkies

[mystifeid]

2 hours ago, Blaarkies said:

so the question is "How much should A be (for the elliptical orbit) to minimize the total required dv?

I think that the total may be minimized as B approaches zero - ie the cheapest inclination change from 0° to 180° is attained by extending the Ap to the edge of the SOI, changing inclination and then circularizing at the Pe.

While it would be nice to be slick at arithmetic so that I could demonstrate mathematically, it is, as usual, easy to test. Here are a few examples of the dV required to flip a 100km Kerbin orbit.

Ap                     A                   B                        Total

100km              0                4492.3m/s        4492.3m/s

1000km        403.3m/s    2318.2m/s        3124.8m/s

5000km        748.7m/s      748.7m/s        2246.1m/s

10000km     830.4m/s       406.3m/s        2067.1m/s

20000km     877.7m/s       212.3m/s        1967.7m/s

40000km     903.3m/s       108.6m/s        1915.2m/s

80000km     916.7m/s          55.0m/s        1888.4m/s

I used Mechjeb for the maneuvers and even after flipping at 80 000km, MJ circularized back at 100km with an inclination of 179.999° - this would probably be a bit trickier without Mechjeb.

[4x4cheesecake]

Well, theoretically, you can perform a fly-by of Eve followed by some aerobreaking at Kerbin to reverse your orbital direction for just ~1300m/s (incl. circularization burn at Kerbin after aerobreaking) but who want's to do that

[mystifeid]

2 minutes ago, 4x4cheesecake said:

Well, theoretically, you can perform a fly-by of Eve followed by some aerobreaking at Kerbin to reverse your orbital direction for just ~1300m/s (incl. circularization burn at Kerbin after aerobreaking) but who want's to do that

I was wondering about the Mun.

[4x4cheesecake]

1 minute ago, mystifeid said:

I was wondering about the Mun.

I don't think that the Mun will help a lot in this case...you can set up a free return trajectory but to flip the orbital direction, you still have to burn a lot of fuel within its SOI which is still pretty close to Kerbin. But its just my gut feeling.

[4x4cheesecake]

@mystifeid I just tried it and actually, the Mun will do a great job^^

852m/s to set up a free return trajectory, 263m/s maneuver at Mun Pe to flip the orbital orientation around Kerbin, some aerobreaking and ~160m/s for minor adjustments and circularization around Kerbin again, which will sum up to ~1275 m/s.

Edited August 16, 2018 by 4x4cheesecake
typo

[Blaarkies]

1 hour ago, 4x4cheesecake said:

@mystifeid I just tried it and actually, the Mun will do a great job^^

852m/s to set up a free return trajectory, 263m/s maneuver at Mun Pe to flip the orbital orientation around Kerbin, some aerobreaking and ~160m/s for minor adjustments and circularization around Kerbin again, which will sum up to ~1275 m/s.

This is awesome! OP just wanted to save their sat contract, and then you guys go find the best solution literally behind the Mun  This ^ is why ksp is so great