Basic maneuver no more working

[Tristan1070]

Hi, I'm trying to achieve the first mission of "Making History" (part 3).

So I've launched from "Woomerang" and I'm in circular orbit (apoapsis ~= periapsis ~= 100 km), but not in equatorial orbit, I have a 52° inclination to the equatorial orbit.

Now, when I try to do this basic maneuver at the ascending node : https://wiki.kerbalspaceprogram.com/wiki/Basic_maneuvers#Match_orbital_inclination, it's getting my orbit inclination to 0, but it's also making it not circular until my ship just leaves orbit.

What is happening here ? Is it a bug ? What should I do ?

Could it be because my orbit is not a perfect circle ? or because my inclination is too important ?

 

[Geonovast]

Changing your inclination also affects your pe and/or ap.  When you're setting the manuever node, you need to also adjust the prograde and retrograde handles to keep your orbit where you want.  So move the normal  or antinormal  handles until your projected orbit gets significantly larger, then pull the prograde  or retrograde  handle to shrink it back down.  You'll have to go back and fourth a few times to get your orbit exactly where you want, then do your burn.

[Tristan1070]

Ok, thanks !

I've added a "NOTE" to the wiki :

"You may need a mix of normal/antinormal + prograde/retrograde to keep the orbit circular."

 

[bewing]

I know you don't have the capability in that MH mission -- but if you have SAS Normal/AntiNormal modes available in your rocket then you can do your inclination burn by hand without affecting your Ap or Pe.

 

[Eriksonn]

I made a small paint diagram to show the correct manuever:

[Tristan1070]

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if you have SAS Normal/AntiNormal modes available in your rocket then you can do your inclination burn by hand without affecting your Ap or Pe.

Well, you can activate SAS in this mission, but maybe it's a not enough precise/advanced SAS ?

In the training scenario about orbital inclination, you can do it with pure normal push for sure.

Quote

I made a small paint diagram to show the correct manuever:

This is a nice diagram, but I'm not sure to understand what the length of the vectors are in this representation ? orbital speed ? The thing is only direction is important in this maneuver, not final speed, so if I 'm ok to get higher equatorial speed then my actual orbit speed, it will look like you diagram with the red line, right ?