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Is there a preferred orientation to approach (or leave) a space station for docking? Is it from below? above? by change in the inclination (port or starboard) from ahead or aft? I know recently the Soyuz has been using a bi-elliptic transfer orbit to get there. That makes me thing from below. I have also read that they have on occasion changed the station attitude for a docking. The pictures show the docking as always being on the earth side. But I don't know if that's because it's the best angle to approach, or because the docking section is the most massive and therefore oriented to earth by tidal forces.

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    $\begingroup$ Surely the best direction to approach from is the side the docking port is on. $\endgroup$ – Russell Borogove Aug 20 '16 at 18:36
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    $\begingroup$ The two common approaches are v-bar (Velocity Vector) and R-bar (radius vector). The wikipedia article on this is quite good, frankly too good for me to be bothered writing up an answer. en.wikipedia.org/wiki/Space_rendezvous Read the section on methods of approach. $\endgroup$ – Organic Marble Aug 20 '16 at 19:21
  • $\begingroup$ If you are approaching too fast, then I suspect approaching from above is better than approaching from below, because you want to slow down first so you should be getting yourself into a higher orbit. :-) Opposite if you are approaching too slowly. $\endgroup$ – a CVn Aug 21 '16 at 15:44
  • $\begingroup$ The V-Bar and R Bar approaches are good information. It appears that the predominant concern is to avoid collision. In both those approaches if the craft stops working it will drift away instead of into the station. The Z-Bar approach is the change of orbital inclination. It doesn't have a drift away by default component, and I assume that is why it is not used. $\endgroup$ – Johnny Robinson Aug 21 '16 at 22:49
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    $\begingroup$ @uhoh You'll have to go there and see for yourself. ;-) $\endgroup$ – a CVn Nov 16 '18 at 8:10
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I don't know about ISS procedures, specifically, but in general it'd be preferred to approach from the orbit normal or anti-normal (from above or below), assuming the station is not tumbling. This means the docking port would appear to rotate as you approach it, so you just need to set up a roll rate to match. If you approach from any other directions the docking port appears to rotate away from you, which means you need to translate as well as rotate to keep up. In practice, this rate is low (one revolution per orbit) and not a difficult problem to solve, but lacking any other factors (such as lines of sight/clearance, not sprayrng the solar phnels with your thruster exhausts, etc) then an approach along the orbit normal would be easiest.

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  • $\begingroup$ Can you expound on this "This means the docking port would appear to rotate as you approach it, so you just need to set up a roll rate to match." It sounds like you are talking about the R-bar (radius vector) but dealing with a "roll rate" is not something I'm familiar with in this case. $\endgroup$ – Organic Marble Aug 24 '16 at 18:22
  • $\begingroup$ I can try :) the orbit normal is the axis which is not the v-bar or the r-bar. The v-bar is on front of the station, or prograde. Now, if the statien is not in a controlled rotation (ie its holding a fixed attitude relatrve to the stars), one quarter of the way around the orbit you'll be 90 degrees off to one side, and half way around the orbit you'll be behind the station.... $\endgroup$ – Innovine Aug 24 '16 at 21:42
  • $\begingroup$ from your point of view, the docking port wiln always be turning away from you, even when you peint directly at it and hold stationary, a few min later it'll have turned to face away. $\endgroup$ – Innovine Aug 24 '16 at 21:43
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    $\begingroup$ I often approach along the normal vector in KSP, for the issues of geometry described here - approaches along other vectors require roll and translation to keep up with changing orientations along the orbit. But if you position one space craft a short distance normal/antinormal of another ship, within a quarter orbit they will collide if given no other thruster input - as they are in mildly different planes. This may be a concern for manned spaceflight operations - it isn't a failsafe vector. $\endgroup$ – Saiboogu Nov 16 '18 at 17:58
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    $\begingroup$ A very good point @saiboogu $\endgroup$ – Innovine Nov 18 '18 at 10:28

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