1
$\begingroup$

I am writing a document about orbital mechanics am trying to depict the different docking approaches in a simple way. Since I am using paint.net to draw the pictures, my drawings are not that great.

The 3 main docking approaches I am trying to depict are:

  • R-Bar
  • V-Bar
  • Z-Bar

The Wikipedia definition of a R-Bar docking sequence is this:

The R-bar approach consists of the chaser moving below or above the target spacecraft, along its radial vector. The motion is orthogonal to the orbital velocity of the passive spacecraft.

From what I understand, it means that after the chaser and the targeted spacecraft have rendezvous, the chaser has to fly higher or lower to catch up with the satellite. If the chaser is in front of the satellite, the chaser will speed up to be in a higher orbit where it will move slower than the satellite so that satellite can catch up. If the chaser is behind the satellite, it will slow down to be in a lower orbit where it can catch up to the satellite.

So I made these 2 drawings depicting an R-Bar approach: R-Bar maneuver if the chaser is behind the satellite R-Bar maneuver if the chaser is ahead the satellite

In the images, the "chaser" is called spaceship.

The definition from Wikipedia about a V-Bar approach is also somewhat understandable.

That is, from behind or from ahead, and in the same direction as the orbital motion of the passive target. The motion is parallel to the target's orbital velocity. In the V-bar approach from behind, the chaser fires small thrusters to increase its velocity in the direction of the target.

I think the V-Bar approach is similar to an R-Bar approach, but instead of using the advantage of flying slightly higher or lower, the chaser fires the RCS in the radial or anti-radial direction to keep it level with the satellite.

V-Bar approach

My question is about a Z-Bar approach, since I don't fully understand the definition of Wikipedia:

An approach of the active, or "chaser", spacecraft horizontally from the side and orthogonal to the orbital plane of the passive spacecraft

What I think Wikipedia is saying that the chaser comes in orthogonal to the orbital plane, meaning in the normal/ anti-normal direction of the satellite. However, I am not sure if I understand this correctly. I tried to depict what I understand a Z-Bar maneuver is below.

enter image description here

For clarification: In the image above, the spaceship is not underneath the satellite flying upwards, even if it might appear so. Instead, the spacecraft is meant to be on a different plane heading towards the satellite in the normal/anti-normal direction.

I drew this second image to demonstrate what I meant in the picture above: enter image description here

What I thought a Z-Bar approach is, is when a spacecraft docks to a satellite, but comes in at a different inclination.

Q: Is this depiction of a Z-Bar approach accurate? Or did I misunderstand the definition of a Z-Bar approach.

$\endgroup$
4
  • $\begingroup$ I suspect the only thing you're misunderstanding is how different the orbital planes are. They're never exactly the same. If you want a 7000km radius circle to end up a couple hundred meters apart out-of-plane so you can start an approach that's a relatively small angle (displacement = radius * sin (angle)). Then you maneuver to gradually make the inclinations match. $\endgroup$
    – Erin Anne
    Dec 27, 2023 at 21:39
  • $\begingroup$ Perhaps you'd be helped by reviewing the diagrams here: jamesoberg.com/… $\endgroup$ Dec 27, 2023 at 23:32
  • $\begingroup$ @ErinAnne, I understand what different orbital planes are. That isn't the issue. My question is rather what is the definition of a Z-Bar approach. Is a Z-Bar approach when the targeted satellite and chaser are on 2 different planes with a very slight inclination in comparison to each other, which have a point of intersection. So from the targeted satellite perspective, the chaser would be coming in from a normal/anti-normal direction. Is this how a Z-Bar docking would happen? Since that is what I tried to draw in paint.net, or is a Z-Bar docking something else. $\endgroup$ Dec 28, 2023 at 8:29
  • $\begingroup$ @ErinAnne In the images about a R-Bar and V-Bar approach, the chaser and satellite were already on the same orbital plane, unlike in the Z-Bar approach image. In the document I am writing, I already showed how the chaser would catch up to a satellite and how it would change its inclination to be on the same orbital plane. In the Z-Bar picture I drew the chaser's trajectory in a different colour to try and show that in the Z-Bar picture, the chaser is not on the same orbital plane, but has a slight inclination. $\endgroup$ Dec 28, 2023 at 8:37

1 Answer 1

1
$\begingroup$

Since your comments indicate that all you want to know is if the chaser would be coming from the normal or anti-normal direction: yes, the Z-bar is directed in the out-of-plane direction from the target, per the citation linked to from that part of the Wikipedia article: Figure from the following link, with the caption saying (among other things) that the Z-bar is defined along the out-of-plane direction

(that citation is a group's Master Thesis about a space fabrication platform design by Bessel et al, 1993)

I'm not professionally familiar with the "Z-bar" nomenclature, perhaps because ISS rendezvous simply didn't exercise it, though PMA-3 was stored along the port side of ISS for a while.

My unsolicited opinion: not only does your Z-bar rendezvous picture initially read as an altitude-crossing (as you note), but even once you understand that they're supposed to be co-altitude the exaggerated plane crossing angle makes it look like a potential-impact scenario, not a rendezvous. Any such rendezvous will in reality be nearly-coplanar, with a small crossing angle, to limit the delta-V necessary to achieve rendezvous. The Z-bar portion is only important in close to the rendezvous target.

$\endgroup$
1
  • $\begingroup$ Thank you so much for this answer!! You do make a good point that in my images, it looks like an impact zone. I could try to change that, making the inclination angle between the 2 orbital planes a little less in the image. $\endgroup$ Dec 28, 2023 at 9:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.