Probably the most common recovery technique for larger (too large for featherlight recovery) Sputnik-style oddrocs is rear ejection. The Moonik uses the more-complicated "ported ejection." I was attracted to the design for the same reason that John was; it's an interesting approach.
I'm sure that there's a formula for calculating the bevels that need to be cut for the branching tube, but I derived it with a little bit of applied logic and a whole lot of trial and error. I can explain it visually with diagrams in a heartbeat with a drafting board, T-square, protractor and a pencil, but doing it with software is much more of a PITA. Let's see if I can draw a picture with words.
Imagine that you have a vertical pipe, and want to merge another one of the same diameter with it at right angles midway along its length. You don't have a T-coupler, so you have to do it the hard way. You can simply butt the second pipe up against the first one, but it only contacts it in two places, leaving gaps everywhere else. you need to be able to push the second pipe into the side of the first one.
For reasons that would be obvious if I could draw it, you want to be able to push the end of pipe B into the side of pipe A until it is halfway through it. Remember that the two pipes are the same diameter. Now once you have done that, you want to trim off the portion of pipe B that is actually inside of pipe A, leaving only the edges where the two pipes touch. After you have done that, you pull pipe B back out. What does the end look like? It has two half-circles cut out of the end at 180° to each other. (We are just going to focus on the changes to pipe B.) If you looked at the cuts in profile, you would see that the pipe actually has a symmetrical double-bevel on that end.
Now go back and imagine that after you push pipe B halfway into the side of pipe A, you twist it down so that it is no longer perpendicular. Then if you trimmed off the part of B that was inside of A, the end of B would still have two opposing bevels on it, but they unlike the first example, they would not be the same size. The upper bevel would be shallower than a perfect semicircle, and the lower one would be elliptical in shape. In fact, the upper bevel decreased by exactly the same amount that the lower one steepened. Their bevel angle would no longer be equal, but the sum of both of them would still be equal to the sum of their angles when B was perpendicular to A, which was 180° (90° + 90°).
So if, say, you angle B down so that it meets A at an (inside) angle of 60°, then the lower (acute angle) side of B is 60° to A and the upper (oblique angle) side is 120° to A. That is because not as much of the upper side is inside of A as it was in the first example, and more of the lower side is in it.
Next post - how to cut the bevels.