right, it's not easy to solve!

I try here to calculate the final number ...but I'm not sure of the result...

at 250 km the orbital circumference is about 41,600 km

with an orbital speed of 28,000 kmh the LSAM/EDS will do a full orbit in about 90 minutes

I think that the "vertical tolerance" (with orbital inclination) may be larger than the "width tolerance" (because, if the CEV will enter the right orbit, it can use its engines to approach the LSAM/EDS), maybe +/- 3000 km (?)

Then, in the 90 minutes of the LSAM/EDS orbit, only 15% of them can be used to have a CEV/LSAM rendez-vous: less than 15 minutes per orbit

Now there is the most difficult number to find: "for how many hours (of the 18 max hours calculated in the previous post) the CEV/CLV will be in the right position (with the right horizontal and vertical/orbital max tolerance) to be launched so it can reach the LSAM/EDS?"

A simple solution may be: 15% of 18 hours, then less than 3 hours in total!

Probably it's not the right figure, probably the launch width and orbital tolerance may be wider (or closer...), etc.

But 3 hours ONLY in TOTAL (on 95 days of LSAM/EDS max orbital loither time) to launch the CEV are a very little time available to have a probability of successful VSE moon missions higher than ZERO!!!

.