What is the best way to find the pad pitch when the design has 27000+ pads? Using the (combine relation1 relation2) operator crashes my session as Ravel attempts to store 729 million+ combinations. Even if I break the design into quadrants to reduce the number of pads being looked at, Ravel still has to deal with 48 million+ combinations. Also, looking only at the pads nearest the design edge doesn't give me the result I'm looking for. I need to find the minimum pad pitch in the design.
Do you have any suggestions?
the issue is that in order to find the minimum distance between any two pads, all combinations of pads have to be checked, which in your case results in a combinatorial explosion. If you can determine a cutoff value -- a value as small as possible but guaranteed to be at least as large as the pad pitch, you can first do a selection of pad pairs where the pads are within the cutoff distance of each other. This would result in comparatively few pad pairs, from which you can then find the minimum pitch. This is more efficient, since in order to find objects within a certain distance to each other Ravel uses a very efficient algorithm which arranges objects according to proximity and therefore does not have to process every possible combination.