I'm going to base this question using the following assumptions:


a) static weights don't mean much in overall ball reaction;

b) x-holes mainly affect ball reaction by altering the core - that is to say that to be the most effective, an x-hole must be deep enough to remove mass from the core;

c) ball plug is a different density than coverstock, filler, and core materials;


So, based on those assumptions, suppose you have a used ball with an existing x-hole. The span is close enough that you are able to just re-slug (or even plug and re-drill into the same general area).

After throwing the ball you are not happy with the reaction, and you suspect it is the x-hole.  You weigh the ball and realize that an x-hole wasn't necessary to make the ball legal, simply there to fine tune the reaction.

If you were to plug the x-hole, would it really make much of a difference in the reaction?  

If you plug and drill a new x-hole, would the previous x-hole factor into the equation, and if so, how much?

Plug it, Put it on a determinator to find the new PSA position.

Use that when laying out your gradient line for a balance hole. For where your current balance hole placement is on the current gradient line, and use that for your basis of the reaction you'll get using the same point on the new gradient line.  

If the old is at P2, don't put the new one at P2 and expect a different reaction. If you want more dynamics, go to P3 or P4, less to P1.   If you don't check for the position of the current PSA after you plug it, you won't have the exact tools necessary to make an informed progression to another hole placement.