Since the piston is not making actual contact with the cheese, I don't understand why the bore size comes into play. On a French press, the shape or size of the ramrod pushing down on the follower doesn't (to me at least) make any difference. It's the follower that's pushing on the cheese.
For simplicity, assume my mold has a surface area of 50 sq. inches. If I put a 50 pound weight on the top of the follower, the FOLLOWER, not the weight, would be transferring 1 psi (50 pounds / 50 sq. inch = psi) onto the surface of the cheese. The shape (or diameter) of the weight doesn't make any difference. The ramrod from the hydraulic unit is just a different source of "weight" i.e. pressure. Without the follower, the ramrod would easily go right thru the cheese, but the follower disperses that weight evenly over its surface area.
The end location mark on the lever of my French Press delivers a 5 to 1 mechanical advantage. If I hang 10 pounds of lead, sand, or whatever, I get 50 pounds at the ramrod. (I have used a very accurate commercial scale to take measurements). So, if I used a hydraulic ram to push down the same 10 pounds at the end of that lever, I would still get the same 5 to 1 advantage with 50 pounds pushing down on the cheese. The bore size doesn't have any effect. Neither does the lever. It's the surface area of the follower as determined by the diameter of the mold. The PSI for 5o pounds applied to the follower of a 4" mold is very different than the PSI on an 8" mold.
It is obvious that recipes need to talk about PSI and not just weight.