Lifter Bore Measurements

Thanks Pishta.
But we aren't talking about the load carrying surfaces. Only the leakage points, right? We're looking at computing the area of a disc that is .906" OD(estimating .002" clearance) and an ID of .904, X 2 (for the top and bottom). I'm assuming no pressure losses at the internals of the lifters.
I did it this way: Area of the bore minus area of the lifter or crank journal
Area of a circle is PixR^2 so (.904/2)^2 x 3.14 = .6415sq" Area of the bore: (.906/2)^2 x 3.14 = .6443sq" Bore area - lifter area: .6443-.6415 = .003sq", x2 for the top and bottom = .0057sq" for oil to leak.

For a typical main bearing with .002" clearance, assuming no taper and a round bearing shell ID (not technically the case but an assumption here):
Area of main bore: (2.50/2)^2 x 3.14 = 4.9062sq" Area of crank journal: (2.498/2)^2 x 3.14 = 4.8984sq" Bore area minus crank area: 4.9062 - 4.8984 = .0078sq", x2 for front and rear edges = .0156sq" for oil to leak.

Now those are both one in a system. So it looks like if you had 16 lifters with the same clearance, you would have the potential for pressure loss of .0912sq".
So now do we add the rod bearing leakages to the main bearing side of the equation? They are "down stream" so I'm not sure if that's a valid add.

Adding the other four mains we have .0780sq" of leakage with only the mains. So lifter leak more per the math I understand. Unless it's totally wrong - which I'd say is a solid 50/50 at this point...lol