As before PG(3,2) gives a BIBD(15,35,7,3,1). The polarity is as before:

The collineation matrix for the x points is given by:

So the collineation matrix for the y points is:

The correspondence between powers of R and the x points is given by:

The correspondence between powers of S and the y points is given by:

Using these collineations, and matching lines according to the polarity, the initial blocks (0,1,4), (0,2,8),(0,5,10) for the x points are to be matched with (2,1,13), (2,0,9),(2,12,7) for the y points, in that order, all to be developed mod 15. (The last block, in each case, being partially cycled.)

This gives rise to the difference squares:

The resulting OBIBD is given by finding a transversal through the difference squares: (0,13; 1,2; 4,1), (0,9; 2,2; 8,0), ((0,2; 5,7; 10,12) developed mod 15, the last block PC(5).