Dear Hans,
I think you may be looking for a schedule like this:
(4 5) (- -) (0 x) (- -) (- -) (8 2) (7 y) (3 6) (- -) (9 1)
(0 2) (5 6) (- -) (1 x) (- -) (- -) (9 3) (8 y) (4 7) (- -)
(- -) (1 3) (6 7) (- -) (2 x) (- -) (- -) (0 4) (9 y) (5 8)
(6 9) (- -) (2 4) (7 8) (- -) (3 x) (- -) (- -) (1 5) (0 y)
(1 y) (7 0) (- -) (3 5) (8 9) (- -) (4 x) (- -) (- -) (2 6)
(3 7) (2 y) (8 1) (- -) (4 6) (9 0) (- -) (5 x) (- -) (- -)
(- -) (4 8) (3 y) (9 2) (- -) (5 7) (0 1) (- -) (6 x) (- -)
(- -) (- -) (5 9) (4 y) (0 3) (- -) (6 8) (1 2) (- -) (7 x)
(8 x) (- -) (- -) (6 0) (5 y) (1 4) (- -) (7 9) (2 3) (- -)
(- -) (9 x) (- -) (- -) (7 1) (6 y) (2 5) (- -) (8 0) (3 4)
I have used 0 to 9 and x and y as the symbols for the 12 players so you will have to convert these back to A-M. All 12 players should appear exactly once in each row and in each column, so rows can be rounds, and columns can be machines, or vice versa. If you look down the diagonals, the cyclic construction of the schedule should be visible.
Ian.