Calculating Stop Combinations, Part II
(con't from Part I)
Here's the formula:
X = 2 - 1 where
X = the number of possibilities
n = the number of stops
Using this formula ( 2 to the nth power minus one ) all the organist has to do to find out how many possible ways the stops of the instrument ( or the couplers, as a group) can be used singly or in combination is to replace the exponent in the formula with the number of stops (or couplers) provided by the builder, and solve.
For example, let's say an organist has at their disposal a 2 manual organ equipped with 7 stops in the Great division, 9 stops in the Swell, and 5 in the Pedal, 21 stops in all. Applying the formula ( 2 to the nth power minus one ) where n is 21, there are over 2 million possibilities, as follows:
X = 2 - 1 = 2,097,152 - 1 = 2, 097,151
Granted, the majority of these random combinations would sound horrid and find no use at all, but presuming that only 0.1% of them were worth storing on piston memory, that still leaves over 2,000 good sounding, useful combos, nearly 40 for every weekly worship service of the year. The point being, even an unpretentious instrument of 21 stops still provides the artist organist an almost unlimited capability of beautiful landscapes of compound tones, all at his or her fingertips.
What if we had in front of us a very large instrument with 4 times as many stops, where would THAT takes us? Let's find out ...
(con't in Part III)