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In a keyboard disvision based on the width of the keys at the rear of the keyboard, let

a = the width of a black key

b = the width of the tail of a white key between a black key and another white key

c = the width of the tail of a white key between two black keys.

Then the width x of all of the keys from C through E is given by

x = 2a + 2b + c,

and the width of the head of a white key from C through E is

f = x/3 = (2a +2b + c)/3

The width y of the keys from F through B is given by

y = 3a + 2b + 2c,

and the width of the head of a white key from F through B is

g = y/4 = (3a +2b + 2c)/4.

The total octave span w of the keyboard is given by

w = 5a + 4b + 3c, which is typically about 6.5 inches or 16.5 cm.

In the example given in the main article, these numbers all fortuitously stand in small integer ratios to one another, but they need not. Since there are three independent variables, we need a system of three equations to solve for the preferred widths of the keys. The three equations most likely to be used are those for b, c and w above: we know what the octave span is to be and how much room we need for the tails of white keys, and we solve for a, the width of the black keys.

[contact John S. Allen by e-mail]

Last revised 13 May 1997