White Key Widths

[Jim]

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Thank you . . .  I now have a headacche . . .
Jim Kinnear

----- Original Message -----=20
  From: alan and carolyn barnard=20
  To: Pianotech=20
  Sent: Thursday, May 05, 2005 4:21 PM
  Subject: White Key Widths



  In case you were wondering, here's the dilemma that keyboard makers =
have  wrestled with for hundreds of years:

  If you've ever looked closely at a piano keyboard you may have noticed =
that the widths of the white keys are not all the same at the back ends =
(where they pass between the black keys).  Of course, if you think about =
it for a minute, it's clear they couldn't possibly all be the same =
width, assuming the black keys are all identical (with non-zero width) =
and the white keys all have equal widths at the front ends, because the =
only simultaneous solution of 3W=3D3w+2b and 4W=3D4w+3b is with b=3D0.

  =20

  After realizing this I started noticing different pianos and how they =
accommodate this little problem in linear programming.  Let W denote the =
widths of the white keys at the front, and let B denote the widths of =
the black keys.  Then let a, b,..., g (assigned to their musical =
equivalents) denote the widths of the white keys at the back.  Assuming =
a perfect fit, it's impossible to have a =3D b =3D ... =3D g.  The best =
you can do is try to minimize the greatest difference between any two of =
these keys.

  =20

  One crude approach would be to set d=3Dg=3Da=3D(W-B) and =
b=3Dc=3De=3Df=3D(W-B/2), which gives a maximum difference of B/2 between =
the widths of any two white keys (at the back ends).  This isn't a very =
good solution, and I've never seen an actual keyboard based on this =
pattern (although some cartoon pianos seems to have this pattern).  A =
better solution is to set a=3Db=3Dc=3De=3Df=3Dg=3D(W-3B/4) and =
d=3D(W-B/2).  With this arrangement, all but one of the white keys have =
the same width at the back end, and the discrepancy of the "odd" key =
(the key of "d") is only B/4.  Some actual keyboards (e.g., the Roland =
HP-70) use this pattern.

  =20

  Another solution is to set c=3Dd=3De=3Df=3Db=3D(W-2B/3) and =
g=3Da=3D(W-5B/6), which results in a maximum discrepancy of just B/6.  =
There are several other combinations that give this same maximum =
discrepancy, and actual keyboards based on this pattern are not =
uncommon.

  =20

  If we set c=3De=3D(W-5B/8) and a=3Db=3Dd=3Df=3Dg=3D(W-3B/4) we have a =
maximum discrepancy of only B/8, and quite a few actual pianos use this =
pattern as well.  However, the absolute optimum arrangement is to set =
c=3Dd=3De=3D(W-2B/3) and f=3Dg=3Da=3Db=3D(W-3B/4), which gives a maximum =
discrepancy of just B/12.  This pattern is used on many keyboards, e.g. =
the Roland PC-100.

  =20

  The "B/12 solution" is best possible, given that all the black keys =
are identical and all the white keys have equal widths at the front =
ends.  For practical manufacturing purposes this is probably the best =
approach.  However, suppose we relax those conditions and allow =
variations in the widths of the black keys and in the widths of the =
white keys at the front ends.  All we require is that the black keys (in =
total) are allocated 5/12 of the octave.  On this basis, what is the =
optimum arrangement, minimizing the maximum discrepancy between any two =
widths of the same type?

  =20

  Let A, B,...G denote the front-end widths of the white keys, and let =
a#, c#, d#, f#, g# denote the widths of the black keys.  I believe the =
optimum arrangement is given by dividing the octave into 878472 units, =
and then setting

  =20

   f=3Dg=3Da=3Db=3D72156 units            c=3Dd=3De=3D74606 units      =
discrepancy=3D2450

  =20

   f#=3Dg#=3Da#=3D72520 units            c#=3Dd#=3D74235 units      =
discrepancy=3D1715

  =20

   F=3DG=3DA=3DB=3D126546 units      C=3DD=3DE=3D124096 units     =
discrepancy=3D2450

  =20

  The maximum discrepancy between any two widths of the same class is =
1/29.88 of the width of the average black key, which is less than half =
the discrepancy for the "B/12 solution". =20

  =20

  The max discrepancy is 1/358.56 of the total octave for the white =
keys, and 1/512.22 for the black keys.  Since an octave is normally =
about 6.5 inches, the max discrepancy is about 0.0181 inches for the =
white keys and 0.0127 inches for the black keys.  (One peculiar fact =
about this optimum arrangement is that the median point of the octave, =
the boundary between f and f#, is exactly 444444 units up from the start =
of the octave.)


  Just thought you'd like to know. And, no, I didn't sit down and write =
this, I ran into it looking for something else on the Internet.

  Alan Barnard
  Salem, Missouri


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