<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML style="FONT-SIZE: x-small; FONT-FAMILY: MS Sans Serif" xmlns:o = = "urn:schemas-microsoft-com:office:office"><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 6.00.2900.2627" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=#ffffff> <DIV>Thank you . . .  I now have a headacche . . = . Jim Kinnear</DIV> <DIV> </DIV> <DIV>----- Original Message ----- </DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black">From: <A title=tune4u@earthlink.net = href="mailto:tune4u@earthlink.net">alan and carolyn barnard</A> </DIV> <DIV style="FONT: 10pt arial">To: <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">Pianotech</A> </DIV> <DIV style="FONT: 10pt arial">Sent: Thursday, May 05, 2005 = 4:21 PM</DIV> <DIV style="FONT: 10pt arial">Subject: White Key Widths</DIV> <DIV> </DIV> <DIV>In case you were wondering, here's the dilemma = that keyboard makers have  wrestled with for hundreds of years:</DIV> <DIV> </DIV> <DIV> 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=3w+2b and 4W=4w+3b is with = b=0. <o:p> </o:p> 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 = b = ... = = g.  The best you can do is try = to minimize the greatest difference between any two of these keys. <o:p> </o:p> One = crude approach would be to set d=g=a=(W-B) and b=c=e=f=(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=b=c=e=f=g=(W-3B/4) and d=(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. <o:p> </o:p> Another solution is to set c=d=e=f=b=(W-2B/3) and g=a=(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. <o:p> </o:p> If = we set c=e=(W-5B/8) and a=b=d=f=g=(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=d=e=(W-2B/3) and = f=g=a=b=(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. <o:p> </o:p> 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? <o:p> </o:p> 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 <o:p> </o:p>  f=g=a=b=72156 = units            </= SPAN>c=d=e=74606 units      discrepancy=2450 <o:p> </o:p>  f#=g#=a#=72520 = units           &n= bsp;c#=d#=74235 units      discrepancy=1715 <o:p> </o:p>  F=G=A=B=126546 = units      = C=D=E=124096 units     discrepancy=2450 <o:p> </o:p> 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".  = <o:p> </o:p> 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.)</DIV> <DIV> </DIV> <DIV>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.</DIV> <DIV> </DIV> <DIV>Alan Barnard</DIV> <DIV>Salem, Missouri</DIV> <DIV> </DIV> </BLOCKQUOTE></BODY></HTML>