White Key Widths

[Avery Todd]

Alan,

I guess I've had too many Vodkas, but I could use a few less letters! :-)

But you're correct. We have 2 keyboards here made by Kluge from a rebuilders
specs and one original one from New York. The Kluge keyboard is app. 3/8" 
wider,
if I remember correctly! But I still don't like the looks of those gaps between
the keys of the older keyboards! JMHO!

Avery

At 03:21 PM 5/5/05, you wrote:

>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=3w+2b and 4W=4w+3b is with b=0.
>
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>
>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.
>
>
>
>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.
>
>
>
>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.
>
>
>
>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.
>
>
>
>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?
>
>
>
>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
>
>
>
>  f=g=a=b=72156 units            c=d=e=74606 units      discrepancy=2450
>
>
>
>  f#=g#=a#=72520 units            c#=d#=74235 units      discrepancy=1715
>
>
>
>  F=G=A=B=126546 units      C=D=E=124096 units     discrepancy=2450
>
>
>
>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".
>
>
>
>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
>