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. > ><?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" /> > >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 >
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