Given below and elseware are very good examples of what I hope will be called "T Math" or "tech math" Which is math that is taken as true if it works out on the calculator. Ok "C math" but that confuses with C the programing language. Take the proposition that halving the diameter of the wire gives an octave. Simple to prove on a calcualtor. I suppose it is also evident when looking at the formula (Taylor's) solving for frequency. F = T^(1/2)/L*d*.0479 That F bears an inverse relation with both L and d. So if d is divided by 2, (or multiplied by 1/2) F is multiplied by 2. Anyhow it works out on the calculator which is the final proof for the technician---ergo, T Math. That the semitone is related by 6% or .06 is evident when we consider the ratio of the semitone (ET) is 1.059463636. Now if we (as technicians) knock off the 1 we are left with .05946 (this flies in T math) which is darn close to .06, or 6%This is of course an "equal semitone" in that multiplying the starting frequency by this value 12 times will end up with twice the frequency or an octave. Or F*1.0594^12 = 2F or 1.5945^12 = 2.00... Way back in the 1500's as reported by Mersenne the lute makers used 18/17 as the proportion to space their frets. Using Steven's semitone formula this gives a.989 semitone. It can assumed (I think) that the pressure resulting from the space between the fret and the finger board made up the difference. How the 18/17 was arrived at I do not know. Mersenne was sometimes vague in his explainations. Perhaps back then the superparticluar ratios had special significance. A superparticular ratio is where the numerator is one greater than the demoninator. "the greater term exceeds the less by a unit".(Web. III) 3/2, 4/3, 5/4, 6/5 are examples. Since the Fifth, Fourth, Third, and Minor Third were superparticulars, I am sure a superparticular was sought for the semitone. For the lute makers at least, such a ratio was vital to their design. Otherwise the frets would not be spaced "evenly", (even proportion) resulting in intervals that are not the same from fret to fret. The notion that cents = semitone/100 is not correct. One cent is actually the 100th root of a semitone, just as the semitone is the 12th root of the octave. (2/1) More correctly a cent is the 1200th root of an octave. or 2^(1/1200) However in the context below it may be a typo if referring to a semitone expressed as a root of 2---then the cents of that would be such a semitone times 100. (instead of /100) ---ric >Two factors are related by 1 > semitone when their ratio is about 6%, related by a cent when their ratio > is about 0.06%. (For tensions or weights you have to halve these numbers). > > Cents or semitones can be used to relate any factors - pitches, string > lengths, gauges, densities, tensions. The simple formula is > > semitones = 12*(log A/B)/(log 2) for lengths, diameters, pitches > and > semitones = 6*(log A/b)/(log 2) for densities or tensions > > cents = semitone/100 > > e.g. two wires of diameters 0.7 and 0.75 are 12*(log 7.5/7)/(log 2) = > 1.19 semitones apart. > > e.g. brass (density 8536 kg/m^3) and steel (density 7769 kg/m^3) are > 6*(log 8536/7769)/(log 2) = 0.8 semitones apart. > Stephen Birkett
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