fretboard relief

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Ken, consider the following. It is so obvious! I'm surprised you didn't see it at first.

(To obtain a C 2 solution u(x, t), it will suffice for f ∈ C 2 (R), g ∈ C 1 (R).) We will separately
analyze the cases g ≡ 0 and f ≡ 0, and then use superposition. Case 1. g ≡ 0. IC  u(x, 0) = f(x)
ut(x, 0) = 0 (x ∈ R). We have u(x, t) = v(x + t) + w(x − t) for some v, w ∈ C 2 (R). By the IC, v(x) + w(x) = u(x, 0) = f(x (x) − w (x) = ut(x, 0) = 0, so v and w differ by a constant. One solution is v(x) = w(x) = 1 2 f(x). Any other solution is v(x) = 1 2 f(x) + c w(x) = 1 2 f(x) − c for some constant c. So the solution in Case 1 is
u(x, t) = 1 2 f(x + t) + 1 2 f(x − t). Remark. For a solution u(x, t) of utt = uxx, v and w are uniquely determined up to a constant.
 
After much thought..I realise that relief is all ready there... we know it as action at the 12th, usualy 2.5-3mm ...if we kept the same action as at the nut lets say 10 thou, all the way down the neck in theory it should work but this would have to be a super accurate fretboard and the string would have to be tuned very high to produce a very high vibration with minimum waveform.

If any body wants to know why there is so much activity going on in my workshop, and on this forum lately, designing new jigs and such..No! I'm not on drugs..It's just that I gave up "The Drink" about ten days ago and my brain is running in top gear at the moment.:rock:
 
I sand about .005" relief (I don't measure it, just look for light shining under a straight edge) into the FB.

Also, I use a CF rod...and a 5mm thick FB.
 
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