Australian/New Zealand Luthiers Forum

Thanks Trevor, I had a couple of looks at k, but I arrive at the same number for a 0.012" string. Not sure where I'm going wrong. From page 4-109 k = E*A I assume E for steel to be 210*10^9 N/m^2 I calculate A for a 0.012" string to 7.29659e-8 m^2 So E becomes 15322.839 N, or the 4877.409600*Pi N gi...

Above is for no nut and saddle compensation

The dN value of 0.6 was actually calculated as 0.6*10^-3 so that part should have been correct. I think I got the straight neck geometry correct since it's resulting in very plausible values. With an 18th fret action of 2*10^-3 and a nut action of 0.1*10^-3, I get a 12th fret action of 0.00156957225...

Thanks for looking Trevor, Unfortunately it's plenty of day left on this side of the pond, so time to get cross-eyed again. I think the last plot might have been misleading since it didn't really start at 0. So here it comes again along with two plots shifting the nut 0.6mm in either direction. Defi...

Hi all, I've been crunching the equations from 4.7.3.2 in Maplesoft Maple, and am getting somewhere. Problem is I'm not getting plausible results and have stared myself blind looking for errors. I wonder if any of you can see where I went wrong? Attaching an image of the plot of the Intonation error...