I need help with saddle compensation equation 4.7-3

[jonwallace]

Hi,

Equation 4.7-3 which calculates the cents tonal shift for an applied saddle compensation, seems to give a result that is different by a factor of 2 when compared to calculating the value using another technique.

For example, at fret 12 for a scale length of 645, nut compensation of 0 and saddle compensation of 1.85, the tonal shift is calculated to approx 10 cents.

However, I've read elsewhere that the approximate saddle compensation can be calculated by multiplying the cents offset at the 12th fret by 1/100 of the distance between the nut and the first fret (Frank Ford formula). This is double what you would expect, but is needed to take effect of the change of string length on the open string. So for my example a tonal shift of 10 cents requires saddle compensation of 10 * 36.2 / 100 = 3.62.

Can anyone help me make sense of this?
Thanks
Jon

[jonwallace]

Also, just noticed that at the bottom of page 4-101 it says that for a scale length of ~650, 1 cent is equivalent to 0.3753mm change in length, but feeding this value into 4.7-3 gives a tonal change of 2 cents.

Can anyone help me make sense of this?
Thanks
Jon

[Trevor Gore]

Hi Jon;

I assume you're referring to Frank's article here:

http://www.frets.com/FretsPages/Luthier ... pcalc.html

We're calculating different things. Frank is figuring out how much compensation is required, I'm giving a formula that tells you how much tonal shift you get by changing the length of a string.

So to explain it in Frank's (actually Greg Neaga's) terminology:

"Let's assume the pitch at the 12th fret is one complete semitone too flat." If you play the 13th fret, you're fine. So you've shortened the string length by the spacing between the 12th and 13th fret, which is half the distance between the nut and first fret. Different from figuring out how much compensation is required.

Also, just noticed that at the bottom of page 4-101 it says that for a scale length of ~650, 1 cent is equivalent to 0.3753mm change in length,...

That's correct. In Frank's terminology it's ~1% of the nut to first fret spacing (but Frank is making a linear interpolation, which is why it doesn't equate to exactly 1%...)

...but feeding this value into 4.7-3 gives a tonal change of 2 cents.

...probably because you're working it out for the 12th fret length rather than the open string.

Works fine when I tried it.

[jonwallace]

Trevor,

Thanks for your helpful reply.

I had thought these two methods were the reverse of each other - I can now see that they're not.
I was trying to use equation 4.7-3 to feed in different saddle compensation values and see the effect in cents change at each fret.

Thanks
Jon