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<blockquote style="border-left: 2px solid blue; padding-left: 3px;">but
will be in balance with string tension. </blockquote>
<div> <font face="Arial, Helvetica, sans-serif">...along with all of the friction points in the string system, as well, particularly the front segments. With string tension, the pin is twisted axially and in a diminishing amount of torque to the bottom of the pin depending on the friction of the system in the block. The friction and stability of the friction system of the counterbearing felt, counterbearing contact, agraffe or capo will be additive to the perceived "loading". Ron is exactly right that there is no way to tell what twist component remains in the pin because of all of the above complications. I think I'm interpreting what you're saying correctly, Ron. <br>
<br>
Paul<br>
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I'd like to clear up a point of basic physics here. You're not
leaving the pin with no twist in it, you're ideally leaving it
with whatever twist it takes to counter the opposing twisting
force from string tension. That's the equilibrium you're
after. You have no way to tell what twist is still in the pin
because you can't measure it directly. The only indications
you get are what you feel, and what you hear (or see on your
ETD) as the string pitch changes. When you leave the pin at a
point where it takes the same or very slightly more force on
the tuning hammer to lower pitch X amount that it does to
raise pitch by X amount, you're in the ball park. At that
balance of forces, the pin will be left with some twist, but
will be in balance with string tension. <br>
Ron N <br>
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