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<DIV> <STRONG><EM> Hi Ric</EM></STRONG></DIV>
<DIV><STRONG><EM> Hey good food for thought but if 1 mm of key movement
raises the hammer 6mm , to me that is a 6 to 1 ratio.
When I use the spurlock tool all I'm doing is using it as a
quick trouble shooting </EM></STRONG><STRONG><EM>indicator as to how much
trouble the action is going to be. Ie will longer knuckle to center be required
or capstan move.. As you stated this morning.</EM></STRONG></DIV>
<DIV><STRONG><EM> Ed ...I like your thoughts too. The use of smaller
than a 6 mm block takes us in the direction of action
acceleration.</EM></STRONG></DIV>
<DIV><STRONG><EM> Dale Erwin</EM></STRONG></DIV>
<BLOCKQUOTE
style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: blue 2px solid"><FONT
style="BACKGROUND-COLOR: transparent" face=Arial color=#000000 size=2>Hi
All<BR><BR>As long as we all remember that the <<ratio>> defined
this way (ala <BR>Spurlock) is not the same ratio relationship as
Stanwoods Strike Weight <BR>ratio we are ok here. Stanwoods ratio is about how
much weight at the <BR>key front is needed to balance whatever radius weight
the hammer has out <BR>there on the end of that shank. More specifically
we are talking about <BR>how much weight would be needed to bring the action
into the half blow <BR>condition in a frictionless world. That is not at
all the same thing as <BR>how much distance the hammer moves for any given
amount of key travel. <BR>In fact, if you were picky enough about your
measurements you would find <BR>that if you measured for a 8 mm key travel
compared to 4 mm key travel <BR>you'd end up with two different figures there
too.<BR><BR>There is an approximate translation that generally works fair
enough... <BR>and in any case if you are shooting for some BW in an action
re-do you <BR>nearly always end up within a couple grams (nearly always
heavier) if <BR>you've calculated with the distance ratio as your figure for
R. Its a <BR>good thing to gain a basic idea of your general leverage
zone by doing <BR>the Spurlock thing. But if you plug that value into
Stanwoods formula <BR>to arrive at a FW figure for some targeted BW, you are
going to have to <BR>go back and add some few grams of FW in the end. On
the other hand... <BR>the error created is consistent... so you can either
live with it and <BR>still have an even result... or try and compensate a bit
by <BR>approximating a translation from one R to the
other.<BR><BR>Cheers<BR>RicB</FONT></BLOCKQUOTE></DIV>
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