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<DIV>In a message dated 12/18/2004 8:43:56 AM Pacific Standard Time, Erwinsp=
iano
writes:</DIV>
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<DIV><FONT size=3>Hi Jon</FONT></DIV>
<DIV><FONT size=3> Nice posting.</FONT></DIV>
<DIV><FONT size=3> I also have sorted shanks by weight but n=
ot to
this extent. I'd like to try this. My thought though is that with the Abel=
shanks, they are already thinned down & the differences in weight coul=
d
be, say more in the flange than at the tip which would not give any differ=
ence
of strike weight.</FONT></DIV>
<DIV><FONT size=3> If you measure strike weights on th=
e
Stanwood scale setup are you finding this method to coincide with the over=
all
weight of the shank & flange & its contribution to a difference in=
strike weight as you describe in the example below. </FONT></DIV>
<DIV><FONT size=3> Very interesting & thoro. As usual
Jon.</FONT></DIV>
<DIV><FONT size=3> Merry Christmas</FONT></DIV>
<DIV><FONT size=3> Dale Erwin</FONT></DIV>
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e=2>My
reasoning for this is the jaggedness of a hammer weight curve and shank
<BR>SW's. Let's say that one were to<BR>simply install shanks right out =
of
the box in order. A 2 g shank could be <BR>place next to a 1.6 g shank. =
Now
what if<BR>an 8g hammer were set on the first and a 7.6 on the second. T=
heir
<BR>respective SW's would be 10 and 9.2; almost<BR>a gram! However, if t=
he
shanks were swapped then each SW would be 9.6. A <BR>much better solutio=
n
than having to<BR>sand down the first and add lead to the second to achi=
eve
the same result.<BR><BR>If one were to really indulge in the inertia cra=
ze
then you'd have to <BR>weight the hammers first to a curve to mate<BR>wi=
th
similar weight shanks. More work than I think is really necessary
<BR>since that inertial effect is probably<BR>lost through the lever
train.<BR><BR>I better get to work...<BR><BR>Jon<BR><BR>PS Merry
Christmas<BR></FONT></BLOCKQUOTE></DIV>
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