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<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>I’m not sure exactly what you mean,
but in the case of an angled capstan, the effective load will not be where the
capstan enters the key, but at a point on the key which is directly below where
the capstan and wippen heal contact point is. In other words, if you want
to straighten an angled capstan and maintain the same key ratio, you should drill
for the new capstan position directly below where the capstan/wippen heal
contact point is. If you straighten the capstan but keep the hole
location the same, you will reduce the key ratio. Don’t know if
that relates to the issue at large as I haven’t been following the thread
enough lately to comment on whether the analogy is a good one or not.</span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'> </span></font></p>
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<p><font size=2 color=navy face="Times New Roman"><span style='font-size:10.0pt;
color:navy'>David Love<br>
davidlovepianos@comcast.net<br>
www.davidlovepianos.com</span></font><font color=navy><span style='color:navy'>
</span></font></p>
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<p class=MsoNormal style='margin-left:.5in'><font size=2 face=Tahoma><span
style='font-size:10.0pt;font-family:Tahoma'>-----Original Message-----<br>
<b><span style='font-weight:bold'>From:</span></b> caut-bounces@ptg.org
[mailto:caut-bounces@ptg.org] <b><span style='font-weight:bold'>On Behalf Of </span></b>Keith
Roberts<br>
<b><span style='font-weight:bold'>Sent:</span></b> Friday, November 02, 2007
2:15 PM<br>
<b><span style='font-weight:bold'>To:</span></b> </span></font><font size=2
face=Tahoma><span style='font-size:10.0pt;font-family:Tahoma'>College and
University Technicians</span></font><font size=2 face=Tahoma><span
style='font-size:10.0pt;font-family:Tahoma'><br>
<b><span style='font-weight:bold'>Subject:</span></b> Re: [CAUT] Digest, Vol
1103, Issue 85 Moving Wippen Rail</span></font></p>
<p class=MsoNormal style='margin-left:.5in'><font size=3 face="Times New Roman"><span
style='font-size:12.0pt'> </span></font></p>
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<p class=MsoNormal style='margin-left:.5in'><font size=3 face="Times New Roman"><span
style='font-size:12.0pt'>A question back to you. A balance type scale. It has a
platform 4 inches across. the weight doesn't change due to where you put the
load on the platform. It is still transfered back to the point on the lever
that the platform is attached. According to you if we move the load farther
from the fulcrum the weight should change. The design of the intermediary
device has a lot to do with it I think. </span></font></p>
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<p class=MsoNormal style='margin-left:.5in'><font size=3 face="Times New Roman"><span
style='font-size:12.0pt'>Keith<br>
<br>
</span></font></p>
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<p class=MsoNormal style='margin-left:.5in'><span class=gmailquote><font
size=3 face="Times New Roman"><span style='font-size:12.0pt'>On 11/2/07, <b><span
style='font-weight:bold'>Richard Brekne</span></b> <<a
href="mailto:ricb@pianostemmer.no">ricb@pianostemmer.no</a>> wrote:</span></font></span>
</p>
<p class=MsoNormal style='margin-right:0in;margin-bottom:12.0pt;margin-left:
.5in'><font size=3 face="Times New Roman"><span style='font-size:12.0pt'>Hi
Keith<br>
<br>
Been thinking on an off about this during the day and it dawned on me<br>
that your idea of the jack center being the determinant for the output <br>
arm length and that the angle the jack takes makes no difference is<br>
analogous to the position of the capstan hole on the key being the<br>
determinant for the output arm length of a key and that it matters
not<br>
which angle you insert the capstan. <br>
<br>
I'll leave it to you to take it from there.<br>
<br>
Cheers<br>
RicB<br>
<br>
</span></font></p>
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style='font-size:12.0pt'> </span></font></p>
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