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<DIV><FONT face=Arial size=2>The two-straightedge method will indeed =
produce a
true and exact circular arc and is an ingenious way to draw large =
curves.
And calculus is not involved; it is simple, high-school geometry.
</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=Arial size=2>If you view the system as an arc just =
from the left
pin to the top pin and draw lines representing the straightedges, from =
these
two pins to any pencil point on the curve, it is more clear what is =
going
on. Drawing lines from one end of a chord to a point on the =
curve and
then back to the other end of the chord makes a triangle. No =
matter what
point on the curve you choose, the angle between the two lines is =
always
the same for a circle. This is a basic theorem of
geometry. It can be proven simply if you note that lines =
drawn from
the <EM>center</EM> of the circle to each chord-end and to the =
pencil
always creates two isosceles triangles.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Locating one straightedge parallel to =
the chord and
one across the two pins is just a clever way to set them up for our
circle. The top straightedge is thus tangent to our =
circle at
the top pin and the second straightedge is simply another line =
from a
point along the circle to the pencil (now literally at the top
pin).</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Now the question is: does the curve =
<EM>need</EM>
to be a true circular arc? It's hard for me to believe that slight =
discrepancies in an arc of such large radius and short length could =
really
cause any noticable difference in the performance of a =
soundboard.
</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT><FONT face=Arial =
size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Don A. Gilmore</FONT></DIV>
<DIV><FONT face=Arial size=2>Mechanical Engineer</FONT></DIV></DIV>
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<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>
<A title=mfarrel2@tampabay.rr.com
href="mailto:mfarrel2@tampabay.rr.com">Farrell</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A =
title=pianotech@ptg.org
href="mailto:pianotech@ptg.org">Pianotech</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, December 11, =
2003 6:00
AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Rib =
dimensions</DIV>
<DIV><BR></DIV>
<DIV><FONT face=Arial>Lots of good methods for calculating the =
radius of an
arc have been provided. But there is also the question of whether the =
curve is
a true arc or some other shape (this assumes you have a specific shape =
as a
target). Most of the provided methods do not address that concern - in =
fact
you could have an obtuse angle with two straight sides rather than an =
arc.
That's why I suggested making a number of measurements along the curve =
-
offsets from a straight line.</FONT></DIV>
<DIV><FONT face=Arial></FONT> </DIV>
<DIV><FONT face=Arial>Terry Farrell</FONT></DIV>
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style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>
<A title=absolutepiano@comcast.net
href="mailto:absolutepiano@comcast.net">Absolute Piano</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A =
title=pianotech@ptg.org
href="mailto:pianotech@ptg.org">pianotech@ptg.org</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Wednesday, December 10, =
2003 4:40
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Rib dimensions</DIV>
<DIV><BR></DIV>
<DIV><FONT face=Arial size=2><FONT face="Times New Roman"
size=3>Hello,<BR><BR>I'm trying to apply some science to my =
soundboard rib
making and I am<BR>looking for "tables of static values for =
the
Resisting Moment (W) and the<BR>Moment of Inertia (I) for all the =
possible
cross sections of sugarpine and<BR>spruce (DIN 1052 Class I will
suffice).<BR><BR>What is the formula for converting pounds/inch =
squared to
kg/cm squared?<BR><BR>Given a right angle connected to the outside =
of an arc
of a circle, how do<BR>you prove the circle is 60'? (I made a jig =
for
crowning ribs that is<BR>adjustable and I want to calibrate
it.<BR><BR>Thanks,<BR><BR>Jude Reveley,
RPT</FONT><BR></FONT></DIV></BLOCKQUOTE></BLOCKQUOTE></BODY></HTML>