<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 6.00.2800.1276" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=#ffffff> <DIV><FONT face=Arial size=2>Hey Jude:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>The area moment of inertia is only = dependent on the crossectional shape and size and not on the material.  So if you = can find a standard set of formulas, they are good for any material.  Most = common crossections are here:</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2><A href="http://www.efunda.com/math/areas/Common_Geometric_Shapes_Index.cf= m?search_string=moment%20of%20inertia">http://www.efunda.com/math/areas= /Common_Geometric_Shapes_Index.cfm?search_string=moment%20of%20inertia<= /A></FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>You can also easily calculate compound = crossections made up of any combination of these fundamental ones.  I can show = you how to do that too, if you're interested.  </FONT><FONT face=Arial = size=2>The units should come out in "length to the fourth power", like inches^4, or = m^4.  Be careful not to use the "<EM>mass</EM>" moment of inertia as that is used for dynamic calculations and is in units like = lbm-ft^2, slug-ft^2, or N-m^2.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>One psi = .0703 kg/sq.cm</FONT></DIV> <DIV><FONT face=Arial size=2>O</FONT><FONT face=Arial size=2>ne = kg/sq.cm = 14.23 psi.</FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>I don't think I understand your angle question.  60' should mean "sixty arc-minutes", which is one = degree.  What does your jig look like?</FONT></DIV> <DIV><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><FONT face=Arial size=2></FONT> </DIV> <DIV>----- Original Message ----- </DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <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 3:40 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></BODY></HTML>