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<DIV>or tan-1 of .625"/16" = 2.24 degrees. </DIV>
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<DIV>Vicente "Vince" Bortone</DIV>
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<BLOCKQUOTE style="PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px solid">-------------- Original message -------------- <BR>From: "Scott Smith" <js.smith@verizon.net> <BR><BR>> Assuming a 16" prop, 2.24 degrees. <BR>> <BR>> Just plug your numbers in here: <BR>> http://www.carbidedepot.com/formulas-trigright.asp <BR>> <BR>> side A: .625 <BR>> side B: 16 <BR>> Click calculate <BR>> Angle A is your answer <BR>> <BR>> -----Original Message----- <BR>> From: nsrca-discussion-bounces@lists.nsrca.org <BR>> [mailto:nsrca-discussion-bounces@lists.nsrca.org] On Behalf Of <BR>> adriancwong@earthlink.net <BR>> Sent: Monday, February 19, 2007 5:13 AM <BR>> To: NSRCA Mailing List <BR>> Subject: Re: [NSRCA-discussion] Thrust line calculation <BR>> <BR>> To all, <BR>> <BR>> Does anyone know how to translate from inches to degrees in thrust line <BR>> calculation. <BR>> <BR>> For example, from the stab to th
e left tip of the prop is 58-11/16", and the <BR>> other side is 58-1/16". Therefore, the difference is 5/8 ", what would be my <BR>> degree of right thrust? <BR>> <BR>> Thanks in advance, <BR>> <BR>> Adrian <BR>> _______________________________________________ <BR>> NSRCA-discussion mailing list <BR>> NSRCA-discussion@lists.nsrca.org <BR>> http://lists.nsrca.org/mailman/listinfo/nsrca-discussion <BR>> <BR>> <BR>> _______________________________________________ <BR>> NSRCA-discussion mailing list <BR>> NSRCA-discussion@lists.nsrca.org <BR>> http://lists.nsrca.org/mailman/listinfo/nsrca-discussion </BLOCKQUOTE></body></html>