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</o:shapelayout></xml><![endif]--></head><body bgcolor=white lang=EN-NZ link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>This is purely from my head and so may be very wrong but seems correct.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Anyway, assuming that by center, you mean the centre of the ellipse / arc id do the following (sorry no code)<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I’d calculate the Tangent Vector at the Start and End Points and then interpolate a set of direction vectors between the two tangents, then scale the direction vectors and apply to the start point as a set of movements.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>No trig required, I believe.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>TTFN,Struan,<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0cm 0cm 0cm'><p class=MsoNormal><b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> delphi-bounces@delphi.org.nz [mailto:delphi-bounces@delphi.org.nz] <b>On Behalf Of </b>Marshland Engineering<br><b>Sent:</b> Tuesday, 18 January 2011 21:03<br><b>To:</b> delphi@delphi.org.nz<br><b>Subject:</b> [DUG] Arc's<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Is there an easy way of drawing an arc ? I have center, start and end co-ordinates as XY. </span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>From what I can see, I need to look at least 16 combinations to work out the perimeter size depending in which quadrant the start and ends are in. </span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>procedure Arc(X1, Y1, X2, Y2, X3, Y3, X4, Y4: Integer);</span><o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Use Arc to draw an elliptically curved line with the current Pen. The arc traverses the perimeter of an ellipse that is bounded by the points (X1,Y1) and (X2,Y2). The arc is drawn following the perimeter of the ellipse, counterclockwise, from the starting point to the ending point. The starting point is defined by the intersection of the ellipse and a line defined by the center of the ellipse and (X3,Y3). The ending point is defined by the intersection of the ellipse and a line defined by the center of the ellipse and (X4, Y4).</span><o:p></o:p></p></div></div>
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