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<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Hi All,</span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>I've been reading the maneuver description from P-05
sequence and have a question regarding the following maneuver:</span></font></p>
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font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>P-O5.13: Half of a reverse Cuban eight, with ½ rolls: Push
to a 45 degree downline and perform a ½ roll, followed by ¾ of an inside loop.
On the second 45 degree downline, perform a ½ roll, then pull to recover
upright.</span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>The previous maneuver, the "Reverse double immelmann,
full roll first, 2/4 pt roll second, EXIT INVERTED, makes total sense. </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Because the previous maneuver "exits inverted,"
and at the top of the box, how do you "push" to a 45 degree downline
and performa a ½ roll, etc of the P-O5.13 maneuver?</span></font></p>
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font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>It seems to me like the description of P-O5.13 should read; "pull
to a 45 degree downline, etc." </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Or, is the intent of P-O5.13 to "push ¾ outside loop
to a 45 degree downlin, etc.????"</span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>TIA,</span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'> </span></font></p>
<p class=MsoNormal><font size=2 face=Arial><span style='font-size:10.0pt;
font-family:Arial'>Jim W.</span></font></p>
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font-family:Arial'> </span></font></p>
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12.0pt'> </span></font></p>
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