X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Sun, 05 Aug 2007 20:10:39 -0400 Message-ID: X-Original-Return-Path: Received: from mail814.megamailservers.com ([69.49.106.24] verified) by logan.com (CommuniGate Pro SMTP 5.1.11) with ESMTPS id 2244147 for lml@lancaironline.net; Sun, 05 Aug 2007 17:04:12 -0400 Received-SPF: none receiver=logan.com; client-ip=69.49.106.24; envelope-from=jhk@clearwire.net X-POP-User: jhk.clearwire.net Received: from hoben59c2e82ed (74-61-136-213.bel.clearwire-dns.net [74.61.136.213]) by mail814.megamailservers.com (8.13.6.20060614/8.13.1) with SMTP id l75K5e13018158 for ; Sun, 5 Aug 2007 16:05:41 -0400 X-Original-Message-ID: <002c01c7d79c$04adaf70$0201a8c0@hoben59c2e82ed> From: "James H. Keyworth" X-Original-To: "Lancair Mailing List" References: Subject: Re: [LML] Re: lancair announcement? X-Original-Date: Sun, 5 Aug 2007 13:05:44 -0700 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0028_01C7D761.557A0C90" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.3138 Disposition-Notification-To: "James H. Keyworth" X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3138 This is a multi-part message in MIME format. ------=_NextPart_000_0028_01C7D761.557A0C90 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit So, what if, instead of "pointing down", you removed all forward speed and kept the aircraft pointed "flat"? Assuming a stable no-yaw, no-bank, no-knot stall, (and no need for distance) would the descent rate be reduced because the aircraft is now a falling non-aerodynamic cross-shaped flat plate? Compared with the "best glide" vertical speed? And could that descent be converted to Vs 1.3 for a contolled landing? I suppose one would have to chuck all loose weight items (maybe even a co-pilot) into the tailcone to help keep a flat descent, and then haul them out again for elevator control on landing. JHK ------=_NextPart_000_0028_01C7D761.557A0C90 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable

So, what if, instead of "pointing = down", you=20 removed all forward speed and kept the aircraft pointed = "flat"? =20 Assuming a stable no-yaw, no-bank, no-knot stall, (and no need for = distance)=20 would the descent rate be reduced because the aircraft is now a falling=20 non-aerodynamic cross-shaped flat plate? Compared with the = "best=20 glide" vertical speed?

And could that descent be converted to = Vs 1.3 for a=20 contolled landing?

I suppose one would have to chuck all = loose weight=20 items (maybe even a co-pilot) into the tailcone to help keep a flat = descent, and then haul them out again for elevator control on=20 landing.

JHK

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