X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Sun, 31 Aug 2008 07:49:38 -0400 Message-ID: X-Original-Return-Path: Received: from imo-d03.mx.aol.com ([205.188.157.35] verified) by logan.com (CommuniGate Pro SMTP 5.2.6) with ESMTP id 3100626 for lml@lancaironline.net; Fri, 29 Aug 2008 21:12:18 -0400 Received: from Sky2high@aol.com by imo-d03.mx.aol.com (mail_out_v38_r10.8.) id q.d4f.370b2186 (42809) for ; Fri, 29 Aug 2008 21:12:09 -0400 (EDT) From: Sky2high@aol.com X-Original-Message-ID: X-Original-Date: Fri, 29 Aug 2008 21:12:09 EDT Subject: Re: [LML] Turn base to final X-Original-To: lml@lancaironline.net MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="-----------------------------1220058729" X-Mailer: Unknown sub 34 X-Spam-Flag:NO -------------------------------1220058729 Content-Type: text/plain; charset="US-ASCII" Content-Transfer-Encoding: 7bit Gary, Lately I have chosen to do curved arc descents to landing from the downwind (i.e. no squared pattern) at my towered airport. Nobody has complained and I find it easier to control both the descent rate and speed in a constant banked turn (assuming I guessed the wind effect properly) and a minor fix is less dramatic than in those 90 degree pattern turns as its need can be recognized earlier. Oh well, fly it the way you like....... Scott Krueger In a message dated 8/29/2008 6:15:45 P.M. Central Daylight Time, glcasey@adelphia.net writes: There was an interesting comment posted the other day that essentially said that a turn to final won't result in an accelerated stall because the aircraft is descending and therefore has a load factor of less than 1 G. Of course, as long as the aircraft isn't accelerating vertically the situation is the same as for level flight - the load factor will be greater than 1 G. But what if one did decide to turn the 90-degree turn while keeping the load factor at 1 G? Certainly any turn at any bank angle can be made at 1 G as long as the aircraft is allowed to accelerate downward. How much, I asked, so I went to my trusty spreadsheet and did some calculations. If you turn 90 degrees at a bank angle of 15 degrees (load factor of 1.0 instead of the normal 1.04) and an airspeed of about 90 knots you will exit the turn at a vertical descent rate of 1,000 ft/min MORE than when you entered. Do it at a bank angle of 30 degrees and the number is about 2,000 ft/min. at first it seemed to me that it is unlikely that doing this type of thing makes sense, although I suppose one could go from a base leg at level flight and use the turn to make the transition to descending flight. A 1,000 to 2,000 ft/min descent rate isn't far off for a normal final approach, so maybe it would be a reasonable thing to do as normal practice. Just a thought. Gary Casey ES #157 -- For archives and unsub http://mail.lancaironline.net:81/lists/lml/List.html **************It's only a deal if it's where you want to go. Find your travel deal here. (http://information.travel.aol.com/deals?ncid=aoltrv00050000000047) -------------------------------1220058729 Content-Type: text/html; charset="US-ASCII" Content-Transfer-Encoding: quoted-printable

Gary,

Lately I have chosen to do curved arc descents to landing from the down= wind=20 (i.e. no squared pattern) at my towered airport. Nobody has complained and I= =20 find it easier to control both the descent rate and speed in a constant= =20 banked turn (assuming I guessed the wind effect properly) and a minor fix is= =20 less dramatic than in those 90 degree pattern turns as its need can be=20 recognized earlier. Oh well, fly it the way you like.......

Scott Krueger

In a message dated 8/29/2008 6:15:45 P.M. Central Daylight Time,=20 glcasey@adelphia.net writes:

<= FONT=20 style=3D"BACKGROUND-COLOR: transparent" face=3DArial color=3D#000000 size= =3D2>There=20 was an interesting comment posted the other day that
essentially= =20 said that a turn to final won't result in an accelerated
stall=20 because the aircraft is descending and therefore has a load
fact= or=20 of less than 1 G. Of course, as long as the aircraft isn't =20
accelerating vertically the situation is the same as for level=20 flight
- the load factor will be greater than 1 G. But wha= t if=20 one did
decide to turn the 90-degree turn while keeping the load= =20 factor at 1
G? Certainly any turn at any bank angle can be= =20 made at 1 G as long
as the aircraft is allowed to accelerate=20 downward. How much, I
asked, so I went to my trusty=20 spreadsheet and did some calculations.
If you turn 90 degr= ees=20 at a bank angle of 15 degrees (load factor of
1.0 instead of the= =20 normal 1.04) and an airspeed of about 90 knots you
will exit the= =20 turn at a vertical descent rate of 1,000 ft/min MORE
than when y= ou=20 entered. Do it at a bank angle of 30 degrees and the
numbe= r is=20 about 2,000 ft/min. at first it seemed to me that it is =20
unlikely that doing this type of thing makes sense, although I =20
suppose one could go from a base leg at level flight and use the=20 turn
to make the transition to descending flight. A 1,000=20= to=20 2,000 ft/min
descent rate isn't far off for a normal final appro= ach,=20 so maybe it
would be a reasonable thing to do as normal=20 practice. Just a thought.

Gary Casey
ES #157

--
= For=20 archives and unsub=20 http://mail.lancaironline.net:81/lists/lml/List.html

It's only a deal if it'= s where you want to go. Find your travel deal here.

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