X-Virus-Scanned: clean according to Sophos on Logan.com Return-Path: Sender: To: lml@lancaironline.net Date: Thu, 18 May 2006 18:02:27 -0400 Message-ID: X-Original-Return-Path: Received: from smtp105.sbc.mail.mud.yahoo.com ([68.142.198.204] verified) by logan.com (CommuniGate Pro SMTP 5.0.9) with SMTP id 1118689 for lml@lancaironline.net; Thu, 18 May 2006 12:14:37 -0400 Received-SPF: none receiver=logan.com; client-ip=68.142.198.204; envelope-from=elippse@sbcglobal.net Received: (qmail 15283 invoked from network); 18 May 2006 16:13:52 -0000 Received: from unknown (HELO Computerroom) (elippse@sbcglobal.net@75.15.124.219 with login) by smtp105.sbc.mail.mud.yahoo.com with SMTP; 18 May 2006 16:13:51 -0000 X-Original-Message-ID: <000901c67a96$11df0770$db7c0f4b@Computerroom> From: "Paul Lipps" X-Original-To: "Marv Kaye" Subject: Descent vs bank angle X-Original-Date: Thu, 18 May 2006 09:13:56 -0700 MIME-Version: 1.0 Content-Type: multipart/alternative; boundary="----=_NextPart_000_0006_01C67A5B.63EAE970" X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 6.00.2900.2869 X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.2869 This is a multi-part message in MIME format. ------=_NextPart_000_0006_01C67A5B.63EAE970 Content-Type: text/plain; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable I couldn't open the attachment myself on LML so here it is: Here=92re some ruminations for the mathematically-inclined to look over = and add any corrections they deem necessary. I tried to solve for the = power-off rate-of-descent and altitude loss at three different bank = angles, 30=B0 , 45=B0 , and 60=B0 . I used the parameters of my 235 with = me and 30 gallons: W-1400 lb, S-24.9=92, A-78 sq.ft., VY-110 mph IAS, = =CE , Oswald efficiency factor, 0.82*. At VY, best L/D, induced drag, = DI, and parasite drag, DP, are equal. V ft/s=3D 22mph/15, r =3D2.37689E-3, Q=3Dr V2/2, CL=3DW/QA, AR=3DS2/A, = DP=3D QAPD CDI=3DCL2/p AR=CE =3DW2/Q2Ap =CE S2 DI=3DQACDI=3DW2/Qp =CE = S2=3DDP APD=3DDP/Q Even though we would not be flying at sea-level, using sea-level values = simplifies the calculations for this mathematical exercise. Using my VY and my plane=92s parameters, Q=3D30.93 at VY so APD=3D1.282 Assuming that APD is a constant, we can rearrange terms and solve for V2 = with different values of g-load-multiplied weight. V2=3D2gW/r S(p =CE APD)=BD This may be simplified to V=3D(26033g)=BD=20 In circular flight at a bank angle of Q , the g load is 1/cosQ and the = centripetal force, CF, g units,=3DtanQ =3D V2/g0r, where g0=3D32.16, so = radius r=3DV2/CFg0 Multiplying the obtained radius by p to get the 180=B0 turn = circumference, then dividing this by V we get the time t for the turn. = Since the total drag is twice the parasite drag or induced drag, = HP=3D2QAPDV. The rate of descent to give the required HP, VS=3D550HP/W, = ft/sec. So for bank angles of 30=B0 , 45=B0 , and 60=B0 , the following values = were obtained: g: 1.155,1.414, 2.0; IAS mph: 118.2,130.7,155.5; radius ft.: 1618.6, = 1142.6, 933.8; t sec.: 29.33, 18.73, 12.86; drag lb.: 80.43, 112.5, = 159.1; HP: 25.34, 39.23, 65.99; VS ft/sec: 9.955, 15.41, 25.92; Alt. = Loss: 292=91, 289=91, 333=91. It appears there may be an optimum bank angle of 45=B0 that gives the = minimum loss of altitude. I=92ve heard that is what sailplane pilots = use. Now the altitude loss figures do not account for the loss entering = the descent nor the loss during recovery, but the higher bank angle = turns result in less distance away from the runway, which has some = merit. Food for thought! *=CE is 0.75+ for square tips with rounded edges, 0.75- for = rounded-planform tips, 0.80 for Hoerner tips with sharp edges, 0.81 for = square tips with sharp edges (Mooney), and 0.82 for raked tips with = sharp edges (mine). ------=_NextPart_000_0006_01C67A5B.63EAE970 Content-Type: text/html; charset="Windows-1252" Content-Transfer-Encoding: quoted-printable

I couldn't open the attachment myself on LML so = here it=20 is:

Here=92re some ruminations for the mathematically-inclined to look = over and add=20 any corrections they deem necessary. I tried to solve for the power-off=20 rate-of-descent and altitude loss at three different bank angles, = 30=B0 , 45=B0 , and = 60=B0 . I used the parameters of my 235 with me and = 30 gallons:=20 W-1400 lb, S-24.9=92, A-78 sq.ft., VY-110 mph IAS, =CE , Oswald = efficiency factor,=20 0.82*. At VY, best L/D,=20 induced drag, DI, and=20 parasite drag, DP, are=20 equal.

V ft/s=3D 22mph/15, r =3D2.37689E-3, = Q=3Dr V2/2, CL=3DW/QA, AR=3DS2/A, = DP=3D QAPD=20 CDI=3DCL2/p AR=CE=20 =3DW2/Q2Ap =CE S2 DI=3DQACDI=3DW2/Qp =CE=20 S2=3DDP APD=3DDP/Q

Even though we would not be flying at sea-level, using sea-level = values=20 simplifies the calculations for this mathematical exercise.

Using my VY and my=20 plane=92s parameters, Q=3D30.93 at VY so APD=3D1.282

Assuming that APD is a=20 constant, we can rearrange terms and solve for V2 with = different=20 values of g-load-multiplied weight.

V2=3D2gW/r S(p=20 =CE APD)=BD This may be simplified to=20 V=3D(26033g)=BD=20

In circular flight at a bank angle of Q , = the g load=20 is 1/cosQ and the centripetal force, CF, g=20 units,=3DtanQ =3D V2/g0r, where g0=3D32.16, so radius=20 r=3DV2/CFg0

Multiplying the obtained radius by p to = get the=20 180=B0 turn circumference, then dividing this = by V we get=20 the time t for the turn. Since the total drag is twice the parasite drag = or=20 induced drag, HP=3D2QAPDV.=20 The rate of descent to give the required HP, VS=3D550HP/W, ft/sec.

So for bank angles of 30=B0 , 45=B0 , and 60=B0 , the = following values=20 were obtained:

g: 1.155,1.414, 2.0; IAS mph: 118.2,130.7,155.5; radius ft.: 1618.6, = 1142.6,=20 933.8; t sec.: 29.33, 18.73, 12.86; drag lb.: 80.43, 112.5, 159.1; HP: = 25.34,=20 39.23, 65.99; VS ft/sec:=20 9.955, 15.41, 25.92; Alt. Loss: 292=91, 289=91, 333=91.

It appears there may be an optimum bank angle of 45=B0=20 that gives the minimum loss of altitude. I=92ve heard that is what = sailplane=20 pilots use. Now the altitude loss figures do not account for the loss = entering=20 the descent nor the loss during recovery, but the higher bank angle = turns result=20 in less distance away from the runway, which has some merit. Food for=20 thought!

*=CE is 0.75+ for square tips with rounded = edges,=20 0.75- for rounded-planform tips, 0.80 for Hoerner tips with sharp edges, = 0.81=20 for square tips with sharp edges (Mooney), and 0.82 for raked tips with = sharp=20 edges (mine).

------=_NextPart_000_0006_01C67A5B.63EAE970--