gliding back to your departure airport

On Tue, 21 Oct 2003 20:02:00 GMT, "Harold" wrote
in Message-Id: :

If a small single engine plane can out-climb its engine-out glide ratio from
take off through the top of climb point, wouldn't it follow that it can
always theoretically make it back to the departure airport in the event of
engine failure ? Assuming straight out departure, no wind, and the altitude
loss in the 180 turnback is offset by the runway portion you didn't use. If
my best glide is 85 KTAS and it loses 700 fpm at that speed, shouldn't I be
guaranteed I can make it back if I climb at 84 KTAS and 701 fpm ?


The mathematics of turning back to the airport have been thoroughly
discussed in the newsgroup a while back. I suggest you do a
www.deja.com search for articles authored by John Lowry on the
subject. Here's an example:

http://groups.google.com/groups?selm...&output=gplain

From: John T. Lowry )
Subject: Min Turnaround Alt. on Single Engine Aircraft-Engine
Failure Question
Newsgroups: rec.aviation.piloting
Date: 1999/02/26


Dear Mike, Henrik, and All:
For the single engine-out return-to-airport maneuver, all the
various parameters (aircraft weight and flaps settings, runway length
and elevation, wind speed and direction) matter. But a crucial
performance number is, instead of just best glide ratio (which is
important once the turn is made) or minimum sink rate, the maximum
turn rate PER altitude lost, dTheta/dh. As close to (banked) stall as
possible. That rate is:

Max(dTheta/dh) =
-g*Rho*S*CLmax*sin(theta)*sqrt(cos^2(theta)+k^2)/(2*W*k)

where g is 32.2 ft/sec^2, Rho is density, S wing area, W weight, and

k = CD0/CLmax + CLmax/(Pi*e*A)

where CD0 is the parasite drag coefficient, e the airplane efficiency
factor, and A the wing aspect ratio. The optimum bank angle is just a
little (except for flamed-out jets) OVER 45 degrees and is given by

cos(phi_bta) = sqrt(2)*sqrt(1-k^2)/2

You'll find a full discussion in Chapter 9, Glide Performance, of my
forthcoming Performance of Light Aircraft published by AIAA.

John.

John T. Lowry, PhD
Flight Physics; Box 20919; Billings MT 59104
Voice: 406-248-2606

--------------------------------------------------------
Here's the formula for best glide

http://groups.google.com/groups?selm...&output=gplain

From: "John T. Lowry"
Subject: Formula for Vbg
Date: 1999/02/07
Message-ID: #1/1
References:

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Newsgroups: rec.aviation.piloting


Dear Phil, and All:
There is a fixed relationship between speed for best glide Vbg and
speed for minimum descent rate Vmd -- Vbg = 1.3161*Vmd -- but (since
you probably don't have Vmd) that won't help you much.
Vbg depends on the drag characteristics of the airplane, depending
on 1) W/sigma (W gross weight), 2) reference wing area S, 3)wing
aspect ratio A, 4)parasite drag coefficient CD0, and 5) airplane
efficiency factor e, according to
Vbg = sqrt(2*W/sigma*S)*(Pi*e*A*CD0)^-1/4

If you're willing to cut the engines and feather the props, to find
Vbg experimentally, here's a rough outline of the procedure. Go to
some nice high altitude and pick a vertical interval of pressure
altitudes, say for purposes of illustration from 14000 ft down to
13000 ft. Time repeated glides down through that interval and record
the product KCAS*delta_t, where delta_t is the time needed for the
glide. When you've found, by trial and error, the speed V which
maximizes that product, that speed is Vbg.

John

----------------------------------------------------

Larry Dighera wrote:



The mathematics of turning back to the airport have been thoroughly
discussed in the newsgroup a while back. I suggest you do a
www.deja.com search for articles authored by John Lowry on the
subject.

Yeah he's a genius. That's why he keeps wrecking aircraft.