gliding back to your departure airport

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 ?

This is a good thing to test at altitude and know ahead of time. There's
going to be a certain height AGL that you just can't get turned around.
You should know what that is for your plane.


"Harold" wrote in message
...
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 ?

Altitude loss to get a 172 turned around after practice is about 600 feet.
If the pilot is surprised, panicked, or out of practice, it will be more.
If the pilot it very surprised, panicked, and out of practice, it may be
exactly equal to the altitude the maneuver was started at when he stalls due
to a combination of too much bank and pull on the yoke.

--
Roger Long

Keep in mind that it will not be a 180 degree turn but more like 210 degrees
(unless you plan to make your power-off landing parallel to runway surface).
There have been many, many studies of the "Impossible Turn." Google that
term and you will learn a lot.

Bob Gardner

"Harold" wrote in message
...
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 ?

Well obviously, but I'm not talking about in the pattern area like the
Impossible Turn is. I'm talking about 10 minutes after departure at 7k feet
where the departure airport, if you can make it, is the best landing option.
Then for all intents and purposes its a 180 degree turn.

Keep in mind that it will not be a 180 degree turn but more like 210
degrees
(unless you plan to make your power-off landing parallel to runway
surface).
There have been many, many studies of the "Impossible Turn." Google that
term and you will learn a lot.

Bob Gardner

"Harold" wrote in message
...
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 ?

"Harold" wrote in message
...
[...] 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 ?

Do you descend at 700fpm gliding at 85 knots (which you should reference as
*indicated* airspeed, not true) with the wings level? Or did you verify
that descent rate in a turn?

Several factors prevent the simplistic analysis you've made from being
valid:

* The turn itself increases descent rate

* You need to turn a net of closer to 270 degrees: 225 to get you on an
intercept course back to the runway, then another 45 degrees the other
direction to align yourself for touchdown.

* Typically you are climbing into a headwind; that becomes a tailwind
halfway through your turn and through the remainder of the descent. The
tailwind will either push you past the runway, or you need to steepen your
descent by increasing the descent rate. Either way, that interferes with
the basic "if I climb at such-and-such a rate, then instantly turn 180
degrees and descend at a different rate, can I make it back to the runway"
simplification. Assuming "no wind" conditions doesn't make sense, because
that assumption is almost never correct and the consequence is significant.

Don't forget the reaction time it takes to start the turn, and the time
spent at something other than best glide airspeed. For the vast majority of
pilots, a large proportion of the post-engine-failure flight will be done
quite a bit away from optimally.

If you have a hard time believing this, it's easy enough to experiment.
Find yourself a nice quiet airport where you can depart straight out. Climb
straight out to 2000' AGL, then cut the power. Wait a second or two (since
you won't be surprised by the power cut), then go ahead and start your turn
back to the runway. Note the altitude loss at the point at which you are
back aligned with the runway. This will give you the absolute *minimum*
altitude you might successfully attempt such a turn-back.

For extra credit, time the post-power-cut flight, noting your airspeed as
well. This will allow you to figure out how far you actually flew during
the descent, which will give you an idea of whether you'd have actually had
enough runway left to land on by the time you got all set up. For extra
accuracy, take someone along to keep track of the actual airspeed, or use a
GPS to track the experiment (to get distance directly, rather than depending
on speed over time).

Finally, keep in mind that not all airplanes have the characteristic yours
does. In fact, I'd say it's unusual to find an airplane that climbs and
descends at exactly the same airspeed and vertical speed. Especially
powerful aircraft will climb more steeply than they descend, while slower,
lower-powered airplanes will climb less steeply than they descend.

Pete

Due to the vertical lift loss during your more than 180 degree turn back to
the airport, you will loose more than 700fpm and your glide range will
decrease if you try to maintain your best glide speed during your turn. I
believe Barry Schiff wrote that this maneuver is best done at a fairly high
rate of turn which involves an approximate 45-50 degree bank to keep the
radius of turn small. This will enable you to then level the wings and then
obtain your best glide speed in the shortest time while getting you back to
the runway in the shortest distance. Not a maneuver for the non-proficient
or the startled and hesitant.
--
Jim Burns III

Remove "nospam" to reply

"Harold" wrote in message
...
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 ?

"Harold" wrote in message
...
Well obviously, but I'm not talking about in the pattern area like the
Impossible Turn is. I'm talking about 10 minutes after departure at 7k
feet
where the departure airport, if you can make it, is the best landing
option.
Then for all intents and purposes its a 180 degree turn.

To my knowledge, no one has ever said you can't. Other than aircraft that
cannot climb as steeply as they descend (and there are a number of aircraft
that are like that), obviously after a 10 minute climb, it would be very
likely you could make it back to your original departure airport.

I would never say "guaranteed". Other than aircraft performance, there are
still other reasons you wouldn't be able to make it. For example, climbing
with a tailwind will hinder your ability to make it back to the airport
(just because you take off into the wind, that doesn't mean you'll be
climbing out into the wind, even if you depart straight out).

So, a literal answer to your question is "no", you cannot guarantee that you
can make it back. However, it's true that for many aircraft, once you've
climbed that far, the original departure airport is often going to be a very
good choice for an emergency landing site, at least shortly after takeoff.

Does *that* answer your question?

Pete

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:

X-MimeOLE: Produced By Microsoft MimeOLE V4.72.2106.4
Organization: Montana Communications Network
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

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

That wasn't implicit in your post (although "top of climb" is a heck of a
hint). Most such questions deal with failure during the initial climb.

Bob Gardner

"Harold" wrote in message
...
Well obviously, but I'm not talking about in the pattern area like the
Impossible Turn is. I'm talking about 10 minutes after departure at 7k
feet
where the departure airport, if you can make it, is the best landing
option.
Then for all intents and purposes its a 180 degree turn.

Keep in mind that it will not be a 180 degree turn but more like 210
degrees
(unless you plan to make your power-off landing parallel to runway
surface).
There have been many, many studies of the "Impossible Turn." Google that
term and you will learn a lot.

Bob Gardner

"Harold" wrote in message
...
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 ?