Subject: RE: Aileron use in stall turn
From: firstname.lastname@example.org(Erik Shilling)
Date: Aug 05 1996
Subject: Aileron use in stall turn - Why?
Date: Thu, 27 Jun 1996 23:27:08 +0100
So far I haven't seen anyone who answered the above question
correctly, so for what it's worth here is my answer.
<Bill@chivcons.demon.co.uk> Bill Chivers wrote:
>Now for something completely different: Who can help me with this
>Why do we need to apply opposite aileron when doing a stall turn?
I think many do it purely out of habit, since they are of little or
>Before you reply and say 'its a prop effect thing' it isn't
>because jets do it too.
There is a gyroscopic force produced in a high speed jet turbine
but not the "P" factor. Any rotating mass is in essence a gyro.
But because Jet do not have any prop blast on the rudder, the
rudder has to kick while the airplane still has considerable
forward speed. At zero airspeed, rudder is absolutely useless on
a jet since there is no airflow.
When a jet does the stall turn, it is not a zero airspeed
maneuver, as it is with the prop driven airplane. If the rudder is
effective on a jet, then so are the ailerons, and under this
condition the outside wing has considerable lift and can be negated
by the use of aileron.
>Anyone got a solution?
Yes Bill I do have an answer, but I first want to ask a couple
question about your statements which may help in the answer.
Why is it, that if you pull power, you won't roll over on your
back, and why is it that on an airplane with a left turning prop,
you have this same problem in a stall turn, but only when doing a
stall turn to the right ? In each case the problem can be
prevented, or stopped by pulling power. (Proving it is the prop.)
Although I've instructed aerobatics for years, after reading your
post, I decided to go out and fly my Skybolt in order to analyze,
and refresh my memory with some of the possible problems in
performing a (stall turn) Hammer head maneuver.
I think basically there are three things that can cause the
problems in executing a Stall Turn. First of all however, I'd like
to describe the maneuver as I understand it. (In a stall turn to
the left) You pull vertical, then when your forward speed is
exactly zero, kick full left rudder, causing the airplane to rotate
on its left wing tip as the center of rotation. Then when going
down vertically. (your down line should equal to your up line).
(Proof that the airplane is not stalled is that it does not spin.)
Word of caution: look out for over rotation, going past the
vertical down line. In a perfectly executed Hammerhead, the left
wing tip follows the exact path that it made in the up line. If
you have ever seen Art Scholl doing an airshow with his wing tip
smoke generators on, you have seen what I have just describe. When
he finished the hammer head, there were only three smoke lines, not
Incidently the fuselage of your airplane may not be vertical,
except when flying an airplane with a symmetrical airfoil. In
negating the lift of the airfoil in attempting to fly vertically,
the body of the airplane is not vertical. This is the very thing
which may cause the airplane go over on its back. What
the judges are looking for is a vertical line, not a vertical
I think the most common problem stems from not being exactly
vertical. Even though ever so slight, if you have passed the
vertical by 5 degrees, when you get to ninety degrees of rotation,
your wings are already 5 degrees past ninety degrees. Torque
forces are rolling the airplane even further. When this happens,
the only way to prevent the airplane from rolling is to either
What causes this added problem is torque. In the case
of a right turning prop the aircraft tries to roll in the opposite
direction which is to the left. A maneuver seen at many airshows
is called a torque roll normally associated with a tail slide. It
is caused by this same force that caused the airplane to roll over
on its back in a Hammer head. Although in the tail slide torque
roll, opposite aileron input is used, the speed of the tail slide
is so slow it will torque even if the ailerons are not reversed.
Have you ever seen a high-powered WW II fighter when doing a power
check. You will see that the left olio is depress considerably and
the right one is extended.
The third problem is the gyroscopic or precessional force. In the
case of a right turning propeller. A force applied to a spinning
propeller the resultant force occurs 90 degrees in the direction of
rotation. This causes the airplane to pitch up (in relation to the
airplane.) Incidently this is the force, when properly executed,
causes an airplane to perform the weird gyrations of Lomcivak (sp).
Also the same force during takeoff in a tail dragger, causes the
airplane to veer to the left, when the tail is brought up.
To illustrate the problem this may impose, fly perpendicular to a
road. When you pull up, your wings will be parallel to the road,
but on your down line, you find your wing are no longer parallel.
This was caused by the precessional force that the nose up (in
relation to the airplane) therefore on the down line the wings are
no longer parallel.
The remedy is to push forward on the stick, because the prop blast
will be enough to keep the nose where it should be. Many pilots
tend to forget this when doing a hammer head or even ignore the
fact that they haven't recover on a heading of 180 from where they
You're are correct when you say that the use of aileron are
absolutely of no value. The speed of the right wing in this case
is extremely small, and since its speed is dependant solely upon
the speed generated by the rotation, it can be discounted.
Certainly not enough to make any aileron input effective.
Erik Shilling Author; Destiny: A Flying Tiger's
Flight Leader Rendezvous With Fate.
3rd Squadron AVG