Your model’s stability hangs on its neutral point
June 2009 89
If It Flies ... Dean Pappas | [email protected]
IN THE LAST column, I waxed nostalgic
about an airplane I had “designed” and built as
a teenager. The word “designed” is in quotes,
because at least one of the airplane’s critical
features was borrowed, or I could call it
copied—all right, stolen! It was the stabilator
pivot point’s design and placement.
A stabilator is an all-moving horizontal tail,
rather than one with a separate movable surface
at the TE. Piper Cherokees and all supersonic
fighter jets have them. I had been warned that
stabilator design was tricky and that they had a
habit of overloading elevator servos, causing a
crash—or that they tended to flutter off of the
airplane altogether. The solution, for me then,
was to copy a proven design.
Many moons ago, as an engineering
student, you would have thought I would have
had plenty of reading to do for my course
work. But being the airplane geek that I am
(still), after much reading of calculus-laden
textbooks and a few nonmathematical books
written for aeromodelers, I learned that the
secret of successful stabilator design lay in
proper counterbalancing and the placement of
the pivot at the flying surface’s aerodynamic
center (AC).
Let’s leave the counterbalancing issue aside
right now. I’ll write about flutter another time,
and it will be important then.
Why does the pivot have to be placed at the
stabilator’s AC? If the stabilator airfoil is
symmetric, or even a flat plate, the torque
around the AC (or pivot point) is zero. It stays
constant at zero throughout almost the entire
range of useful control-surface deflections.
That means that a servo could provide the
necessary control, no matter how fast the
airplane were flying. That’s a big deal. When
your model is in a dive and accelerating
earthward, the last thing you want is an
elevator control that gets harder and harder to
move, especially when up-elevator seems like
such a good idea.
Our goal, in this column, is to extend the
AC concept to the entire airframe. Sometimes
the entire airframe’s AC called the “neutral
point” (NP); to avoid confusion, I will use that
terminology.
Why do we want to know about the entire
airframe’s AC? I’ll describe (or define) the AC
in a different way. If you mounted the wing
like a stabilator, with the pivot at its AC, you
could measure the lift the airfoil makes and the
pitching torque at the same time, and you
would find that:
• For a fixed angle of attack (AOA), both the
lift and nose-down pitching torque would
increase as a function of the airspeed squared.
So would the drag.
• For a fixed airspeed, the lift is nearly
proportional to the AOA above the zero-lift
angle.
06sig3.QXD 4/22/09 1:04 PM Page 89
• For a fixed airspeed, the nose-down pitching
torque is nearly constant at every AOA, except
in or near stall. For symmetric and flat-plate
airfoils, this nose-down torque equals zero and
increases with airfoil camber.
The fact that it is possible to pick a point
on the wing where you can imagine the lift as
acting through that point, and that the nosedown
torque does not change as the AOA and
lift change, means that you can analyze what
happens as the airplane’s nose rises and falls.
And as long as the aircraft’s airspeed does not
change much, only the lift that is provided by
the wing and horizontal tail change.
That simplifies the analysis. That’s why
aerodynamicists find the AC to be a useful
concept.
Now that I have established that having a
known AC is good, let’s extend the idea to the
entire airplane. The problem we are trying to
investigate is, how does the balance of forces
in pitch change as the airplane changes attitude
(nose up or down) as it flies through the air?
Does the balance of forces change to
restore the airplane’s original attitude? Does it
have no effect? Will the initial disturbance
worsen? That can’t be good!
If we assume that the fuselage is a skinny
object that contributes negligible lift (not
always true), we can make the following
simplification about the problem before us.
1. Because the wing’s nose-down torque,
taken by itself, is constant throughout the
range of AOA, and …
2. Because the stabilizer’s nose-down
torque, on its own, is constant for all flying
AOA …
3. Taken together, the nose-down torques
(caused by lift) have an effect that does not
change with AOA. Just the lift of the two
surfaces change, and then …
4. Even though the wing’s nose-down
torque (its pitching moment) must be
counteracted with some elevator trim or
stabilizer incidence compared to the wing …
5. We can analyze the airplane’s stability
in pitch by looking only at the changes in lift
made by the wing and stabilizer as the
attitude of the aircraft changes. And as we
know now, the lift of each surface acts at the
AC of each surface, and …
6. Because a flying surface’s lift depends
on its area, the problem becomes a simple
seesaw balance problem …
7. With the wing area representing an
adult’s weight on one side and the stabilizer
area representing a child’s weight at the end
of the other arm of the seesaw, we find
where the fulcrum would have to be to get a
balance.
This fulcrum location is the airframe’s
NP. It’s the same thing as the AC of the
entire airplane. I will show that if the
airplane is balanced at the NP, it is unstable;
it is going to be almost unflyable.
Pilots who fly 3-D aim for just in front of
this point for maximum maneuverability.
Stability results when the CG is placed in
front of the NP.
Let’s look at the seesaw. In the drawings,
the adult on the left end is sitting closer to
the fulcrum or pivot than the child at the
right end. This stands to reason, but we need
a way to figure out exactly where that pivot
is.
You might recognize the simple balance
equation. The weight on the left side
multiplied by the distance from the fulcrum
equals the weight on the right side multiplied
by its distance from the fulcrum. Or:
Weightadult x Distanceadult = Weightchild x
Distancechild
This requires that you know where the
fulcrum is, so I’ll present the result of a bit
of light algebra. High school algebra
students may try to derive this as a useful
exercise. The class of problem is called
“translation of axes.” The best derivation by
a student gets an attaboy or attagirl in print.
Measuring all distances from the
unoccupied leftmost end of the seesaw as a
starting point, the result is:
Fulcrum distance from left end = [(Weight adult x
Distance adult from left end) + (Weight child x
Distance child from left end)] ÷ (Weight adult +
Weight child)
This formula gives the same result and
does not require you to even guess where the
fulcrum will be.
Let’s apply this to the airplane. The “left
end” of the seesaw board could be anywhere,
but let’s use the back face of the spinner; it’s
as good a place as anywhere and easy from
which to measure.
Now we need to know the distance from
the propeller face to the wing’s AC (for a
simple, constant-chord wing, it’s the quarterchord
point) and the wing area. We also
need to know the distance from the propeller
face to the AC of the stabilizer and its area.
The same formula looks like:
NP location from propeller face = [(Wing Area x
Wing AC location from propeller face) + (Stabilizer
Area x Stabilizer AC location from propeller face)]
÷ (Wing Area + Stabilizer Area)
I’ll “plug in” the numbers from my
Goldberg Tiger 60. The wing has 900 square
inches of area, and the stabilizer has 195
square inches. The wing’s quarter-chord point
is 133/8 inches behind the propeller face, and
the stabilizer’s AC is 473/8 inches behind the
propeller face.
NP location from propeller face = [(900 x 133/8) +
(195 x 473/8)] ÷ (900 + 195), or 197/16 inches
behind the propeller face
On the Tiger, that puts the NP 93/16 inches
behind the LE. That’s 72% of the wing chord
from the LE. That’s unusually far back for
most sport models, because the Tiger has a
large stabilizer and a long tail moment.
Most designs have an NP that is close to
55% or 60% of the wing chord back from the
LE. Short-tailed designs such as some Scale
models will end up even farther forward.
Right now, the Tiger is balanced
approximately 43/4 inches behind the LE on a
125/8-inch chord. That puts the CG at 38% of
the chord, and the difference between it and
the NP is called the “margin of static
stability.”
In this case, it equals 34% of the wing’s
average chord. That leads to a nice, stable
airplane. A 10% static margin will be twitchy
but flyable, assuming that you don’t have too
much elevator throw.
So why does balancing an airplane at or
behind the NP make it unstable? Let’s go
back to the seesaw example.
Assume that because of some disturbance,
gust, elevator-servo twitch, or whatever, that
the airplane has a slightly more nose-up
attitude than it had just one-tenth of a second
ago. Both the wing and the stabilizer are
going to make changes in lift, in proportion to
their areas.
The increase in the wing’s lift tends to lift
the nose, and the tail will tend to lift the tail
and drop the nose. Because the NP represents
that seesaw’s balance point, you could say
that the total lift was happening at the NP. If
the CG is in front of that, the added lift will
be behind the CG, and the tendency will be
for the tail to rise and the original attitude will
be restored.
If the CG is behind the NP, that increase
in AOA will tend to cause the nose-up
attitude and AOA to increase further and
further and further. That’s what you see when
you try to balance a broomstick on end; it can
be done but requires constant attention—or a
flight computer. Now you can see that having
the CG far in front of the NP gives the
airplane a greater mechanical advantage with
which to restore itself.
The Tiger was a simple example, with a
skinny fuselage and a constant-chord wing. In
the next column, we will look at how to
calculate the NP of a blended wing/fuselage
airplane such as an F-16.
Until then, have fun and do take care of
yourself. MA
Edition: Model Aviation - 2009/06
Page Numbers: 89,90
Edition: Model Aviation - 2009/06
Page Numbers: 89,90
Your model’s stability hangs on its neutral point
June 2009 89
If It Flies ... Dean Pappas | [email protected]
IN THE LAST column, I waxed nostalgic
about an airplane I had “designed” and built as
a teenager. The word “designed” is in quotes,
because at least one of the airplane’s critical
features was borrowed, or I could call it
copied—all right, stolen! It was the stabilator
pivot point’s design and placement.
A stabilator is an all-moving horizontal tail,
rather than one with a separate movable surface
at the TE. Piper Cherokees and all supersonic
fighter jets have them. I had been warned that
stabilator design was tricky and that they had a
habit of overloading elevator servos, causing a
crash—or that they tended to flutter off of the
airplane altogether. The solution, for me then,
was to copy a proven design.
Many moons ago, as an engineering
student, you would have thought I would have
had plenty of reading to do for my course
work. But being the airplane geek that I am
(still), after much reading of calculus-laden
textbooks and a few nonmathematical books
written for aeromodelers, I learned that the
secret of successful stabilator design lay in
proper counterbalancing and the placement of
the pivot at the flying surface’s aerodynamic
center (AC).
Let’s leave the counterbalancing issue aside
right now. I’ll write about flutter another time,
and it will be important then.
Why does the pivot have to be placed at the
stabilator’s AC? If the stabilator airfoil is
symmetric, or even a flat plate, the torque
around the AC (or pivot point) is zero. It stays
constant at zero throughout almost the entire
range of useful control-surface deflections.
That means that a servo could provide the
necessary control, no matter how fast the
airplane were flying. That’s a big deal. When
your model is in a dive and accelerating
earthward, the last thing you want is an
elevator control that gets harder and harder to
move, especially when up-elevator seems like
such a good idea.
Our goal, in this column, is to extend the
AC concept to the entire airframe. Sometimes
the entire airframe’s AC called the “neutral
point” (NP); to avoid confusion, I will use that
terminology.
Why do we want to know about the entire
airframe’s AC? I’ll describe (or define) the AC
in a different way. If you mounted the wing
like a stabilator, with the pivot at its AC, you
could measure the lift the airfoil makes and the
pitching torque at the same time, and you
would find that:
• For a fixed angle of attack (AOA), both the
lift and nose-down pitching torque would
increase as a function of the airspeed squared.
So would the drag.
• For a fixed airspeed, the lift is nearly
proportional to the AOA above the zero-lift
angle.
06sig3.QXD 4/22/09 1:04 PM Page 89
• For a fixed airspeed, the nose-down pitching
torque is nearly constant at every AOA, except
in or near stall. For symmetric and flat-plate
airfoils, this nose-down torque equals zero and
increases with airfoil camber.
The fact that it is possible to pick a point
on the wing where you can imagine the lift as
acting through that point, and that the nosedown
torque does not change as the AOA and
lift change, means that you can analyze what
happens as the airplane’s nose rises and falls.
And as long as the aircraft’s airspeed does not
change much, only the lift that is provided by
the wing and horizontal tail change.
That simplifies the analysis. That’s why
aerodynamicists find the AC to be a useful
concept.
Now that I have established that having a
known AC is good, let’s extend the idea to the
entire airplane. The problem we are trying to
investigate is, how does the balance of forces
in pitch change as the airplane changes attitude
(nose up or down) as it flies through the air?
Does the balance of forces change to
restore the airplane’s original attitude? Does it
have no effect? Will the initial disturbance
worsen? That can’t be good!
If we assume that the fuselage is a skinny
object that contributes negligible lift (not
always true), we can make the following
simplification about the problem before us.
1. Because the wing’s nose-down torque,
taken by itself, is constant throughout the
range of AOA, and …
2. Because the stabilizer’s nose-down
torque, on its own, is constant for all flying
AOA …
3. Taken together, the nose-down torques
(caused by lift) have an effect that does not
change with AOA. Just the lift of the two
surfaces change, and then …
4. Even though the wing’s nose-down
torque (its pitching moment) must be
counteracted with some elevator trim or
stabilizer incidence compared to the wing …
5. We can analyze the airplane’s stability
in pitch by looking only at the changes in lift
made by the wing and stabilizer as the
attitude of the aircraft changes. And as we
know now, the lift of each surface acts at the
AC of each surface, and …
6. Because a flying surface’s lift depends
on its area, the problem becomes a simple
seesaw balance problem …
7. With the wing area representing an
adult’s weight on one side and the stabilizer
area representing a child’s weight at the end
of the other arm of the seesaw, we find
where the fulcrum would have to be to get a
balance.
This fulcrum location is the airframe’s
NP. It’s the same thing as the AC of the
entire airplane. I will show that if the
airplane is balanced at the NP, it is unstable;
it is going to be almost unflyable.
Pilots who fly 3-D aim for just in front of
this point for maximum maneuverability.
Stability results when the CG is placed in
front of the NP.
Let’s look at the seesaw. In the drawings,
the adult on the left end is sitting closer to
the fulcrum or pivot than the child at the
right end. This stands to reason, but we need
a way to figure out exactly where that pivot
is.
You might recognize the simple balance
equation. The weight on the left side
multiplied by the distance from the fulcrum
equals the weight on the right side multiplied
by its distance from the fulcrum. Or:
Weightadult x Distanceadult = Weightchild x
Distancechild
This requires that you know where the
fulcrum is, so I’ll present the result of a bit
of light algebra. High school algebra
students may try to derive this as a useful
exercise. The class of problem is called
“translation of axes.” The best derivation by
a student gets an attaboy or attagirl in print.
Measuring all distances from the
unoccupied leftmost end of the seesaw as a
starting point, the result is:
Fulcrum distance from left end = [(Weight adult x
Distance adult from left end) + (Weight child x
Distance child from left end)] ÷ (Weight adult +
Weight child)
This formula gives the same result and
does not require you to even guess where the
fulcrum will be.
Let’s apply this to the airplane. The “left
end” of the seesaw board could be anywhere,
but let’s use the back face of the spinner; it’s
as good a place as anywhere and easy from
which to measure.
Now we need to know the distance from
the propeller face to the wing’s AC (for a
simple, constant-chord wing, it’s the quarterchord
point) and the wing area. We also
need to know the distance from the propeller
face to the AC of the stabilizer and its area.
The same formula looks like:
NP location from propeller face = [(Wing Area x
Wing AC location from propeller face) + (Stabilizer
Area x Stabilizer AC location from propeller face)]
÷ (Wing Area + Stabilizer Area)
I’ll “plug in” the numbers from my
Goldberg Tiger 60. The wing has 900 square
inches of area, and the stabilizer has 195
square inches. The wing’s quarter-chord point
is 133/8 inches behind the propeller face, and
the stabilizer’s AC is 473/8 inches behind the
propeller face.
NP location from propeller face = [(900 x 133/8) +
(195 x 473/8)] ÷ (900 + 195), or 197/16 inches
behind the propeller face
On the Tiger, that puts the NP 93/16 inches
behind the LE. That’s 72% of the wing chord
from the LE. That’s unusually far back for
most sport models, because the Tiger has a
large stabilizer and a long tail moment.
Most designs have an NP that is close to
55% or 60% of the wing chord back from the
LE. Short-tailed designs such as some Scale
models will end up even farther forward.
Right now, the Tiger is balanced
approximately 43/4 inches behind the LE on a
125/8-inch chord. That puts the CG at 38% of
the chord, and the difference between it and
the NP is called the “margin of static
stability.”
In this case, it equals 34% of the wing’s
average chord. That leads to a nice, stable
airplane. A 10% static margin will be twitchy
but flyable, assuming that you don’t have too
much elevator throw.
So why does balancing an airplane at or
behind the NP make it unstable? Let’s go
back to the seesaw example.
Assume that because of some disturbance,
gust, elevator-servo twitch, or whatever, that
the airplane has a slightly more nose-up
attitude than it had just one-tenth of a second
ago. Both the wing and the stabilizer are
going to make changes in lift, in proportion to
their areas.
The increase in the wing’s lift tends to lift
the nose, and the tail will tend to lift the tail
and drop the nose. Because the NP represents
that seesaw’s balance point, you could say
that the total lift was happening at the NP. If
the CG is in front of that, the added lift will
be behind the CG, and the tendency will be
for the tail to rise and the original attitude will
be restored.
If the CG is behind the NP, that increase
in AOA will tend to cause the nose-up
attitude and AOA to increase further and
further and further. That’s what you see when
you try to balance a broomstick on end; it can
be done but requires constant attention—or a
flight computer. Now you can see that having
the CG far in front of the NP gives the
airplane a greater mechanical advantage with
which to restore itself.
The Tiger was a simple example, with a
skinny fuselage and a constant-chord wing. In
the next column, we will look at how to
calculate the NP of a blended wing/fuselage
airplane such as an F-16.
Until then, have fun and do take care of
yourself. MA