SME
Flight Mechanics, Stability & Control
Static stability
Use this for “An airplane has hac = 0.25, lₜ = 4 m, Sₜ = 3.2 m², c̄ = 1.6 m, S = 16 m² and tail effectiveness 0.33. Find the neutral point and the static margin with the CG at 0.3c̄.”
Diagram
SM = hₙ − h = 0.415 − 0.3 = 0.115, 11.5% of c̄ behind the CG
A nose-up gust makes a nose-down moment: the airplane returns.
This is an example. Type your own values to start a new problem.
VH
=
lₜ (m)
×
Sₜ (m²)
c̄ (m)
×
S (m²)
hₙ
=
hac
+
VH
×
ηₜ
SM
=
hₙ
−
h
P
Pitch stability
worked out (not typed)
Formulas
VH
=
lₜ
×
Sₜ
c̄
×
S
0.5
=
4
×
3.2
1.6 × 16
hₙ = hac + VH × ηₜ
0.415 = 0.25 + 0.5 × 0.33
SM
=
hₙ
−
h
0.115 = 0.415 − 0.3
P
=
stable
if
SM
>
0,
neutral
if
SM
=
0,
else
unstable
0.115 > 0, so stable
Assumptions
•
Stick-fixed: the elevator is held, and the tail’s lift slope and the downwash are constants.
•
Positions hac, h and hₙ are fractions of the mean chord c̄, from its leading edge.
•
Tail effectiveness ηₜ = (aₜ ÷ a)(1 − dε/dα): the tail-to-wing slope ratio times the downwash loss.
•
Stable needs the CG ahead of the neutral point, a positive static margin SM.
Step-by-step
WE KNOW
hac = 0.25, lₜ = 4 m, Sₜ = 3.2 m², c̄ = 1.6 m, S = 16 m², ηₜ = 0.33, h = 0.3
FIND
tail volume (VH), neutral point (hₙ), static margin (SM), pitch stability (P)
Step 1 · Find tail volume (VH)
Use VH = lₜ × Sₜ ÷ (c̄ × S)
The tail volume: the tail’s area times its arm, over the wing’s area times its chord.
VH
=
4
×
3.2
1.6 × 16
VH
=
4
×
3.2
25.6
VH
=
12.8
25.6
VH = 0.5
Step 2 · Find neutral point (hₙ)
Use hₙ = hac + VH × ηₜ
The tail moves the neutral point aft of the wing’s aerodynamic center by VHηₜ.
hₙ = 0.25 + 0.5 × 0.33
hₙ = 0.25 + 0.165
hₙ = 0.415
Step 3 · Find static margin (SM)
Use SM = hₙ − h
How far the CG sits ahead of the neutral point, in mean chords.
SM = 0.415 − 0.3
SM = 0.115
Step 4 · Find pitch stability (P)
Use P = stable if SM > 0, neutral if SM = 0, else unstable
With the CG ahead of the neutral point, a nose-up gust makes a nose-down moment.
0.115 > 0: stable
P
=
stable
if
0.115
>
0,
neutral
if
0.115
=
0,
else
unstable
P
=
stable
Check
Put the numbers back in. Both sides should match.
0.5
=
4
×
3.2
1.6 × 16
✓
0.415 = 0.25 + 0.5 × 0.33 ✓
0.115 = 0.415 − 0.3 ✓
0.115 > 0, so stable ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.