SME
Flight Mechanics, Stability & Control · Problem type
Trim angle of attack
Use this for “An airplane has Cₘ₀ = 0.05 and Cmα = −0.012 per degree. At what angle of attack does it trim, with Cₘ = 0?”
Chart
Cₘ = −0.012α + 0.05
Slope Cmα = −0.012: down 0.012 for every 1 across
Intercept Cₘ₀ = 0.05: it crosses the y-axis at (0, 0.05)
Cₘ = −0.012 × 25/6 + 0.05 = 0 when α = 25/6
This is an example. Type your own values to start a new problem.
Cₘ
=
Cₘ₀
+
(
Cmα
)
×
α
°
P
Pitch stability
worked out (not typed)
Formulas
Cₘ
=
Cₘ₀
+
Cmα
×
α
0 = 0.05 + (−0.012) × 4.167
P
=
stable
if
Cmα
<
0,
neutral
if
Cmα
=
0,
else
unstable
−0.012 < 0, so stable
Assumptions
•
The pitching-moment coefficient Cₘ about the CG is a straight line in α below the stall.
•
Trim means no pitching moment, Cₘ = 0, so the airplane holds its angle of attack.
•
Static stability needs Cmα < 0; trimming at positive lift then needs Cₘ₀ > 0.
Step-by-step
WE KNOW
Cₘ₀ = 0.05, Cmα = −0.012, Cₘ = 0
FIND
angle of attack (α), pitch stability (P)
Step 1 · Find angle of attack (α)
Use Cₘ = Cₘ₀ + Cmα × α
Take Cₘ₀ from Cₘ, then divide by the slope; with Cₘ = 0 this is the trim angle.
α
=
Cₘ − Cₘ₀
Cmα
α
=
0 − 0.05
−0.012
α
=
−
0.05
−0.012
α
=
4.167°
Step 2 · Find pitch stability (P)
Use P = stable if Cmα < 0, neutral if Cmα = 0, else unstable
A
falling
Cₘ
line:
more
α
brings
a
nose-down
moment
that
pushes
back
to
trim.
−0.012 < 0: stable
P
=
stable
if
−0.012
<
0,
neutral
if
−0.012
=
0,
else
unstable
P
=
stable
Check
Put the numbers back in. Both sides should match.
0 = 0.05 + (−0.012) × 4.167 ✓
−0.012 < 0, so stable ✓
Topic check
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