SME
Flight Mechanics, Stability & Control
Flight control
Use this for “Cₘ₀ = 0.05, Cmα = −0.012 per degree, Cmδe = −0.02 per degree: what elevator angle trims at α = 6°?”
Diagram
The relative wind meets the fuselage at α = 6°.
δe = −1.1°: trailing edge up, the tail pushes down and the nose rises.
This is an example. Type your own values to start a new problem.
Cₘ₀
+
(
Cmα
)
×
α
°
+
(
Cmδe
)
×
(
δe
°
)
=
0
Formulas
Cₘ₀
+
Cmα
×
α
+
Cmδe
×
δe
=
0
0.05 + (−0.012) × 6 + (−0.02) × (−1.1) = 0
Assumptions
•
Linear pitching-moment coefficients about the CG, per degree; nose-up moments are positive.
•
Trailing edge up is negative δe and pitches the nose up; elevators move about 30° each way, past which the linear model fails.
•
A forward CG makes Cmα more negative, so trim needs more up elevator.
Step-by-step
WE KNOW
Cₘ₀ = 0.05, Cmα = −0.012, Cmδe = −0.02, α = 6°
FIND
elevator angle (δe)
Step 1 · Find elevator angle (δe)
Use Cₘ₀ + Cmα × α + Cmδe × δe = 0
Trimmed, the three pitching moments add to zero; solve for δe.
δe
=
−
Cₘ₀ + Cmα ×
α
Cmδe
δe
=
−
0.05 + (−0.012) × 6
−0.02
δe
=
−
0.05 + (−0.072)
−0.02
δe
=
−
−0.022
−0.02
δe
=
0.022
−0.02
δe = −1.1°
Check
Put the numbers back in. Both sides should match.
0.05 + (−0.012) × 6 + (−0.02) × (−1.1) = 0 ✓
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Bank angle for a coordinated turn
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