1 of 6

Bombardier Challenger 650

Static Stability Analysis

Technical Report

Stability & Control

Group - 3

Ganjare Aavishkar (22AE30010)

2 of 6

Weight & Center of Gravity

Balance points define aircraft stability and control characteristics.

Empty C.G.

10.10

m

Operating Empty: 12,315 kg

MTOW C.G.

10.79

m

MTOW: 21,315 kg

0.69

m

Aft Shift

Ganjare Aavishkar (22AE30010)

3 of 6

MAC & Static Margin

Mean Aerodynamic Chord

Geometric reference for aerodynamic balancing

MAC (c̄)

2.595m

meters

Aerodynamic Center

Wing AC location from aircraft nose. It’s the point on the airfoil about which the aerodynamic moment is independent of AoA.

X_ac

11.5m

meters

Static Margin

At MTOW

27.3%

percent

Natural Longitudinal Stability

Positive static margin ensures the aircraft naturally returns to equilibrium after disturbances

Priyangshu (22AE10048)

Static Margin is a measure of natural longitudinal stability,

4 of 6

Longitudinal Static Stability (Cmα)

Quantifies the aircraft's natural tendency to pitch down and recover when disturbed.

Wing contribution provides baseline, enhanced by T-tail configuration with high tail volume coefficient

Lift Curve Slope (CLα)

from airfoil data

0.1

per degree

Static Margin

CG to AC distance

27

percent

Wing Contribution

baseline estimate

-0.027

per degree

Total Aircraft Cmα

with T-tail enhancement

-0.02 ~ -0.04

per degree

Phani Surya (22AE30006)

5 of 6

Pitching Moment Calculation

Equilibrium Condition: ΣCm = 0

Wing

Moment Arm

11.5 - 10.79 = 0.71 m

Normalized (Static Margin)

0.71 / 2.595 = 0.273

Lift Component

-0.48 × 0.273 = -0.131

Airfoil Component

Cm,ac ≈ 0 (negligible)

Tail

Tail Volume Coefficient

VHT = 0.735

Tail Lift Coefficient

CL,tail ≈ -0.12

Incidence Setting

-2° to -3° (Raymer)

Restoring Moment

-0.735 × (-0.12) = +0.088

Fuselage

Equilibrium Equation

0 = -0.131 + 0.088 + Cm,fuselage

Solve for Cm,fuselage

Cm,f = 0.131 - 0.088

Moment Equation

Cm,f = 0.043

Phani Surya (22AE30006)

6 of 6

Trim Angle Calculation

Cruise Condition

Mach 0.80 @ 37,000 ft

Cruise Weight

199864 N

Pressure

21.7 kPa

Density

0.364 Kg/m3

Required CL to sustain Cruise Flight

Lift Curve Equation

Symmetrical Airfoils

NACA 0011-64 & 0010-64

CL0 = 0

Simplified Equation

α = CL / CLα

Trim Angle Derivation

CLα ≈ 0.1 per degree

α = 0.45 / 0.1

4.5

°

Dheeraj (22AE30009)