Spacecraft and Aircraft Dynamics Matthew M. Peet Illinois Institute of Technology Lecture 4: Contributions to Longitudinal Stability
Aircraft Dynamics Lecture 4 In this lecture, we will discuss Airfoils: The wing Contribution to pitching moment Rotate Lift and Drag vectors Calculate moment about CG ( M = r F) Use approximations to simplify expression. M. Peet Lecture 4: 2 / 26
Two Confusing Figures Confusion: For an airfoil, angle of attack is measured to the zero-lift-line. Thus C M0 = 0 for an un-inclined airfoil. Confusion: We assume that the aerodynamic center is on the FRL. Thus as measure from the CG, r ac = X cg X ac 0. Z c g If there is any confusion on a problem, ask me to clarify. M. Peet Lecture 4: 3 / 26
Next Subject: Wing Contribution Lift and Drag The wing is a lifting surface which produces both Lift and Drag. Lift is perpendicular to free-stream velocity. If αfrl > 0, then lift vector pitches forward (nose-down direction) in the body-fixed frame. Drag is parallel to free-stream velocity. If αfrl > 0, then drag also rotates in the nose-down direction. To determine the contributions of Lift and drag in the body-fixed frame, these forces must be rotated by the angle of attack and any additional wing inclination. M. Peet Lecture 4: Wing Contribution 4 / 26
Rotating the Lift and Drag Rotation Matrices Lift and Drag vectors rotate with angle of attack and wing inclination. Example: Given a vector v = x y and a pitch-up rotation, θ, z cosθ 0 sinθ v = 0 1 0 x xcosθ +zsinθ y = y sinθ 0 cosθ z xsinθ +zcosθ The matrix is called a rotation matrix. M. Peet Lecture 4: Wing Contribution 5 / 26
Rotating Vectors Rotation Matrices Rotation matrices can be used to calculate the effect of ANY rotation. Roll Angle φ : v = R 1 (φ) v Pitch Angle θ : v = R 2 (θ) v Yaw Angle ψ : v = R 3 (ψ) v Remember to use the right-hand rule to determine what is a positive rotation. M. Peet Lecture 4: Wing Contribution 6 / 26
Rotating Vectors Rotation Matrices The rotation matrices are (for reference): Roll Rotation (φ) : R 1 (φ) = 1 0 0 0 cosφ sinφ 0 sinφ cosφ Pitch Rotation (θ): R 2 (θ) cosθ 0 sinθ = 0 1 0 sinθ 0 cosθ Yaw Rotation (ψ): R 3 (ψ) cosψ sinψ 0 = sinψ cosψ 0 0 0 1 M. Peet Lecture 4: Wing Contribution 7 / 26
Rotating Vectors Rotation Matrices: Multiple Rotations Rotation matrices, can be used to calculate a sequence of rotations: Roll-Pitch-Yaw: v RPY = R 3 (ψ)r 2 (θ)r 1 (φ) v Note the order of multiplication is critical. ( ( ( ) ) ) v RPY = R 3 (ψ) R 2 (θ) R 1 (φ) v 1 2 3 M. Peet Lecture 4: Wing Contribution 8 / 26
Rotating Vectors Rotation Matrices: Example Consider a pure lift force of 10MN after a pitch up of 10deg and a yaw of 20 deg. L = 0 0 10 cos20 sin20 0 cos10 0 sin10 L PY = sin 20 cos 20 0 0 1 0 0 0 0 0 1 sin10 0 cos10 10 0.9397 0.3420 0 0.9848 0 0.1736 = 0.3420 0.9397 0 0 1.0000 0 0 0 0 0 1.0000 0.1736 0 0.9848 10 0.9254 0.3420 0.1632 = 0.3368 0.9397 0.0594 0 0 = 1.6318 0.5939 0.1736 0 0.9848 10 9.8481 M. Peet Lecture 4: Wing Contribution 9 / 26
Rotating Vectors Rotation Matrices: Example Compare this to the same rotations in the reverse order (Yaw, then Pitch) cos10 0 sin10 cos20 sin20 0 L YP = 0 1 0 sin 20 cos 20 0 0 0 sin10 0 cos10 0 0 1 10 = 1.7365 0 9.8481 Which is still in the x z plane!!! Why? Compare to Pitch, Yaw value: L PY = 1.6318 0.5939 9.8481 M. Peet Lecture 4: Wing Contribution 10 / 26
Next Subject: Wing Contribution Lift and Drag The magnitude of lift and drag are given by: L wing = C L,w QS and D wing = C D,w QS The two important angles are angle of attack of the aircraft, α FRL and inclination i w of the wing. Most wings are rotated up - this increases the angle of attack of the wing. C L can be determined as C L = C Lα α w = C Lα (α FRL +i w ) where α w = α FRL +i w is the angle of the wing with the free-stream. M. Peet Lecture 4: Wing Contribution 11 / 26
Next Subject: Wing Contribution Lift and Drag C D is usually determined using C Dα and C D0 from the tables. C D = C Dα α w +C D0 M. Peet Lecture 4: Wing Contribution 12 / 26
Wing Contribution Moment Contribution The wing makes three contributions to the moment about the Center of Gravity (CG) of the aircraft. The lift force, multiplied by lever arm The drag force, multiplied by a shorter lever arm The moment exerted on the wing itself (due to C M0) As a reference point, we choose the leading edge of the wing. X ac is the distance from the leading edge to the aerodynamic center of the wing X cg is the distance from the leading edge to the center of mass of the aircraft Z cg is the height of the CG above the Fuselage Reference Line (FRL) M. Peet Lecture 4: Wing Contribution 13 / 26
Wing Contribution Moment Contribution The moment produced by a force about point is M = r F For example, F = F x 0 F z and r = r x 0, which implies r z M = r F = det ˆx ŷ ẑ r x 0 r z = (r x F z F x r z )ŷ. F x 0 F z M. Peet Lecture 4: Wing Contribution 14 / 26
Wing Contribution: Simple Case Zero Angle of Attack Consider when α = 0. Then the force in the body-fixed frame, the lift and drag forces are D w X cg X ac F w,α=0 = 0, applied at r w = 0 L w Z cg Note that we assume the aerodynamic center is on the FRL. Now we have ˆx ŷ ẑ M w,α=0 = M 0,wŷ + r F = M wŷ +det X cg X ac 0 Z cg D w 0 L w = (M 0,w +L w(x cg X ac) D wz cg)ŷ. M. Peet Lecture 4: Wing Contribution 15 / 26
Wing Contribution: Simple Case Zero Angle of Attack Thus the pitching moment produced by the lift and Drag at zero angle of attack (α FRL = 0) is M w,α=0 = M 0,w +L w (X cg X ac ) D w Z cg. Converting to non-dimensional form, the pitching moment is C Mw,αFRL=0 = C M,w +C L,w (X cg X ac ) C D,w Z cg. M. Peet Lecture 4: Wing Contribution 16 / 26
Wing Contribution Moment Contribution: Angle of Attack Because Lift is perpendicular to free-stream and Drag is parallel, a positive angle of attack will rotate F in the negative pitch direction (nose down-direction). In the body-fixed frame, these forces become cos( α) F w = 0 sin( α) D w D wcosα+l wsinα 0 = 0 L w D wsinα L wcosα 0 1 0 sin( α) 0 cos( α) where α = α FRL and recall sin( α) = sinα, cos( α) = cosα Now we can calculate the moment contribution as before from M w = M 0,w + r w F w M. Peet Lecture 4: Wing Contribution 17 / 26
Wing Contribution Moment Contribution: Angle of Attack This contribution is readily calculated as M w = M 0,w + r w F w That is, M w = M 0,w +F x Z cg F z (X cg X ac ) = M 0,w +(L w sinα D w cosα)z cg +(D w sinα+l w cosα)(x cg X ac ) or, in non-dimensional form, divide by QS to get Recall that C M,w = C M,0,w +(C L,w sinα C D,w cosα) Z cg +(C D,w sinα+c L,w cosα)( X cg X ac ) C L,w = C Lα α w = C Lα (α FRL +i w ) and C D = C Dα α w +C D0 At this point we decide the equation are too complicated. We need to make some Approximations. M. Peet Lecture 4: Wing Contribution 18 / 26
Wing Contribution: Simplification Small Angle Approximations Our moment equation is now C M,w = C M,0,w +(C L,wsinα C D,wcosα) Zcg +(C D,wsinα+C L,wcosα) ( Xcg ) Xac But this is nonlinear and I promised linear equations. Therefore we use the Small Angle Approximations to linearize. If α = 0, then sinα = α cosα = 1 M. Peet Lecture 4: Wing Contribution 19 / 26
Wing Contribution: Simplification Other Approximations Replacing sinα with α and cosα with 1, we get C M,w = CM,0,w +(C L,w α C D,w ) Z cg +(C D,w α+c L,w ) ( Xcg X ) ac. The equation is linear, but still complicated. We now make 2 more assumptions 1. Z cg is small compared to X cg X ac. CD,wZ cg +C L,w(X cg X ac) = C L,w(X cg X ac) 2. α is small. CD,wα+C L,w = CL,w What we have left is C M,w = CM,0,w +C L,w ( Xcg X ) ac. M. Peet Lecture 4: Wing Contribution 20 / 26
Wing Contribution: Approximations Exceptions M. Peet Lecture 4: Wing Contribution 21 / 26
Wing Contribution: Approximations Exceptions M. Peet Lecture 4: Wing Contribution 22 / 26
Wing Contribution: Continued We now include the expression for Lift: this yields C M,w = C M,0,w +C Lα,w (α FRL +i w ) C L,w = C Lα,w α w = C Lα,w (α FRL +i w ) = C M,0,w +C Lα,w i w ( Xcg X ac ( Xcg X ) ac ) ( Xcg +C Lα,w α FRL X ac ) α FRL Thus we have where C M,total = C M,total,0 +C M,total,α α FRL C M,total,0 = C M,0,w +C Lα,w i w ( Xcg C M,total,α = C Lα,w ( Xcg X ac X ) ac ) M. Peet Lecture 4: Wing Contribution 23 / 26
Wing Contribution: Stability Recall for nose-up steady flight, we need C M,total,0 = C M,0,w +C Lα,w i w ( Xcg X ) ac 0 and C M,total,α = C Lα,w ( Xcg X ) ac 0 Since C Lα,w > 0, this can only be achieved when we have both CG ahead of wing: Negative Camber: X cg X ac < 0 which is unusual. Thus we need a tail contribution. C M,0,w > 0 which would be odd. M. Peet Lecture 4: Wing Contribution 24 / 26
Conclusion In this lecture, we have Calculated the moment about the CG produced by Lift and Drag on a Wing. Made small-angle approximations to get a linear expression. Neglected some Moment contributions. Developed impractical conditions for stability of a wing-only aircraft M. Peet Lecture 4: Wing Contribution 25 / 26
Next Lecture Do the same thing for the tail. Discuss our new design variables. Discuss equilibrium and stability conditions. M. Peet Lecture 4: Wing Contribution 26 / 26