Dynamics and Control of Rotorcraft - Helicopter ... · Dynamics and Control of Rotorcraft Helicopter Aerodynamics and Dynamics Abhishek Department of Aerospace Engineering Indian

Dynamics and Control of RotorcraftHelicopter Aerodynamics and Dynamics

Abhishek

Department of Aerospace EngineeringIndian Institute of Technology, Kanpur

February 3, 2018

Rotor LoadsTrim and Rotor Response

Overview

1 Rotor LoadsRotor Loads Calculation

2 Trim and Rotor ResponseTrim


Rotor Loads Calculation

Section 1

Rotor Loads



Subsection 1




Rotor Forces and Moments - 1

H-force: in-plane alongfreestream rotor drag,positive towards tail

Y-force: in-plane, positivetowards right

T: rotor thrust,perpendicular to rotor disk

Mx : roll moment

My : pitch moment

Q: rotor torque

NOTE: these are in fixed-frame !




T =

Nb∑i=1

∫ R

0dFzi Q =

Nb∑i=1

∫ R

0ydFxi

H =

Nb∑i=1

∫ R

0(dFxi sinψi + dFyi cosψi )

Y =

Nb∑i=1

∫ R

0(−dFxi cosψi + dFyi sinψi )

Mx =

Nb∑i=1

∫ R

0dFzi y cosψi

My =

Nb∑i=1

∫ R

0dFzi y sinψi




The periodic nature of rotor: ψi = ψ + (i − 1) 2πNb

. The sectionalforces in rotating frame are given by:

dFx = dLφ+ dD

dFy = −dLβ + dFR

dFz = dL

To obtain steady forces in fixed frame:

T =1

2π

∫ 2π

0T (ψ)dψ




In non-dimensional form it is given as:

CT

σClα

=1

2

∫ 1

0(u2

T θ − uPuT )dy

The non-dimensional velocities are given by:

uT = r + µ sinψ

uP = r∗β + λ+ βµ cosψ

uR = µ cosψ

Similarly non-dimensional H-force, Y-force, torque Q and roll andpitch moments can be obtained.




Consider a rigid rotor (no flapping), with three blades (Nb = 3)

Fz =Fz

ρcClalpha(ΩR)2R=

1

2

∫ 1

0(u2

T − uPuT )dy

Assuming the cyclic pitch variation and assuming uniform inflow:

Fz = F0 + F1c cosψ + F1s sinψ + +F2c cos 2ψ + F2s sin 2ψ

+F3c cos 3ψ + F3s sin 3ψ







The total force due to all three blades is:

Rotor acts like a filter, allowing only Nb/rev harmonics to fixedframe. Rest all harmonics are blocked. Nb/rev originates fromNb ± 1/rev







For rotor loads, inertial forces also need to be included. Let mo beblade mass/length.

Blade root shear forces (rotatingframe)

sx =

∫ R

e[dFx −mo(y − e)ζ]dy

sy =

∫ R

e[dFy + moyΩ2]dy

sz =

∫ R

e[dFz −mo(y − e)β]dy

Hub forces in rotating frame are:fx = sx , fy = sy , and fz = sz




Blade root bending moments (rotating frame)

nf =

∫ R

e[dFz −mo(y − e)β −moyΩ2](y − e)dy

nl =

∫ R

e[dFx −mo(y − e)ζ](y − e)dy

nt =

∫ R

edMxdy

Hub moments (rotating frame) are given by: mx = nf + esz ,my = nt , and mz = −nl − esx



Numerical Integration of Loads

Aerodynamic forces and moments are calculated for eachblade element and integrated over the length of the blade toobtain root loads (shears and bending moment)

Root loads are integrated over the whole rotation for eachblade to get mean loads for trim analysis



Numerical Integration of Loads

Aerodynamic forces and moments are calculated for eachblade element and integrated over the length of the blade toobtain root loads (shears and bending moment)

Root loads are integrated over the whole rotation for eachblade to get mean loads for trim analysis


Trim

Section 2

Trim and Rotor Response


Trim

Subsection 1

Trim


Trim

Helicopter Trim - 1

Trim means equilibrium: all forces and moments are balanced.Need to draw FBD of the helicopter in free flight!


Trim

Helicopter Trim - 2

xcg , ycg ≡Cg location wrt hub.

h ≡ height of hub.

lt ≡ tail location.

T = Rotor thrust, D = airframe drag, H= rotor drag, Y =rotor side force

YF = Tail rotor thrust

αs = longitudinal shaft tilt

φs = lateral shaft tilt.

Helicopter forces are in body frame and need to be transformedfrom xbybzb to xEyE zE . This involves rotation about xb followedby φs and then about yi by αs .


Trim

Helicopter Trim - 3

The transformation matrix for the two rotations are given by:

[T1] =

1 0 00 cosφs sinφs0 − sinφs cosφs

[T2] =

cosαs 0 − sinαs

0 1 0sinαs 0 cosαs

Net transformation from body axis to earth axis is:

[T1] = [T1][T2] =

cosαs 0 − sinαs

sinαs sinφs cosφs cosαs sinφscosφs sinαs − sinφs cosαs cosφs


Trim

Helicopter Trim - 4

So,FxeFyeFze

= [T ]

HYT

=

cosαsH − sinαsTsinαs sinφsH + cosφsY + cosαs sinφsTcosφs sinαsH − sinφsY + cosαs cosφsT


Trim

Helicopter Trim - 5

So, vertical force equilibrium equation is:W − T cosαs cosφs + Y sinφs − H sinαs cosφs + Yf sinφs +D sin θfp = 0Similarly:

D cos θfp + H cosαs − T sinαs = 0

(Y + Yf ) cosφs + T cosαs sinφs + H sinαs sinφs = 0

My + Myf −W (xcg cosαs − h sinαs)− D(xcg sinαs + h cosαs) = 0

Mx + Mxf + Yf h + W (h sinφs − ycg cosφs) = 0

Q − Yf lt = 0

In addition we have a equation of inflow (if using uniform inflow):

λ = µ tanαs +kCT

2√µ2 + λ2


Trim

Trim Variables

Inputs are: (θ75, θ1c , θ1s , αs , φs , θtr , λ0)Outputs are: (T ,Y ,H,Mx ,My ,Q, λ)Question is how do we solve non-linear trim equations?Popular approach is Jacobian based i.e using Newton Raphson.


Trim

Trim Procedure

Define residual for each equation:R1 = res(1) = T −WR2 = res(2) = D + H − Tαs

R3 = res(3) = Y + Yf − TφsR4 = res(4) = My + Myf + w(hαs − xcg )− hDR5 = res(5) = Mx + Mxf + w(hφs − ycg ) + Yf hR6 = res(6) = Q − Yf ltR7 = res(7) = λ− µ tanαs + kCT

2√µ2+λ2

R1 through R7 should be zero.We can do Taylor’s series expansion for same:R1 + ∂R1

∂θ1∆θ1 + ∂R1

∂θ2∆θ2 + ....+ ∂R1

∂θ7∆θ7+HOT = 0

R2 + ∂R2∂θ2

∆θ2 + ....+ ∂R1∂θ7

∆θ7+HOT = 0


Trim

Newton Raphson Iteration

The iterative procedure that NR-algorithm is given as :

θn+1 = θn − J−1R

where J - Jacobian of the system which is calculated by

Jij =∂R i

∂θj

≈Ri(θj+∆θj ) − Ri(θj )

∆θj

where θj is the j th element of input control θR i is the i th element of residual R∆θj is perturbation in θj

and n - is the number of iteration


Trim

Coupled Trim Algorithm


Trim

Numerical Integration Methods

Trapezoidal rule:∫ ba f (x)dx ≈ (b − a) (f (a)+f (b))

2Gaussian quadrature:∫ b

af (x)dx ≈ b − a

2

n∑i=1

wi f

(b − a

2xi +

a + b

2

)

i wi xi1 0.3607615730481386 0.6612093864662645

2 0.3607615730481386 -0.6612093864662645

3 0.4679139345726910 -0.2386191860831969

4 0.4679139345726910 0.2386191860831969

5 0.1713244923791704 -0.9324695142031521

6 0.1713244923791704 0.9324695142031521


Trim

Thank You

Documents

Dynamics and Control of Rotorcraft - Helicopter ... · Dynamics and Control of Rotorcraft Helicopter Aerodynamics and Dynamics Abhishek Department of Aerospace Engineering Indian