Estimating Aerodynamic Properties

The aerodynamic properties of a given aircraft are critically dependent on the aerodynamic properties ofits various lifting surfaces. For example, recall that the slope of the pitch moment coefficient Cm , a termof primary importance in determining longitudinal stability, takes the form

Cm = CLwb (h hnwb) VHCLt(1 d

d

)+ Cmp .

The value of this expression depends on CLwb , CLt , hnwb , anddd , all of which are determined by the

geometric properties of the wing and horizontal tail and parameters describing the flow field (such as theMach number and Reynolds number). In this lecture, we discuss various tools available for estimatingaerodynamic properties of lifting surfaces. These notes are adapted from [1].

y

c y( )

0

cr

ct

fuselagecenterline

b2

Figure 1: Sketch of a half-wing with constant sweep and taper.

Geometric properties of lifting surfaces. Consider the half-wing shown in Figure 1. Note that thewing has been extended to the centerline of the fuselage; the root chord cr is defined there. The meangeometric chord of a wing is defined as

c =2

b

b2

0c(y)dy

=2

b

S

2

or

c =S

b.

The wing aspect ratio is

1

AR =b

c=

b2

S.

The mean aerodynamic chord of a wing is defined as

c =2

S

b2

0c(y)2dy.

(Note that this definition is purely geometric. The mean aerodynamic chord length does not depend onthe wing aerodynamics.)

For the wing depicted in Figure 1, the local chord c(y) varies linearly from the root chord cr to the tipchord ct, so we say that this wing has constant taper. We define the taper ratio

=ct

cr.

For a wing with constant taper, the mean geometric chord is

c =

(1 +

2

)cr

and the mean aerodynamic chord is

c =2

3

(1 + + 2

1 +

)cr.

Note that, if = 1 (i.e., if the wing is untapered), then c = c = cr = ct.

The nth percent chord line of a wing is the curve connecting each point nc(y) measured aft of the localleading edge (where 0 n 1). (For example, the 0th percent chord line is the leading edge of the wing.)For the wing depicted, the angle of incidence of each chord line is constant, so we say that this wing hasconstant sweep. For a wing with constant sweep, one defines the sweep angle n of the n

th percent chordline as shown for n = 0. Given the sweep angle m of the m

th percent chord line for a wing with constantsweep and taper, one may determine n according to the formula

tann = tanm 4(nm)AR

(1 1 +

).

The mean aerodynamic center (x, y, z) is defined as that point about which the total aerodynamic momentof the wing does not vary with angle of attack. The location of the mean aerodynamic chord (for thehalf-wing) is determined relative to the mean aerodynamic center. While the mean aerodynamic chordc is defined purely by the wing geometry, the location of the mean aerodynamic center depends on thewing loading. Under certain assumptions about the form of the load distribution, the location of the meanaerodynamic center may be computed explicitly. See Appendix C of [1] and, in particular, Figure C.3.

2

mc

Figure 2: Sketch of the mean aerodynamic chord for a wing with constant sweep and taper.

Because the airplane is symmetric about the vertical plane through the centerline, the mean aerodynamiccenter of the complete wing lies in this plane. For a wing with constant sweep and taper whose loaddistribution is proportional to the local chord length, the distance from the leading edge of the root chordto the leading edge of the mean aerodynamic chord is

m =b

6

(1 + 2

1 +

)tan0.

Such a load distribution would result, for example, if the local lift coefficient of the airfoil sections consti-tuting the wing does not vary with y. (See Appendix C of [1] for information on other load distributionsand planform shapes.)

Aerodynamic properties of airfoils (Sectional properties). The aerodynamic properties of liftingsurfaces can be estimated in terms of the aerodynamic properties of the 2-D airfoil sections which makeup the complete lifting surface. For a given lifting surface we generally need three 2-D properties

Cl = the 2-D lift-curve slope

0L2D = the 2-D zero-lift angle of attack, and

Cm0L2D = the 2-D pitch moment at zero lift (i.e., the moment coefficient about the aerodynamic center).

0.09c

t0.99

t0.90

TE

12

Figure 3: Definition of the trailing edge angle TE.

We first consider the problem of estimating the 2D lift-curve slope Cl for a given section. The methodoutlined in Section 1 of Appendix B in [1] applies to wings with constant sweep and taper and without

3

twist moving at Mach numbers less than the critical Mach number. (The critical Mach number is thatvalue of the free stream Mach number at which the local flow is sonic at some point on the aircraft; thecritical Mach number is always less than one.) The method proceeds as follows

1. Given the wing thickness ratio tc , estimate the theoretical 2-D lift-curve slope (Cl)theory from FigureB.1,1(b). The curve appears to be well-approximated by the formula

(Cl)theory = 2pi + 4.9t

c.

2. Measure or estimate the trailing edge angle TE. This angle is defined

TE = 2arctan

(12

(tc

)90%

12(tc

)99%

0.09

)

where(tc

)90%

is the wing thickness ratio at the 90% chord line and(tc

)99%

is the wing thickness ratioat the 99% chord line. See Figure 3.

3. Determine the factor K which corrects for trailing edge angle and Reynolds number from FigureB.1,1(a). (Interpolate for Reynolds numbers between 106 and 108.)

4. Correct for trailing edge angle, Reynolds number, and Mach number to obtain the 2-D lift-curveslope

Cl =1.051M2K (Cl)theory.

If detailed airfoil information is unavailable and the wing is thin, then an approximate value of the 2-Dlift-curve slope is

Cl =2pi

1M2 .

The parameters 0L2D and Cm0L2D must be determined by some other means, such as a wind tunnel testor a computational fluid dynamics (CFD) model.

Aerodynamic properties of lifting surfaces. First, we determine the lift-curve slope for an untwistedwing with constant sweep and taper moving at a subcritical Mach number. Let

=

1M22pi

Cl .

Note that = 1 for a thin wing (i.e., a flat plate). Using the sectional lift-curve slope from above, andcorrecting for the finite wing-span and the sweep angle, we obtain

4

CL =2piAR

2 +

(AR2(1M2)

2

)(1 +

tan2 1/2(1M2)

)+ 4

.

Next, we determine the zero-lift angle of attack 0L. We assume that the sectional zero-lift angle 0L2D isgiven. Following are two cases for which 0L can be easily determined.

1. Constant sweep angle and constant airfoil section, where the cross sections are taken normal to thenth percent chord line. (No twist.)

0L = arctan

(tan0L2Dcosn

).

Note that, in the special case of an unswept wing, 0L = 0L2D .

2. Geometric twist and zero sweep. (Note: A wing with geometric twist has constant airfoil sectionsacross the span, however the zero-lift line of the sections varies from root to tip. A wing withaerodynamic twist achieves a similar effect by varying the airfoil section from root to tip.) Refer thewing angle of attack to the root chord: w := root. Then

(y) = root +(y)

where (y) is the wing twist and (0) = 0. The wing zero-lift angle of attack is

0L =2

S

b/20

(0L2D (y)) c(y)dy.

We next compute the wing zero-lift pitch moment Cm0L , given Cm0L2D across the span. This term canbe easily estimated in the special case of an untwisted wing with constant sweep angle, where the airfoilsections are taken parallel to the free stream:

Cm0L =

(AR cos2 1/4

AR+ 2 cos1/4

)(Cm0L2D |root + Cm0L2D |tip

2

).

Note that, if the wing section does not vary across the span, the latter factor becomes simply Cm0L2D .

Wing downwash parameter dd . The wing downwash parameter can be crudely estimated by assuminga thin, finite wing with an elliptic load distribution, which gives

d

d=

2CLpiAR

.

A more accurate method applicable to wings with constant sweep and taper is described in Appendix B.5of [1]. The estimate takes the form

d

d=

4.44[KAKKH

cos1/4

]1.191M2

5

where KA is a correction for the aspect ratio of the wing, K is a correction for the taper ratio of thewing, and KH is a correction for the location of the horizontal tail. The correction factors are determinedaccording to the following formulas:

KA =1

AR 11 +AR1.7

K =10 3

7

KH =1

hHb 3

2lHb

where, in the last formula,

hH = the (signed) orthogonal distance from the extended root chord line to the horizontal tail a.c.

lH = the longitudinal distance from the wing a.c. to the tail a.c.

b = wingspan.

Tutorial Example

wac

V

FuselageReference

Line

-it

V

( - )

tac

NACA2412

NACA0009

Figure 4: Sketch for the example.

Consider an aircraft with a trapezoidal wing and horizontal tail. Following are the geometric properties.

Wing properties: NACA 2412

b = 15 m, AR = 6, 1/4 =pi

6rad, =

1

4,

0L2D = 2.0, cm0L2D = 0.047.

Tail properties: NACA 0009

bt = 6 m, ARt = 4, 1/4t = 0 rad, t = 1,

6

0L2Dt = 0.0, cm0L2Dt

= 0.0, it = 1.0, hH = 0 m, lH = 15 m.

Compute the following:

1. The mean aerodynamic chord length and location for the wing and tail and the longitudinal locationof the wing and tail mean aerodynamic centers.

2. The lift coefficient of the wing as a function of fuselage angle of attack at Re = 106 and M = 0.5.(This requires computing both the slope of the lift curve and the lift coefficient at = 0.)

3. The zero-lift pitch moment of the wing Cm0Lw = Cmacw .

4. The lift coefficient of the tail as a function of fuselage angle of attack at Re = 106 and M = 0.5.(In addition to the lift slope, this requires computing the downwash parameter dd .)

Item #1.)

Wing m. a. chord length. The mean geometric chord is

c =S

b=

b

AR= 2.5 m.

Because the wing is trapezoidal, we compute

cr =2

1 + c = 4 m.

The mean aerodynamic chord is

c =2

3

(1 + + 2

1 +

)cr = 2.8 m.

Wing m. a. chord location. Assuming a uniform load distribution, the distance from the leading edgeof the root chord to the leading edge of the mean aerodynamic chord is

m =b

6

(1 + 2

1 +

)tan0

where

tan0 = tan1/4 4(0 14)

AR

(1 1 +

)= 0.68

which corresponds to a leading edge sweep angle of about 34. We thus compute

m =b

6

(1 + 2

1 +

)tan0 = 2.03 m.

The leading edge of the mean aerodynamic chord is roughly two meters aft of the wing apex.

Wing m. a. center. We may estimate the location of the wing mean aerodynamic center from FigureC.3 in [1]. For a taper ratio = 0 and aspect ratio AR = 6, we would find that xw 0.32c. For a taperratio = 12 and aspect ratio AR = 6, we would find that xw 0.26c. Interpolating for = 14 , we find that

xw 0.29c = 0.81 m

aft of the leading edge of the mean aerodynamic chord, which means the wing aerodynamic center is 2.84meters aft of the wing apex.

7

Tail m. a. chord length and location. The mean geometric chord of the tail is

ct =St

bt=

bt

ARt= 1.5 m.

Because the wing is not tapered ( = 1),

ct = c

t = 1.5 m.

Tail m. a. center. From Figure C.3 in [1], we estimate that the tail mean aerodynamic center is locatedat

xt 0.20c = 0.30 maft of the leading edge of the tail.

Item #2.)

Wing lift-curve slope. Based on the definition of the NACA 4-digit airfoil series, the thickness ratio forthe NACA 2412 is tc = 0.12. (See [2] for details about the definitions of the 4, 5, and 6 digit series.) FromFigure B.1,1 (b) in [1], we find that

(Cl)theory 6.87 rad1.Referring again to the definition of the NACA 4-digit series, one may compute

TEw = 2arctan

(12

(tc

)90%

12(tc

)99%

0.09

)= 0.26 rad

which is around 15. From Figure B.1,1 (a), we find that

K 0.77.The 2-D lift-curve slope, corrected for wing thickness, trailing edge angle, Reynolds number and Machnumber, is

Cl =1.051M2K (Cl)theory 6.41 rad

1.

We next compute

=

1M22pi

Cl = 0.88.

The formula for the lift-curve slope CLw requires the sweep angle of the mid-chord line. We compute

1/2 = arctan

{tan1/4

4(12 14)AR

(1 1 +

)}= 0.45 rad

or about 26. We thus find that

CLw =2piAR

2 +

(AR2(1M2)

2

)(1 +

tan2 1/2(1M2)

)+ 4

= 4.2 rad1

or about 0.07 per degree.

Wing zero-lift angle of attack. Now, the zero-lift angle of attack of the airfoil is given to be 0L2D =2.0. Thus, the zero-lift angle of attack of the wing is

0Lw = arctan

(tan0L2Dcos1/4

)= 0.040 rad

8

or about 2.3. We thus find that

CLw = C

Lw(w 0Lw)

= CLw ( 0Lw)= 0.07( (2.3))= 0.16 + 0.07,

where CLw has units of deg1 and is given in degrees.

Item #3.)

Using the formula for Cm0L presented earlier, we find that

Cm0Lw =

(AR cos2 1/4

AR+ 2 cos1/4

)Cm0L2D = 0.027.

Item #4.)

Tail lift-curve slope. The tail lift coefficient is

CLt = CLt ( (w) it)

= CLt

(

(0 +

d

d

) it

)

= {CLt (0 + it)}+{CLt

(1 d

d

)}.

The critical parameters which we must compute are at, 0, anddd .

Based on the definition of the NACA 4-digit airfoil series, the thickness ratio is tc = 0.09. From FigureB.1,1 (b) in [1], we find that

(Cl)theory 6.75 rad1.(The subscript t is omitted with the understanding that all of the following computations relate to thehorizontal tail.) Referring again to the definition of the NACA 4-digit series, one may compute

TEt = 2arctan

(12

(tc

)90%

12(tc

)99%

0.09

)= 0.20 rad

which is around 11. From Figure B.1,1 (a), we find that

K 0.79.

The 2-D lift slope, corrected for wing thickness, trailing edge angle, Reynolds number and Mach number,is

Cl =1.051M2K (Cl)theory 6.47 rad

1.

We next compute

=

1M22pi

Cl = 0.89.

9

Because the horizontal tail is unswept, we compute

CLt =2piAR

2 +

(AR2(1M2)

2

)(1 +

tan2 1/2(1M2)

)+ 4

=2piAR

2 +

(AR2(1M2)

2

)+ 4

= 3.9 rad1

or about 0.07 per degree. Since the horizontal tail is symmetric, the zero-lift angle of attack of the airfoil,and of the entire tail, is zero.

Downwash at the tail. Next, we estimate the downwash parameter dd using the relation

d

d=

4.44[KAKKH

cos1/4

]1.191M2 .

The correction factors are:

KA =1

AR 11 +AR1.7

= 0.12

K =10 3

7= 1.32

KH =1

hHb 3

2lHb

= 0.79.

We thus computed

d= 0.40.

To determine 0, we observe that no downwash is generated when the wing generates no lift. The zero-liftangle of attack of the wing is 0Lw = 2.3. We thus compute

(w) = 0 = 0 +d

d0Lw

= 0 +d

d(2.3)

which tells us that 0 = 0.92 = 0.016 rad.

In the end, we obtain

CLt = {CLt (0 + it)

}+

{CLt

(1 d

d

)}

= 0.14 + 0.04where is measured in degrees. Notice that CLt < 0 when = 0. At zero angle of attack, the tailgenerates a downward force which generates a nose-up moment about the center of gravity, as desired.

References

[1] B. Etkin and L. D. Reid. Dynamics of Flight: Stability and Control. John Wiley and Sons, New York,NY, third edition, 1996.

[2] W. H. Mason. Geometry for aerodynamicists. Notes available on-line at:http://www.aoe.vt.edu/aoe/faculty/Mason f/CAtxtAppA.pdf.

10

Review of Linear, Time-Invariant Ordinary Differential Equations

First order, homogeneous, LTI ODEs. Following is the normal form for a homogeneous, linear,time-invariant ordinary differential equation of first order.

x+ ax = 0, x(t0) = x0. (1)

The dot indicates differentiation with respect to time. Suppose we wish to solve (1) for x(t).

Approach #1 : Use an integrating factor. That is, multiply by a factor that will make the left hand sidean exact differential.

eat (x+ ax) = 0

xeat + axeat = 0d

dt

(xeat

)= 0

Integrating both sides from t0 to t gives tt0

d

d(x()ea ) d =

tt0

0 d x(t)eat x(t0)eat0 = 0 x(t) = x0ea(tt0)

Comments:

The integrating factor also works when there is forcing on the right hand side and when the equationis linear time varying (i.e., when a depends on t).

As stated, the approach only works for first order linear equations. However, it can be generalizedto systems of first order linear equations. (Recall that any number of linear equations of any ordercan be rewritten as a set of first order linear equations.)

Approach #2 : Motivated by divine insight, assume a solution of the form

x(t) = cet. (2)

(Actually, this form seems quite reasonable given the solution in Approach #1.) Substituting into (1) gives

x+ ax = 0d

dt

(cet

)+ a

(cet

)= 0

(+ a)cet = 0

(+ a)x(t) = 0.

Now, x(t) will not be zero, in general; only in the special case that x0 = 0 will x(t) be zero. Since theidentity above must hold for any initial state, it follows that

= a.To determine c, we evaluate x(t) given in (2) at t0 and use the initial condition:

x(t0) = ceat0 c = x0eat0 .

Substituting the values of and c into (2) gives the solution

x(t) = x0ea(tt0).

1

Approach #3 : Since this is a linear, time-invariant system we may use the Laplace transform to convert thedifferential equation in time into an algebraic equation in s. Having expanded the solution X(s) in a seriesof partial fractions, we may then transform back to the time-domain. Assume that t0 = 0. There is noloss of generality in doing this; because the system is time-invariant we simply change the time coordinate,shifting the origin of time such that t0 = 0. Taking the Laplace transform of the equation gives

L{x+ ax = 0} (sX(s) x0) + aX(s) = 0 X(s) = x0s+ a

.

Taking the inverse Laplace transform gives

x(t) = L1{

x0

s+ a

}= x0e

at.

To better understand the nature of the solution, define the time constant

T =1

a.

Dividing through by the initial state (which we assume is nonzero), we have

x(t)

x0= e

t

T .

If a > 0 (so that T > 0), then the solution decays with time. In this case, at time t = T (that is, after onetime constant has elapsed), x(t) is only 37% of its initial value. Slightly before this time, x(t) is exactlyone-half its initial value. To determine this time to half-amplitude thalf , observe that

x(thalf)

x0=

1

2= e

thalf

T thalf = ln(1

2

)T 0.69T.

Suppose that a < 0, so that T < 0. Then x(t) grows without bound. The time required for x(t) to growto twice its initial amplitude is obtained as follows:

x(tdouble)

x0= 2 = e

tdouble

T tdouble = ln (2)T 0.69|T |.

Now suppose x0 = 0. Then x(t) = 0 for all time, regardless of the value of a. Thus, x(t) = 0 is anequilibrium. Stability of this equilibrium depends on the sign of the constant a. If a > 0, then trajectoriesstarting near the equilibrium approach it asymptotically and we say that the equilibrium is stable. If a < 0,then trajectories starting near the equilibrium diverge from it and we say that the equilibrium is unstable.

First order, heterogeneous, LTI ODEs. Following is the normal form for a heterogeneous, linear,constant coefficient ordinary differential equation of first order.

x+ ax = f(t), x(t0) = x0. (3)

Suppose we wish to solve (1) for x(t). (Assume that a > 0.)

Approach #1 : Use an integrating factor.

eat (x+ ax) = eatf(t) ddt

(xeat

)= eatf(t).

Integrating both sides from t0 to t gives

x(t)eat x(t0)eat0 = tt0

eaf()d x(t) = x0ea(tt0) + tt0

ea(t)f()d (4)

2

Note that x(t) is the sum of a term involving the initial condition and a term involving the force f(t). Thisobservation reflects the principle of superposition for solutions to linear equations. This principle is usefulin solving more general problems (i.e., heterogeneous equations of arbitrary order).

Approach #2 : The method of undetermined coefficients applies to forced LTI equations where f(t) solvessome linear differential equation. Equivalently, the method applies to forced LTI equations where a finiteset S contains f(t) and its derivatives of all orders. For example, the forcing function

f(t) = t sin(t)

is admissible because the finite set

S = {t sin(t), t cos(t), sin(t), cos(t)}

contains f(t) and its derivatives of all orders. (Or rather, f(t) and its derivatives of all orders maybe constructed by linear combinations of the elements of S.) Note that each element of S is linearlyindependent ; no element can be formed by a linear combination of the remaining elements. (You maycheck this by computing the Wronskian determinant and making sure it is not zero for all t; see yourtextbook on differential equations.)

The method of undetermined coefficients proceeds as follows:

1. Solve the associated homogeneous problem, leaving the constants arbitrary. The general solution tothe homogeneous problem is termed the complementary solution.

2. Assume a particular solution as a weighted sum of all the elements of S. Leave the coefficients ofthe various terms undetermined.

3. Sum the particular and complementary solutions, substitute into the original equation, and matchthe coefficients of like terms. This will determine values for some of the coefficients.

4. Impose the initial conditions on the resulting solution to determine the remaining coefficients.

Approach #3 : Use the Laplace transform.

L{x+ ax = f(t)} (sX(s) x0) + aX(s) = F (s) X(s) = x0s+ a

+1

s+ aF (s).

Taking the inverse Laplace transform, we find

x(t) = x0eat + eat f(t)

= x0eat +

t0ea(t)f()d.

Compare this solution with (4).

Second order, homogeneous, LTI ODEs. We now turn our attention to the equation

x+ a1x+ a0x = 0, x(t0) = x0, x(t0) = x0 (5)

where a1 and a0 are scalar constants.

Approach: Again, assume thatx(t) = cet (6)

3

Substituting the assumed form of x(t) into the equation gives

0 = 2(cet) + a1(cet) + a0(ce

t)

= (2 + a1+ a0)x(t).

We assume the initial condition is such that x(t) 6= 0 for all time. Then it must be true that

2 + a1+ a0 = 0.

This quadratic equation has two distinct solutions, in general,

1,2 =1

2

(a1

a21 4a0

).

Thus, x(t) generally contains two components of the form (6). By the principle of superposition, thegeneral solution is a sum of the two:

x(t) = c1e1t + c2e

2t. (7)

Evaluating at the initial conditions determines the parameters c1 and c2. In the case that a214a0 = 0, the

two candidate solutions are linearly dependent. In that case, we obtain a secular term in the solution:

x(t) = (c1 + c2t)e

a1

2t.

Generically, the initial condition response of a second order system can be described either as exponentialconvergence or exponential divergence. The former case occurs whenever both eigenvalues have negativereal part. The latter case occurs if either eigenvalue has a positive real part. The boundary whichseparates these two generic behaviors occurs when one or both of the eigenvalues lie on the imaginary axis.In this case, and assuming both eigenvalues are not zero, the initial condition response is a sinusoid whosefrequency is the magnitude of the imaginary part.

If the initial condition response involves convergence to zero, then we say that the equilibrium at (x, x) =(0, 0) is stable. If the initial condition response involves divergence from zero, then we say that theequilibrium at (x, x) = (0, 0) is unstable. Otherwise, we say that the equilibrium at (x, x) = (0, 0) isneutrally stable.

We will be most concerned with the case where a0 > 0 (which is necessary for static stability). Definethe natural frequency n and the damping ratio as follows:

n =a0 and =

a1

2a0.

With these definitions, the equation of motion becomes

x+ 2nx+ 2nx = 0, x(t0) = x0, x(t0) = x0

and the characteristic equation is2 + 2n+

2n = 0.

The characteristic values are

1,2 =1

2

(2n

422n 42n

)=

(

2 1

)n.

Consider the following two cases:

4

Re

Im

Re

Im

Re

Im

-!n

!n

!d

increasingthrough 1 decreasingthrough-1-1 < < 1

Figure 1: Characteristic Values of a Second Order System as the damping coefficient varies.

1. 0 2 < 1: In this case, the characteristic values are a complex conjugate pair. We may write1,2 = n j

1 2n

= n jd.The initial condition response takes the form

x(t) = ent(x0 cosdt+

1

d(x0 + nx0) sindt

)

If > 0, this response is a damped oscillation and we say that the equilibrium at the origin isdynamically stable, as well as statically stable. If < 0, the response diverges and we say that theequilibrium at the origin is dynamically unstable. If = 0, the response oscillates forever and we saythat the equilibrium at the origin is neutrally stable.

Referring to the factor ent in the expression above, the real part n plays a role similar to thereciprocal of a first order time constant. As in the case of a first order system, we may compute thetime to half-amplitude (if 0 < < 1) or the time to double amplitude (if 1 < < 0):

thalf or tdouble 0.69||n .

Since the period of oscillation is Td =2pid, we may also compute the number of oscillations to half or

double amplitude as

Nhalf/double =thalf/double

Td(0.69

||n

)(d2pi

)= 0.11

1 22.

2. 2 > 1: In this case, the characteristic values are real. If > 1, then both characteristic valuesare negative. The response is an overdamped convergence to the equilibrium which is dominated bythe larger (less negative) characteristic value. Thus, the response is essentially a first order responseand one computes the time to half amplitude as one would for a first order system (using the largercharacteristic value). As grows larger positive, one characteristic value approaches zero from the leftwhile the other moves toward . As the larger characteristic value approaches zero, convergencebecomes slower and slower.

If < 1, then both characteristic values are positive. The response is a non-oscillatory divergencewhich is dominated by the larger (more positive) characteristic value. One computes the time todouble amplitude as for a first order system, using the larger characteristic value. As grows largernegative, the dominant characteristic value moves toward and the other approaches zero from theright.

5

Second order, heterogeneous, LTI ODEs. Second order LTI ODEs with forcing can be solved usingthe method of undetermined coefficients or Laplace transforms exactly as in the case of first order ODEs.1

Higher order systems. We have seen that the the response of a first order system is (generically)exponential growth or decay. The response of a second order system is either a combination of first orderresponses (in the case that both characteristic values are real) or is an oscillatory exponential growthor decay (in the case that the characteristic values are complex conjugates). In fact, the roots of apolynomial with real coefficients are always real or complex conjugate pairs. Thus, with the exception ofthe degenerate case of repeated characteristic values, the general response of an nth order linear systemcan be understood as a superposition of responses of first and second order responses.

1Even the technique of an integrating factor extends to second order equations, provided they are re-written as a system

of first order equations; since this can always be done, the technique applies in general. The generalized integrating factor

method is related to the method of variation of parameters which you may have seen in your course on differential equations.

6

Evans Rules for Sketching the Root Locus

Absolute and Relative Stability. A control system is called absolutely stable if the controlled transferfunction Hd(s) from the reference signal yd(s) to the output signal y(s) has all of its poles in the openleft half plane. One technique for determining absolute stability of a control system is the Routh-Hurwitzstability analysis technique. This very useful technique is presented in Section 5-7 of [1].

Absolute stability is an essential quality for a control system, but it says nothing about the performancecharacteristics of the system, i.e., the transient response. Two absolutely stable systems can respond toa step input in very different ways; one might exhibit a very slow, overdamped response while the otherexhibits a very fast, underdamped response.

To compare the performance of two absolutely stable systems, it is useful to consider the notion of relativestability or degree of stability. Degree of stability can be rather narrowly defined as the horizontaldistance between the imaginary axis and the nearest pole. This distance will typically determine the speedof response of the system, however it tells you nothing more about the nature of that response (e.g., ifit the system is overdamped, critically damped, or underdamped). More generally, one may examine thespecific locations of the closed-loop poles. Knowing these pole locations gives a good sense of the natureof the systems transient response.

The Root Locus Method. The root locus method, also known as Evans rules in honor of W. R.Evans, is a technique for determining how the poles of a feedback control system move in the complex planeas a parameter is varied. Typically, the parameter is a control gain, although any parameter of interestcan be used. (For this reason, the root locus method is useful in dynamical system theory, where one isoften interested in sudden changes in a systems qualitative behavior, called bifurcations, as a parametervaries.)

yd

uyFd+ - P

Figure 1: One degree of freedom closed-loop control structure.

Consider the simple feedback control system shown in Figure 1. The closed-loop transfer function is

Hd(s) =y(s)

yd(s)=

P (s)Fd(s)

1 + P (s)Fd(s).

Closed-loop poles are values of s for which

1 + P (s)Fd(s) = 0.

Since P (s)Fd(s) is a function of a complex variable, the equation P (s)Fd(s) = 1 can be expressed interms of the magnitude and phase of P (s)Fd(s):

|P (s)Fd(s)| = 1 and P (s)Fd(s) = (2k + 1)pi k = 0,1,2, . . .

In words, the magnitude of the loop gain is always one and the phase is an odd power of pi.

Suppose that P (s)Fd(s) can be written in the form

P (s)Fd(s) = Kb(s)

a(s).

1

This would be the case, for example, if P (s) = b(s)/a(s) and Fd = K, as for a simple proportionalcontroller. The control structure might be more complicated than this, however we assume that a themultiplicative factor K appears and that this parameter may vary.

The root locus is the locus of possible roots of the closed-loop transfer function as the multiplicativeparameter K is varied. In fact, the entire root locus can be determined from the angle condition alone. Themagnitude condition is then used to determine which value of K corresponds to which set of closed-looppoles along the locus of all possible closed-loop poles.

Rather than learn Evans rules to begin with, it is more illustrative to consider a series of increasinglycomplicated examples.

Example 1. To begin, we consider the very simple example

P (s)Fd(s) = K1

s(s+ 2).

We will compute the closed-loop poles as explicit functions of K. In general, this is a tedious, anduninformative exercise, but for this simple system it serves to illustrate how closed-loop poles vary as thegain K is varied. The closed-loop transfer function is

Hd(s) =

Ks(s+2)

1 + Ks(s+2)

.

The closed-loop poles are obtained from

0 = 1 +K

s(s+ 2)

= s2 + 2s+K.

They are

s =1

2(24 4K)

= 11K.

Re

Im

xx

Figure 2: Root locus example #1.

2

When 0 < K < 1, there are two distinct poles which are located on the real axis between 0 and 2. WhenK = 1, the poles coalesce at s = 1. As K continues to increase, the poles split apart and move in oppositedirections parallel to the imaginary axis.

To see that the locus of closed-loop poles shown in Figure 2 can be obtained from the angle condition

1

s(s+ 2)= (2k + 1)pi k = 0,1,2, . . . ,

we first recall some facts about complex numbers. First, a complex number can be represented in polarform, for example z = rei where r is the radial distance from the origin to the point z and is the angleto z measured counter-clockwise from the positive real axis. Consider the complex function

C(s) =(s z1) (s zm)(s p1) (s pn) .

Each term in the numerator can be considered a vector from the zero zi to the point s. Similarly, eachterm in the denominator can be considered a vector from the pole pi to the point s. Each of these vectorshas a magnitude and an angle, so we may equivalently write

C(s) =

(rz1e

iz1) (rzmeizm)(

rp1eip1) (rpneipn )

=

(rz1 rzmrp1 rpn

)ei(z1++zmp1pn)

where rzi (or rpi) is the magnitude of the vector from zi (or pi) to s and zi (or pi) is the angle of thevector from zi (or pi) to s.

Re

Im

xx

s

qp2

qp1

Figure 3: Angle condition for root locus example #1.

Applying these observations to the current example, we find that

1

s(s+ 2)= s (s+ 2). (1)

Now, for any point on the real axis to the right of p1 = 0, equation (1) gives zero, which is not an oddnumber times pi. Similarly, for any point on the real axis to the left of p2 = 2, equation (1) gives 2pi,

3

which is also not an odd number times pi. Thus, the real axis to the left of p2 and to the right of p1 is notpart of the root locus. However, for points between p2 and p1, equation (1) gives

1

s(s+ 2)= s (s+ 2) = pi 0,

which is an odd number times pi. Thus, points on the real axis between p2 and p1 are part of the rootlocus.

Considering next the points on the vertical line s = 1, we choose a point and determine 1s(s+2) . The

vectors from p1 and p2 to any such point form an isosceles triangle. The sum of the two angles is pi forpoints above the real axis and 3pi for points below the real axis, giving 1

s(s+2) = pi or 3pi, respectively.Thus, the line s = 1 is part of the root locus.To find the value of K which corresponds to a particular pair of closed-loop poles, we use the magnitudecondition. For example, suppose we would like to choose K so that the closed-loop system has a damping

ratio =22 . Any pole lying on the radius =

3pi4 in the complex plane has damping ratio =

22 . Thus,

we would like to choose K to give closed-loop poles at

s = 1 i tan pi4= 1 i.

Choosing a particular pole, say s = 1 + i, we substitute into the magnitude condition to obtain K(1 + i)((1 + i) + 2) = 1

orK = |(1 + i)(1 + i)| = | 2| = 2.

Thus, choosing the gain K = 2 gives the closed-loop poles s = 1 i. We have assumed that P (s)Fd(s) can be written in the form

P (s)Fd(s) = Kb(s)

a(s).

where b(s) has degree m, a(s) has degree n m and where K > 0 is a parameter (e.g., a control gain)which may vary.

An important observation is that, as K 0, the closed-loop poles approach the poles of the loop gain. Tosee this, write the closed-loop characteristic equation as

a(s) +Kb(s) = 0.

Clearly, as K 0 the roots of the polynomial on the left approach the roots of a(s).One may also observe that, as K , the closed-loop poles must either diverge to or approach a zeroof the loop gain. To see this, recognize that as K , b(s)

a(s) must become very small so that the product

is 1. There are two ways that b(s)a(s) can become very small. First, b(s) can go to zero (which happens

when s approaches a zero of the loop gain). Second, a(s) can go to infinity (which can only happen when|s| goes to infinity.) In general, m branches of the root locus approach the zeros of the loop gain while theremaining nm branches go to infinity.Example 2. Consider the following example from [1]:

P (s)Fd(s) =K

s(s+ 1)(s+ 2).

4

This system has poles at p1 = 0, p2 = 1, and p3 = 2. Recalling that

P (s)Fd(s) =mi=1

(s zi)n

j=1

(s pj),

we first consider which, if any, points on the real axis are part of the root locus. For any point to theright of s = 0, P (s)Fd(s) = 0, so the positive real axis is not part of the root locus. For any point1 < s < 0, P (s)Fd(s) = pi, so these points are part of the root locus. For any point 2 < s < 1,P (s)Fd(s) = 2pi, so these points are not part of the root locus. Finally, for any point s < 2,P (s)Fd(s) = 3pi, so these points are part of the root locus.Next, we consider what happens to the root locus as s grows large. In the limit that s grows large, we have

lim|s|

P (s)Fd(s) = lim|s|K

s(s+ 1)(s+ 2)= lim

|s|K

s3.

Now, no matter how large |s| is, the angle condition must be satisfied, so we must havelim|s|

P (s)Fd(s) = limr P (re

i)Fd(rei)

limr

K

(rei)3

= ei3

= (2n+ 1)pi n = 0,1,2, . . .or

= 2n+ 13

pi.

Trying n = 0 gives = pi3 . Trying n = 1 gives = pi. Trying n = 2 gives = 5pi3 . Other choices of ngive repeated angles. In the limit that |s| , the three closed-loop poles follow asymptotes that extendradially in the directions pi3 and pi.A logical question is How large must |s| be before the root locus converges to these asymptotes? Rephras-ing the question, How does the root locus look for smaller values of |s| given that it converges to theseasymptotes as |s| grows large? A partial answer can be obtained by determining where the three asymp-totes are centered. A simple way to determine this is to notice that a slightly more precise approximationfor P (s)Fd(s) for large |s| is

lim|s|

P (s)Fd(s) = lim|s|K

(s+ 1)3.

To see this, compare the polynomials

(s+ 1)3 = s3 + 3s2 + 3s+ 1 and s(s+ 1)(s+ 2) = s3 + 3s2 + 2s.

The two polynomials agree to next-to-highest order. The root locus for the large-|s| approximation is thethree asymptotes computed previously, centered at the point s = 1. For the true system, the root locuswill behave a bit differently for small |s|, but we have at least located the origin of the asymptotes whichdescribe the large |s| behavior.Two of the three asymptotes extend into the right half complex plane, while the third follows the negativereal axis. Intuitively, the closed-loop pole which starts (for small K) at s = 2 will follow the negative realaxis asymptote as K increases. Therefore, the two closed-loop poles which rest on the real axis betweens = 1 and s = 0 must coalesce and split off to follow the asymptotes at pi3 .1

1They must first coalesce because poles must be either real numbers or complex conjugate pairs and because the closed-loop

pole locations vary continuously with K.

5

To determine precisely where this split occurs, known as a breakaway point, we recognize that theremust be a double pole between s = 1 and s = 0 for some value of K. Recall from System Dynamics thata double-root s = s of a polynomial C(s) satisfies not only C(s) = 0, but also C (s) = 0. You can see thisby recognizing that C(s) = (s s)2n2i=1 (s pi) if there is a double pole at s.For the feedback control system, we therefore have not only a(s) +Kb(s) = 0, when the closed-loop polescoalesce, but also

d

ds(a(s) +Kb(s)) = 0.

Solving for K, the value of the parameter for which the double-pole occurs, we find

K = a(s)b(s)

.

Substituting back into the condition for a closed-loop pole, we have

a(s) a(s)b(s)

b(s) = 0

orb(s)a(s) a(s)b(s) = 0 (2)

Re

Im

xx x

Figure 4: Root locus example #2.

For our system, we have b(s) = 1 and a(s) = s(s+ 1)(s+ 2). Solving (2) for the multiple-pole, we obtain

(3s2 + 6s+ 2) = 0.

Two solutions are

s =612

6= 1

3

3

The solution is not on the root locus, so the breakaway point must be s = 1 +33 0.4.

Note: You can find the value of the gain K at which the root locus passes into the right half plane byperforming a Routh-Hurwitz stability analysis and finding conditions on K for stability.

Example 3. Next consider a system for which

P (s)Fd(s) = Ks2 + 3s+ 2

s2 + 2s+ 4.

6

The loop gain has two zeros at z1 = 1 and z2 = 2 and two poles at p1,2 = 1 i3.

First, determine if any closed-loop poles lie on the real axis. The vectors from the two poles to points onthe real axis form an isosceles triangle. The sum of their contribution to PFd is 2pi. Similarly, a pair ofcomplex conjugate zeros would contribute 2pi to the angle calculation. Thus, we see that complex conjugatepoles and zeros have no impact on whether a given segment of the real axis is part of the root locus. Thereal axis poles and zeros alone determine which portion of the real axis is part of the root locus. In fact,one can easily verify that the following statement is true:

The portion of the real axis which is part of the root locus lies to the left of an odd number of polesand zeros.

For the present example, we observe that the only part of the real axis which contributes to the root locuslies between the zeros at z1 and z2. Recall that for small values of K, the closed-loop poles are close to thepoles of the loop gain P (s)Fd(s). To determine how the closed-loop poles leave p1 and p2 and approachthe real axis locus, we first compute the angle at which the locus departs from p1. (Of course, the locusnear p2 will simply be a mirror image about the real axis.) Taking a test point s very near p1, it is easy tosee that the contributions of z1, z2, and p2 to P (s)Fd(s) will remain more or less constant as we move sin a small circle around p1. Choosing s = p1, a point near p1, we must have

P (p1)Fd(p1) = (2k + 1)pi k = 0,1,2, . . .= (p1 z1) + (p1 z2) (p1 p1) (p1 p2) (p1 z1) + (p1 z2) d (p1 p2)

where d is the departure angle from p2. Taking k = 0, we therefore have

d = pi + (p1 z1) + (p1 z2) (p1 p2)

= pi + arctan(

3

0

)+ arctan

(3

1

) arctan

(23

0

)

= pi + pi2+

pi

3 pi

2

= 2pi3.

By mirror symmetry, the departure angle from p2 must be2pi3 .

Next, we compute the breakin point exactly as we did before, i.e., by using the condition for the existenceof a double-pole. From the coalescence condition

b(s)a(s) a(s)b(s) = 0,

we require that

0 = (2s+ 3)(s2 + 2s+ 4

) (2s+ 2) (s2 + 3s+ 2)= s2 + 4s+ 8.

Two solutions are s = 2 23; only the solution lies on the root locus, so the breakin point iss = 2 23 1.46.

A total of m closed-loop poles approach the zeros of the loop gain P (s)Fd(s) as K . Theremaining nm closed-loop poles follow asymptotes outward to infinity.

7

Re

Im

x

x

oo

2p/3

Figure 5: Root locus example #3.

Following is the general procedure for constructing a root locus plot, as adapted from [1].

Step 1. Locate the poles and zeros of the loop gain P (s)Fd(s). First, compute the zeros of the loopgain (the roots of b(s)) and place an o at their location in the complex plane. Next, compute the polesof the loop gain (the roots of a(s)) and place an x at their location in the complex plane.

Step 2. Determine what, if any, portion of the real axis is part of the root locus. The anglecondition requires that the real axis portion of the root locus lies to the left of an odd number of poles andzeros. Equivalently, since complex poles and zeros must occur in conjugate pairs, the real axis portion ofthe root locus lies to the left of an odd number of real poles and zeros.

Step 3. Determine the asymptotes of the root locus. Given that there are m zeros and n mpoles, m of the closed-loop poles will approach the loop gain zeros as K and the remaining n mwill converge to asymptotes which extend radially to infinity from some starting point on the real axis.The asymptote angles are

(2k + 1)pi

nm k = 0,1,2, . . . ,

which can be proved by approximating the loop gain with Ksnm

for large values of |s|.The center of the asymptotes can be computed from a slightly better approximation obtained as follows.Write

b(s)

a(s)=

(s z1) (s zm)(s p1) (s pn)

=sm + (z1 zm)sm1 + sn + (p1 pn)sn1 +

Dividing both the numerator and the denominator by the numerator gives

b(s)

a(s)=

1

snm + ((z1 + + zm) (p1 + + pn)) snm1 + (3)

For large values of |s|, this ratio of polynomials can be approximated by(s+

(z1 + + zm) (p1 + + pn)nm

)(nm). (4)

8

That is, (3) matches (4) to order snm1. The root locus for this (approximate) loop gain consists of nmrays extending radially from the point

=(p1 + + pn) (z1 + + zm)

nm .

The real number is the center of the asymptotes for root locus corresponding to the true loop gain.

Step 4. Find the breakaway and breakin points. Recall that these points correspond to values ofthe gain K for which the closed-loop system has multiple closed-loop poles at a particular point. For adouble-pole, the condition

b(s)a(s) a(s)b(s) = 0must be satisfied. The roots of this algebraic equation give possible breakaway or breakin points. Todetermine whether these are, in fact, breakaway or breakin points, one must check whether these pointsare actually on the root locus.

Step 5. Determine the angles of departure from the loop gain poles and the angles of arrival

at the loop gain zeros. Recall that as K 0, the root locus approaches the poles of the loop gain andas K , m branches of the root locus approach the zeros of the loop gain. The angle of departure fromthe kth loop gain pole pk can be obtained from the angle condition as

d = pi +i

(pk zi)j 6=k

(pk pj).

That is, the departure angle is pi plus the sum of all the angles of vectors pointing from the loop gain zerosto pk minus the sum of all the angles of vectors pointing from the remaining loop gain poles to pk.

Similarly, one can use the angle condition to show that the angle of arrival at the kth loop gain zero zk is

a = pi i6=k

(zk zi) +j

(zk pj).

In general, it is a good idea to also compute the value of K at which the root locus crosses into the righthalf of the complex plane for the first time. This can be done using the Routh-Hurwitz procedure. Thegain value at which the root locus first crosses into the right half plane generally serves as an upper limiton acceptable choices of the parameter K.

To determine the value of K corresponding to a particular closed-loop pole on the root locus, one mustuse the magnitude condition. Recognizing thatK b(s)a(s)

= 1 K =a(s)b(s)

,we have, for a particular closed-loop pole s,

K =

j |(s pj)|i |(s zi)|

.

Example: Stabilizing an Inverted Pendulum. The nondimensional equation for an inverted pendulumis

2n = 2nu,so the transfer function from torque to angle is

G(s) =2n

s2 2n.

9

10 8 6 4 2 0 2 4 6 8 1010

8

6

4

2

0

2

4

6

8

10Root Locus

Real Axis

Imag

inar

y Ax

is

Figure 6: Root locus for proportional feedback.

The system has no zeros and two real-conjugate poles.

We start by applying proportional feedback Fd(s) = kp. The loop gain becomes

PFd = kp2n

s2 2n.

The root locus is shown, for particular parameter values, in Figure 7. Clearly the feedback control lawonly marginally stabilizes the system, provided kp is large enough.

40 35 30 25 20 15 10 5 0 5 1020

15

10

5

0

5

10

15

20Root Locus

Real Axis

Imag

inar

y Ax

is

Figure 7: Root locus for proportional-derivative feedback.

Next, we apply proportional-derivative feedback

Fd = kds+ kp = kp

(kdkps+ 1

).

The loop gain becomes

PFd = kp

(1s+ 1

)2n

s2 2n

10

where =kpkd. The compensator introduces a new loop-gain zero at . We will assume that remains

constant as kp is varied. (That is, kd varies in direct proportion to kp.) The root locus is shown in Figure 7.Clearly, the feedback controller stabilizes the system. The independent freedom in kp and allows one toobtain any desired closed-loop pole locations (provided the poles are both real or are complex conjugates).

Example: Longitudinal Autopilot. The longitudinal dynamics of the airplane shown in Figure 8 aredescribed by the following equations

(

V

)q

=

(1 00 1

V

)(( DL)+RT ()

(u1u2

)+RT ()R()

(01

))qu2

,

where R() is the proper rotation matrix

R() =

(cos sinsin cos

)

and the aerodynamic forces satisfy

D = KDV 22 (5)L = KLV 2(1 + ). (6)

(Note: These equations have been normalized. All quantities are dimensionless.) The inputs are u1,representing thrust, and u2, representing the elevator deflection. The small positive scalar accounts fora slight downward force due to positive deflections of the elevator.

V

vu

g

a

q

Figure 8: Longitudinal aircraft motion.

Define the state vector x = [V, , , q]T . Assuming that KL = 1/V2, we linearize the dynamics about

the equilibrium x = [V , 0, 0, 0]T . A somewhat tedious series of calculations gives the linearized dynamics

d

dtx =

0 1 1 0 2

V 2 1

V0 0

0 0 0 10 0 0 0

x+

1 00

V

0 00 1

u.

Suppose we take the output of interest to be flight path angle :

y = = [0, 1, 1, 0]x.The transfer function from the elevator u2 to the flight path angle is

G(s) = [0, 1, 1, 0]

s 1 1 02V 2

s+ 1V

0 0

0 0 s 10 0 0 s

1

0

V

01

= s

(

Vs2 + s+ 1

V

)s2(s2 + 1

Vs+ 2

V 2

)

11

=

(

Vs2 s 1

V

)s(s2 + 1

Vs+ 2

V 2

) .Note that this transfer function is nonminimum phase for all > 0. Recall that a stable, non-minimumphase system initially responds to a positive step input in the negative direction. Physically, the downwardforce on the elevator and tail fin due to a pitch-up command cause the airplane, initially, to acceleratedownward. This results in a negative flight path angle until the integrated effect of the tail momentincreases the aircraft pitch angle sufficiently to provide upward lift.

2 0 2 4 6 8 10 126

4

2

0

2

4

6Root Locus

Real Axis

Imag

inar

y Ax

is

Figure 9: Root locus for proportional feedback.

We would like to design an autopilot to regulate the flight path angle for an aircraft with flight conditionsand aircraft parameter values given by

V = 1, and = 0.25.

(In fact, the aircraft is already stable, but its motion is very lightly damped. We start by applyingproportional feedback Fd(s) = kp. The loop gain becomes

PFd = kp

(

Vs2 s 1

V

)s(s2 + 1

Vs+ 2

V 2

)Consider the root locus shown in Figure 9. Observe that straight proportional feedback drives this slightlystable system unstable! The right half plane zero draws the branches of the locus into the right half plane.

Notice that the root locus in Figure 9 appears not to satisfy Evans rules. For example, the real axis locusdoes not lie to the left of an odd number of poles and zeros. This is a consequence of the negative sign inthe numerator, the very term which makes the system nonminimum phase. Essentially, it is as if we havechanged the sign of the proportional gain and are now applying positive feedback. For positive feedback,the angle condition changes to

PFd = 2kpi k = 0,1,2, . . . ,and Evans rules change accordingly.

To stabilize the system, suppose we introduce an additional left half plane zero in the loop gain by applyingproportional derivative feedback. This additional zero will have the effect of drawing the branches of the

12

15 10 5 0 5 105

4

3

2

1

0

1

2

3

4

5Root Locus

Real Axis

Imag

inar

y Ax

is

Figure 10: Root locus for proportional-derivative feedback.

root locus into the left half plane. We choose

Fd = kds+ kp = kp

(kdkps+ 1

).

The loop gain becomes

PFd = kp

(1s+ 1

)(

Vs2 s 1

V

)s(s2 + 1

Vs+ 2

V 2

) .where, once again, =

kpkd.

Figure 10 shows the root locus for this system. The branch coming into the RHP zero from the right isan extension of the branch moving the left. It is merely an artifact. (Root loci for nonminimum phasesystems are just weird.) Thus, we see that proportional derivative control provides stable regulation of theflight path angle.

Example: Parametric Stability Analysis for Watts Regulator. Watts regulator is an earlyexample of mechanical feedback control. This apparatus can be simply modeled as a planar pendulumwhich is made to rotate about its vertical axis by some device, such as a steam engine, whose speed is tobe controlled. The speed of the engine is governed by the regulator through a mechanical linkage betweenthe pendulum and the engine throttle. A simple dynamic model is

q

v

Figure 11: A simple model for Watts regulator.

13

+ b + sin (2c v cos

)= 0 (7)

v + av = a(v k( )

), (8)

where is the elevation angle between the pendulum and the vertical axis and v represents the (square ofthe) angular speed of the regulator shaft. The nominal value of v is v and the nominal value of is

= arccos

(2cv

).

The parameter a > 0 determines the response time of the engine. The constant b > 0 is a dampingparameter for the regulator and c > 0 is the undamped natural frequency of planar swinging motion(when v 0). The parameter k is a control gain.As manufacturing processes improved in the latter half of the nineteenth century, friction in mechanicallinkages was dramatically reduced. Coincidentally, new machines whose speed was controlled by Wattregulators began to exhibit undesired speed oscillations due to oscillations in the elevation angle . Thisbehavior was referred to as hunting. Lets use a root locus plot to investigate this phenomenon. To doso, suppose that, for a particular regulator,

a =61

81and k = 80

3

and that the regulators nominal (equilibrium) state is

(, , v)e = (, 0, v) =(pi3, 0, 80

).

Linearizing the dynamics about this equilibrium and computing the characteristic equation for the statematrix gives

0 = 813 + (61 + 81b)2 + (4860 + 61b)+ 4880

=((+ 1)(812 20+ 4880))+ b(812 + 61)

or

0 = 1 + b812 + 61

(+ 1)(812 20+ 4880) .

The key observation is that the eigenvalue equation can be written in a way which resembles the charac-teristic equation for a feedback control system with loop gain

812 + 61

(+ 1)(812 20+ 4880) .

Evans rules apply equally well to this problem. Note that the loop gain has two zeros at = 0 and at = 6181 . There are also three loop gain poles: at s = 1 and at

s =20400 81 4880

162 0.123 i3.879.

Note that the closed-loop poles, that is the characteristic values, have positive real part for small valuesof b. As b is increased, two branches of the root locus approach the two loop gain zeros and the thirdfollows the asymptote along the negative real axis. The portion of the root locus which lies on the realaxis lies between the zeros at s = 0 and s = 6181 and to the left of the pole at s = 1.We know that the three branches of the root locus leave the poles of the loop gain and that two of thesebranches somehow approach the two zeros while the third diverges to infinity along the negative real axis.

14

One way this could happen is that the two branches originating at the complex conjugate poles couldcoalesce between s = 0 and s = 6181 . However, there is another possibility which becomes apparent whenwe attempt to compute the break-in point. To compute the break-in point, we define

b(s) = 81s2 + 61s

a(s) = 81s3 + 61s2 + 4860s+ 4880

and compute

b(s)a(s) a(s)b(s) = (162s+ 61) (81s3 + 61s2 + 4860s+ 4880) (243s2 + 122s+ 4860) (81s2 + 61s)= 6561s4 9882s3 + 389939s2 + 790560s+ 297680

which has roots at

s = 7.432, s = 1.529, s = 0.502, and s = 7.957.

Note that all but the last point lie on the root locus. The root locus is clearly more complicated thanwe first imagined. A simple exercise in logic shows that the only feasible shape for the root locus is theone depicted in Figure 12. Because branches of the root locus leave loop gain poles as b increases, thepoint s = 1.529 must be a break-away point rather than a break-in point. The remaining two pointsare obviously break-in points. Suppose the two branches which leave the complex conjugate poles were tobreak in at s = 0.502 and converge to the zeros, as we originally hypothesized. Then we would have acontradiction between the fact that the real axis to the left of s = 1 is part of the root locus and the factthat there is a break-away and break-in point to the left of the pole at s = 1. (Note: a single pole cannotsuddenly split and become two poles!)

The angle of departure for the pole at 0.123 + i3.879 is

d = pi + 1.539 + 1.349 1.289 pi2

= 3.170

and, by mirror symmetry, the angle of departure for its complex conjugate is 3.170 radians.

Re

Im

oo

x

x

x

Figure 12: Root locus as b varies from 0 to . Note that as b 0, the closed loop eigenvalues move intothe right half plane.

The hunting phenomenon is a consequence of the fact that, while the damping coefficient had previouslybeen sufficiently large to ensure stability of the closed-loop system, improvements in machining lowered b

15

below a critical value for stability. The oscillatory behavior referred to as hunting is a consequence ofwhat dynamical systems theorists call a Hopf bifurcation in which a complex conjugate pair of eigenvaluescrosses over the imaginary axis as a bifurcation parameter is varied. The equilibrium about which thedynamics are linearized becomes unstable and the nonlinear system begins to exhibit a periodic oscillationabout the equilibrium state. Note that this oscillation is not predicted by the linearized dynamics; theright half plane poles of the linearized system would suggest that the state diverges exponentially, but thisdoes not happen in reality. The oscillation is a nonlinear phenomenon referred to as a limit cycle.

References

[1] K. Ogata. Modern Control Engineering, Fourth Ed. Prentice Hall, Upper Saddle River, NJ, 2002.

16

Lecture 1: Introductory Remarks

Aerospace engineering encompass a broad range of challenging topics which must be mastered in order todesign atmospheric and space flight vehicles. Topics of fundamental importance include

aerodynamics

propulsion

structures and materials, and

vehicle dynamics and control.

A fifth topic, vehicle design, envelops all of these more basic areas. It involves integrating knowledge in eachof the four subject areas in order to synthesize a complete vehicle which satisfies prescribed performancerequirements.

This course focuses on dynamics and control of atmospheric flight vehicles, particularly fixed-wing aircraft.This topic is also referred to as flight mechanics. Flight mechanics comprises three major subtopics:

performance,

stability and control, and

aeroelasticity.

Conventionally, each of these subtopics is studied individually although the three are very much related.In studying aircraft performance, one considers issues such as range, take-off and landing distance, andtrajectory planning for a given aircraft. This involves determining the forces necessary to achieve a givenpath of motion, assuming that these desired forces can be generated. Thus, one typically models theaircraft as a point mass subject to three control forces: lift, side force, and thrust. Performance isconcerned with the large-scale aircraft motions associated with takeoff, landing, turning, etc.

In studying stability and control, one takes a closer look at the aircraft and recognizes that lift and side forceare not true control forces. Rather, these forces are a consequence of the aircrafts orientation with respectto the local air flow. To generate a desired lift force, for example, the vehicle must effect a particularangle of attack. Thus, in stability and control, one is concerned with how the vehicles orientation, orattitude, changes under the influence of moments generated by the actuators. These moments are typicallygenerated by a pilot through a suitably designed interface (such as a stick and pedals).

In studying performance, one assumes that the aircraft is a point mass. In studying stability and control,one typically assumes that the aircraft is a rigid body. In studying aeroelasticity, one recognizes thatno aircraft is truly rigid and, moreover, that changes in the vehicle shape due to varying load conditionscan have dramatic effects on the vehicles motion. Aeroelastic phenomena that can arise for real aircraftinclude wing or control surface flutter, roll control reversal, and other effects.

In this course, we will consider only the second sub-topic: stability and control. As the course title suggests,there are two issues of primary importance. Stability relates to the intrinsic flying qualities of the aircraft.Stability is a characteristic of the vehicle dynamics which is independent of the pilots actions.

Control relates to a pilots interaction with the aircraft. Of interest are the following two questions:

How effective are the various actuators at forcing the aircraft into a desired motion?

1

How much effort is required of the pilot to generate the necessary actuator commands?

Issues related to these two questions include actuator placement and sizing, stick motion cues for aircraftwith power-assisted actuators, and the use of feedback control to enhance handling qualities and rejectdisturbances.

We will begin by considering stability. To discuss stability of a steady motion, we must first introducesome terminology to describe the motion. Suppose we fix a reference frame to some point in the aircraft,as shown in Figure 1. We denote by xB the unit vector pointing through the nose of the aircraft. Thisaxis is often referred to as the longitudinal axis. We let zB represent the unit vector pointing through thebelly of the aircraft; this is often called the directional axis. Finally, we define the lateral axis in terms ofthe unit vector yB = zB xB. Viewing the aircraft from behind, yB points to the right.

xI

yI

zI

xB

yB

zB

V

!

Figure 1: Inertial and body-fixed reference frames.

To describe the orientation of the aircraft, we define an inertial reference frame, which is denoted by thefixed unit vectors xI, yI, and zI. The reason we choose to describe the aircrafts orientation with respectto an inertial frame is that Newtons laws of motion only hold in an inertially fixed frame. In this course,we will typically consider an earth-fixed frame to be an inertial frame. Although the resulting equationsof motion will technically be incorrect, the error due to the earths rotation, its revolution about the sun,etc. will be small over the time periods of interest in studying stability and control.

As the aircraft is assumed to be rigid, the location of any point in the airplane is uniquely determinedby the position and orientation of the body-fixed reference frame. Therefore, we will often represent theaircraft simply by its body-fixed reference frame. Suppose that the aircraft (i.e., the body frame) translatesat some velocity with respect to the inertial frame. We let

V =

uv

w

denote the translational velocity of the body with respect to the inertial frame, but expressed in the bodyframe.1 Also, suppose that the aircraft rotates at some angular velocity with respect to the the inertial

1Note the distinction, here! While any given vector can be expressed in any given reference frame, derivatives are always

taken with respect to a specific frame.

2

frame. We let

=

pq

r

denote the angular velocity of the body with respect to the inertial frame, but expressed in the body frame.

Axis Linear Aerodynamic Angular Angular AerodynamicVelocity Force Displacement Velocity Moment

Longitudinal (xB) u X p L

Lateral (yB) v Y q M

Directional (zB) w Z r N

The angular displacement variables , , and do not generally represent angles about the body-fixedaxes. These angles, referred to as the Euler angles, define a series of three rotations which transformvectors from the inertial frame to the body frame, and vice versa. We will discuss the parameterizationof vehicle attitude in more detail later in the course. Until then, we will only consider simple motions inwhich, for example, the pitch angle is truly a rotation about the lateral (yB) axis.

The aerodynamic forces and moments are conventionally denoted in terms of dimensionless coefficients.Let V = V be the airspeed, let S denote a reference area, and let l denote a reference length. Then onewrites

X = CX

(1

2V 2

)S

Y = CY

(1

2V 2

)S

Z = CZ

(1

2V 2

)S

L = Cl

(1

2V 2

)Sl

M = Cm

(1

2V 2

)Sl

N = Cn

(1

2V 2

)Sl

Note: The reference area is typically chosen to be the wing planform area. Reference lengths may differdepending on the context. For the pitch moment coefficient, for example, one typically takes l = c, themean aerodynamic chord. For the roll and yaw moment coefficients, one takes l = b, the wing span.

The use of upper-case subscripts in the force coefficients is consistent with the notation for aerodynamicforces. The apparently inconsistent use of lower-case subscripts in the moment coefficients avoids a potentialambiguity between roll moment coefficient and lift coefficient.

The dimensionless coefficients CX , CY , CZ , Cl, Cm, and Cn, are primarily functions of the Mach numberM = V/a (where a is the speed of sound), the Reynolds number Re = (V l)/ (where is the fluid densityand is the dynamic viscosity), and the aerodynamic angles and . Recall that the aerodynamic anglesare defined solely in terms of the body translational velocity:

= arctanw

uand = arcsin

v

V.

3

vu

w

V

Figure 2: Aerodynamic angles.

These angles are shown in Figure 2. In normal flight conditions, the dimensionless coefficients dependprimarily on the variables and parameters mentioned above. They also depend, to a lesser extent, on thebody angular rate and the aerodynamic angle rates and .

An often-used simplifying assumption is that an aircraft is symmetric about the xB-zB plane. Motionswhich are restricted to this plane of symmetry, such as wings-level climbs and loops, are called symmetric orlongitudinal motions. Motions out of the plane of symmetry, such as banked turns, are called asymmetricor lateral-directional motions. Accordingly, the components of velocity and aerodynamic force and momentare often decomposed into these two groups:

Longitudinal (or symmetric) quantities: u, w, q, X, Z, M

Lateral-directional (or asymmetric) quantities: v, p, r, Y , L, N

Static longitudinal stability. When discussing flight of atmospheric vehicles, the term stability refersto a property of a special class of motion known as steady motion. For a vehicle in steady motion, allcomponents of body translational velocity V and body angular velocity are constant. A special caseof steady motion is equilibrium flight, in which the vehicle acceleration is zero. Note that these twodefinitions are distinct. Steady, wings-level flight at constant altitude is equilibrium flight. A horizontalturn at constant radius and velocity is not equilibrium flight; the constant yaw rate turn requires a constantcentripetal acceleration. Equilibrium flight is a steady motion for which = 0.

Stability (or instability) is a property corresponding to a steady motion. Loosely speaking, if a vehiclewhich is slightly perturbed from a steady motion returns to that steady motion, the motion is stable. Ifthe vehicle motion diverges in response to a small perturbation, the motion is unstable.

The flight mechanics literature distinguishes between two finer notions of stability: static and dynamicstability. The term static stability is somewhat of a misnomer because, by definition, stability (or instabil-ity) refers to a systems motion in response to a disturbance. Static stability refers to the initial tendencyof a vehicle, if displaced from a given steady motion, to return to that motion. No information about thevehicles subsequent motion is required, only its initial tendency. Thus, one may determine static stabilitywithout solving the differential equations that describe the airplanes motion.

Even though a given steady motion may be statically stable, the vehicle may diverge from the given motionwith time. To characterize this latter phenomenon, one must consider dynamic stability in which thecomplete vehicle motion, not just its initial motion, is important. A given steady motion is dynamicallystable if, after a small displacement, the aircraft returns to the steady motion asymptotically in time.Dynamic stability is stronger than static stability:

Dynamic stability Static stability but Static stability 6 Dynamic stability

4

0 1 2 3 4 5 6 7 8 9 105

4

3

2

1

0

1

2

3

4

5

Nondimensional Time

Non

dim

ensio

nal P

itch

Angl

e

Reference value for steady motionStatically and dynamically stableStatically stable and dynamically unstableStatically and dynamically unstable

Figure 3: Sketches depicting static and dynamic pitch stability.

For example, a vehicles state may undergo diverging oscillations about a statically stable steady motion.See Figure 3.

We will begin by investigating the conditions under which steady wings level flight is statically stable inpitch. Consider a rigid aircraft with a reference frame fixed in the body at its center of gravity (CG) asshown in Figure 1. If the xB-zB plane is a plane of symmetry, then the pitch rate equation is

q =1

Iy

((Iz Ix)pr + Ixz(r

2 p2) +M), (1)

where Ii is the moment of inertia about the ith coordinate axis and Ixz is a product of inertia. This is one

of three first order ODEs for the body angular rate; there are also equations for p and r. In addition, thereare three first order ODEs for the components u, v, and w of translational velocity. These six equationsdescribe the aircraft dynamics. Six more first order ODEs describe the variation of position and attitudedue to changes in velocity, that is, the aircraft kinematics.

Equation (1) is a nonlinear ordinary differential equation; the dependent variables p and r appear quadrat-ically. We consider the case of steady, wings-level flight. In this case, p = r = 0. Also, v = 0. In theabsence of asymmetric disturbances, the lateral-directional variables remain zero; the motion is purelylongitudinal. The pitch rate equation becomes simply

q =1

IyM =

1

Iy

((1

2V 2

)Sc

)Cm.

The dimensionless coefficient Cm depends on a number of variables and parameters. For pure longitudinalflight, the primary influences are angle of attack , Reynolds number Re, and Mach number M. To alesser extent, Cm also depends on q and . For now, we will ignore all dependencies save and writeCm = Cm(). Formally expanding this expression in a Taylor series about zero angle of attack gives

Cm = Cm0 + Cm+ h.o.t. (2)

5

If we assume that (given in radians) remains fairly small, then we may neglect the higher order termsin (2). The term Cm0 is the pitch moment coefficient at zero angle of attack and

Cm =Cm

=0

is the slope of the pitch moment coefficient curve. Because of its critical role in determining both staticand dynamic stability, Cm is referred to as a stability derivative.

One should keep in mind that (2) is only an approximation. It ignores the dependency of Cm on Re andM, as well as higher order terms in . Moreover, for fast, asymmetric maneuvers, Cm also depends on, , , p, q, and r and possibly other variables and parameters. For now, we consider only the case ofsymmetric (wings-level) equilibrium flight

Cm

3. 0Cm

2. = 0Cm

CL

eqCL

eq eq

Figure 4: Pitch coefficient possibilities.

From a force balance in the zI direction, we see that the lift which the aircraft generates must perfectlybalance its weight for equilibrium flight. Suppose that we measure the angle of attack from the airplaneszero-lift line, so that

CL = CL.

Given that CL > 0 and that lift must act in the upward direction to balance the airplanes weight, apositive angle of attack is required at equilibrium, say = eq > 0.

ASIDE: If the angle of attack (call it , for now) is given in reference to some other line in the body sothat

CL = CL0 + CL,

we may shift the origin by defining = 0L

where

0L = CL0CL

.

Doing so givesCL = CL,

as we have assumed.

We know that must be positive and constant for equilibrium flight. Moreover, equilibrium flight requiresthat the angular rate be zero, i.e., that q = 0. It follows that Cm() must be zero at the precise value of > 0 for which lift balances weight. Referring to Figure 4, there are three possibilities to consider:

6

1. Cm0 < 0 and Cm > 0

2. Cm0 = 0 and Cm = 0

3. Cm0 > 0 and Cm < 0

In each case, the pitch coefficient is zero when = eq. What distinguishes the three cases is what happenswhen the equilibrium is disturbed.

Case 1. If an impulsive pitch disturbance causes the angle of attack to decrease ( < eq), then the pitchmoment coefficient becomes negative. This results in a nose-down pitching moment which drives the angleof attack even lower. Alternatively, if a pitch disturbance causes the angle of attack to increase ( > eq),then the pitch moment coefficient becomes positive. This results in a nose-up pitching moment whichdrives the angle of attack even higher. Thus, if Cmcg > 0, steady wings-level flight is statically unstable.

Case 2. In this case, variations in the angle of attack have no effect on the pitch moment coefficient. Thusno additional moment is developed which would either drive the aircraft away from the equilibrium motionor cause equilibrium to be restored. If Cmcg = 0 (with Cmcg0 = 0), steady wings-level flight is calledneutrally stable.

Case 3. If an impulsive pitch disturbance causes the angle of attack to decrease, then the pitch momentcoefficient becomes positive, resulting in a nose-up pitching moment which drives the angle of attack backup toward eq. Alternatively, if a pitch disturbance causes the angle of attack to become positive, thenthe pitch moment coefficient becomes negative, resulting in a nose-down pitching moment which drives theangle of attack back down toward eq. Thus, if Cmcg < 0 (with Cmcg0 > 0), steady wings-level flight isstatically stable.

Static pitch stability clearly requires that Cm() have a negative slope when it crosses the -axis at theequilibrium angle of attack eq. Moreover:

Static longitudinal stability requires Cm < 0 and Cm0 > 0.

7

Lecture 2: Introduction to Static Longitudinal Stability

Transferring moments. Recall that in the previous lecture we began discussing static longitudinalstability. We obtained requirements on the dimensionless pitch moment coefficient as a function of theangle of attack . Specifically, we found that static longitudinal stability requires Cm < 0 and Cm0 > 0.Before we discuss the various aircraft components and their contributions to Cm, we should review thebasic notion of equivalent representations of forces and moments.

z

x

a

F

M

O

z

x

F

( - )M Fx

=x

a

Figure 1: Equivalent force and moment diagrams.

Consider the planar rigid body shown on the left in Figure 1. The body is subject to a force F acting atthe point xa and a moment M , which is a pure couple. For this system, we have

Fz = F and

MO =M Fxa.

One may easily transfer a set of forces and moments acting at a given point to any other point. Forexample, one may transfer the force and moment above to the origin O, as shown at the right in Figure 1.

L

D

M

V

c

a

a

a

a

L

D

M

V

c

b

b

b

b

x x

Figure 2: Equivalent force and moment diagram for a wing.

Now consider a rectangular wing. We assume that lift force, drag force, and aerodynamic moment areknown, as functions of angle of attack, at the point xa. (In keeping with aerodynamics convention, thesigned distance x is measured positive aft from the wing leading edge.)

Suppose we wish to transfer the forces and moment from the point xa to another point xb along the chord.The forces are equal at either point:

Lb = La = L and Db = Da = D.

It remains to determine the moment Mb given Ma, La, and Da. First, compute the moment of the systemon the left about a particular point, say the leading edge:

Ml.e. =Ma L(xa cos)D(xa sin).

Next, compute the moment of the system on the right about the same point:

Ml.e. =Mb L(xb cos)D(xb sin).

1

Equating the two expressions for Ml.e. and solving for Mb gives

Mb =Ma + (L cos+D sin)(xb xa).

Dividing through by(12V

2)Sc gives

Cmb = Cma + (CL cos+ CD sin)(xbc xa

c

).

(Note: Since we are considering a rectangular wing, the mean aerodynamic chord c is simply the constantchord length c.) Define the nondimensional distances

ha =xac

and hb =xbc.

For small angles of attack1, we have

Cmb = Cma + (CL cos+ CD sin) (hb ha) Cma + CL

(1 +

CDCL

)(hb ha) .

For a well-designed wing operating below stall , CDCL

1. And since we have already assumed that is small, the product CD

CL may be neglected. We therefore have the following approximate equation for

transferring an aerodynamic moment between points on a wing:

Cmb Cma + CL (hb ha) . (1)

Aerodynamic reference points. A common reference point for the aerodynamic forces and the pitchmoment on a wing is the aerodynamic center. The aerodynamic center is that point about which thepitching moment does not vary with angle of attack. To find this point, note that by definition

Cmac

= 0.

Thus, if one knows CL and Cma , about some point ha, as functions of (from wind tunnel tests, forexample), one may obtain Cmac through the following procedure:

1. Let hb in equation (1) denote the aerodynamic center.

2. Set the derivative of equation (1) with respect to equal to zero:

0 =Cmac

=Cma

+CL

(hac ha) .

Notice that if CL and Cm are linear in (and we will generally assume that they are), all terms inthe equation above are constants.

3. Solve for the location of the aerodynamic center:

hac = ha (CL

)1 Cma

. (2)

4. Substitute hac back into equation (1) to obtain Cmac .

1The error in this approximation is less than 5% for || pi12

radians 15.

2

Given a set of wind tunnel data, there is often a simpler approach to find Cmac . If = 0L, that is, theangle of attack corresponding to zero lift, then CL is zero in equation (1) and we have

Cmb = Cma = Cmac .

Thus, the zero-lift pitching moment has the same value as the (constant) moment about the aerodynamiccenter:

Cmac = Cm0L .

Letting xa = xac, we may re-write the moment transfer formula (1) as

Cmb Cmac + CL (hb hac)= Cm0L + CL (hb hac)

Aerodynamic data for wings are typically referenced to the wing aerodynamic center or some other wing-related reference point. When writing the equations of motion for an entire aircraft, however, it is mostconvenient to sum moments about the aircraft center of gravity. Thus, a typical application of the formulaabove will be to transfer the wing aerodynamic moment to the aircraft center of gravity.

To this point, we have only discussed the aerodynamic center for a rectangular wing. For a more generalwing, one introduces the notion of mean aerodynamic center xac. Appendix C in [1] presents techniques fordetermining (or approximating) this point, as well as the mean aerodynamic chord c, for wings of generalshape. In subsonic flight, the aerodynamic center is located roughly one-quarter chord aft of the wingsleading edge. In supersonic flight, the aerodynamic center shifts aftward to roughly the half-chord point.

Another reference point which is sometimes important is the point at which the moment generated by thewing vanishes entirely. The center of pressure is the point about which the moment due to the aerodynamicforce generated by the wing (i.e., the vector sum of lift and drag) precisely balances the pure aerodynamiccouple generated by the wing. To find the center of pressure, we solve

0 = Cmcp = Cm0L + CL (hcp hac)to obtain

hcp = hac Cm0LCL

.

Note that the center of pressure varies with because CL varies with . For this reason, the center ofpressure is generally a less useful reference point in aircraft dynamic modeling.

Conditions for Static Longitudinal Stability. Lets return now to the problem of static longitudinalstability. The two requirements we obtained are that the pitch moment coefficient Cm about the centerof gravity (CG) must have a negative slope and be positive at the zero lift angle of attack 0L. Thefirst condition ensures that a restoring moment is generated in response to small perturbations from eq.The second condition ensures that an equilibrium angle of attack exists for which the wing generates thepositive lift necessary to balance the airplanes weight.

Given xac and Cmac , we may write

Cmcg = Cmac + CL (hcg hac) .The first condition for static longitudinal stability is that Cmcg < 0, where

Cmcg =Cmcg

=Cmac

+CL

(hcg hac)= CL (hcg hac) .

3

Since CL > 0 and x is measured positive aft of the leading edge, this condition says that the center ofgravity must be forward of the aerodynamic center.

Now consider the second condition for static longitudinal stability, that Cmcg |0L > 0. This conditionensures that the pitch coefficient curve passes through zero at an angle of attack eq for which CL(eq) ispositive. Thus, positive lift will be generated when the pitch moment is zero. Assuming that the dynamicpressure is appropriate, the aircrafts weight will be perfectly balanced by the lift that it generates. Wecan express this condition as a condition on Cmac as follows:

0 0 and Cmcg < 0 or, equivalently, Cmac = Cm0L > 0 and (hcg hac) < 0.

Cm

CL

CL

Cm

CL

eqCL

0L

0LCm , C acm

eq

eq eq

eqeq

0LCm , C acm

Figure 3: Generic lift and pitch moment coefficient curves. The bottom and top graphs are equivalentexcept that, in the lower graphs, is measured from the zero-lift line while.

Shown in Figure 3 are representative lift and pitch moment coefficient curves, where Cm = Cmcg . At thetop, CL and Cm are plotted versus where is not measured from the zero-lift line. Below these plots,CL and Cm are plotted versus , which is measured from the zero-lift line (meaning 0L = 0). Note thatthe the pitch moment coefficient curve represents a statically stable wing for which the balanced angle ofattack corresponds to positive lift. That is, Cmac = Cm0L > 0 and Cmcg < 0.

Example. (Courtesy of Dr. F. Lutze) Suppose lift force and pitch moment data have been obtained fora rectangular flying wing. Moment data are referred to the one-third chord point x = c3 where x ismeasured aft from the leading edge. The data are given in the table below.

4

(deg) CL Cm1/30.5 0.2 -0.02

3.0 0.4 0.00

5.5 0.6 0.02

8.0 0.8 0.04

Clearly CL and Cm1/3 vary linearly over this range of because each 2.5 increment in gives an equal

increment in CL and Cm1/3 . A least squares fit (though unnecessary for these data!) gives

CL =CL

= 0.08 deg1 = 4.6 rad1

Cm(1/3) =Cm1/3

= 0.008 deg1 = 0.46 rad1

We have

CL = CL0 + CL

= 0.16 + 0.08 deg

= 0.16 + 4.6 rad.

and

Cm1/3 = Cm(1/3)0 + Cm(1/3)

= 0.024 + 0.008 deg= 0.024 + 0.46 rad.

Let us compute the aerodynamic center of the wing from these data. Equation (2) gives

hac = ha (CL

)1 Cma

=1

3 0.008

0.08 0.23

which says that the aerodynamic center is at roughly 23% chord. Recall that Cmac = Cm0L , and note fromthe tabulated data that 0L = 2. Substituting into the above equation for Cm1/3 gives

Cmac = Cm1/3 |0L = 0.04

Next, we investigate static stability. Notice from the data that the slope of the Cm1/3 curve is positive.If the CG were located at x = c3 , then the flying wing would be statically unstable because small pitchdisturbances would drive the state away from the balanced flight condition. The critical CG location atwhich the slope of the Cm curve becomes zero is the aerodynamic center, which we have computed. Recallthat a necessary condition for a flying wing of this design to be statically stable is that the CG be locatedforward of the aerodynamic center, i.e.,

(hcg hac) < 0,for only then do we have

Cmcg

< 0.

5

This condition alone is not sufficient for static stability, however. The aircraft must also be capable ofgenerating positive lift at the balanced flight condition, in order to balance its weight. Thus, we also requireCm0L > 0. (See Figure 3.) In our example, Cm0L = Cmac = 0.04; this flying wing is statically unstableregardless of the CG location. If the CG is forward of