# Mode Shapes

Let us return to the longitudinal equations of motion for the Cherokee 180 (Equations 9.36, 9.42, and 9.45). The stability of the motion is determined from the transient solution. For this, it is assumed that u, a, and в are of the

(9.57a)

(9.51b)

(9.57c)

The set of equations then become, after dividing out e01,

(173<r + 0.185)и, + (-0.0637)a,+ 0.5430, = 0 (9.58a)

1.09a, + (175<r + 4.68)a,+ (- 17Oo-)0, = 0 (9.58 b)

(3.32 а + 0.741)a, + (210<r2 + 7.42<r)0, = 0 (9.58c)

One could solve Equation 9.58a to 9.58c for each variable (uu ab and 0i) in turn. However, to obtain cr, it is more direct to use the fact that for uu «ь and 0i to have nonzero values, the determinant formed from the coefficients of these terms must vanish. This leads to the following quartic for a. This is the characteristic equation for the system.

tr4 + 7.84 x 10 V + 4.80 x 10“V + 5.40 x lO’V + 7.55 x 10"8 = 0 (9.59)

The problem is to extract the roots of this quartic. Since all of the coefficients are positive, there will be no positive real roots. However, there may be a positive real part of a complex root. This can be determined without actually finding the roots. We write the preceding polynomial in the following general form.

a* + a3<r3 + a2a2 + a, a + a0 = 0 (9.60)

It can be shown (Ref. 8.2) that the system for which this is the characteristic equation will be stable if a quantity R, known as Routh’s discriminant, is positive. R is defined by

R = a3a2at – a,2- a2a0 (9.61)

For a cubic, R takes the form (with a3 = 1)

R = a2ai – a0 (9.62)

In addition to satisfying the requirement that R be positive, it is emphasized that each of the coefficients must also be positive.

In the case of Equation 9.59, each of the coefficients is positive, and

R =1.539 x 10"9

Hence we now know that the stick-fixed longitudinal dynamic motion of the Cherokee 180 is stable. Although this in itself is valuable, we need to know more before we can assess the flying qualities.

Unless one has a canned computer routine handy, the extraction of complex roots from a quartic can be formidable. One method, which involves some trial and error or graphical procedures, begins by writing Equation 9.60 as the product of two quadratics.

(cr2 + Ba + C)(<r2 + D<r + E) = 0 (9.63)

Expanding Equation 9.63 and equating coefficients Equation 9.60 gives

a3 = В + D a2 = C + BD + E a, = DC + BE a0= CE

These can be reduced to give

)’-“і+с+й” <9-65)

_ fliC a3C2

a0~C2

Equations 9.65 and 9.66 are two implicit relationships involving В and C. They can be solved graphically by calculating В from each equation over a range of C values and noting the value of C graphically that results in the same В value from each equation. Numerically, one can subtract Equation 9.66 from Equation 9.65. Calling this difference /(C), the problem reduces to finding the value of C for which /(C) = 0. Having С, В is obtained immediately from either Equation 9.65 or Equation 9.66. Because of the symmetry of the problem, D and E can then be calculated from

Using a programmable calculator, the foregoing procedure was applied to

Equations 9.65 through 9.67 with the following result.

В = 0.0775 (9.68a)

C = 0.00472 (9.68b)

D = 8.84 x 10~4 (9.68c)

£ = 1.60 x 10-5 (9.68d)

Substituting these values into Equation 9.63 and solving the two resulting quadratic equations gives the following four roots for cr.

a = -0.0388 ±0.0567/ (9.69a)

(t = -0.000442 ± 0.00397 і (9.69 b)

As we concluded earlier, there are no real positive roots, so the motion is stable. Having these roots we can now write the transient solution for, say й, as

й = e-0.0388T(M|e0.0567,V + 0.0567,+ ^-4.42X10 ^^0.00397ir + ^-0.00397,(9 Щ

This equation is in terms of the dimensionless time, r. It can be expressed in terms of t by substituting Equation 9.29 for r.

In this example, t* = 0.016 sec. Hence, in terms of t, Equation 9.70 becomes

й = [c-2 43,(u, c3 54" + m2c 3 54")] + [<Г0 0265′(м3е0’24*" + a4c 0248")] (9.71)

The two separate functions of time within the brackets are referred to as normal modes. Ui, u2, щ, and u4 are arbitrary complex constants to be determined by the initial conditions when the particular solution is included. It is generally true, as in this specific example, that the homogeneous solution can be expressed as the sum of the normal modes.

These two normal modes are typical of the stick-fixed longitudinal motion of most airplanes. The shapes of both modes are similar, but have different time constants. The real parts of these functions are presented graphically in Figure 9.4a and 9.4b. Notice the different time scales used in the two figures. The first mode, in Figure 9.4a, has a much shorter period than the other mode and is heavily damped. The damping of the amplitudes are shown by the dashed lines on each figure.

A measure of damping is provided by the time to damp to half-amplitude. The amplitude of either function is of the form

If the amplitude is A, at time fb the increment in time, T1/2, to give an amplitude, A2, equal to half of A, is found from

or, the time to damp to half-amplitude, Tl/2, is calculated from

Г№ = ^ (9-72)

The period of the mode, T, is the time required to complete one cycle. Thus,

u)T = 2тг

or

(9.73)

For the mode shown in Figure 9.4a,

a = 2.43

ш = 3.54rad/sec

Thus,

Г = 1.77 sec

TW2 = 0.285 sec

For obvious reasons, this mode is referred to as the short-period mode. As noted, it is heavily damped. Because of its nature, this mode is not normally discernable to a pilot. For the other mode,

a = 0.0265 ш = 0.248 rad/sec

Thus, for this mode,

T = 25.3 sec Tm = 26.2 sec

This more lightly damped mode is called the long-period, or phugoid, mode. It is easily discerned by the pilot.

The shapes of these modes, that is, the relative magnitudes and phase between the three displacements а, в, and m, can be found from Equation 9.58 with the known values of a. Using one of the displacements, say в, as the reference, Equation 9.58 can be solved for the ratios ujdi and ai/0,.

From Equation 9.58c, which involves only a and 0, we can write

a, 210tr2 +7.420- 0, -3.32o–0.81

For the real part of the phugoid, a = -0.000442 + 0.00397i. Combining these relationships gives

^ = 0.0364e‘* (9.75)

01

where ф = -78.1°.

Substituting а 1Ю1 into Equation 9.58a or 9.58b and solving for ujd, gives

£ = 0.78e“* (9.76)

“i

where ф = 99.1°.

For the short-period mode, <r = —0.0388 + 0.0567/. Substituting this root into Equation 9.74 gives

—■ = 1.33еіф (9.77)

bi

where в = 2.18°.

Substituting this into Equation 9.58a gives

!h = 0.0407е’ф (9.78)

where ф = 52.8°.

Using в as the reference, we are now in a position to describe the two modes by referring to Equations 9.75 to 9.78. For the phugoid, the amplitude of a is seen to be small compared to the amplitude of the velocity increment. The latter is seen to lead the pitch angle by approximately 99°, while the angle-of-attack increment lags в by approximately 78°. For the short-period mode the velocity increment is small compared to the angle of attack. In this mode both a and и lead в by angles less than 90°.

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