# Category AERODYNAMICS, AERONAUTICS, AND FLIGHT MECHANICS

## I Derivatives

Equation 10.17b governs primarily the angular acceleration about the roll axis. The stability derivatives Clft, C(., Clf, the control derivatives Ch, C)s, and the parameters ix and ixz are needed to evaluate this equation.

From the definition of ix given by Equation 10.16a, for the Cherokee 180,

ix — 24.6

The product of inertia Ixz is unavailable for the example airplane. If the jc-axis is a principle axis, Ixz is equal to zero, so it is reasonable to assume that Ixz is small. Therefore, it will be assumed that

ixz = 0

я

This derivative, known as “dihedral effect,” has been previously covered in some depth. Ccan be calculated on the basis of Equations 8.108 to 8.112. The Cherokee 180 is a low-wing airplane having a dihedral angle of 7.5°. The

wing is unswept, untapered, and has an aspect ratio of 5.625, From Figure 8.39, the contribution to Clfi from the wing equals -0.0895/rad. To this we add 0.0092 to account for the presence of the fuselage. Thus, for the wing-fuselage combination,

The vertical tail placed above the center of gravity also contributes to Ctf). At a sideslip angle of /3, the vertical tail develops a side force in the negative у direction equal to

А У = ~VtqSvavp

Acting at a height above the center of gravity of Zv, this increment in the у force gives rise to a rolling moment equal to

Д/ = – r),qSvavAvf}

Thus, in coefficient form,

Qfv = – V,^fav (10.23)

For the Cherokee 180,

Adding the contribution of the vertical tail to that of the wing-fuselage combination gives a total dihedral effect of

An effect that is not included here, but that can be important, particularly at low speeds with flaps down, is illustrated in Figure 10.5. As a propeller – driven airplane slips to the right, the slipstream trails to the left, causing an increased lift on the left wing. This can result in a significant increase in Q. This interaction is difficult to predict in a general way. It is best obtained by means of wind tunnel testing with a powered model.

C’p

The roll damping coefficient can be calculated on the basis of Figure 8.36 for both the wing and horizontal tail. For the Cherokee’s wing, Ct. = -0.420. For the horizontal tail, A, = 4.10 and A, = 1.0. Thus, from Figure 8.36, Ctf = -0.335. However, this value is based on the tail area and span. To base С/р on the wing’s dimension, we note that

/, = V, qS, b, || C, p:

 Figure 10.5 Effect of slipstream on dihedral effect.

For the Cherokee,

Thus the total Ct/ due to the wing and horizontal tail is equal to -0.426.

The vertical tail also contributes to C(_. As recommended in Chapter Eight, Figure 8.36 can also be applied to the vertical tail as if it extended below the fuselage to the same extent as above. The value obtained is then halved and corrected according to Equation 10.24. For the Cherokee the increment to Ci – from the vertical tail is approximately half of that from the horizontal tail. Thus, for the Cherokee 180,

Ci. = -0.429

*P

Ci,

The rate of change of rolling moment with yawing velocity was con­sidered in Chapter Eight. Ctf is composed of two contributions, one from the wing and the other from the vertical tail. According to Equation 8.103 for the

Cherokee 180, the wing’s contribution to Qr will equal

= 0.181

The contribution to С/, from the vertical tail is given by Equation 8.105.

Hence, the total Q. for the Cherokee 180 is estimated to equal

Clf = 0.198

Aileron roll control was covered in Chapter Eight. C, e can be estimated on the basis of Figure 8.30b. For the Cherokee 180,

A = 1.0 A = 5.625 Xi = 0.603 jc2= 1.0

— = 0.193

c

Using these equations and a linear interpolation for aspect ratio for values of A between 4 and 6 gives a value for C,. from Figure 8.30b of

°a

°a

Since the maximum aileron deflection for the Cherokee is 30° up and 15° down, the total Sa of 45° gives a predicted

C,, = 0.0417

dmax

C/8 can be found from Equations 8.101 and 8.102. For the Cherokee 180, CYs was estimated to equal 0.117/rad. Therefore,

N Derivatives

cNf>

The change of yawing moment coefficient was covered in Chapter Eight. Equation 8.90 expresses the contribution from the vertical tail as

CNvp = і?(Кас(1 – єр)

for ле Cherokee.

The contributions to CNfl from both the propeller and fuselage are obtained from the values estimated in Chapter Nine for CM„- There, the total Cua from the fuselage and propeller was estimated to equal 0.153/rad. To obtain CNfl, the sign must be reversed and multiplied by c/b. Thus, due to the fuselage and propeller,

Thus, for the Cherokee 180,

CN is given by Equation 8.107. For the Cherokee 180 at a trim lift coefficient of 0.543, this equation reduces to

CN – = -0.0905

this contribution comes from the wing. There will also be a small contribution from the vertical tail, given by

к

Y* b or, for the Cherokee,

CN. =-0.017

Pv

Thus, the total becomes

CN – = -0.0735

iyP

Сц}

CNf can be obtained directly from the vertical tail and propeller con­tributions to CYf. It is left to you to show that

CNf = – CYf+CYf± (10.25)

or, for the Cherokee,

CNf = -0.188(0.435) – 0.0232(0.236)

= -0.0873

The adverse aileron yaw is difficult to estimate. In view of the differential aileron deflections used on the Cherokee, it will be assumed that CNs – 0.

cNgr

Cn, is given by

<Ч = – Сч£ (10.26)

Thus, for the Cherokee,

In addition to the stability derivatives, the parameter iz is needed to evaluate Equation 10.17c. From Equation 10.16b, for the Cherokee 180,

iz — 1.18

## A SUMMARY LOOK AT THE STABILITY DERIVATIVES AND OTHER PARAMETERS AFFECTING LATERAL-DIRECTIONAL DYNAMIC MOTION

Y Derivatives

Equation 10.17a is the equation governing primarily the acceleration of the airplane’s mass in the у direction. It contains the stability derivatives, CYf), CYf, and CYp the control derivative CYs, and the parameters, Сц and fi. is the trim lift coefficient and needs no further explanation. The dimensionless mass /м is defined by Equation 10.16d. For the Cherokee 180 example used previously at an altitude of 1500 m,

M pSb

= 15.2 Cl, = 0.543

The side force derivative, CYfl, consists primarily of contributions from the fuselage, the vertical tail, and the normal force resulting from the propulsion system.

Referring to Figure 8.25, a positive sideslip angle of /3 results in a negative side force on the fuselage and vertical tail. The force on the fuselage can be estimated on the basis of Equation 8.71. The у force on the vertical tail will be given by

Yv = – ri, qSvavp~ ee) or

Cy„ = -7j(^a„/3(l-^) (10.18)

For the Cherokee 180, the effective aspect ratio of the vertical tail (see discussion following Equation 8.90) is estimated to equal 2.84. This value of A substituted into Equation 3.70 gives an estimated lift curve slope of 3.04/rad. The ratio SJS equals 0.0713. ep is taken to be zero, and tj, to be unity so

The fuselage and propeller contributions to CYfi are assumed to equal their contributions to CZa that were previously estimated to total -0.18/rad. Thus the total side force derivative for the Cherokee 180 is estimated to equal

The side force resulting from the rolling velocity is normally small. It results primarily from the vertical tail lying above the longitudinal axis through the center of gravity. If the aerodynamic center of the vertical tail lies a distance of Z„ above the center of gravity, then it will experience, as shown in Figure 10.3, an angle of attack resulting from a roll rate, P, equal to

PZV

Uo

The direction of Да is such as to produce a negative Y force, given by

v – PZv

Yv—- Vtqbvav

Uo

In coefficient form,

n – _•? Sy Zy,, – Су, s ^vP

where

Pb

For the Cherokee 180, ZJb = 0.09. Thus,

CY – = -0.039

1 P

Cy.

The side force due to yaw rate results from the damping force on the vertical tail and on the propulsor.

If the vertical tail is aft of the center of gravity a distance of /„, a yaw rate of R will produce on increment in the angle of attack, as shown in Figure

10.4, equal to

U0

A side force in the у direction results, given by

У„ = VlqSvav yj*

b’O

In coefficient form, this becomes

Cy„ = 2q, avVvr (10.20)

Vv is the vertical tail volume defined previously, and r is the dimension­less yaw rate.

For the Cherokee 180, Vv = 0.031. Thus,

CY =0.188

vr

In a similar manner, a propeller experiences an angle-of-attack change

Aa-*’

Rlv

Figure 10.4 Angle-of-attack increment at vertical tail due to yaw rate. Top view, due to R, given by

where lp represents the distance of the propeller ahead of the center of gravity. The direction of Да is opposite to that shown in Figure 10.4, so the Y force on the propeller is

Rip

U0

In dimensionless form this becomes

For the Cherokee 180, the following quantities were estimated previously for the trim condition of 50 m/s at an altitude of 1500 m and a gross weight of 10,680 N.

—jb – = 0.8/rad Сщ = 0.0615

lplb equals 0.236. Hence

rprop

The total CYf for the Cherokee is therefore estimated to be

Cy? = 0.165

The control derivative giving the rate of change of side force with rudder deflection is found from

Ду = t),qSvavTr 8r

or, in coefficient form,

CySr = V, avrr ^ (10.22)

For the Cherokee 180, i> is estimated, using Figures 3.32 and 3.33 to equal 0.54. Thus,

## REDUCTION OF THE LATERAL-DIRECTIONAL EQUATIONS OF MOTION

The lateral-directional equations of motion are linearized and nondimen – sionalized in a manner similar to that followed for longitudinal motion. The resultant velocity is taken to be nearly constant and equal to U0. V and W are assumed to be small by comparison to U0. The angles ф, в, and ф and their derivatives are also taken to be small. Orders higher than the first for any of these small quantities are neglected. Also, the velocity V is replaced by Uafi. Hence the lateral-directional equations reduce to

у + mgф = mUdP+ R) (10.12 a)

Ь = Іхф-ІхД (10.12b)

N = IZR — Іхгф (10.12c)

The aerodynamic force, У, and moments, L and N, are expressed in coefficient form using S as the reference area and b as the reference length. The coefficients are then expanded in a Taylor series, assuming them to be a function of /3, p, r, Sa, and Sn p and r are dimensionless rates similar to q and

(10.13a)

(10.13b)

Again a characteristic time t* is defined. However, in the lateral-direc­tional case, the reference length is taken as Ы2.

The dimensionless time r is given by

The mass and moments of inertia are expressed by

Substituting Equations 10.13 to 10.16 into Equation 10.12 results in the final nondimensional, linearized equations of motion governing lateral-direc­tional motion.

In Equation 10.17, all derivatives indicated by a dot are with respect to the dimensionless time r.

## EULER ANGLES

Although we will not be concerned with them to any extent, it might be well at this point to define the so-called Euler angles. Finite angular rotations of an airplane about its own body axes are not commutative. The final orientation of the airplane will depend on the order in which the rotations are performed. To illustrate this, hold a small model in front of you, heading directly away from you with wings level. Now rotate the model 90° about its x-axis, then 90° about its у-axis, and then 90° around its 2-axis, all directions positive in accordance with the right-hand rule. The model will now be pointing nose-down with its top toward you. Now reverse the order. Rotate it 90° about its 2-axis, then the у-axis, and then the x-axis. In this case, the model’s final orientation will be nose-up with its top toward you.

The Euler angles, starting with a given airplane orientation, are rotations denoted by ф, в, and ф about the x, y, and 2-axes, respectively. However, the order of rotation is first about the 2-axis, then about the у-axis, and then about the x-axis. In other words, the airplane is first yawed, then pitched, and then rolled.

## LA TERAL-DIRECTIONAL DYNAMIC STABILITY AND CONTROL

EQUATIONS OF MOTION

In the previous chapter we dealt with motion in the plane of symmetry. Motion of the plane of symmetry will now be considered. This motion consists of transition in the у direction called sideslip; rotation about the x-axis referred to as rolling; and rotation about the z-axis, or yawing. To begin, the equations of motion will once again be derived, but in a somewhat more basic and complete manner than that which was followed in Chapter Nine. The following derivation closely parallels a similar development presented by Seckel (Ref. 10.4).

Figure 10.1 illustrates a particle of mass of the airplane located at the point x, y, z, in the moving-body axis system. As shown, the axis system is translating with instantaneous velocity components of U, V, and W in the x, y, and z directions while rotating about these axes at angular rates of P, Q, and R.

The linear velocity components of AM in the x, y, and z directions are, obviously,

x = U + Qz – Ry у = V-Pz + Rx z = W + Py – Qx

The accelerations are obtained directly by differentiating the above velocities.

x = U + Qz + Qz – Ry – Ry у – V – Pz – Pz + Rx + Rx z = W + Py + Py – Qx – Qx

 Figure 10.1 Motion of a particle of mass in a moving reference system.

x, y, and z are obtained from Equations 10.1, so Equation 10.2 becomes:

x = U + QW + QPy – Q2x – RV + RPz – R2x + Qz-Ry (10.3a)

у = V – PW – P2y – PQx+ RU + RQz – R2y – Pz + Rx (10.3b)

z = W + PV – P2z + PRx – QU – Q2z + QRy + Py – Qx (10.3c)

It is convenient at this point to use the concept of inertia forces and moments. Briefly, this concept allows one to treat a dynamic system as a static one by employing pseudoforces acting on the system equal in mag­nitude to the product of each mass and its acceleration. The forces are directed opposite to the accelerations. As an example, consider Figure 10.2. The dynamic equation of motion is, obviously,

F = Mx (10.4)

Now, however, add a force on the mass opposite in direction to x equal to Mx. As a problem in statics, the sum of the forces on the mass equals zero.

F – Mx = 0

This, of course, is equal to Equation 10.4.

The inertia forces on the mass element can be written as

FXi = – mx (10.5a)

Fy. = – my (10.5b)

Fz. = – mz (10.5c)

Figure 10.2 Concept of inertia force.

with x, y, and z given by Equation 10.3. These inertia forces give rise to inertia moments. The moment arms can be seen in Figure 10.1. For example, an x force on Am gives rise to a moment about the у-axis equal to FXZ. Therefore, the three inertia moments become:

Mx. = Fz. y – Fyz (10.6a)

Mn = FXiz – Fzx (10.66)

MZI = Fyx – Fx. y (10.6c)

This can be written in vector form as

M, = Rx F,

The total inertia forces and moments acting on the airplane are obtained by summing Equations 10.5 and 10.6 over the total airplane mass. In evaluat­ing these sums, the following sums vanish since the origin of the coordinate system is at the center of gravity

2 A mx = 0 2 bmy = 0 2Amz = 0

Also, because of symmetry,

2 A mxy = 0

Not included in these equations are angular momentum vectors that may be present because of rotors or other rotating components. If h denotes such an angular momentum vector, then ы x h must be added to the above momentum equations. Thus Equation 10.8d, 10.8c, and 10.8/ becomes,

L = IXP – IXZR + Qhz – Rhy

M = IyQ – Phy + Rhx N = IZR – IXZP + Phy – Rhz

Equations 10.8a, 10.8c, and 10.8c were derived previously in Chapter Nine and were used to examine longitudinal dynamic stability and control. Equa­tions 10.8b, 10.8d, and 10.8/ will be used for the analysis of lateral-directional stability and control.

## Flight Path Stability

Another criterion relating to longitudinal motion discussed in Reference

9.1 is that of flight path stability. This refers to the flight-path-angle change effected by elevator control at a constant power setting. For the landing approach flight phase, this angle as a function of true airspeed should have a slope at the minimum operational speed, which is negative or less positive than:

Level 1: 0.06°/kt (9.99a)

Level 2: 0.15°/kt (9.99b)

Level 3: 0.24°/kt (9.99c)

In effect, these are tantamount to stating that one should not design an airplane to operate too far into the backside of the power-required curve. In this region, the flight-path-angle (positive for climb) will increase as the speed is increased for a constant power setting.

Short-Period Mode

Both the short-period frequency and damping ratio are important to achieving satisfactory flying qualities. When the damping is too low, the short-period response can produce an annoying oscillation. When the damping

Table 9.1 Short-Period Damping Ratio Limits

 Categories A and C Category В Level Minimum Maximum Minimum Maximum 1 0.35 1.30 0.3 2.00 2 0.25 2.00 0.2 2.00 3 0.15 — 0.15 —

is too high, the response to control input can be sluggish. Therefore, upper and lower limits to short-period damping are recommended in Reference 9.1 in order to achieve a given Cooper-Harper level. These limits are given in Table 9.1 according to the following flight categories.

Category A These are nonterminal flight phases that require rapid maneuver­ing, precision tracking, or precise flight path control such as air-to-air combat, in-flight refueling (receiver), terrain-following, and close for­mation flying.

Category В These are nonterminal flight phases that are normally accom­plished using gradual maneuvers and no precision tracking, although accurate flight path control may be required. This category includes climb, cruise, and descent, in-flight refueling (tanker), and aerial delivery.

Category C These are terminal flight phases requiring gradual maneuvers but precise flight path control. These include takeoff, catapult takeoff, ap­proach, wave-off/go-around, and landing.

Specifying the damping ratios alone, as in Table 9.1, is not necessarily sufficient to assure adequate flying qualities. As mentioned earlier, the short – period frequency is also important. Indeed, it appears as if the values of frequency and damping ratio for satisfying flying qualities are interdependent.

Reference 9.1 and other sources present a number of graphs similar to Figure 9.11. These are sometimes referred to as target plots. Since the establishment of these boundaries is subjective in nature, the contours defining the Cooper-Harper levels should not be taken as hard and fast. However, an airplane that falls to the left of level 3 in Figure 9.11 can be dangerous to fly.

For the Cherokee 180, the roots of the characteristic equation for the short-period mode were calculated previously. In real time these were

<T i2 = — 2.43 ± 3.54І

 Figure 9.11 Short-period handling qualities criteria.

Thus, from Equation 9.97, for this mode

£ = 0.566 шп – 4.29 rad/sec = 0.68 Hz

These values are seen to lie well within the level 1 region of Figure 9.11.

## LONGITUDINAL FLYING QUALITIES

The evaluation of an airplane’s flying quality is one of subjectiveness. It is difficult to quantify how an airplane feels to a pilot. An airplane may even have an unstable mode (as we will see in the next chapter) and yet feel fine to the pilot if the time to double amplitude is sufficiently long.

Figure 9.10 (taken from Ref. 9.1) presents the latest revision of the so-called Cooper-Harper Scale for evaluating airplane flying qualities Obviously, with adjectives such as “excellent,” “good,” “fair,” “moderate,” “considerable,” and “extensive,” different pilots will give the same airplane different ratings. Nevertheless, this system does provide a rational and somewhat objective base for measuring an airplane’s flying quality. Because of the subjective nature of the Cooper-Harper Scale, Reference 9.1 mainly emphasizes three levels within the scale.

Level 1: Cooper-Harper scale = 1-3.5 Level 2: Cooper-Harper scale = 3.5 – 6.5 Level 3: Cooper-Harper scale = 6.5 – 9+

In order to assure that an airplane lies within one of these levels (level 3 is really undesirable but flyable), Reference 9.1 specifies definite dynamic characteristics that the airplane should possess.

 Figure 9.10 Cooper-Harper scale for rating airplane handling qualities. Phugold Mode

First, with regard to the phugoid mode,

Level 1: £>0.04 (9.96a)

Level 2: f > 0 (9.96b)

Level 3: Г2> 55 sec (9.96c)

C is the damping ratio, and T2 is the time to double amplitude.

The damping ratio, £, for a particular mode, is related to the roots of the characteristic equation as follows. Let the roots defining an oscillatory mode be given by

(t12 = — а ±іш

Now consider the product

 (a – a,)(cr – a2) = 0 Expanded, it is cr2 + 2acr + a2 + ш2 = 0 (9.97) Comparing Equation 9.97 to Equation 9.50, it is obvious that (o„2= a2 + co2 (9.98a) i = — Шп (9.98 b)

For the example of the Cherokee 180, in real time, a = 0.0265 and (o = 0.248 for the phugoid. Hence, the damping ratio, £, equals 0.106 and the undamped natural frequency, ton, equals 0.249 rad/sec. Thus, according to the criteria of Equation 9.96, the Cherokee falls within level 1.

## SOLUTION OF THE LONGITUDINAL EQUATIONS OF MOTION USING AN ANALOG COMPUTER

An analog computer is used primarily for solving differential equations. Although it is not as accurate as the digital computer, for certain applications it is more valuable. For example, one can examine the effect of varying a parameter on the solution of a differential equation simply by turning a potentiometer on the analog computer.

The heart of an analog computer is the operational amplifier, which is a high-gain, high-impedance amplifier. The symbol for such an amplifier is shown in Figure 9.6a. Consider what happens when this amplifier is connected into the circuit shown in Figure 9.6b. eu ег, and e3 represent three time – dependent input voltages, which are applied to resistance Ru R2, and R3, respectively. The input resistors are tied to a feedback resistor, RF, at a point SJ known as the summing junction. As shown, RF is across the operational amplifier. All voltages are relative to ground. If ix denotes current flow

id)

Figure 9.6 Analog computer elements, (a) Operational amplifier. (b) Summer, (c) Integrator. (d) Potentiometer.

through the x-circuit component then, from Kirchoff’s theorem, the net flow into the junction SJ must equal zero.

£sj ~ gp
Z

In this equation, Z is the impedance of the amplifier, which is assumed to be high. Since all voltages are limited in order to stay within the linear range of

the amplifier, it follows that

і A =0

eSJ and e0 are related by the gain of the amplifier.

or

G is also assumed to be high, so that approximately

Csj — 0

SJ is referred to as a virtual ground point.

A good operational amplifier will have a DC gain of approximately 107 and an impedance of 10’°fl. Its frequency response (gain versus frequency) will be flat from zero to 106 Hz with feedback. Obviously, the approximation, given by Equation 9.93, is a very good one.

The voltage drop across a resistance is given by Ohm’s law.

e = iR

Combining Equations 9.83 to 9.86, it follows that

— / Rf, Rf, Rf

е°-~ж;ех+1с2ег+1г, ег)

Thus, the patching shown in Figure 9.6b results in an output voltage e0 equal to the negative sum of the input voltages multiplied by the ratio of the input resistors to the feedback resistor. Most standard analog computers provide for ratios (referred to as gains) of 1.0 or 10.0. Taken as a whole, the combination of operational amplifier, input resistors, and feedback resistor is known as a summer. The symbol for a summer is shown in Figure 9.6b.

It is left to you to show that an integrator is formed by using a feedback capacitance, CF, in place of RF. In this case the patching shown in Figure 9.6c gives an output voltage equal to

By putting an initial charge on the feedback capacitor, one is also able to allow for an initial condition on the quantity being represented by the output voltage.

The schematic for a potentiometer is also included in Figure 9.6. The input voltage is supplied to one end of a resistance coil, and the other end of the coil is grounded. The output voltage is taken from a wiper. When the wiper contacts the input end of the coil, the output voltage is equal to the
input voltage. As the “pot” is turned, the wiper moves toward the ground, so that the output voltage goes to zero. Thus, one can effectively multiply a voltage signal by a factor from zero to unity by feeding the voltage signal through a pot. Combined with an input gain of 10 into a summer or integrator, the pot allows a programmer to multiply a signal by a factor from 0 to 10. In practice one tries to avoid setting a pot below a value of 0.1, since accuracy is lost. There are times, however, when this is unavoidable if a term represents a small contribution to the solution.

Generally, problems to be solved with an analog computer have to be both time-scaled and magnitude-scaled. This is best explained by means of an example for which we refer to Equation 9.50. The example will also illustrate the general procedure that is followed to construct the patching diagram to solve a given differential equation. We begin by solving for the highest derivative. In this case,

x = f(t) – 2£w„x – a>n2x (9.89)

To be more specific, suppose x represents the displacement of a light mass supported on a stiff spring with a damping ratio less than critical. Let

o>„ = 600 rad/sec £ = 0.5

fit) = 0 x(0) = 0.85 mm x(0) = 0

Equation 9.89 becomes

x = – 600X – 360,000* (9.90)

We now “guesstimate” the maximum values to be expected for x and x. In this case, it is not too difficult. Since it is a damped oscillation, the initial value of x should be the maximum. We will therefore pick a convenient maximum value for x, known as a “rounded up maximum” (RUM), of 1.0 to which to reference x. RUM values should be higher than any expected values of a particular variable, but not too high, or accuracy will be lost.

The maximum value of x should equal approximately the maximum value of x multiplied by шп. A RUM value for x of 1000 should therefore suffice. We now rewrite Eqn 9.90 expressing x and x in terms of their RUM values.

x = – 600,000( – 360,000(j)

If we now divide through by 1 x 106,

(шу)–°-6Ш-°-“(т) <9-92>

The problem appears to be magnitude-scaled satisfactorily at this point, since the coefficients of the scaled values of x and x lie between 0.1 and 10. However, to integrate x/( 1 x 106) and get x/1000, we would need to multiply the former by 1000. Thus we need to time-scale the problem. In this case, we need to “slow down” the action in order to plot the motion. Denoting the problem time by т and the real time with which the computer works as t, let

r = /31 (9.93)

Now consider

=|ооо/Ш1‘"

= 1оо°0/(шіоо)л

Thus, multiplying x by /3 and integrating with respect to real time results in x as a function of r.

The RUM values correspond to the rated voltage of the computer, which is referred to as the machine unit. Thus, for example, suppose the voltage signal at a given instant equals З V for a 10-V machine, or 0.3 of a machine unit. If this signal represents a variable having a RUM value of 1000, the value of the variable at that instant would be 300.

Now consider Figure 9.7a. To begin, it is assumed that the output from summer 1 equals -(x/1 x 10*). This signal is fed through pot 1, which is set at 1000/3, and then into integrator 2. The output of this integrator equals (x/1000). This output is fed through pot 2, which is set again at 1000/3, and then into integrator 3. The output of this integrator is equal to — (jc/I). Thus we easily obtain x and x by simply integrating x twice. Having these quantities we can now satisfy the differential equation, Equation 9.86, as illustrated in Figure 9.7b. Since jc/1 x 106 equals the sum of -0.6x/1000 and 0.36x/l, we take the voltage from amplifier 2, which is proportional to x/1000, and run it through summer 4 to change the sign. The output of this summer is then fed into pot 3 set at 0.6 and from the pot into amplifier 1. Also, the voltage from amplifier 3 is fed into pot 4 set at 0.36 and from the pot into summer 1.

The initial condition on the problem is provided by patching from a + 1 machine unit reference into pot 5. Since x is initially equal to 0.85 of its RUM value, this pot is set at 0.85. The output of this pot is then fed into the IC (initial condition) connection for the integrator module patched into amplifier 3.

For this relatively simple example, it is obvious that the RUM values chosen for x and x are sufficiently high to assure that the amplifier outputs will not exceed the rated machine voltage. However, if an amplifier overflows (as indicated by some kind of marker, usually a light), one simply rescales, choosing higher RUM values.

Let us now turn to the solution of the set of simultaneous, linear differential equations, Equations 9.36, 9.42, and 9.45. In particular consider the following experiment. At an altitude of 1500 m the airplane is trimmed to fly hands-off at a true airspeed of 50 m/s. Without changing the power setting, the stick is held back to a position that provides a steady speed of 40 m/s. Thus, relative to the trim airspeed, м = -0.2. Now return the stick suddenly to its trimmed position and hold it fixed (in the manner of a step function). Thus disturbed, the airplane will exhibit an unsteady longitudinal motion, which we will attempt to predict using the analog computer

Expressing all derivatives with respect to real time, all angles in degrees, and the incremental velocity, u, in meters per second, the governing equations become:

й =-0.0668м+ 0.0201 a-0.1710 (9.94a)

a = -0.449и – 1.69a + 0.9830 — 0.3375 (9.94b)

0 = – O.988d – 15.1a -2.21^^65 (9.94c)

Unsteady aerodynamic terms in a are included in Equation 9.94

From the experiment we are given that и equals —10 m/s initially. Also, at t = 0, all derivatives are zero. Thus Equation 9.94 can be solved to give the following initial values.

«(0) = 0.0570 rad = 3.26°

0(0) = 0.0848 rad = 4.86°

Since 5 is zero for t > 0, the problem can be run on the computer in one of two ways. First, S can be patched into the problem and the airplane “flown” through the pot representing 8. This pot is adjusted until и = -10 m/s, at which condition the pot setting should correspond to the preceding 5(0) value. a and 0 will also assume their correct initial values. A function switch can then be thrown to disconnect the 5 input, simulating the return of the stick to its trimmed position. The ensuing analog behavior should represent the unsteady behavior of the airplane. The second, and simpler, way to run the problem is to impose the proper initial values on u, a, and 0, letting 5 equal zero. In scaling this problem one might assume that the initial values of u, a, and 0 represent maximum upper bounds on these variables. However, if one chooses the RUM values on this basis, the amplifier representing 0 will overflow. By trial and error, a RUM value for 0 of 20° was found to be satisfactory. Five degrees and 10 m/s were chosen for a and u, respectively, based on their initial values. Magnitude-scaled; Equation 9.94 becomes:

 (?)= _|Ш4(ш) +00201 (: t)-ft684(s) (9.95a) (s)- -0.420(f) f 0.976(1 )-°223(тЬ)- 0И5(й) (9.95b) -0.442(|)-0.ТО(| )-°’l99(l>)- 8’92(i>) . (9.95c)

These equations do not have to be time-scaled in order to examine the phugoid mode. For the short-period mode, however, the action has to be slowed down. Figure 9.8 presents the patching diagram to solve these equa­tions. The top row of amplifiers sums and integrates U to obtain u. The second row performs the same function for a. The third row of amplifiers sums and integrates 0 twice to obtain 0 and 0.

Observe how the equations are tied together. For example, to satisfy the second term on the right-hand side of Equation 9.94b, the output from amplifier 6 is fed into pot 11 set at 0.976 and is then fed into amplifier 1, which is summing a. The numbering of the amplifiers and pots may seem somewhat haphazard, but it was chosen to be convenient for the analog computerThat was available to me. /

The outputs for amplifiers 2, 3, and 9, representing (м/10), (a/5), and

 Figure 9.9 Longitudinal motion of a Cherokee 180 as predicted by an analog computer.

(в/20) as a function of real time, are presented in Figure 9.9. These results were obtained directly from the traces of an x-y plotter with a built-in time base. As reflected by the trace, the short-period mode damps out quickly, leaving only the phugoid. You should satisfy yourself that the frequency, damping, and phase relationships shown here are close to those calculated previously for the phugoid.

This experiment, the results of which are predicted in Figure 9.9, was flown by me with some of my students using only the standard equipment in the airplane. Flying at a density altitude of 1500 m, the airplane was trimmed, hands-off, to fly straight and level at a true airspeed of 50m/s. A small telescoping rod was then placed between the back of the control wheel and the instrument panel in order to reference the trim stick position. The wheel was pulled back and held at the position to maintain a steady true airspeed of 40 m/s. Simultaneously, the wheel was pushed forward and held fixed at its trim position, and a stopwatch was started. The times when the airspeed attained maximum and minimum values were then visually noted. The values themselves were also noted simply by reading the airspeed indicator. In this manner, the experimental points shown in Figure 9.9 were obtained. The

agreement between the analog results and the experimental points is good considering that the stick motion only approximates a step function, that there is some lag in the airspeed indicator, and the errors inherent in the simplistic manner in which the data were obtained.

Predictions of longitudinal dynamic motion for relatively simple configurations such as the Cherokee are not too difficult to accomplish and can be done so with some degree of confidence. This is not necessarily so, however, for high-performance aircraft operating at high Mach numbers or for V/STOL aircraft, in which the lift and propulsion systems may be integrated. Because this is an introductory textbook, the details of treating these types of aircraft are beyond its scope. In the case of high-speed aircraft, the aerodynamic stability derivatives must also include the dependence of the forces and moments on speed. For example, suppose an aircraft is operating just below its critical Mach number. If it is disturbed by some type of input, the perturbations in the aircraft’s velocity and angle of attack can produce compressibility effects that drastically alter the lift, drag, and pitching moment coefficients.

## Phugoid (Long-Period Mode)

The phugoid represents motion at a nearly constant angle of attack. Although the pitch angle is varying periodically, the altitude is also changing so as to maintain a nearly constant a. This is depicted in Figure 9.5. The displacements are exaggerated for clarity. Beginning at the top of one cycle, the airplane has slowed down to its minimum airspeed, and its attitude is nearly level. It then begins to lose altitude. As it does so, its speed increases, followed by a nose-down attitude. At the bottom of the cycle its airspeed is a maximum and its attitude is again nearly level. It then begins to climb. The

 O’ as constant Figure 9.5 The phugoid mode; oscillating, longitudinal motion at a constant angle of attack.

airspeed begins to drop off, followed by a nose-up attitude, and the cycle is repeated.

Since a, and hence CL, is approximately constant for the phugoid, an approximate value for the period of the mode can be obtained as follows. At any instant of time, the net force on the airplane in the vertical direction can be written approximately as

Fz=W-p{U02 + 2Uau)C^S But

w = Pu02cus

Thus, the unsteady force accelerating the airplane vertically downward is

Fz = — pUoCi^Su

The approximate equation of motion in this direction is

Fz = mZ

Integrating this equation, assuming и to be of the form

и = мтах sin wt

gives

_ P 1-А) Sll max _

Z =——– 5—- Sin bit

ты

Thus the maximum change in the potential energy (PE) of the airplane is

2gpU0CLoSu

max

APE =———- 5—-

a>

This must equal the maximum change in the kinetic energy (KE) of the airplane, given by

AKE = 2mU0umax

It follows that ы is given approximately by

C’o

This approximate result is seen to be independent of airplane geometry. It can be found in other references, but is frequently written in a somewhat disguised form as

As written here, the units of ы are radians per air seconds. Expressed in real time, this equation reduces to Equation 9.79. For the example case of the Cherokee 180 at 50m/s, Equation 9.79 predicts an ы of 0.278 rad/sec, or a

period of 22.6 sec. This period is 2.7 sec, or approximately 10%, shorter than the exact value calculated previously.

Short-Period Mode

A very close approximation to the short-period mode is obtained by assuming that и is constant for this mode. With this approximation, Equation 9.33 for the control-fixed case reduces to

(2ijl – CzJd – Cza – (2fi + С2.)в + Сц tan в0в = 0 (9.80a)

— Смаос — Cmjh + іуд — См .в = 0 (9.80 b)

Both CZa and Cz. are normally small compared to 2ft. These two terms will therefore be neglected. If we let в0 = 0, the following quadratic is obtained for the characteristic equation.

2/i/ycr2 — [CzJy + 2(і(См^ + См.)](г + СгСмц ~ 2и-Сма = 0 (9.81)

or

Cz CM’ + CM<i T /Cz См’ + СмЛ2 /2цСма-СгСмЛЛт

a 4n 2iy ~ l 4/i 2iy / 2fiiy /J

For the Cherokee 180 example, this approximation to <r becomes

<t = -0.0391 ± 0.0544/

Thus, with this approximation, the damping is predicted within 1% and the frequency within 4% of the previously calculated values.

## 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

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°.