# 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°

5(0) =-0.0193 rad = -1.11°

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.

## Leave a reply