MDC G2968

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SIMULATION OF A FLEXIBLESPINNING VEHICLE

Final ReportQ

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https://ntrs.nasa.gov/search.jsp?R=19720023205 2020-03-11T19:31:46+00:00Z

MICDONNELL.DOUGLAS

SIMULATION OF AFLEXIBLE SPINNING VEHICLE

Final Report

JULY 1972

PREPARED BY

W. A. BEAUDRYGUIDANCE AND ADVANCEFLIGHT MECHANICS DEPARTMENT

WCDONNELL DOUGLAS ASTRONAUTICS COIMPANY-WEST

5301 Bolsa Avenue, Huntington Beach, CA 92647

C

MDC G2968

PRECeThTiG PAGE BLANK NOT FILMPD

FOREWORD

This report was prepared by McDonnell Douglas Astronautics

Company (MDAC), under National Aeronautics and Space

Administration Contract No. NAS8-26700. It covers work

conducted from 3 May 1971 to 2 July 1972.

Preceding page blank iii

PRE.CEING PAGE BLANK NOT F7,M1T7,"

CONTENTS

INTRODUCTION

DESCRIPTION OF SIMULATION SYSTEM

2. 1 Flexible Booms2.2 CMG Wobble Damping2. 3 Active Mass Balance Wobble Damping

Control System2.4 Reaction Jet System

SIMULATION RESULTS

3. 1 Rigid Body Spin-up and Spin-down3.2 Flexible Vehicle3. 3 Active Mass Balance Wobble Damper3.4 Active Mass Balance Wobble Damper

with Flexible Booms3. 5 CMG Wobble Damper3.6 CMG Wobble Damper with Flexible

Booms

Section 4

Section 5

Appendix A

DISCUSSION OF RESULTS

CONCLUSIONS

DERIVATION OF EQUATIONS OF MOTION

Appendix B COMPARISONEQUATIONS

WITH CANTILEVER BEAM

Appendix C SOLUTION TO EQUATIONS

Preceding page blank

Section 1

Section Z

Section 3

V

PrBC, DWTG PAGE BLAt1K NOT FILET

FIGURES

2-1 Flexible Vehicle Simulator Model

2-2 Boom Mechanism

2-3 Close-up of Mass Balance Wobble DamperWeight

2-4 Close-up of Body Axis Angular VelocitySensor Assembly

2-5 Inertia Augmentation Boom Assembly

2-6 Close-up of Inertia Augmentation BoomAs sembly

2-7 Model Coordinate System

2-8 Flexible Boom Equation Diagram

2-9 Approximate Equation Diagram for FlexibleBoom

2-10 Manipulation of Flexible Boom Equations

2-11 Twin CMG Wobble Damper

2-12 CMG Gimbal Servo Drives

2-13 Simplified CMG Wobble Damper

2-14 Spin-Up System

2-15 Reaction Jet Wobble Damper

2-16 Schematic Diagram-Reaction Jet System forSpin-Up and Wobble Control

3-1 Rigid Body Spin-up and Down

3-2 Rigid Body Spin-up and Down Euler Angles

3-3 Flexible Body Spin-Up Run

3-4 Caged Spin-Up to 0.64 rad/sec (36.8 deg/sec)

Preceding page blank| vii

3-5 Caged Spin-Up to 0. 59 rad/sec (34 deg/sec)

3-6 Caged Spin-Up to 0. 51 rad/sec (29. 1 deg/sec)

3-7 Rigid Body Active Mass Balance Runs

3-8 Active Mass Balance Wobble Damper withFlexible Booms

3-9 Rigid CMG Wobble Damper-Configuration 2

3-10 CMG Wobble Damper with Flexible Booms-Configuration 2

3-11 CMG Wobble Damper with Flexible Booms-Configuration 2

4-1 Nutation Period vs Spin Rate-Configuration 2

4-2 Configuration 1 Flexible Booms-X Componentof Angular Velocity

4-3 Configuration 1 Flexible Booms-Y Componentof Angular Velocity

4-4 Configuration 1 Flexible Booms-Boom AngleTheta

4-5 Configuration 1 Flexible Booms-Z Componentof Angular Velocity

4-6 Configuration 1 Flexible Booms-Euler AngleBeta 1

4-7 Configuration 1 Flexible Booms-Euler AngleBeta 2

4-8 Configuration 1 Flexible Booms-Euler AngleBeta 3

4-9 Configuration 1 RCS Wobble Damping-XComponent of Angular Velocity

4-10 Configuration 1 RCS-Y Component of AngularVelocity

4-11 Configuration 1 RCS-Z Component of AngularVelocity

4-12 Configuration 1 RCS Damping-Boom AngleTheta

Viii

4-13 Configuration 1 RCS Damping-Euler AngleBeta 1

4-14 Configuration 1 RCS Damping-Euler AngleBeta 2

4-15 Configuration 1 RCS Damping-Euler AngleBeta 3

4-16 Configuration 2 CMG Damping-X Component ofAngular Velocity

4-17 Configuration 2 CMG Damping-Y Component ofAngular Velocity

4-18 Configuration 2 CMG-Z Component of AngularVelocity

4-19 Configuration 2 CMG-Boom Angle Theta

4-20 Configuration 2 CMG-Gimbal Angle

4-21 Configuration 2 CMG-Euler Angle Beta 1

4-22 Configuration 2 CMG-Euler Angle Beta 2

4-23 Configuration 2 CMG-Euler Angle Beta 3

ix

Section 1

INTRODUCTION

This report presents the results of an experimental investigation of the con-

trolled and uncontrolled dynamical behavior of a rotating or artificial-gravity

space station including flexible-body effects. A dynamically scaled model

was supported by a spherical air bearing which provided a nearly moment-

free environment. Reaction jet systems were provided for spin-up and spin-

down and for damping of wobble motion. Two single-gimbal gyros were

arranged as a control-moment gyro wobble damping system. Remotely con-

trollable movable masses were provided to simulate mass-shift disturbances

such as arise from crew motions. An active mass-balance wobble damping

system which acted to minimize the wobble motions induced by crew motions

was also installed. Flexible-body effects were provided by a pair of inertia

augmentation booms. Inertia augmentation booms are contemplated for use

on rotating space stations to cause the spin axis moment of inertia to be the

largest of the three moments of inertia as is necessary to assure gyroscopic

stability. Test runs were made with each of the control systems with the

booms locked (rigid body) and unlocked (flexible body).

1

3 AFS BLOW W0~NT RE~siiD

Section 2

DESCRIPTION OF SIMULATION SYSTEM

Figure 2-1 shows the general arrangement of the flexible spinning vehicle

simulation model. Disk-shaped weights mounted on threaded rods visible at

the right end of the model (a similar set is at the left end) provide means for

adjustment of the static and dynamic (principal axis alignment) balance, The

pedestal of the air bearing support is visible beneath the model. The model

is floated on a film of air beneath a 0. 2 5 4 m (10 in. ) diameter beryllium

sphere. One of the two inertia augmentation booms may be seen protruding

from an opening in the front side of the model. Figure 2-2 shows the booms

vibrating. The gas supply (nitrogen) for the reaction jet system (RCS) is

stored in two gas bottles symmetrically located on the centerline of the

model and ball to preserve the static balance as gas is used. To the right of

the boom arm on the front of the model are the pressure gages and pressure

regulator of the RCS. The jets are visible on brackets at the right end of the

model, a similar set (not visible) is located at the left end.

The square box on top of the model near the left end provides mounting for

the two control moment gyros (CMG's) of the CMG wobble damping system.

The assembly of electronic circuit cards of the control system is in the box

on the top of the model toward the right end. To the right of this are mounted

the solenoid valves of the RCS .

The disturbance mass system is composed of two masses on vertical guide

rods located at diagonally opposite corners of the model. The two masses

are connected by a cable which is wrapped around and fastened to the pulley

of a dc servomotor. On command from a potentiometer on the control panel,

the servomotor will rotate to the commanded angle causing the two masses

to move equally and in opposite directions generating a product of inertia

disturbance. The active mass balance wobble damper system is also

Preceding page blank 3

CR71

Figure 2-1. Flexible Vehicle Simulator Model

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c .8 u

E

o

o

CQ

CM-

«<l <u 3

composed of two vertically guided masses connected by a cable and moved

by a servomotor. The active mass balance system position is commanded

to be proportional to the sensed angular velocity about the long (Y) axis of

the model and thus acts to minimize the product of inertia disturbance. To

the right of the pressure regulator in Figure 2-1 is a vertical square guide

for one of the two disturbance masses. A similar guide rod and mass may

be seen near the left end of the model and is one of the two masses of the

active mass balance system. Figure 2-3 is a close-up view of one of the

two active mass balance wobble damper weights and its guide rod. The

weights have ball bushings to minimize friction. Above the rod is the servo-

drive assembly of the disturbance mass system. A similar drive system,

not visible but directly behind on the opposite side of the model, drives the

active mass balance system.

Figure 2-4 is a close-up of the body axis angular velocity sensor assembly,

composed of three orthogonally mounted rate gyros and located on the rear

side of the model.

Figure 2-5 is an overall view of the inertia augmentation boom assembly

prior to mounting in the model. The positions of the tip masses can be

changed to provide different moments of inertia. Figure 2-6 is a close-up of

the mechanism. The band at the top constrains the booms to vibrate in the

antisymmetric mode. The springs at the bottom can be adjusted to obtain

various boom vibration frequencies. A potentiometer on one pivot provides

a readout of the boom angle.

The signal and power-wire follow-up servo is shown in Figure 2-1. Immedi-

ately below the junction box attached to the ceiling of the insulated room in

which the model is located is the follow-up servomotor which drives the

attached slip ring assembly in synchronism with the spin of the table. The

follow-up servosystem functions in a rate command mode with the rate com-

mand updated once per revolution based on a comparison of the time differ-

ence with which two light beams are broken. The upper light beam is in the

servomotor housing and is broken once per revolution by a vane attached to

6

CR71

Figure 2-3. Close-up of Mass Balance Wobble Damper Weight

f ! 3 C

< ( a a 3

Psl

t c i ( . . ( . . < i i ( j ( i ( ( ( ( ( r (

CR71

Figure 2-5. Inertia Augmentation Boom Assembly

i Q.

3

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the motor-shaft, and is sensed by a silicon photosensor. The other light

beam is interrupted once per revolution by a vane attached to the bottom of

the table.

Figure 2-7 shows the coordinate system used throughout this report. The

model body axes, X, Y, and Z, are related to inertial axes, XI, YI, and Z I

by the Euler angles, p I, p2, and P33 . Equation (2-1) gives the transformation

from body axis rates, wXPy, and wZ, to Euler angle rates, P1,' 2' and 3'

The rate signals from the body axis rate sensor assembly are fed to an analog

computer which performs the transformation to Euler angle rates and sub-

sequent integration to yield the Euler angles:

EULER ANGLE TRANSFORMATION- - ~~~ r- -I_

I~z

p36

CP33'c D3cP 2

s P3

Cp 3 Spz

CP 2

S p3

CD 2

c~3C 133

Sp 2 CP3

CD 2

0

0

1

Wy

oZ

(2-1 )

CR71z

ZI

Y

XIYI

Figure 2-7. Model Coordinate System

11

2. 1 FLEXIBLE BOOMS

The equations of motion of the spinning vehicle with flexible booms are

derived in Appendix A (Equations A-39, A-41 and A-42). It is also shown in

the appendix that the hinged booms have the same dynamical behavior (natural

frequency and variation of frequency with spin angular velocity) as tip

masses supported by cantilever beam structural members. The equations

can be diagrammed as shown in Figure 2-8. At the frequency range of

interest, the nutation frequency squared is approximately Y 0. 01 rad /2 2 2~ 2.

sec , much less than the spinning boom natural squaredw -2. 8 rad /sec;

therefore, Figure 2-8 can be approximated by Figure 2-9 which in turn can

be manipulated into Figure 2-10. The spinning vehicle with flexible booms

can therefore be approximated by a spinning rigid vehicle with I Z and II~~~~~~~~~~~~~ T

replaced by I Z and Iy wherein the boom inertia contribution is reduced

by a factor depending on boom flexibility and spin speed. Based on the

approximation of Figure 2-10, the nutation frequency of the flexible vehicle

is given by

(I~ - IX I I(Z y] (2-2)

Figure 2-10 is convenient for investigating the controlled characteristics of

the system.

It will be noted in Figure 2-1 that the centerline of the booms is displaced

from the center of the air bearing ball along the Y axis a distance Ry which

differs from the assumptions of Appendix A. It may be shown that the

model still behaves according to the equations of motion of Appendix A pro-

vided that 2mRy is added to IX and I z , the moments of inertia of the rigid

center body. This has been done in the reported results.

12

CR71

Figure 2-8. Flexible Boom Equation Diagram

CR71

Figure 2-9. Approximate Equation Diagram for Flexible Boom

13

CR71

1IIxS

wx

I = Iz + 2m (d + 9)2 (1 - ) Iy' = Iy + 2m (d + 2)2 (1_ -- )S 2 WS2

Figure 2-10. Manipulation of Flexible Boom Equations

2. 2 CMG WOBBLE DAMPING

The CMG wobble damper on the simulation model is composed of two single-

gimbal CMG (SGCMG) which have their gimbal axes parallel with the table

spin axis (Figure 2- 11). The gimbals are driven so that the gimbal angle of

one CMG is the negative of the gimbal angle of the other CMG.

The torques acting on the transverse axes due to the CMG's are

TX= H11 Sal -Wz H1 Sa1 + H 2 &2 2 2 + Z H2 Sa 2

Y = H1 1 Z 1 Cal - H2 2 Ca2 - Z H2 C2 2

T =.' X H Sa - y H Ca -W X H Sa2

+ Wy H Ca2Z XlI 1 1 1 X 2 2 'Y 2 2

(2- 3)

(2-4)

(2- 5)

14

a 2

H 2

zY

X

H1

Figure 2-11. Twin CMAG Wobble Damper

A block diagram of the wobble-damping control system is shown in Fig-

ure 2-12 where cZ

= f = constant. The gimbal angle control law is

al = -aZc = - Kw (2-6)= Ky

A tracking loop (KTR is the tracking loop gain) is included to maintain the

r elationship

a1 ~~~~~~~~~~(2-7)al = - a? = a (2- 7)

Substituting this relationship into the torque equations and letting H 1 = H2

= H yields

T X " - 2HQ2Sa

Ty 2HaSa

(2-8)

(2-9)

(2- 10)Tz = - 2HwxSa

15

CR71

a-cc,

C'J

CD

0C,Z..

0co -a.?N &C)

CL

LLT

16

C)3

If we assume small gimbal angles (Ca z 1, Sa = a), the system equations

can be linearized and conventional control analyses can be applied.

Figure 2_-13 is a simplified block diagram of the wobble damper. The gimbal

angle control loop will have a much higher bandwidth than the rest of the

system; therefore, if we assume that a = a , the simplified system equations

are

IX X + (IZ

- Iy)QWy + 2H2a = 0 (2- 1)

Iy &y + (IX - Iz ) Q cWX - 2H6 = 0

The characteristic equation is

S2 + 2HKQJI(IZ - Iy)(IZ - IX)2

IX IyixY

Figure 2-13. Simplified CMG Wobble Damper

17

(2- 1 2)

(2-13)

CR71

x

Y

Ix)

The damping factor is

KHQ (Iy IX ) (2-14)~y +

and nutation frequency is

[Iz -)( - IX)1 1/2[y X Y ] (2-15)IXI Y

The influence of the flexible booms may be determined by using the reduced

moments of inertia I' and Iy in the above equations. Since both Y and Iz yare reduced from the rigid body values in this case, increased damping is

to be expected with flexible booms.

2. 3 ACTIVE MASS BALANCE WOBBLE DAMPINGCONTROL SYSTEM

The theory and computed results of operation of an active mass-balance

control system are fully presented in Reference 1.' The principle of opera-

tion is to move a mass parallel to the Z axis an amount proportional to wy

thus generating an Ixz(t) product of inertia which acts to damp the wobble

motion including that due to mass shifts. The main equation of Reference 1

is reproduced here for completeness.

Cm I 2 x2 31Z coy+ Cm Icl D >b + (a + b6a) bQ2C = b23 YZ ( 0 ) (2-16)

where mc is the control mass, rX

and ry are the position components of the

control mass guides, and C is the gain constant in the commanded position

of the control mass

*Reference 1. D. W. Childs. A Movable-Mass Attitude-Stabilization Systemfor Artificial-g Space Stations. Journal of Spacecraft and Rockets, Vol. 8,No. 8, August 1971.

18

IZ -Iya IZ Y (2-17)

a = =- B of this report

Iz - IXb = -A of this report (2- 18)

I Y

= (ab)I / 2 (2- 19)

Iyz(0) is the product of inertia disturbance arising from a mass shift such

as crew motion. The term 6ba which increases the wobble natural frequency

is not significant. The ratio of effective damping to critical damping is given

by

Cm Jrx QC c X (2-20)

21 XY

The influence of flexible booms may be examined by use of the reduced values

of I z and Iy and since they are snmaller than for a rigid body, indicate that

increased damping would be obtained.

The model wobble damping system is composed of two movable masses

located at diagonally opposite corners of the model. The two masses are

connected by a cable and operated by a dc torquemotor so that they move in

equal and opposite directions thus preserving the model static balance. It

may be shown that the attitude equations of motion are identical with those

of Reference 1 provided that the control mass, m , is understood to include

both model masses.

2.4 REACTION JET SYSTEM

The reaction jet system located in the air bearing table has two functions:

A. To spin up and spin down the table (about Z axis).

B. To provide wobble damping during spinup and spindown (about

X axis).

19

To accomplish these functions a two-axis rejection jet system with four pairs

of nozzles is used. Functionally, the jets on the Z axis operate independently

from the jets on the X axis.

2.4. 1 Spin-up System

The spin-up system is shown in Figure 2-14. Upon command, the gas jets

are turned on and are left on until the desired spin rate is reached. During

the course of a run, air drag on the table acts against the table spin. When

the rate decreases below the commanded rate, the jets are activated to keep

the spin rate within a given tolerance of the nominal. This is accomplished

by putting a dead zone in the reaction jets as shown in the figure. To spin

down the table, the command, WZC, is set equal to zero.

2. 4. 2 Reaction-Jet Wobble Damping

During spin-up or spin-down, it is not convenient to use the CMG's for

wobble damping. The X-axis reaction jet system is used for wobble damp-

ing during these times. This system can also be manually commanded by

CR71

JETS

uWZC I1K Z ~ t L Iz s

I -

Figure 2-14. Spin-Up System

20

means of a switch on the control console to introduce initial conditions or

disturbances about the X axis.

A block diagram of the system is shown in Figure 2-15.

2. 4. 3 Reaction Jet Control System Hardware

The system, as shown schematically in Figure 2-16, consists of four pairs

of cold gas (nitrogen) reactionjets. The reaction jets are driven by two

solenoid valves which are controlled by the spinup/wobble-control

logic.

The nitrogen is carried in two symmetrically located tanks (to reduce any cg

shift due to mass expulsion). The volume of the tanks is approximately3

400 in.. This sizing of the tanks allows a number of simulation runs to be

conducted before recharging is required.

Each of the jets is 0. 66 mm (0. 026 in. ) in diameter and develops a thrust of

0.059 N (0. 013 lb) when the supply pressure is 17.24 N/cm (25 psi). The

CR71

JETS

Figure 2-15. Reaction Jet Wobble Damper

21

CR71

FILL VALVE200 IN3 200 IN3

FILL VALVE

WOBBLE CONTROL

Figure 2-16. Schematic Diagram - Reaction Jet System for Spin-Up and Wobble Control

distance between jets is 2. 286 m (6. 94 ft). The pair of on-jets develop a

nmoment of 0. 124 N m (0. 091 ft/lb) at this pressure. The spin up jets accel-

erate the model to a speed of 0. 63 rad/sec (6 rpm) in approximately 20 min.

The wobble damper dead zone was set at ±0. 00524 rad/sec (0. 3 deg/sec).

22

Section 3

SIMULATION RESULTS

Two model configurations with slightly different parameters were used in

the simulation runs. The experimentally determined parameters for the

two configurations were the following:

Center Body

Configuration 1 Configuration 2

148.46 kg m (111.16 slug ft2

)

22.70 kg m ( 17.0 slug ft2

)

144.10 kg m (107.90 slug ft 2)

g 2 2148.16 kg m (111.16 slug ft )

22.90 kg m ( 17.15 slug ft2

)

144.20 kg m (107.97 slug ft2

)

4.354 kg m ( 3.26 slug ft2

) 4. 260 kg m ( 3. 19 slug ft 2)

For both configurations the boom parameters were

2I = 5.107 Kg m (3.824 slug ft2

)B

WB = 1.636 rad/sec (0.260 cps)

d+!I = 1. 2115

R = 0.9929

2 2Rcs = 2.658 + 1.2115 Q

The rigid body parameters are obtained by adding 2I B to the IZ

and Iy

values for the center body.

23

IX

Iy

IX- Iz

I z

I1

3.1 RIGID BODY SPIN-UP AND SPIN-DOWN

Large angular excursions can occur during the low spin portion of a run

since the system does not have a nonspinning attitude control system. To

prevent excessive attitude excursions due to the spin-up jet misalignment,

the wobble-damping jets, which provide torques about the X axis, were oper-

ated manually by means of a toggle switch on the control console to minimize

the attitude excursions until spin speed is built up sufficiently to allow con-

trol by the RCS wobble damping system.

in this run the model was spun up and down with the booms locked (rigid

body). At time zero, the spin-up jets were turned on and attitude was con-

trolled manually as described above.

At 568 sec when the speed was 0.489 rad/sec (28 deg/sec 4.67 rpm), con-

trol was turned over to the RCS wobble damper which allowed wX to move to

the edge of the RCS dead zone 0.0052 rad/sec (0.3 deg/sec) and remain there.

At 926 sec, the speed had reached the set speed of 0.62 rad/sec (35.6 deg/

sec). At 1, 100 sec the RCS wobble damper was turned off and the spin-up

jets allowed to maintain a constant speed while the steady-state nutation

motion was observed. The nutation period observed of 68.4 sec is in excel-

lent agreement with that calculated from the experimentally determined rigid-

body parameters (68.0 sec). The offset from zero on wx indicates a

significant IXZ dynamic imbalance on this run. At 1,428 sec, the spin-up

jets were turned off and the model allowed to spin down under air resistance.

At 1,453 sec, the RCS wobble damper was turned on again and damped the

oscillation to the edge of its dead zone in about half a cycle (43 sec). The

length of this run precludes inclusion in this report; however, Figure 3-1

shows the constant speed and RCS wobble damping portion of this run from

time 1,113 to 1,513 sec.

Figure 3-1 has coZ as the spin rate on channel 1, the scale is 0.035 rad/sec

(2 deg/sec) per division, and the time scale is 1 sec per division. X and coy

are on channels 2 and 3, with scales of 0. 0017 rad/sec (0. 1 deg/sec) per

division. Channels 4-5 and 6 are not in use on this run; however, Channel 4

is the boom channel. The boom is locked in place (rigid body). The signals

24

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allowed to coast so that the free motion could be observed unperturbed by

the RCS system. Figure 3-4 shows the result of a spin-up and release at a

spin of 0. 64 rad/sec (36. 8 deg/sec). wX and the boom angle immediately

diverged indicating that the stability boundary has been approached in good

agreement with the calculated value. (Stability factor = 1. 0 at 2 = 0. 69 rad/

sec, 39.4 deg/sec.)

Figure 3-5 shows the result of a spin-up and release at 0.59 rad/sec

(34 deg/sec). A significant boom oscillation was excited, indicating opera-

tion not far from the stability boundary. Again the nutation period is close

to that of the rigid body. The jerky movement of the booms is again evident.

Figure 3-6 is the record from a spin-up and release at 0.51 rad/sec (29.1

deg/sec). A relatively small boom amplitude is excited indicating operation

relatively far from the stability boundary.

3.3 ACTIVE MASS BALANCE WOBBLE DAMPER

Figure 3-7 gives simulation results obtained with the active mass balance

system with the booms locked (rigid body). The model was spun up to 0.626

rad/sec (36 deg/sec) and released and the free motion observed for about one

cycle. The wX offset from zero indicated a negative IXZ product of inertia.

The wobble damper damped the oscillation in about one cycle. The disturbance

mass was then commanded to move 0.6 inch in a direction to relieve the pro-

duct of inertia, again the active mass system damped the motion in one cycle,

although the spin had decreased considerably. The broad wy trace is due to

a low-amplitude, high-frequency oscillation of the active masses. It does not

show in the active mass trace because the signal has been put through a low-

pass filter with a break frequency of 20 rad/sec. The locked boom structural

frequency appeared to be participating in this and is possibly related to the

natural frequency of the cable connecting the two masses. It is possible that

additional low-pass filtering on the input signal wy could eliminate the low-

amplitude oscillation.

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Figure 3-11. CMG Wobble Damper with Flexible Booms-Configuration 2

38

4

1 DEG

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CR71

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Section 4

DISCUSSION OF RESULTS

The system with flexible booms tended to exhibit the rigid-body nutation

period. A number of runs were made with configuration 2 at various spin

speeds, and the nutation periods were measured to investigate the phenomenon

in greater detail. The results are presented in Figure 4-1. The upper

curve is the flexible-body nutation period and the lower curve is the rigid-

body nutation period. The crosses are experimentally determined points.

It will be noted that the observed points generally fall between the two

curves. At 0.419 rad/sec (24 deg/sec) the experimentally observed points

lie close to the rigid-body curve. The points deverge from the rigid-body

curve as the spin rate is increased. Near the stability boundary, the points

jump up closer to the flexible body curve. The dashed lines indicate the

general trend. It is believed that the observed behavior is caused by slip-

stick friction in the boom pivot bearings. The simulation run records show

that the boom angle tends to move in small pumps as is typical of slip-stick

friction. At the lower spin-rates where only small boom amplitudes occur,

the system tends to behave as a rigid body. At the higher spin-rates where

large boom amplitudes tend to' occur, the system behaves more nearly

according to flexible-body predictions.

In view of the foregoing, it is somewhat difficult to assess the influence of

the flexible booms on the active and disturbance mass systems. The chara-

teristics of the active mass balance system are C = 16m/rad/sec (53 ft/

rad/sec), mc = 0. 777 kg (1. 72 lb) each, rX = 0. 254 m (0. 833 ft),

X = 0. 148. For the conditions (configuration 1) of the rigid active mass

balance run (Figure 3-7), 2 = 0.628 rad/sec (36 deg/sec), yield 5 = 0.5,

according to Equation (2-20); and this appears to agree with the experimental

result. Computed on a rigid-body basis we would expect 5 = 0.37 to be

observed in the flexible boom run (Figure 3-8) since the spin speed is lower,

0.489 rad/sec (28 deg/sec). Actually the damping appears nearly the same

39

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in both runs. If we compute the damping for the flexible boom case by using

the reduced inertia we obtain , = 0.55. This indicates that on this run,

the flexible effects are acting according to theory. The nutation period

observed on this record, 120 sec, is closer to the theoretical flexible boom

value of 128 sec and distant from the rigid body value of 87 sec. A somewhat

similar result is obtained with the CMG wobble damper (configuration 2).

The angular momentum of the two CMG's is 0.922 kg m /sec (0.68 slug ft /

sec), and the input gain is 6 (dimensionless). On a rigid body basis,, Equation

(2- 14), these parameters would produce a damping ratio, I = 0. 7, which is

about the value obtained in the simulation run (Figure 3-9). The' flexible boom

CMG run yields more nearly critical damping (Figure 3- 10). If the CMG

damping for the flexible run is computed from the reduced inertias (Figure

2-9), a damping ratio, T = *0.9, is obtained in reasonable agreement with

the observed value.

Figures 4-2 through 4-8 present the results of a digital computer simulation

using the data of configuration 1 and a boom viscous damping.ratio = 0. 06

for the flexible booms at a spin rate of 0.628 rad/sec (36 deg/sec). This

run exhibits the theoretical nutation period of 122 sec. Figures 4-9'through.

4-15 present digital simulation results for configuration lwith RCS wobble

damping. The wobble damping dead zone was 0. 00524 rad/sec (0.3 deg/sec)

and the pair of on jets developed a moment of 0. 149 Nm (0. 11 ft lb).

Figures 4-16 through 4-23 give digital simulation results for configuration 2-

with the CMG wobble damper. The gimbal angle command was a = OwX.'C' x.

The integral of the gimbal angle with a time constant of 400 sec-was sub-

tra'cted from the gimbal angle command to cause the gimbal angle to' slowly

return to its nominal position.

41

CR71

z L

a-w 1

-3

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TIME SECONDS

Figure 4-2. Configuration 1 Flexible Booms - X Component of Angular Velocity

42

0 o00 2oo

TIME SECONDS300

Figure 4-3. Configuration 1 Flexible Booms - Y Component of Angular Velocity

43

CR71

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Figure 4-4. Configuration 1 Flexible Booms - Boom Angle Theta-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ . _-

44

CR71

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45

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46

CR71

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47

CR71

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TIME SECONDS

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48

CR71

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49

CR71

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Figure 4-10. Configuration 1 RCS - Y Component of Angular Velocity

50

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51

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Figure 4-12. Configuration 1 RCS Damping - Boom Angle Theta

52

CR71

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.o i111 111111111

II 1 1 X 1111 X

-.. I' -I'11111111111

CR71

w . . ' I . . II Iw

IIIm II II i II II II I I I |

TIME SECONDS

Figure 4-13. Configuration 1 RCS Damping - Euler Angle Beta 1

53

Figure 4~~~,,13.Cniuain1 RSDmigElrAgeBtIIIII11111111 II IIIII3

CR71

IL I I I I 1I1 1

" - -- -,I ll ll .1 1 Ii l A :(A , , I,.llll, r "T11 '1 I vi I Illl'IlI in i

0

w

0

I 10 200 300 11 00 01TIME SECONDS

Figure 4-14. Configuration 1 RCS Damping - Euler Angle Beta 2

lull ~ I~ R A I~ ~U~_~ ~ III~I~IRI:I LA~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~hk .IA~~~~~~~~~~nl~~~~~l~~~l~

54

CR71

18000

16o00

o1000

12000

10000

(A

aal

-iCDz<C

8000

6000

q000

2000

-o 200

TIME SECONDSo00

Figure 4-15. Configuration 1 RCS Damping - Euler Angle Beta 3

55

300 4oo0 sow

~oi

.0 _ _…-'-

LB2 _…t___t-_____ __ 1J\…Io

1.8 _ ____ ---- __ _ __- __ _-… _ _

.'2- X -_ __--_--

_ '°I ---=I=…

.8

._i %

L

II

_ I I I-1 --ii- i i -;

o10I I I

9oo 300

TIME SECONDSo09

Figure 4-16. Configuration 2 CMG Damping - X Component of Angular Velocity

56

CR71

z0UwoU)

0:

U.

U)ww

CDw0

-)0I-

-4

(aZ

-. 29O0

= -. I T-9 - I 1 I 1 1 I 1i

I I I I I I 1 1 11 1 I11 - 1 i l II

.08 -i

.0-

-. 02 t

LI-. - I I-

. 1 __ __

'I _ - - --

e Io0 00 ao00

TIME SECONDS1Qo

Figure 4-17. Configuration 2 CMG Damping - Y Component of Angular Velocity

57

.10.

CR71

L

azo0

JU)

Z

oul

0

U

w

Cz_1

So0 o00

-TI -T -T -T -T -F -T -T -F -T-T

35.9

35.90

35.89

35.8

35.89

3l. T_IF o00 ; 3100O

TIME SECONDSe0s gOt

Figure 4-18. Configuration 2 CMG - Z Component of Angular Velocity

58

CR71

z0Uww

Ix-

U

0

co

-Jw

C

z'C

Ibi

n

DJ

OLi

BOOM ANGLE THETA,e

.7 ir

i I~~~~~~~~~~~~~~~~I

. _ _____ _ ___ ____ -__

.. … _ _ _ __ _ _ -_ __ _ _ __ _ __ _ _-_- -…

FJ

.2 _ _ _

LL

%~~~~~~~~~~.1

t-1 I I I. I E | i | 1 -- - - 4Th--- - ¢

.i eeo I TI I aeeI 1 p t IT r-! it'0 200l 0 0 I ME SECONO

TIME SECONDS

Figure 4-19. Configuration 2 CMG - Boom Angle Theta

59

CR71

.

-

-1

-2

-5

-6

-o

6

1I

X

i00 eoo 300

TIME SECONDS

Figure 4-20. Configuration 2 CMG - Gimbal Angle

60

CR71

()w

CDco20

w

z<

/7-8

-9

-I0

-ll

,.'7

7,

o00 S0e 600

1.

l

w --0 t -f-i-f-

to

- ii ij ;I i I I II I

-3 __ __

-S00 Zoo00TINE SE

Figure 4-21. Configuration 2 CMG - Euler Angle Beta 1

CR71

;CONOS

61

.1 1o ''i¾ ' - l l l l l l l l

D :i;ii l'

T IME SE(

Figure 4-22. Configuration 2 CMG - Euler Angle Beta 2

CONOS

CR71

62

,,fOO

10000

16000 ,1

1000

I0000

2000 _ _ - - - - _…

// aeU9 1i0 M 10Soo

TIME SECONDSq00 roe

Figure 4-23. Configuration 2 CMG - Euler Angle Beta 3

63

CR71

WW

wa

J

z

600

PRECEDRIG P4GE BLANK NOT FILMBD

Section 5

CONCLUSIONS

The rigid body behavior of the simulation system was in good agreement with

theoretical expectations. The flexible boom system results were in good

agreement with the predicted stability boundary and in reasonable agree-

ment as to predicted nutation frequency when operating near the stability

boundary. The RCS wobble damper functioned very well under all conditions

simulated. The spin-up RCS functioned satisfactorily but was not as well

aligned as is desirable. The disturbance mass and active mass systems

performed excellently. The active mass system showed itself as a simple

and effective wobble damper under both rigid and flexible body conditions.

Simulation runs with the CMB wobble damper exhibited the increased damping

ratio predicted for operation with the flexible booms as 'did the active mass

wobble damper.

On the whole the air-bearing simulation model appears to be a very effective

tool for the investigation of the interaction of control systems with spinning

vehicles with flexible booms. The pressure of time and funds did not allow

as much effort to be devoted to precise adjustment of the model as is desir-

able to exploit its full potential as a simulation tool.

Movement of center of gravity with use of RCS gas upset the static balance

and resulted in the model being operated in a less well-balanced condition

than would otherwise be necessary. The small misalignments in the spin-up

thrusters tended to obscure the phenomena under observation. In addition

these extraneous moments often, but not consistently, appear similar exper-

imentally to product of inertia disturbances and interfere with precisely

dynamically balancing the model.

Preceding page blank 65

It is recommended that additional effort be devoted to

A. Improving the model static and dynamic balance, a straightforward

but time-consuming task.

B. Change the boom pivot bearings to flexural pivots which would then

exhibit structural rather than friction damping.

C. Devise means to eliminate the small amplitude high-frequency

oscillation of the active mass system.

66

Appendix A

DERIVATION OF EQUATIONS OF MOTION

Figure A- 1 illustrates the configuration under consideration. A rigid central

body of mass M and principal moments of inertia Ix, Iy, I is spinning with

angular velocities , wo and ,' = Q + wo where' is much greater than o,x' y z z x

a, and aZ. Stability augmentation booms with tip masses m 1 and m 2 and of

length f are hinged at the points of connection to the central body a distance d

from its center of gravity. The hinges restrict motion of the booms to the

x-z plane, such mnotion being opposed by a torsion spring of constant KH

at

the hinge so that in the unperturbed state the booms are aligned with the x

principal axis of the center body. The x, y, z axes form an orthogonal coord-

inate system with axes parallel to the principal axis of the center body and

origin at the center of gravity of the system. The small displacements ql and

q2

of the tip masses are thus restricted to be parallel to the z axis. (It may

CR71

z /Z m2 /

/

X

Figure A-1. Rotating Body with Flexible Stability Augmentation Booms

67

be shown that in-plane motions produce only second-order attitude effects.

The center body small displacement from equilibrium is Q. The displace-

ments of the tip masses relative to the center body are qlr and q 2

Although the equations of motion are developed for hinged booms, it will be

subsequently shown that with minor modifications the same equations apply to

tip nmasses supported by elastic cantilever beams.

For each of the three masses, one can write

In. a. F. (A-l)11 1

Where a. is the inertial acceleration of m. and F. is the sum of the internal1 1 1

connection forces and external forces. In terms of quantities defined in the

rotating coordinate system xyz,

ai = A + cxr. + 2wxi. + wxwxr. +r i. , (A-2)

where A is the acceleration of the origin o (system CG) expressed in xyz0

components, J is the inertial angular velocity vector of the center body in the

xyz system, and r. is the position vector of body i relative to o. The com-

ponents of the position vectors of the bodies are;

body mi, r= [d + f, 0, ql

n 2 , r 2 = -(d + ), o, q 2

M, Qxf Q y I Q z I (A-3)

Inserting Equation (A-3) in Equation (A-2) and neglecting second-order terms

gives the component equations, where the subscript e denotes external forces,

for mi

1na = mlA ox Fix iex

68

mlaly = rmlAoy + ml(d + Q)cz = Fly + Fley (A-5)

mlalz = mlA - ml(d + f)Ly + ml(d + e)Zx + m

= F. + Fe (A-6)lz lez

for m2

2m2aax = m2Aox + rn2(d + F2x + F2ex (A7)

2a2y m2Aoy - m2(d + )z = F + F2ey (A-8)

m2a2z = m2A + m2d + f) - n 2 (d + w + m)W + 2

= F2 + F2ez (A-9)

for M

M + MA = _-F F -~ F7 + F. (A- 10)MQx + Mox lx - x Mex

MQy + MA = F - F2y + ey (A-ll)Qy oy 1ly 2y Mey

M + MA = _ F - F + F (A-12)Qz oz lz 2z Mez

If we let m 1 = m 2 = m and add the corresponding component equations we

obtain for the x direction,

MQx + (M + 2m) Aox = Fe (A- 13)Qx ox ex

Since M + 2m is the total mass of the system and A is the x component ofox

acceleration of the center of gravity, the second term of Equation (A-13) is

equal to the right-hand side and therefore Qx 0.

Similarly for the y direction, we obtain

Q =0y

69

In the z direction, the result is

MQZ + mql + m 2 + (M + 2m) Aoz(A-14)

ez

Reasoning as above, Equation (A- 14) becomes

MQz + mql + mq 2 = 0

Equation (A-15) implies

ql +q2 m z

(A-15)

(A-16)

In the following, it will be assumed that no external forces are present (i. e.,

FMe' F e F2e' = 0), so that A = 0. The internal connection forces

can be expressed in terms of the small deflections of the boom relative to the

center body. Figure A-2 illustrates the forces acting on a mass and the cor-

responding reaction forces on the boom.

CR71

Z

M (d + Q) C2

FI Z

mm (d + Q) 522

KH FIZ q1r

d I

Figure A-2. Forces Acting on Mass and Boom

70

!

Summing moments about the hinge point and solving for Flz we obtain,

F z + m(d + f)2 ]. (A-17)

and similarly

FZ -2H + m(d + )Q2] 2r (A- 18)2z 1~ .e

From the geometry we have, letting Q = Q,

ql qlr

q 2= qr + Q (A- 19)

Inserting Equations (A-17) and (A-18) in Equation (A-12) and making use of

Equations (A-16) and (A-19) yields

Q. + (M -+ 2m)m(d +~ ) 2 +H Q 0 (A-20)z\m/L M

Now

1(d + ) 2 KH 2

c s f +- 2

is the spinning vibration frequency of the booms if they were attached to a

rigid spinning shaft, thus Equation (A-20) can be written,

+ ( +2m 2 Q = 0 (A-21)

Equation (A-21) indicates that the system can perform vibrations parallel to

the z axis with m 1 and m2

moving in the same phase and opposite in phase to

M with frequency given by

(= M + 2m)2 (A-22)

71

The absence of attitude variables in Equation (A-21) indicates that this trans-

lational vibration is uncoupled with the attitude motion of the system.

To arrive at the equations of angular motion of the system, we can equate the

rate of change of angular momentum I to the torque T acting for each of the

bodies.

R = m. r.xa. and . = r.xF. (A-23)1 1 1 1 1 1 1

The component equations are, for ml

ilx = 0 = Tix (A- 24)H ~OZT (A-24)

2.It = -m(d + f)l ql - m(d + f)ql + m(d + f) Cy

- m(d + f)Qwx=Tly (A-25)

T -m(d + ) 2Q 2 q + (d + KH9r + m(d + f) 2 q (A-26)ly 2 l26

or

Tly = m(d + f) ((Ws- 2 ) ql 2 Q (A-27)

2.HA1 = m(d+) = Tlz (A-28)

Similarly for mass 2,

HZ2x = 0 = T2x (A-29)

2 22H 2y = m(d + f)2 q 2 + m(d + f)q2 + m(d + .e) y

- m(d + e) Qx = T2y (A-30)

72

T n(d + 2) [ 2Q)q 2 r 2Q (A-31)

2.2z m(d + ) T 2 z (A-32)

For the center body, the corresponding H equations are the usual Euler

equations for Q >> wx Cy

I cx +(I - I)Qw = -T - T (A-33)xx z y y lx 2x

I o + -(I I) = )- T 2-y -M (A-34)yy Ix z x ly 2y c

I = -r - T (A-35)z z lz 2z

where T appears with a negative sign since the connection moments occur in

equal and opposite pairs.

Adding Equations (A-25) and (A-30), the equation of motion for the booms is

obtained

m(d + ))22 (qz

- ql) + m(d + Q)(q2 - q1) + 2m(d + af)2 J

2- 2m(d + f) Qw = M (A-36)x c

M = -m(d + )( 2 )(q 2rqlr) (A-37)c s

Equation (A- 19) shows that (q2r - q 1 r ) = q2

- ql' if we let

? -9 q1 20? (A-38)

Equation (A-36) may be written

2mQ(d + f) 6 + 2m(d + E) 2. + ZmQ(d + f)lQ0 - 2m(d + P) Q2o

= M = -2m(d + )(2 _ 2) (A-39)c s

73

dividing through by the coefficient of the first term Equation (A-39) can be

written

220+ d+ d+ w s03 + os - y x (A-40)

Inserting the appropriate expressions for T, Equations (A-33) through (A-35)

become

I &x + (I z- I )QY = 0 (A-41)

I + (I - I )Qo = 2mT(d + P)(w - 2)0 (A-42)y y x z x 5

I z -2m(d + ) 2z (A-43)zz z

Equations (A-40), (A-42), and (A-43) then are the equations of free motion of

the system. (Equation (A-43) merely indicates that ;0 = 0. ) It will be noted

that M , the connection moment, occurs on the right side of Equations (A-39)

and (A-42) (with opposite signs). In order to determine the equation of motion

for a systenl with cantilever booms, it is necessary to insert the correspond-

ing connection nnmoment expression.

Figure A-3 illustrates a massless uniform beam of length P mounted a

distance d from the spin axis, (i. e., the portion d is rigid) with a concen-

trated mass m mounted at the tip and spinning with angular velocity Q. The

beam is subject to a deflecting force F and a tension P = m(d + 9)Q22

The basic beam equation is

EI Y" = M(x) . (A-44)

where El is the flexural rigidity of the beam,Y" = d2y/dx2

, M(x) is the bending

moment applied at point x (origin taken at d), and Y is the lateral deflection.

The moment is given by

M(x) = F( - x) - P[Y() - Y(x)]. (A-45)

74

CR71

SPIN Q(J AXIS

d

Y(X P

F

. ~~ xX X '

| Z v~~~~~~

Figure A-3. Stiffness of a Spinning Boom

Differentiating and combining Equations (A-44) and (A-45) yields,

IP =FE1 E1

with end conditions

Y(o) = Y'(o) = 0; Y"(e) = 0

(A-46)

(A-47)

Let a2

= P/El. Laplace transforming Equation (A-46) and using the end

conditions at x = 0 gives the transform of the solution as,

Y(S) -F + Y"(o)

EI S2(S2- a 2 ) S(S2 -a

2)

The inverse transform of Equation (A-48) is

=_F 1 a )Y"(o)cosh x - 1)Y (x) - 2 ( sinhax - x) + (cosh-x - 1)-EIa a

75

(A-48)

(A-49)

Differentiating Equation (A-49) twice and using the condition that Y"(e) = 0

enables determination of Y"(o) as

FY"(o) = tanh a . (A-5(

Inserting Equation (A-50) in Equation (A-49) produces the complete solution

Y(x) F 3 [ax - sinh ax + tanh at (cosh ax - 1)] . (A-5EIa

The tip deflection at x = P is given by

0)

1)

Y(P) = F (aP - tanh aP) = (EIa

- i tanh a) .a (A-52)

The spring constant is given by

FK =

YP

(A-53)1P - - tanh aga

Vibration frequency of spinning cantilever.

given by

The spinning frequency cs, is

P1

m(f- - tanh aP)

(d + ±) Q2 11

(1 - -tanh acf)wue

This can be expanded in a series for purposes of comparison with the hinged

boom solution as follows;

Let 3 = aP then Equation (A-54) can be written

2 EIs m 3

mP(A-55)P 3

\p - tanh

76

2 Ks m

(A-54)

Let

3 1 a-ntan hpf =_3-_____ 1and g- a(A-56)p-tanhp ' f P3

then using the series expansion for tanh p

1 2 2 17 4 62 6 + (A-57)g =3 5 1 315 2835

f can be expanded in a Taylor series in

f,,(o) 2f(p) f f(o) + (o) + *21 + (A-58)

and

1 g'(o)f(o) g(o) f'(o) etc

g(o) ()

as a result of these operations we obtain

6 2 1 4f(p) 3 + 2 17 1 +. (A-59)

and

2 3EI 6 (d + e) Q2 1 me(d + )24 + A-60

s m3 5 e 175 EI

2the first term is coB , the nonspinning natural frequency.

Accuracy of Approximation

If we use the first two terms of the series as an approximation, the accuracy

can be estimated by a representative numerical example.

Let Q = 0. 63 rad/sec, wB = 0. 55 rad/sec, m = 189. 17 kg (13 slugs),

d = 0, - = 30. 48m (100 ft),

77

Q2 3(d + C)9ae =: 1B = 1. 98393

2 2 1

[d - 2 Q tn Z = 0. 87817

I 6 ( 2tanh a d

CO ( 2 5 f = 0. 88249 rad/secs B 5 e

the difference is 0. 00432 rad/sec

central body about the cg is

Mc = F(d + P) - PYc

or 0. 49 percent. The moment exerted on the

(A-63)

Let 0 = Y/P then since

2 F Fa) = K/m =

s Ym me6(A-65)

and

P = m(d + 2)n2

2 _ 2)0M = m2(d + )(co - )c s

(A-65)

(A-66)

It will be noted that the expression for M c in Equation (A-66) is identical to

that in Equations (A-39) and (A-42) for hinged booms so that for equal values

of X the spinning boom frequency the equations of motion are identical.

However, the expressions for cs are different functions of the spin rate -2.

For comparison

2w cantilever =

s(d + 2) Q2

P(1 - e tanh ae)7RP

2 6 (d + 2) 2= CB +5 e

78

then

(A-61)

(A- 62)

(A-67)

2 2 d+e2co2 hinged = + d (A-68)

If d + P/e hinged = 6/5 (d + P)/P cantilever as for instance in a dynamic model

correct tracking of Xs with spin rate will be obtained.

79

Appendix B

COMPARISON WITH CANTILEVER BEAM EQUATIONS

Hinged boom vibration frequency with boom arm mass included

qJ m . p=m(d+QKZ 2

K4

the potential and kinetic energy for

2V = KHO2 + me(d + )QF2 0 +

= KH2 + 22 [(m + B )

2T =m [ + (d +

= m{E6)

2

'+ (d + R 22]

a small displacement O' are

CQ 2 0 2 f x (d + x) dxo

mBd] 2(d + 2) + m-d]

+f dm x 2 62 + (d + x)Z2]

whe re

c is the mass per unit length of the boom arm and m B = cP is the total mass

of the boom arm

then

d 8_0. + ae - (m +- 3) e' + IKH + [(m + 3 ) (d + e)

+ m-Bd]Q 2 } 0' = 0

Preceding page blank j 81

(B-4)

(B-l)

(B-2)

(B-3)

whence

2 2 d + 2 Qs B + C R (B-5)

where

mB [3d + 2P]m m + 2 [d + d ]] (B-6)

and

m + B/3R m (B-7)m'

where

2 KHcob =

3

The moment applied to the center body about the cg is

M = F(d + e) - PY - C2 f (d + x) x dxo

which can be put in the form

(d + e)21 [ 2 2] (I (B-8)c d+JM = ml (d + e AS R - Q

The object of this discussion is to demonstrate the relationship between the

equations of motion and stability criteria for flexible cantilever booms and

hinged booms on a rotating space station. It was shown in Appendix A that

the angular equations of motion for the configuration under consideration are

I C1

+ (I3

- I2)Qc 2 0 (B-9)

I2'2 + (I1 - I3)2c 1 = 2 M (B- 10)

82

2me (d + e) 0 + 2m (d + e)22 + 2m #(d + e) Q2 - 2m (d + e)2Qwol

=- _2M (B-11)C

where M is the internal connection moment between the central body and thec

two booms, I1 , I2, 13 W, Z1 , 2Q are the principal moments of inertia and

angular velocities of the central body, m is the boom tip mass, and 0 is the

boom angle. It was shown that for massless hinged booms and a massless

cantilever boom M is given by

2 e 2 2M m (d + e)

2-2 2) O (B-12)

c

and w s for the massless hinged boom is

2 2 (d + P) 22S = a,2 + ( 2 (B-13)os :oB + £

2 KH2oB 2 (B- 14)

me

for the massless cantilever booms it was shown also

2 (d + P) 2Wws -1 (B-15)P 1

(1 - tanh oa)

or the approximation

2 2 6 (d +- e) 2Ws aB) + (B- 16)

whe re

2 3 E1 2 m(d + e) 2'B 3 and a E (B- 17)

83

If the mass of the arm of the hinged boom is accounted for, it was shown

that

M = ml (d + ) 2 e+ [ s2R ]- Q ]oc ~~~~(d + e) I

~2O (B- 18)

and the spinning vibration frequency is

2 2 d + P s22's = CB 2+ R

mB 3d -+ 2e]m -- m + - [J+3 [2 (d + f)]

m + /3R = m

m I

(B-19)

(B-20)

(B-2 1)

Equation (B-ll) takes a different form due to the distributed mass of the boom

arms as follows;

dt = 2m (d + ) + 2c (d + x)x dx] ' + m (d + )0

+ 2c f (d +x)2

dx]W 2 + [I2m (d + e) + 2cO

- [2m (d + I) + 2c f (d + x) 2 dx2w 1l0

eJ (d + x) x dx]2Q2O

= -2Mc

The integrals involved evaluate to

c 9d x'd mB 3d + 2)]c d X) 2 dx =3 (d + d(d+ )

0 L-

cf0

(B-22)

(B-23)

(d +x) 2= ([3dd + )+ 2(d + x) dx (d + e)

3 (d (d 4 )2O (B-24)

84

and

The last expression is the moment of inertia of the boom arm about an axis

through the center of the station (not including the tip mass).

The first and third bracketed expressions on the right-hand side of Equa-

tion (B-22) then become

2£ (d + m) /3 d + + B/3 2 (d + f) m' (B-25)

the second and fourth expressions become

m B21 2m (d + 2 (d + )2 [3d (d + ) + 6)

(d + e)2

This last is the total moment of inertia of the boom (arm and tip mass) about

an axis through the center of the station and would directly add to the total

spin inertia if the booms were rigidly connected. Equation (B-22) can then

be written

2P (d + f) m' 8' + 2I6 c2 2 + 22 (d + P) m! 8' - 2IB = -2M (B-27)

with definitions

I 11 3 1 3 - 2A B

12 I 1

and

c' 2 m' (Rd + ( 2 2) (B-28)

2 s

the equations of motion of a rotating space station with hinged massy booms

are

+1 + BQ2 = 0 (B-29)

[r 85

w+2 + AQ2w d +c' 0 = (B-30)

IB IBf26' + R 0 + $ 01-0 (B-31)+ Rs (d + 9)m' W2 (d + f)m' (B-31

The equations of motion of the system with cantilever booms with

c = 2m ((d + s

Q (B-32)2

are

W1 + BQ2 Z = (B-33)

2 + A)wl d C = (B-34)

2 d-e + d + e 0 (B-3=)s 5 + 2 - -- (B-35

The simplest way to arrive at t:he characteristic equation for these systems

is to differentiate the last two and use the first to eliminate c01. It should

perhaps be noted at this point that I B/(d + e)m' - d+e/£ and R- 1 for the

dynamic model, so that the two sets of equations have nearly equal coefficients

as they stand. However to demonstrate the variations with spin rate Q further

discussion is advantageous.

Carrying out the suggested operations and inserting the associated values of

Ws, the spinning natural frequency, yields for the cantilever boom system,

2 2 _2mg(d + ) 2 + (6 d 2] BaB 2+ 2 B 5

+ [ 2 + 6 (d + e)29 (d + e) + (d + e)2 Bco 0 (B-37)B 5 e 2 P 2 =

86

The associated determinant is

S4 + (S2 - AB 2 + c)S 2 + BQ2(c - Awo 2 ) = 0 (B-38)

It may be shown that for the type of spacecraft under consideration with

A < B < 1 the requirement for stability is

2c > Ac (B-39)

or

2m(d + P)I

1-I3

2s

2 2s

2 6 (d + P) z 2CB + 5

2 6d 1 2cB + ±¥ +

(B-40)

for a boom geometry similar to Skylab d = 0 and Equations (B-36), (B-37) and

inequality (B-40) become

2 m AB2 [B2 [ABQ B 2 - IL

2me2

I1-I3

1 2] + IQ j 0

+ Q2 B2 = 0

(B-41)

(B-42)

(B-43)

2 6 Q2

2 12ob 5+

For the massy hinged booms the equations corresponding to Equations (B-41)

and (B-42) are

2 AB 2 2m'e (d + f) 2 d 2 '2 - AB ( w2 - I B P (B-44)

... 1 2 d P ~221 I_ I_",B PB +B .m l B d2 20 + Rw + f f2 ' +BRoB od $ P~m' 2 T(d + B cM2+ 7 f~~~d + Om' 2~ +2~m

= 0 (B-44)

87

The corresponding characteristic equation is

I4 2 2 B 2 2S + (Rw - AB2 + C')S +BQ

(d + ) mn

JB _C' - ARCw 2 0 (B-45)

(d + #) m '

The condition for gyroscopic stability is

21 Rw 2 Rw + 2-B_ >, S - B P -(B-46)

I1

- I3 R2 Q2 R 2 + d 22s B P

Comparing stability criteria Equations (B-43) and B-46) , it was noted that

IB

is the total moment of inertia of the hinged massy boom about an axis

normal to the spin axis and therefore is directly analagous to 2me2

of

Equation (B-43).

If we choose

IB 2 RwB hinged -B B cantilever (B-47)

and

d + 1 + d 6=1+ (B-48)

P 5 e e 5

the stability conditions of the hinged boom model will be identical to those of

a cantilever boom vehicle at all spin speeds.

The equations of motion of the massy hinged boom system Equations (B-44)

and (B-45) would differ from Equations (B-41) and (B-42) of the cantilever

system by the presence of the coefficients IB/2(d + P)m', in the last two terms

of Equation (B-45) and a difference of the coefficient of the last term of

88

Equation (B-44). This condition can be corrected by making a linear trans-

formation of the boom angle variable

0- (d + )m' o' (B-49)B

in Equations (B-44 and (B-45). Each term of Equation (B-45) will then contain

the factor I B/(d + e)m' as part of its coefficient, and can be cancelled out.

Equations (B-44) and (B-45) then become

;2 AB Q 2W 2 - 2IB [Bz +Q2IE (B-50)

+ [ 2 + d + + + BQ 2 W2 0 (B-51)+B c02 e 2

Thus with the appropriate choice of parameters, the coefficients of Equa-

tions (B-50) and (B-51) can be made identical to those of Equations (B-41) and

(B-42). The stability criteria are unchanged by the transformation. The canti-

lever boom angle 0 is obtained from the model boom angle 0' by means of

relation (B-49).

The air bearing model parameters actually used result in the following

d = 0. 1397 m (0. 4583 ft), e = o. 6604 m (2. 1667 ft)

m = 3. 634 kg (0. 2491 Slugs), m B 0. 9819 kg (0. 0673 Slugs)

m + mB/3 = 3. 9617 kg (0. 2715 Slugs)

m' = 3. 9889 kg (0. 2734 Slugs)

m + m B/3R = . 9932

I 2. 5785 kg m2

(1. 9023 Slug ft2

)B

P(d + Om1 I#

d + = 1. 2115e

89

. ... G PPAG3 'BLAN NOT FILM

Appendix C

SOLUTION TO EQUATIONS

The equations of free motion of the hinged boom model including the boom

arm mass are;

&1 + BRQw2 = (C- 1)

(o2+ Af21_2m'e (d + ) (R2

I2

s2

2 IB

6 + Rw 0 + I 2m e(d + e)

Q )0 = 0

IB

m '(d + I)

I 1 -I3A I

I2

I- 123 2B m B [3d + 2f]

= m + 3 2(d + Q)

m +mB

3 2R rn

2 d + e Q2= eoB+ R

To obtain the equations for the cantilever beam booms substitute;

d + efor , m for m, and w2 cantilever for Rw2m' e(d + e)

where

2 2 6 (d + e) 2Ws cantilever WB c 5 +-

Preceding page blank-- .. .. 91

where

(c-2)

(C-3)

(C-4)

(C-5)

Some additional definitions are convenient.

Let

21B

1 3

which may be called the inertia margin and

IlI (1 S2 £22 SF (C-7)

1 1 \ Rw/

which may be called the stability factor, since the types of space stations

under consideration (I1

> 13 > I2), the requirement for gyroscopic stability

is SF > 1.

Equation (C-2) may be solved for a)1

and substituted in Equation (C-3) to

eliminate wol and Equation (C-2) may be differentiated and 1I substituted

from Equation (C-1) to eliminate ol from Equation (C-2). The resulting

equations are;

+ Rx (1 - SF) e + 1 + A) w2 = 0 (C-8)~~s ~ 2m' #(d + e)

Z 2mme(d + 2) ( _2 2 _ 2m e(d + e (R 2 -2)e 0 (C-9)2z - 2 ¥Z I2

s

where

2 AB (C-10)'1 = AB£2 (C-10)

92

Taking the Laplace transform of Equations (C-8) and (C-9) we obtain in matrix

form;

s I2 IM (1+A)2m' Q (d+P)

2 2S -Y

S + Ro2

(1 - SF)S

2 m' Q (d+P) (Rw 2 _I 52

92) S

S + 6 + I2 IM(+A)o o 2 m' (d+ ) 22o

S Wo20 +2m'e (d+) (R2

20 I2 s2

Q2) 0 -

O

The terms on the right with zero subscript are initial conditions.

minate A of these equations is

-A = S + S2 [R) (1 + SFA) - y ] + R (SF- 1) y = -a= 0

Equation (C-12) is quadratic in S2

and can be written

- = (S3

+ a2) (s2 b

where a and b are given by

a - R Rc2 (1 + SFA)- ¥ --- R (I + SFA)2 1 s 4

+ Rw,) (I -SF) 1+5

b2

1/2 [Ra(, (1 + SFA) -

The deter-

(C- 12)

(C-13)

(C- 14)

2Y] + I [RcW0 (1 + SFA)- _-y2

2 1- S+ R (1 - SF) 2 (C- 15)

From Equation (C-14) it may be noted that in order that a > 0 and real the

requirement is for SF > 1, as noted earlier.

93

z2 (S)

0,(S)

(C-ll)

The solution to Equation (C- 1) is

2w2(S) -2m'(d + sL(R -

I I2I1 1S'A L(S( 2 20(s) (s

2-2)Ii [

I IM(1 + A)2

sO + 00 + 2m'--(d + ) C2o

2+ -. 2m'f(d + f)(RwS2o + 20 - I 2

Q 22) - Is2 + R (1 - sF)I]S I2IM(1 + A)

2m'f(d + f)

Q 2) 0o0

With initial conditions 0 = 0 = = o o 20o lo = cl(t = 0)

From Equation (C-2), W2o = -AQ2lo

_s2

wz(s) = 2-(s

+ Rw (1

+ a 2)(s2

- sF)]

+ b 2 ) 2o

Inserting in Equation (C-16) yields

Is2 + Rw (1 - sF)| AQolo

-(s2

+ a2 )(s + b 2 )

s__ __2_o I2IM(1 +A)

(s + a 2 + b2

) 2m'Q(d + f)

2Ib(l +A)2 Swlo

2m'L(d + L) (s2

+ a2)(s 2 + b 2

With the aid of the transform pair,

(s2 + a 2 )(s 2 + b 2 )

cos at - cos bt2 2

b - a

the time solution to Equation (C-19) is,

IB(1 + A)Qwlo(t) = 2 2 (cos at - cos bt)

m'Q(d + f)(b - a )

94

(C- 16)

(C- 17)

O(s) =

(C- 18)

(C- 19)

(C-20)

(C-21)

Equation (C-18) maybe written

[s2 + Rwi 2 (1- S-F)] AB2Q 2

_s 0o (+ 2B 100Z(s) = 2 2 2 2

-(S + a ) (S + b ) BO

Since ABQ 2 2 and a 2 b 2 = R 2 (1 - SF)Ys

Equation (C-22) becomes

(S 2 y + a2b2)

W2(S) =-_ (S

2+ a

2) (S 2 + b 2

)

10

BO(C-24)

with transform pairs

Y-1I

and

c-1 i

S 2

(S2 + a2 ) (S2 +b2 )

I(S + a )(Sz

1=I

+ b2)

- a sin at + b sin bt

b 2 _ a 2

1 1 1

2 (a sin at - sin bt)b - a

the time solution to Equation (C-24) is

W10 2 2 2W2 (t) = (

22 a) [a(y2 _ b

2 ) sinat + b(a 2

BQ (b _ a )_ y2) sinbt]

With initial conditions 0 =0 w = 0o, 2 0 = w2 (t = o)

We obtain from Equation (C-16)

S 2S Rw SF (1 + A) w20 S [L2 + Rw 2 (1 -_-F)],0o

(S 2 + a2 ) (Sz + b2)

95

(C-22)

(C-23)

(C-25)

(C-26)

(C-27)

W2 (s) =(S 2 + a

2) (S

2 + b 2)

(C-28)

r

S[RS ( + SFA)] W2 0 + S 3 2 0

2 (s ) = (s2 + a2) (S2 + b2)

2 2 2 2since a + b = Row (1 + SFA) - 2Y the time solution is

S

)20W2 (t) = 2 2

b a(C-30)(b2 + y2) cos at - (a 2 + y2) cos bt|

The solution for initial conditions on both co1 and w2 is obtained by adding

Equations (C-27) and (C-30) (superposition).

The transform of the boom angle 0 to an initial condition on wc2 only is given

by Equation (A-16) as,

O(s) -I B (1 + A) BQ wo20

m'(d + ) (S2

+a2

) (S 2 + b 2)

(C-31)

with time solution,

0(t)IB(1 + A) BQ 2oZ=-2 2 (cos at - cos bt)

m' (d + 9) (b - a )(C-32)

Equation (C-32) may be combined with Equation (C-21) to give the response to

initial conditions on both c1 an 1d o2'

IB(1 + A)2Q

2(t) = 2 2 0 P- 2c 2 0 ) (cos at - cos bt)m'P(d+e) (b - a )(C-33)

From Equation (C- 1)

(C-34)l (t ) = 0 - BQ2 W2 (T) dT

96

(C-29)

Inserting Equations (C-27) and (C-30) in Equation (C-34) and performing the

indicated operations yields

xl(t) - 0 (Y2 - b ) cos at + (a2

- ) cos bt]

20 2+ Y a + Y 2 b2 (b

2 2

a sin at - b sin bt (C-35)

97