Spin decay of a flexible satellite by structural dissipation of energy

12 JMt

MARC:a 1965

SPIN DECAY OF A FLEXIBLE SA TELLITE BY STRUCTURAL DISSIPATION OF ENERGY

by

Bernard Etkin

..

-MARCH~1965

SPIN DECAY OF A FLEXIBLE SATELLITE BY STRUCTURAL DISSIPATION OF ENERGY

by

Bernard Etkin

ACKNOWLEDGMENT

The work reported herein was supported by Air Force Office of Scientific Research Grant No. AFOSR 222-64.

Thanks are due to Mr. J. Mar of the Defence Research Telecommunications Establishment, Ottawa, for supplying the required data on the Alouette 1 satellite.

SUMMARY

This report analyzes the coupling of the rigid-body and elastic modes of motion of a spinning satellite under two conditions of motion: (i) Wobbling, in which the direction of the spin vector is

appreciably different from that of a principal axis of inertia and (ii) Quasi

-steady spin, in which the spin vector nearly coincides with the principal axis of maximum inertia. In the former case, elastic modes are excited into periodic motion, with consequent energy dissipation, even in the absence of an external force field. In the latter case the elastic mode excitation is caused by, and requires the presence of, the external gravity-gradient field. Order-of-magnitude estimates of the rate of energy dissipation and spin decay are presented for both cases for the Alouette 1 satellite. It is concluded that the mechanism studied is not the principal one causing the spin decay of Alouette 1 satellite.

iii

I T ABLE OF CONTENTS SYMBOLS INTRODUCTION TORQUE-FREE MOTION 1. 1 Rigid-Body Wobbling

1. 2 Behaviour of Çl,n Elastic System 1.3 Ca1culation of Energy Dissipation

1. 4 A-posteriori Check of 'Small'. Elastic Effect

Page No. v 1 2 2 6 7 13

II DISSIPATION ASSOCIATED WITH GRAVITY-GRADIENT 14

2. 1 Torque Acting on Satellite 2. 2 Elastic Motion

2.3 Numerical Value for Alouette 1 REFERENCES

FIGURES 1-13

APPENDIX A: Calculation of Normal Modes

iv 14 16 18 19 20

a a (A,B,C) E EI f f h J k K L M p

(J>

(P, Q, R) r T

u

SYMBOLS

satellite body radius acceleration of element principal moments of inertia constants (see Eqs. 1. 3. 10, 11) energy

effective bending stiffness of booms frequency

part of gravity force vector angular momentum vector an integral (see Eq. 1. 3.6) modulus of elliptic functions

complete elliptic integral of the first kind

semi-lengths of long and short antennae (see Fig. 5) torque (rolling moment)

bending moment elliptic frequency

period of elliptic functions

components of

W

-

in body-fixed axes position vector of element

kinetic energy of rigid body

kinetic energy associated with elastic motion strain energy

(X, Y, Z) body-axis coordinates

(x, y, z) trajectory-fixed coordinates (see Fig . . 9)

amplitudes of elliptic functions (see Eq. 1. 1. 4) an angle (see Fig. 1)

dissipation constant (see Eq. 1. 3. 13) angle of roU

orbital angular velocity angular velocity vector

INTROD UCTION

Since it was launched in 1962. the Alouette 1 satellite has exhibited a reduction in its rate of spin fr om the initial 1. 4 RPM to 0.15 RPM at the end of 1964. (See Fig. 1.) Several physical mechanisms have been considered which could contribute to this effect. These involve principally the interaction of the vehic1e with the Eartht _{s gravitational and} magnetic fields. resulting in loss of both spin energy and angular momenturn. Another mechanism that can occur without the participation of an external field is loss of energy at constant momentum if the vehic1e is not initially spinning about a principal axis.

This investigation was undertaken to study the m echanism and magnitude of the spin decay associated with hysteretic structural

energy dissipation during periodic flexural distortion of the long flexible antenna booms under two conditions:

(i) When there is no external field and the spin axis is not .coincident with a principal axis.

(ii) When spinning about the axis of maximum inertia in the presence of the gravity-gradient field.

The analysis as originally formulated allowed for a general orientation of the ~ and

i]

vectors in space. and for dynamic elastic motions. However. it was found that this generality served only to complicate the analysis without providing significant additional insight. Consequently. a quasi-statie elastic analysis was finally used in both cases. and moreover. the gravity-gradient case selected for treatment is an especially sirnple one. Nevertheless it retains all the essential features of the mechanism. and gives a result of the same order of magnitude as would be found for more complicated cases.

As to the solutions themselves. it should be pointed out that the energy method is not exact. involving as it does the averaging of periodic quantities over a cycle. and making a basic assumption about the dissipation (Eq. 1. 3.13). However. the use of the detailed equations of motion (a more complicated procedure) while avoiding the averaging process. would still require a basic assumption about the form of the damping force associated with hysteresis in the material. and hence would propably be no more accurate in the long run.

I TORQUE-FREE MOTION 1. 1 Rigid - Body Wobbling

When a general rigid body spins in the absence of external forces, the motion is a complex wobbling. The complete solution is given in Ref. 1 and is also treated in Ref. 2. The relevant results are reproduced in summary form below for convenience.

The principal moments of inertia are

A) B)t:

(1.1.1)

and the constant momentum and energy are h, T respectively. A critical condition is given by

(1. 1. 2)

which gives the h, T relation for steady spin about the Y axis, (the axis of intermediate inertia). Two different solutions are obtained for values of h greater and less than that of Eq. (1. 1. 2)

Case A

The initial condition is angular velocity

;)0'

lying in the XZ plane, at angle

d

to OX, see Fig. 2a, such th at

-Po

=Wocosd

Qo

=

0

Ro

=

-dJ

oSin6

The subsequent motion is given by the elliptic functions , where P =

0<

dn pt Q =

~sn

pt R =

Ó

cn pt (1.1.3) (1. 1. 4) C(A-C) B(A-B) (1. 1. 5)

ó

=

-ti)

oSind

and _P

₌

Wocosd (A-C) (A-B) (1. 1. 6)

BC

tan~ ~

.

)

k

₌

B-C

A-B

k is the modulus of the elliptic functions, defined by

u

=

[1>

d 9 , = F

<cp.

k) (1-k 2 sin 2 9)"2 IJ sn u

=

sin '/) (1.1.7) 1( K

=

F (

2"'

k) is the complete elliptic integral of

the first kind.

The functions sn pt, etc., are ,periodiç', with periodO'

to A-. = ~ thus

't'

2 '

pdi-and = F

(~,

k)

=

K (k)

~=

4K P

Note that as

k~O, K...,~

and

(p-'"

21"t/

p

1/

4f'

iCor:r..esp@nds

(1. 1. 8)

For the Alouette, the booms are long enough that the

satellite approximates a lamina, i. e.,

o

A

=

B

+

C

In that case, we have approximate~y:

/>

=

~

osin

cf

= -

ó

p

=

0( (1. 1. 9)

k =

tand!

B-C '

From the known values of P, Q, R in body axes, the orientation in inertial space can in principle be found by integration of the equations for Euler angle rates. This leads to complicated relationships, not needed for the present analysis, and hence not presented here.

Case B (Fig. 2b)

h2

<

2 B T,

The solution for case B is analytically very similar to that for A:

where P

=

0 Qo

=

R

=

0 P

₌

Q

₌

R

₌

p

W

ocosI

0 ini tial val ue s -Wosind ~cn pt

fl

sn pt

o

dn pt A(A-C) B(B-C) (A-C)(B-C) "AB k

=

A(A-B) C(B-C) For the approximation A ~ B

+

C we get

,d

,0

~

~cos5j

A • .. " " : " B-C

.

"

p

=

k

=

(1. 1. 10) (1. 1. 11) (1. 1. 12) (1. 1. 14)

.. 'I'

Variation of

Wo

and

cf

with T for const. h

For both cases(A)and(B) we find that

tan

f ::

-.

A 2AT - h 2 C h 2 - 2CT or withTcrit = h2_{/2B, and ?:=T/Tcrit '}

tanJ

=/

~.

A'L-

B

B-Ct'

We can also show that

h 2 '- 2CT 2AT - h 2

=

+

-AB BC or when A = B

+

C,

tij

2

=

2(A+C) T - h2 o AC

Alternatively, in terrns of

Z'

and T .

ent

tiJ

2

=

o 2 Tcrit AC

(A+C)l' -

B ) (1. 1. 15) (1. 1. 16) (1.1.17) (1. 1. 18) When

T

=

1, the values of k given by Eqs. (1.1. 6) and (1.1.13) m~t~çs equal; . whence it follows that kcrit = 1 and

d

crit = tan

-lJ

A .'

-C B-C

For A == B

+

C,

tan

Numerical Values for Alouette 1

-1~

.

J

s:c

(1. 1. 19)

The values for the Alóuette 1 satellite are: A == 681 slug. ft2

B == 604

C == A-B == 77 (actually, C == 82)

. I

It follows from (1. 1. 16) th at

tan

J

=

and from (1. 1. 18) that

JJ

_o

=

8.857- 7.85 .886 - .113

T

A 2T crit (1. 1. 20) 7.84 (1. 1. 21)

For

cl

to be real, we must have. 887

~

t

~

7. 85, and for

Wo

to be real,

?:~.

796.

Hence the limits on

?:'

corresponding to

cf

= 0 and 900 are respectively . 887 and 7. 85.

From Eq. (1. 1. 19),

~

_crit

=

48.70 •

Equations (1.1.20) and (1.1.21) are plotted on Fig. 3. If the satellite loses energy at an infinitesimal rate while h remains constant, its state would 'move' down these curves in the direction of decreasing

't .

1. 2 Behaviour of an Elastic System

When the system is elastic, the above solutions are not in general exactly applicable. One must then inc1ude the elastic degrees of freedom with their associated additional equations of motion, and the variable inertia terms to allow for deformation in the three main moment

equations. Such a complete system of equations would reveal all the . details of the mechanism of coupling of the rigid and elastic modes. Solutions to them, with dissipation terms present in the elastic degrees of freedom, would yield the rate of slowingdown of the motion (the secular terms).

In the case of Alouette I, the following argument shows that the conditions are such that a quasi-static analysis can give a good estimate of the dissipation rate. The various undamped normal modes of this satellite that can be excited in this motion (the rotational modes) have been calculated (Appendix 1) and found to have the shapes and frequencies shown in Fig. 4.

Now the spin rates we are interested in are 1. 4 RPM at launch, decreasing to very small values ultimately. It will be shown below that the period of the 'driving force' for bending of the booms is

is taken to correspond to 1.4 RPM, the driving frequency for the elastic

modes is

f

~

2.8 c.p.m. Hence the frequency ratio for the lowest mode is

J:...

~ ~

_

=

.49 f₁ 5.75

Reference to Fig. 8. 18a of Rei. 3 shows. that for this frequency ratio the dynamic amplification of the motion can be at most about 30%, and

that it will be less than that for all times after launch. It is concluded that

a quasi-static analysis wiU give a good approximation to the actual motion. This means that at each moment the instantaneous elastic configuration is the same as though the stress distribution associated with it had existed indefinitely.

We now make the further assumption that the rigid-body solutions for P, Q, R given in Sec. 1. 1 are valid for the elastic case as

weU. This implies that the elastic deformation and the energy associated with it are 'smaU' in sorne suitable s.~se. This assumption is checked a posteriori.

1. 3 Calculation of Energy Dissipation

As stated above, we neglect the relative elastic motion

and assume that the values of P, Q,. Rare those given in Sec. 1. 1. The

acc.eleration field corresponding to this motion produces 'inertia forces'.

on the rods, which in term induce deformation and strain energy. We

calculate these as follows:

The accelerations of mass elements on the long booms (1)

and the short booms (2) are indicated in Fig. 5. From. Rei. 3, Appendix A. 7 the expresf:jions for the accelerations are found to be

{ aX(O, 0, Z)

=

Z (PR

+

Q)

on (1):

•

a_y (0, 0, Z)

=

Z (QR P) ti { a

_x

(0, Y, 0)

=

Y (PQ - R) on (2): a_Z (O,Y, 0)

=

Y (QR

+

pr

(1. 3. 1)

These accelerations, and hence also the associated bending loads, increase linearly along the booms from zero at the origin.

From Eqs. 1. 1. 4, with the approximations of 1. 1. 9 we get for the accelerations:

ax (0, 0, Z)

=

Z(0<5 +Jp) cnpt dnpt

=

0

a_y (0,'0, Z) =

z</O'

+ O<pk 2 ) snpt cnpt = Z(-52 +

ex.

2k 2)snpt cnpt

a

X (0, Y, 0)

=

y(o<j5 +(fp) snpt dnpt

=

0

. 2

a

Z

(~,

y, 0)

=

Y(/t

-O<pk2 ) snpt cnpt

=

_Y(J2 +0(2k )snpt cnpt Hence the accelerations, the bending loads, and the

deflections alllie in the plane of the booms. The loads, with boom mass

f

per unit length for both booms are

on (1): on (2): where w 1 (Z)

=

gl(t)Z w 2 (Y) = g2(t) Y

.

gl (t)

=

f(QR - .P) . (1. 3. 2) (1.3.3)

The product snpt cnpt has twice the frequency of snpt, and hence the period of the loads on booms (1) and (2) is

~&.

Thus, as anticipated in Sec. 1. 2, the period of the fluctuating load is

(f /2.

The bending moment distributions obtained from Eqs. 1. 3. 2 are on (1): M 1 (Z) .= gl (t)

(L

1 2 _{(L1 + Z)} - Z) 3 6 _{(1. 3. 4)} M 2(Y) = g2(t) (

~

_ y)2 ( / 2 + Y) _{3 6} on (2):

The strain energy associated with these bending moments (applied quasi-statically) is for all four booms.

dZ ₌ J 1 _{EI gl} 2 _(t)

f~

M 2 2 J 2 2 U_{2 = 2} - - dY = g2 (t) 2EI EI Y=a

e

where J₁ =

f<l -

Z)4Jl

+

~)2

dZ 1 3 6 a

~

(1.3.6)

f(~

_ y)4 (f2

y2

J_{2 =} - ) dY 3 6 a

Thus the total strain energy is

(1.3.7)

J 1 J

2 For Alouette 1 the numerical values are :lEI = 2.79 x 10

9 ; EI = 1. 94 x 107 • The values of gl and g2 are different for the two cases (A) and (B) of Sec. 1. We shall treat here only the case

d<J

crit'

i. e. case (A), on the assumption that the initial angular error of the

spin vector of Alouette 1 was less than 48.7°. In that case from Eqs. 1. 4, and the differentiation rules for elliptic functions, we get

g1 2

f

2 2

=

<_([2 +0<.2 k2) sn2 pt en pt <1. 3.8)

and sim ilar Iy

It follows that (1. 3.9) where

.,e

=

f

~i

<-r

+0(2 - k2)2 +

f

2

~;

<t

+0(2 k2)2 (1. 3. 10). 9

-t

=

b 10 4

₊

_{brp<4 - b} 30(2lf2 (1.3.11) and bI

=

.f2 _EI (J ₁

+

J₂) b₂

=

k4b 1 b 3

=

k 2

E

(J 1 - J 2) EI or alternatively, using (1. 1. 5)

-t

=

Wo

4(b₁ sin4

d

+

b₂cos4

cf

-

_{b 3sin}2

~

cos2

d)

(1. 3.12) The function (snpt cnpt)2 has period

!

f'

and a maximum value of

-i-.

Thus it flillows. that the strain energy oscillates between 0 and

.

J:...&,

with

4 period

~

!

.,

_

.. '

,

as' iildicated on Fig. 6.

We now assume that the energy dissipated per strain cycle is a fixed fraction of the maximum strain energy, i. e.

L1

E

= -

e

U

= -

J:...

€

.

~

max 4

the time for this loss is

LI

t

=

-ft?

and the ave rage rate of loss is

É

=

~

=

_e~

Llt

p

e8-p

4 K (k)

(1.3.13)

where

t

(Wo'

d)

is given by Eq. (1. 3.10). Upon using Eq. (1.-1. 9) for

p

and (1. 1. 5) for 0( we get

•

.

(J

E = -

f

.

tiJ

cosd (1. 3.14)

,

4

K(k) 0

The tot al energy of the system is

E

=

T

+

Te

+

U (1.3.15)

where T is the energy associated with the rigid-body motion, given

by E q. (1. 1. 17) 1 2

T

=

(AC

tJ)

+

h2) (1. 3. 16)

and Te is the kinetic energy associated with the elastic vibration. (The frame of reference must be correctly chosen. so that the angular

momentum of the elastic motion is zero relative to it. Otherwise there

would be a cross-term in the energy equation containing both rigid-body and elastic velocities.) The total elastic energy is given nearly enough by

(1.3.17)

during any one cycle. in which the energy decreases only negligibly. :

Hence E :: T

+

U_max 1 :: 2(A+C) and

•

_•

(1. 3. 18) or. from (1. 3. 14.)-

-_~8p

::

AC

W

~

+

~ 0 0 4K A+C

~e

4 (l.3.19)

From Eq. (1. 3. 10). we get "

•

~::

(1. 3. 20) where

_•

0(

• (1. 3.21)

o

The derivative of

cf

can be found using Eqs. (1. 1. 15) and (1. 3.16). and gives 1

L

A2

cotá

+

C2

tan8]

Wo

•

(1. 3. 22) 11

It follows tha t

•

•

_C2

0(=

-w

₀ 1 A 2 _ C 2 cosd (1. 3. 23) •

•

A2

0=

-ti)

1 0 A 2 - C2

sin~

Thus it appear§. that the axial spin component P

=

0( dnpt actually increases as

Wo

decreases due to energy loss. Af ter substituting for

•

•

o<:Ö.o(..

t ..

in Eq. (1. 3. 20) weget

-8

=

fi)~3 ~

S

sin2J(4b 1A2

+

2b3C 2 ) - cos 2 J(4b 2C 2

₊

2b 3A2

)1

A - C

t

(1. 3. 24)

"

The differ.ential equation for

W

0 is finally obtained by substitutingofor

.e

and

-t

in Eq. (1. 3.18): ,

~

d)o[AC

+

~202

2' fSin2d(4b1A2

+

_{2b 3C}2 ) - cos 2J(4b_2C2

+

2b3A2~

lA

+C 4 (A .. C ) [

j

= -

~o

4

,

~ COS~(b1

sin4

d

+

_{b 2cos}4

d -

b₃sin2Jcos2J)

(1.3.25) We can also show. by eliminating T between (1. 1. 15) and (1. 1. 17) that

(' A 2W02- h2 tan "

=

h 2 _ C2W 2

'0

(1. '3.26)

which is needed in conjunction with (1. 3. 25) to eliminate

d .

•

Values

ofW

have been calculated from Eqs. (1. 30 25 and

1 . 0 ,

1. 3. 26). and

-iJ

0 is pJptted vs.

d)

0 on Fig. 7,# for the initial conditions ~

The

Wo.

t

WO

i

=

1. 4 cycles per min. at di

=

45 0

. relation is then obtained from

f

W4t

d~o

t'

6

t

= ~

il>ole.

IÀ).

and is plotted on Fig. 8.

From Fig. 8 it can be se en that the final state has been closely approached by the time

ê

t

=

10- 5 yrs. Thus, if

é

=

10- 4

for example (a reasonable value), the corresponding time is t = 10- 1 years. It appears that if the initia 1 conditions after launch were such that

wobbling occurred, then the process discussed above may well have been the controlling one in itially , and fairly rapidly (within a month say) reduced the energy and spin rate to those corresponding to steady

rotation about the x-axis.

1. 4 A-posteriori Check of 'SmalP Elastic Effect

If the strain energy U is small compared to the kinetic energy T, or if the deflections at the tips of the booms are small compared to

their lengths, we consider the neglect of distortional motions to be

justified in deterrnining the main motion P, Q, R.

The maximum strain energy is, from Eq. (1. 3. 9)

U

=..l-J,

max "4

and T is most conveniently given for this calculation by

Hence, using Eq. (1. 3.10) we get

U max _ .1

b

~

ttSin4d

+

_{b 2cos}4

cf

-

h.3

sin2[ cos 2

J"

T

-"8

A cos 2

d'

+

C s in2

d"

1;)

2 o

This ratio is evidently largest at or nearthe beginning of the motion when

;Jo

is largest. At the initial conditions given in Sec. 1. 3,

,;) 2

=

.0215 o sin

J

=

cos

$

=

/2'

k -. .878, k 2 = • 770 bl = 1. 265 x 104 b 2 = .592b 1 = .750 x 104 b₃ = . 960 x 104 13

1 ) Umax 1

'4

(b 1

+

b 2 . - b 3

=

(.0215)

₌

0.0187 T 8

_-

1 (A

+

C) 2-_

Thus U;ax

«

1, and the approxim a tion is justified.

II DISSIPA TION ASSOCIA TED WITH GRA VITY -GRADIENT

As the satellite rotates. the gravity-gradient field in general induces periodic bending of the booms. As in the analysis of part I, the frequency separation between the elastic normal modes and the rigid body motion justifies a quasi-statie analysi~, in which elastic equilibrium is assumed at all times.

The gravity-gradient effect is always present. The deflections of the booms produced by it are additive to those that occur in the torque -free motion studied in Sec. I. and hence the dissipation and spin decay ca1culated there are to that extent underestimated.

When one considers the most general case of a wobbling satellite. the exact computation of the gravity-gradient-induced bending moments is severely complicated by the involved nature of the

expressions for the Euler angles. which give the orientation in space relative to the local gravity vector. In order to obtain an estimate of the possible magnitude of this effect. without unnecessary complication. we consider the special case illustrated in Fig. 9.

It illustrates a simple rotation. in which the satellite spins about a horizontal axis that coincides with the satellite's X axis. In this condition. the gravity-gradient torque vector G is also on the X axis. and there is no precession or wobble. At the same time. the bending of the booms is a maximum. since they lie in a vertical plane. It is assum ed that

W

is enough larger than the orbital frequency tha.t the situation pictured holds constant for at least one revolution of the satellite about its spin axis.

(If

X were perpendicular to the orbital plane. it would remain so except for orbit-plane precession. )

2. 1 Torque Acting on Satellite

The torque-producing part of the gravity-field per unit mass (Ref. 4) is,given by (see Fig. 10)

fx

=

_Jl

2x

f

₌

-.ll2y (2.1.1)

Y

or

(2. 1. 2)

where

The first term of Eq. (2. 1. 2) is a radial force and can produce no torque on the satellite. and no bending moment on the booms . . Only the last term then is required for our ca1culations. It leads to the torque about OX. and since we get y = y cos<P - Z sin</' z = Z cos

+

+

Y

Sincp

.-(2.1.3) (2.1.4)

L

=

3.Q.?

[~

Sin2cp S<Y2 - Z2) dm

+

cos2tj>fXY dmJ

Since X Y Z are principal axes. the last term is zero. and

whence

3 1"'\2 . ~

L = 2~~ (C - B) sm2't' (2.1.5)

Thus the equation of motion for rotation is

••

i-

..n

2 (C - B) sin2

cp

=

A

1>

(2.1.6) w'hence

••

~=

~ 1"\ 2 2 ~~ • C-B A . sin2' "

_r

15

2.2 Elastic Motion

In the quasi-static approximation, we neglect the accelerations and inertia forces associated with relative elastic motion.

, Thus the effective force system for 'dynamic equilibrium' is as shown in Fig. 11. The terms 3-'l.2z are

th~.effective

gravity forces from Eq. (2.1.2), and th~ terms containing

cp

are the inertia forces.

The distribution of bending load on the two boom s is then

.-w1

=

Z~gdZ

+

3 mz sin1>fdZ on boom (1)

...

w2

=

Y,,~dY

3.o:z

cos tffdZ on boom (2) or, by virtue of Eqs. (2. 1. 4)

..

cf.

3 .

(

+

_ f l 2

sin 2

"J.)'Z

.• _ 2

'r

(<;-

i-1l.

2 sin

2r})Y

(2.2. 1)

The bending moments are therefore (cf. Eqs. 1. 3. 4)

I

2.t Z

on (1) MI

=

g3(t) (..(1 - Z)

(3

+

6)

on (2) M ₂

=

g4(t)

(~

- y)2

(~2

+

i)

(2.2.2)

••

=

f (

t.

+

i-

n3

sin

2~)

=

f

(1) -

i-

~

sin 2

rj;)

3 ,f"\2 .

A..

C-B

=

"'2

f~1.. sm

2't'(A

+

1) 3 2 . ~ C-B

=

2:fJL.

sm 2)"( A - I ) where

The strain energy associated with the above bending moments, (cf. Eqs. 1. 3.5) for all four booms, is

u

= _UI

₊

_{U 2} J1 2 (2.2.3) UI = g3 (t) EI J 2 2 U 2 = _EI g4 (t) U =

~~

_2..fl.. 4 _sin2 _{2CPC 1 (C-B}

₊

₁₎

₊

J 2 ( C-B 1j

-

-4 EI A EI A (2.2.4)

This periodic strain energy has the maximum value

Umax

=

'49 c f '.2 4 4

where

c

=

~

( C-B

+

1).,

+

J 2 (C-B _ 1)

EI A EI A

and a period corresponding to

</>=

-z- '

te which is approximately

L1

t

=

1 21'1;:

4

~

(2.2. 5)

Following the same line of argument as in Sec. 1. 3, we assume the average rate of energy loss to be

T

= _

e

Umax

~t

The energy rate equation is then

-

(')

•

= _

€

~ cf2.J2.4 2tJ

4 ~

-

•

E

=

T

+

U where 1 T

=

AW

2, 2

•

•

T

=

AWW

and from Eq. (2. 2.4) o

~

=

·: .c

f2.JL4 .

2 sin 2 1 . 2 cos

2tP.<f

=

9

cf

2..12. 4 sin 2rfCOS 2

tf·

{i)

"

Whence the mean value of U is

and hence or

-

.

,

U

=

0 ,. .tI

E

=

T

17 (2. 2. ö) (2.2.7) (2.2. 7) (2.2.8')

whence

(2.2.9)

2.3 Numerical Value for A10uette 1 A

=

681

If

=

.00212 .n..~ .001 rad/sec. c

=

2. 7 9 x 10 9 (1- . 7 7

4~

+

1. 94 x 107 (-1 -. 7 74)

=

59.7 x 107

•

. , . 9

-w

=

€

-2 7 59.7 x 10 681 =

~

5.65 x 10- 12 rad/sec. 2 = ~ 1. 70 x 10- 3 rpm /year

•

For

tE

=

10- 4, - fA)

=

1. 7 x 10- 7 rpm /yr. This mechanism of despin is thus seen to be very weak. In order to produce a 10ss of order 1 rpm per

year

€

would have to be of order 600. The maxim urn value of

e

is undoubted1y less than unity, so th is process must be ruled out as the principa1 cause of the Alouette spin reduction.

1 2 3 4 Synge, J. L. Griffith, B. A. Thomson, W. Etkin, B. Etkin, B. REFERENCES

Principles of Mechanics, Sec. 14.1, McGraw HilI Book Co., N. Y.

Space Dynamics, Sec. 5.11, J. Wileyand Sons, N.Y.

Dynamics of Flight, J. Wiley and Sons, N. Y.

Dynamics of Gravity-Oriented Orbiting Systems with Application to Passive

Stabilization. AIAA Jour. 2, 6, June, 1964.

APPENDIX A

Calculation of Normal Modes

The coordinate systePl and notation for the analysis is

shown in Fig. A. 1. The differeni-ial equations for the maximum

deflection of uniform elastic be.ams in a simple harmonic vibration are

where d4g _{/\ 4g} '1

₌

0

d~

d4 h d

'7

4

.À

4h = 0 ) 4 =

!W

2 B

f

= mass per unit length of each rod

2()

= circular frequency of vibration

B = bending stiffness, EI.

(A. 1)

The general solutions of (A. 1) are

= al

COShÀ~

+

a₂ sinh

AS

= bI cosh

A

7

+

_{b 2} sinhÀ

>;

We have now to apply sufficient conditions to evaluate the 10 unknowns,

al' .. ; . _{b 4}

,À,

Sb' The boup.dary conditions available are: at

~

=

0: g(O) = -a Sb

(1)

g'(O)

₌

_{- Bb} (2)

at

'7

= 0: h(O) = a Bb (3 )

h' (0) = _Bb (4) (A.3)

g<it)

=

1 (5) (normalized mo~e)

gil

<.t:.)

=

0 (6) (zero bending moment)

h"(~)

= 0

h'

,,(,t;)

= 0

(8) (zero bending moment)

(9) (zero shear) The above yield nine of the required ten conditions . The last is

provided by the equation of motion of the centre body. Figure A. 2 shows

the forces and moments applied to the body when g"(O), g'''(O), h"(O), h"'(O) are all

>

0 and when the displacement is a maximum.

The equation of motion then follows as

(A.4) where MI

=

B g"(O) M₂

=

B h"(O) VI

=

B g"'(O) V 2

=

B h"'(O) whence Eq. (A. 4) becomes

h"(O) - g"(O)

+

a g"'(O) - a h"'(O)

₌

(A. 3)(10)

Equations A. 3, (1) to (10) then provide the necessary conditions. From A. 3, ( 1 ). . . . (4) and (6). . .. (9), we get

al Dll a2 D12 a3 al =

- -

= = a -eb Dl eb Dl ' e _b eb a4 1 a2 - ₌ eb eb b 1 D 21 b 2 D 22 b 3 b 1 =

- -

= = a -eb D 2 eb D 2 eb eb b4 1 b2 = eb

.Ä

eb 21

where

ba

sin').L

1

+

~

cos

À!i>

(COShÀ.t

₁

+

COsÀ.t:'>

(A.6)

(cosh:l

Rl

+

cos

A

~l)

(a cos

A.t;.

+

~

sin

À

1'1)

)

a· b·

Substitution of the above solutions for _1 , .=..l.. into boundary condition (5)

then yields eb explicitly, eb Qb

eb ::

(:..!.

cosh;l

~

+

a2

sinhl~

+

a3

COSilel

+

él.4

sin.À~

1

'

-1

eb Qb eb e b ; t

and substitution in copdition (10) leads to the characteristic equation for the eigen values.""-, i' This turns out to be:

/

Ib.,1 2

=

A(l) + B

<A)

(A.8)

2af

where

A

2 a 2 +1 l2 2

coshA.

~1

sinA.

Rl

+ 2 sinhA.

~

+ : t a -1

sinhÀ.ticos~é'l

A<À) = _.a~a

_________

-=--_:--::;....--~~!::...--_ _ _ _ _ _

( 1

+

cosh,A

el

cosÄ.

t

1)

B~)

"

-The normal fre~uencies are then obtained by solving Eq. (A. 8) numerically

for the values Ai that satisfy it. The mode shapes are then found boy substituting for theA i into Eqs. A.5, 6, 7, and thence for the Ai and Bj

into (A. 2). " a

₌

1. 5 ft.

1

1

=

73. 5 ft.

~

=

36.0 ft. Ib

=

7. 66 slug ft2

.f

=

.00212 slug/ft. EI

=

350 lb. ft. 2

These data, and the above equations, were used to calculate the frequencies and shapes of the modes shown in Fig. 4.

c 1.4 0

I

0 0 0 1.2 0 0 ₀ RP}! 0 0 1.0 0 0 0 0 0 -v 0.8 0 0.6 tY'I 0 0 o 00«0 0 0.4 0 o 0 0.2 -~ - o n 0

,

100 2CO JOO 400 500 600 700 800

G /

~

y

y

FIG. 2. M (a) h2

>

2BT

x

1

(b) h2

<

2BT F SPIN

•

&

80

0 7.8$

60

0

40

0

20°

cJ

o

=

Wo

~

2

~'jt

o

I I

o

I

I

I

7.0

8.0

1.0

I

2.0

3.0

4.0

5.0

6.0

0.887

Energy Ratio.

'r

,..

Wo

10

B

6

tt

2.

o

f,

=5.75

cpm

f3

=

10.15

cpm

~Z

f

~

~

~_2

y

~Z

f

2

=

I tt.29

cpm

~y

FIG. 4. ROTATIONAL NORMAL MODES OF ALOUETTE 1

><

f'f

-"...

N

..

0

..

0

' -) (

ctS

•

>-

..

0

~ )C (0

N

/

,...

~

\~

Cl ~ ~ >-I

>-m

~ Z 0 >-I ~

<t:

~ Çi1 ~ ~ U U

<t:

.

l i j

.

ö

-

>-I

-..

~

\>-u

=

b

sn

2

_pt.

_cn

2

_pt

u

b

-

_'t

t

-(~)

RPMIrR 1.6 1.2

0.8

o.tt

X 10" 18 16 Ilt 12 10 8 6 4 2

o

I RPM/YR

for

€ =IO-S

1·0

t

WO MIN &=

0

\.\ \.2.

I.r

FIG. 7. RATE OF SPIN DECAY AT CONST. MOMENTUM

0.2.

0.4

0.6 0.8 1.0 1.2 I.~ X 10-1"

et

t YEARS

Horizontal

_p_'aL_e

_____

0

_,c~ar=!d==,==h==.==Q==3>

-

...

x,X

FIG. 9. CONFIGURATION STUDlED FOR GRAVITY-INDUCED SPIN DECAY

~---.

y

.

---~~

y

3.n.2Zd~

y

3n'zdm

z

>-z 0 ... E-< <!! E-< 0 Z Çl Z <!! Cf.l ~ E-< <!! Z ... Çl ti

a

~

I I )-. N 0 u

"

11 C""

,.,..

... <ti ti ... rx. - : - -

-..