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KUNOE SPIN DECAY OF A CLASS OF SATELLITES 'iLIEGTUlGBOlJWQ\~lI01\11ttK u u CAUSED BY SOLAR RADIA['ION
PART 11: ARBI~RARY ATTITUDE OF THE SPIN AXIS
by
D. B. Cherchas
'.'
SPIN DECAY OF A CLASS OF SATELLITES CAUSED BY SOLAR RADIATION
PART- 11: ARBITRARY ATTITUDE OF THE SPIN AXIS
by
D. B. Cherchas
Manuscript received April,
1969.
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ACKNOWLEDGEMENT
The material presented herein is based on work done under the super-vision of Dr. P. C. Rughes of the Institute for Aerospace Studies at the University of Toronto. This research was sponsored in part by the Air Force of Scientific Research, Office of Aerospace Research, United States Air Force under AFOSR Grant No. AFOSR-68-l490, and in part by the National Research Council under Grant No. 67-3.
Data on the spin rate, spin axis atti tudlè.~~and percent sun histories for the Alouette 1 and Explorer XX were obtainedl from SPAR Aero space Company of Canada and the Defense Research Telecommunications Establishment (D.R.T.E) of D.R.B •
•
SUMMARY
The analysis studies the marmer in which solar radiation heating and pressure acting on the long flexible antermae of a spin stabilized satellite can cause the spin rate to significantly decrease. The problem is formulated by finding a differential equation for the shape of a boom antenna under heating
effects and inertial farces and deriving the expression for torque acting on the satellite due to salar radiation pressure on booms of known shape. By assuming small boom deflections and incorporating, for the first time, tpe spin axis attitude and salar exposure histories, the spin rate versus time histories of two example satellites, Alouette 1 and Explorer XX, are found, and excellent agreement obtained with actual flight data .
1. 2.
3.
4.
5.
6.
TABLE OF CONTENTS NOTATION INTRODOCTIONCALCULATION OF TORQUES THAT INFLUENCE BOOM SHAPE
2.1 Torques Due to Inertia1 Forces 2.2 Torques Due to Solar Heating
2.3 Differentia1 Equation for Boom Shape SOLAR RADIATION TORQUE
SOLurION OF EQUATIONS OF SHAPE AND MOTION
4.1 Simp1ifying Assumptions
4.2 Details of Torque Calcu1ations 4.3 Ca1culations App1ied to A10uette 1 4.4 Ca1culations App1ied to Exp10rer XX
DISCUSSION C ONCLUS I ONS REFERENCES APPENDIX A B C D j-"E F G FIGURES 1-16 1 1 2
4
7
8
9
9
1315
16 1619
20D d d e E G !,J.,~
X,Y,Z
!* ,.si*
,.!?-x, y, z I ,I ,I x Y z I I P L M p r s b.S n Te
T a: NarAT10l\TAll symbols not listed below are defined in the text. unit vector in direction of thermal torque
position vectcr of point on boom,written in satellite co-ordinate system
diameter of boom
emissivity of boom surface modulus of rigidity
shear modulus of boom
axes and parameters of a co-ordinate system fixed in space
axes and parameters of a co-ordinate system fixed to the satellite
non-twisting axes located at a boom cross-section
moments of inertia of satellite about satellite axes
!
*
,
.si*,
~*bending moment of inertia of boom x-section polar moment of inertia of boom cross section
length of boom
torque acting on satellite due to solar radiation pressure
on one boom
solar radiation pressure reflectivity of boom
distance along boom axis measured from boom base
unit vector, normal to boom surface, pointing into the boom unit vector tangent to boom axis
temperature at surface of boom thermal time constant
absorptivity of boom
angle of a diameter lin~ at any given cross-section measured
counter-clockwise from .si or
.si*
)', n
E p.!.l
'. T 1*
T wangle to diameter line having greatest positive temperature difference across it
two angles which make up ~; )' is fixed for any given diameter,
n
varies with the amount of boom twisting
unit vector in direction of solar radiation
unit vector in direction of force due to solar radiation reflection
solar radiation vector
thermal curvature
mass per unit length of boom
moment at a point on the boom due to inertial forces
component of !l along boom axis
thermal moment at a point on the boom
total moment acting at a point on the boom
total bending moment acting at a point on the boom
Euler angles relating
!*,
~*, ~ to!'
~, ~•
"
1. INTRODUCTION
This note continues the study of the interaction of a spinning
sate-llite with long flexible booms (Fig.l) with solar radiation. It was shown in
Ref.l that this interaction can result in a torque on the satellite which, when
taken over a long time period, will effect a significant change in spin rate.
The original work on this mechanism was prompted by the strange
spin-down of the Alouette 1, l~nched in 1962, which exhibited a spin decay of
approximately one rpm per year. The study (Ref.l) was done by B. Etkin and P.
C. Hughes at the Institute for Aerospace Studies (University of Toronto) in
1965. In this analysis, the spin axis of the satellite was held a~ ninety
degrees to the solar radiation vector and radiation was assumed to fall on the
satellite twenty-four hours per day; then an exposure factor was introd~ced to
fit the theoretical spin decay with the data. This fitting did not invalidate
the theory since it was the only theoretical decay which, with rational fitting,
could be made to coincide in magnitude and curve shape with the actllal decay. The work described herein is aimed at removing the exposMTe factor,
i.e. at finding the boom shape and satellite torque using the actual solar
exposure history ~nd for an arbitrary orientation of the spin axis relative to
the solar radiation vector. Therefore, much of the original work is left as it
was and an application of it is made to a more general case.
A differential equation for the shape of a boom subject to heating
effects and inertial forces is found first. Then, on the assumption that the
shape solution could be found, an analysis is made to find the torque acting on the satellite due to the interception of the solar radiation momentum flux by
the booms. The equations for boom shape and torque which have evolved to this
point in the study are complicated integral-differential equations. In order
to make the problem tractable, the equations are simplified by assuming small
boom deflections. The simplified equations are solved for an arbitrary direction
of spin axis and the solution applied to two example satellites using their
spin axis attitude and solar exposure histories. The Canadian Alouette 1 and
U.S. Explorer XX are the example satellites chosen.
2. CALCULA'tION OF TORQ,UES THAT INFLUENCE BOOM SHAPE
T~e following material will derive the torques which determine the
boom shape and the differential equation for shape. No attempt will be made
to solve the equation in this section.
In this analysis, both the boom bending and the mass of the booms are
taken to be small enough so that the satellite centre of mass is and stays at
the origin of the
!*, J*,
~ or body axes system (Fig.2). Also the booms aretaken to start at the origin of the above axis system. These simplifying
assumptions ca~ qe justified by the small boom mass/body mass ratio and the
body radius/boom length ratio for the example satellite studied. For example,
for Alouette 1,
boom mass
body maRS 0.0188 (.--for longest boom)
body radius
boom length = 0.04 (for shortest boom)
The i*, ~*, ~* system has its ~*, ~* axes coincide~t with the unbent
boom directio~s for the Alouette 1 configuration (Fig.3) and for small boom
deflections can be taken to be the princip~l axes of that configuration. For
the Explorer XX, the body axes are as shown in Fig.4 and also for small boom
deflections can be taken to be principal axes of that satelli te. The assumption
of the body axes to be principal axes is made throughout the remainder of this
section and in all other sections. The principal moments of inertia are assumed
to be constant. The~,~, ~ or space axes system is a system ofaxes fiNed in
space with the ~ axis pointing ~owards the sun. The origin cf the body axes is
at the origi~ of the space axes and the body axes are located with respect to
the space axes by three Euler a~gles and their derivatives as shown in Fig.2. The boom shape is determined by the torques which exist along the
length of the boom. In this case the important torques are the thermal torque
due to a temperature difference across the boom cross-section, the inertial
torque due to the acceleration of the boom and the elastic torque due to the
structural rigidity of the boom itself. The next two sections will derive
ex-pressions for the inertial torq e and the thermal torque.
2.1 Torques Due to Inertial Forces
The first moment to be evaluated is ~l' the moment at an arbitrary
point
P
due to inertial forces on the boom. The motion of the satellite aboutits center of mass is the only motion we shall consider in finding the inertial
force on a boom element. Referring to Fig.2, we shall find the acceleration of
an element of the boom at P' in order to obtain the inertial force at that
point. The acceleraticn of P in t,he ~,
;1,
~ ref'erence frame can be calculatedfr om
a = a* + 2w x V* + w x ~ + ~ x(~ x ~) (II.l)
a is the acceleration in the i, j , k reference frame and V*and a* are the velocity
and acceleration in the i*,
i*,
k*reference frame respectively.- By writing theright hand side of Eq. (11.1) in-body axes system co-ordinates, ~ will be written
in the axes set in which we want to work. The right hand side terms are written
in the
!*,
J*, ~ system as follows:d
=
xi* + y;i* + zk*.
•
•V-'I.-
=
xi* + y/J.* + zk*..
y~-*
..
a*=
xi* + + zk* w w i* + w j* + wk* x- y-z-.
w = ~ x-i* + ~ y-j* + ~ z-k*where wx,w ,wz' ~x'~y and ~z are as given in Appendix A. Therefore, expanding
a
=
[
x + 2 (w Z -w y) + (~ z-~ y)+(w W - W2X_w2x +w W z)] i* Y z Y z x y y z x y -+ [y
+ 2 (w x-wz)
+(t
x-~ z)~(w w X _w2y _w2y + W W z)] J* z x zx
x y x z y z . + LZ + 2 (wY
-w x)+(t y +t
x)+(w W x_w2z_w2z + W W y)] k* x Y x Y xz x Y yz-The inertial reaction force on the boom due to this acceleration is
f -P!:;
f i* + fj* + f k*
x-
r
z-where p
=
mass per unit length of boom, and!
is the inertial force per unitlength at a pomnt P on the boom. The point P' will be identified by co-ordinates
x', y', x'. If these co-ordinates are used in f, the force per unit length at
P', identified as
!', is obtained. The distance vector between Pand
P'
is~sep = (x' - x)!* + (y' - y)J* + (z' - z)~
The cross produc~ of the inertial force at P' with this distance vector gives
the moment at P due to the inertial force per unit length at P' ie.
T. d x f' -ln -sep = [(y'-y)f' (z'-z) f' ] i* z Y + [(z'-z)f' - (x'-x) f' ] J* x z + [(x'-x)f' - (y'-y) f' ] k* Y x
T. is the moment at P due ~o the inertial force per unit length at P'.
Mult--ln
iplying by ds' will give the moment at P due to inertial force on an element
at P' and integration of this torque for P' going from P to the end of the
boom will give the total inertial moment at P, i.e.
2:1 (s)
~
lL
2:in(s' ,s) ds' (II.2)Later it will qe necessary to know the component of ~l along the boom axis.
~his component can be fcund by taking the dot prcduct of ~~ with a unit vector
in the direction of the boom axis at the particular point ln question. The
unit vector tangent to the boom is
T
=
d
.2: i* +
ds
-and the axial component of ~l at P is
d d
.J.
.* + ~ k*ds
.J.
ds-r*'
=
T • T·1 -1
-Thus3 we have found the inertial torque and its axial component at P.
2.2 Torques Due to Solar Heating
The other torque which acts to bend the boom is athermal torque
generated by a temperature difference across the boom cross section as effected by solar heating. We will derive an expression of this torque in the material
below.
Reference 1 proposes th at a first order differential equation closely
describes the temperature difference established across a diameter, ie.
where
è
~e
(t ) + ~e
(t,s, 1)dt
'
s,1
T
e
1
e
2.
(see Fig.3)Referring to Fig.5,
e ,
e~ are temperatures of elements~l and ~2 respectively, T is a thermal time cOnstant, ~l is the rate of heat absorption from solarradiation per unit area at 68 1 and q2 is the same quantity at ~2; cl is a
constant based on the mass density and specific heat of the boom.
Now we define E as the solar radiation vector, E
= -
j E, whereE = solar energy flux in thermal units. Then q 1 and q'2 iiiay be -wri tten as
<1'1
=
P (ex ~. 68 1) where p(x) = x x>
O.=
0 x<
O. now 'a.. - ~2 = P (0: E· 6S ) - p( 0: E • ~ ) ~.L - - 1 - - 2 and, since ~2 = -ts
l ' =ex
.§.. 68 1Thus the differential equation for the temperature difference across a diameter of a given cross section can be written as
d
A 0. (t ) 6 El (t,s,r)~ D Ö ,s,r + T
(11. 3)
It wi11 be usefu1 to expressL~ in the i*, j*, k* system. 681 can
easi1y be wri tten in terms of the
1,;,"
r,
!-
system (see Fig.5) ,
T,
T,
rareaxes at the cross-sectioQ which move with the boom but do not rotate as the
boom twists and if P was at the base of the boom,
'1,
T,
'k
wou1d be coincidentwi th
.?:.
*, .J.*,.!?
We can write the inward normal to-.6Sî
as68 1
A ~ '~A-k
COSf-'
..J -
Sluf-' _To write ~ in the i*, j*, k* system we need the trans~IDrmation between the
untwisted i,
T,
'k
system and-the i*, j* , k*' .system. This transformation canbe effected by two rotations. ReferrIng to Fig.6 and noting that
-1 dx TJ 1 cos ds -1 dz TJ 3
=
cos ds cosTJ 1=
cosCP1 cos(90 -TJ ) 3we can easi1y write matrices to describe the rotations. The first rotation
abqut the
!
axis by (90-TJ3) i.e.sinTJ3 0 -cosTJ
3 0 sinÇ3 cosTJ3
0 1 0 -cost) -cost)
cosTJ
3 0 sinTJ3 -sint) -sint) sinTJ3
-The next rotation is about the k axis by - CP1 and transforming 68 1 from the
previous stage lIS 1 = = 68 2 is just the
o
o
o
1sir$ cosTJ3_ COSCP1 + cost)
sint) cosTJ 3 sinCP1
-
cost)-
sinÇ3 sinTJ3 m i*+ m j* + m k* x- J z-negative of this.5
sint) cosTJ 3 -cost) (11.4 ) isIt is important to note that ~ is not constant for a given diameter
through changes in the Euler angles. It will be helpful to write
~ (s,t)
=
/
+n
(s,t)n=o@s=o
n
is due to torque acting along the boom axis and boom is not subjected to a torque along its axis.an axial component which is, as found earlier,
Using
dn
dsT!
=.!l . T
=
~ ',-I GI p/ is the value of ~ if the
Only the inertial moment has
and ignoring any time lag in twisting (see Ap~ed~ix G) we can write
n
Therefore, sJ
* (s, t)=
.Tl ds 0 GI~
(s,t)= /
+
r
o
p T*
ds 1 GI pEq. (11.3) is the differential equation for t:,
e
across a given diameter at across-section. We are interested in the maximum t:,
e
which will'occur at agiven point P. The value of ~ which gives t:,
e
max is denoted ~ and determinesthe direction of thermal torque. The thermal curvature vector is then KTQ
where
K
=c
t:,e
T
4
max (11. 5)and D is a unit vector at right angles to the diameter of~ in the plane of
the cross-section of interest (see Fig.7). In the
r,
r,
~ set it isQ
=
sin~
r -
cos~
r
and transforming to the ~*, ~*, k* axis set
cos r:J>1 cos 1)3 cos ~ sill<pl sin ~
D
=
sin 1>1 cos 1)3 cos ~ cos</Jl sin ~
-sin1)3 cos ~
D i* + D .* + D k*
x - y.J. z
For small boom deflections it can be shown that (Ref.4) where M EI 1 r
=
K r=
radius of curvature K = curvatu;re M=
bending momentThus the thermal bending moment is given by
T
=
EI-2
The tota1 torque at a point
P
isT
= inertia1 torque
+ thermal torquewhere ~1 and ~2 are given by (11.2) and (11.6) respective1y. 2.3 Differential Equation for Boom Shape
(n.6)
We now wish to use the tota1 torque va1ue at
P
to find a differantia1 equation for boom shape. A curve in space can be described by three unitvectors which change their direction as one moves a10ng the curve (e.g. Ref.
5,
pp. 311, 312), Fig.8 i1lustrates these vectors. It is true thatbut K can be written in
K
Tb EI
terms of the bending moment bending moment
EI
By taking ~ x
!
we have (see Fig.8)therefore T X T = K N EI EI and T X
T
=CIS
EI 7 i.e. (11. 7) !Equation (11.7) is the differential equation for the boom shape.
3.
SOLAR RADIATION TORQUENow that the differential equation for boom shape has been derived,
the next step is the calculation of the torque on the satellite due to solar
radiation flux being intercepted by the satellite booms. Due to the time lag
in the temperature difference across a boom cross-section as effected by
solar radiation heating (see Eq. (11.3)), there will be a time lag in the boom
bending relative to the solar radiation input history. It is expected that this
time lag in boom bending will lead to a net solar radiation torque per spin
cycle on the satellite.
The force due to solar radiation pressure can be calculated from
(Ref.l). where dF A := ds
.
d ex p.
cos v dF R4
ds d 2 - 3,
. .
r.
p.
cos vdF is the differential element of force due to solar radiation momentum flux,
p, striking a boom element with lengths ds and diameter d. ex and rare the
absorptivity and reflectivity respectively of the boom and
21
and.
Q2
are unitvectors in the direction of the force due to absorption and ~he force due to
reflection respectively as shown in the sketch below,
9.
2T
E
~l
-
-
si.
and2.
2 can be found from
T x
C!:.
x.J. )
5 :=-2
I
!x,Si.I
and cosv can be found fromThe torque on the satellite due to radiation forces on all elements of the boom is
L L
!i=
J
~=J
o
0where ~ = x!* ~ YJ* + z~ is the position vector of a point on the boom written
in the satellite co-ordinate system. For the tot al torque on the satellite a
similar integration would have to be performed for each satellite boom and the sum of the torques from each boom taken.
The equations of motion for the satellite can now be written. They
are L n I w - w w (I - 1 ) =I
J
dM :. x x y Z Y Z Xl i=l 0 n L • (I - 1 ) =IJ
1 w - w w dM . Y Y Z x Z X yl i=l 0 IW
- w w (I - 1 )=t
.
J
L dM . Z Z x y x Y Zl i=l 0for an n-boomed satellite.
4.
SOLUTION OF EQUAT-IONS OF SHAPE AND MOTION4.1 Simplifying Assumptions
In this section the equations of shape and motion are simplified to
tractable farm and solved. In thëssimplification the following assumptions are
made:
(i )
e
and ~ and ~ are quasi-steady. Forces more significantthan solar radiation pressure influence the attitude of the spin axis, i.e. gravity gradient and magnetic moment effects. Appendix D gives the spin axis attitude histories
for the two example satellites chosen. It is clear from
these histories that for short intervals of timeC9 and ~
can be assumed const~t. Figures 15 and 16 show the slow
spin decay of Alouette 1 and Explorer XX respectively.
(ii) Terms which are second orde~ or higher in boom deflection
or slope are negligible in comparison to first or zero
orderi3terms provided j;hat other factors do not cause terms
in which the higher order terms appear to fall in the same magnitude range as.lower order terms.
(iii) The component of torque along the boom axis is neglected. i.e., the boom is considered to be untwisted. For boom
deflections which are small, the axial component ofT~l' the
inertial torque, will be small and ~2 contributes no ~orque
along the boom axis.
As a result of assumptions (ii) and (iii) it can be shown that, referring to Eq. (11.4) cos epl 1 ~ sin ep
=
0 1 cos TJ3
o
sin TJ3
= 1i.e. ~l has the same components in the][, ~ ~axes system as it has in the
bodyaxes set. This means that for purposes of calculating the thermal torque
on the boom, the boom is considered to lie along its nominal line.
In connection with assumption (ii), it should be noted that the small
deflection assumption is only useful for purposes of simplifying the boom shape
and solar radiation equations if the boom lies nominally along i*, j* or k*.
For any booms which do not lie nominally along one of these axes thë procëdure
is to change the !*, j*, ~* system temporarily so that the simplification can
be made and then transform the shape and torque values to the principal
!*, J*,
.!?
system.This procedure is particularly easy for our example satellites since
the booms are all nominally in the same plane normal to the spin axis and the
co-ordinate transformation is a rotation about the k* axis until the i* axis
is collinear with the nominal boom line. Since we are analyzing the motion
about the k* axis only, k* is the only axis which has to remain as a principal
axis and the return transformation would not have to be made. This procedure
is implicit in the analysis which follows where any boom analyzed is taken
nominally along the i*,
Using these assumptions, the equations for findin~ boom shape become:
(i) ~hermal bending equation
C
3 [-cos~sin~sinep + cose cosep cos~ cos~ - sin~sine cosep ]
C
3
F( 8,c:p,
~, ~) (IV.l)(ii) total torque (acting on boom) components T x ==
9
L T YpJ[
T Z == s PJ[
L sz
(s'-s)~ '2 s'(z'-z)ds' + EI KTsinp ~ '2 (' ·2(iii) boom shape equations
o
EI
d2y T ds2 zEI
d2z ds2 = T YS~nce
dx/ds = constant and(~)s=o=l,
it is clear that xtl.ons.
(IV.3)
s under these
assump-The boom shape equations are solved for the steady-state y and z deflection values by assuming the following forms for y and z:
for 8,~,7/J constant. That is
z 0
y _ ~2 Y
This choice of solution forms is compatible with the sinusoidally varying thermal
torque component leading to be~ding in the ffi*, j* plane and the constant component
leading to bending out of this plane. For further discussion of this reasoning
see section
5.
become:
Using the assumed solution forms, the equations for ~inging boom shape
(i) thermal bending
~ (~8)
+i
(~
8)=
C3
F(8,~,7/J,~
)(ii) boom torques
T
=
0 X T=
Y LJ
'2 p 7/J s L T Z PJ
[_~2
s'(y'-y)+ 2?Î;2Y'(s'-s)]dS:!: slJ.
(IV.5)
(Iv.6)
(Hi) boom shape EI d 2 x 0
2
== ds d2 EI J T (IV.7) ds2 z EI d 2 z ds2 -T yIt should be noted that T depends on, in the inertial term, and effect's only
the z deflection and similarly with Tand the y deflection. In other words
z
the y and z deflections are uncoupled when second order and higher deflection
terms are dropped.
The validity of the assumed solutioa form is established in the
following way. The thermal bending equation is solved (see Appendix F) for
given ~,8,~ and ~ to find the ~ value
(€)
which gives 68max from which KTcan be calculated. First values for Tand TZ are found by neglecting the
integral terms in the boom torque equations i.e. using only the thermal
torque terms. These values for Tand T .are used in the boom shape
equa-tions to give the shape for the bgom. Th~t shape is used to evaluate the
inertial, i. e. integral, terms in T and T . These, added to the thermal
torque terms, give new values for T
Y
and TZ• These new values are used tofind a new shape. The iterative prgcess or
fin~ing
better values of TandT from each new shape, and then finding a better shape, is repeated
u~til
s~cceeding shapes agree within a specified limit. The above described
con-vergence on a shape is performed for values of ~ varying between 0 and 2 H.
Th~s a history of y and z for one satellite revolution is obtained and y is
indeed found to vary sinusoidally with ~ and z is found to be independent of
~(see Fig.9). Figure 10 shows a typical boom shape. The assumed forms for
y and z in this way are shown to be solutions of Eqs. (IV.l), (IV.2) and
(IV.3) and since they are compatible with the radiation input which drives
the shape, we take them to be the desired steady-state solutions for shape.
Therefore, for given e,~ and ~, the y and z values at any point in
the revolution could be found by converging to a shape at some value of ~
and then establishing Y(s) from
Y(s) == sin(~ y( s, 7jJ)
+
'l/J )o
and Z(s) from the found values. Y(sj, 'l/J and Z(s) are all that are needed
o
to give y and z at any value of ~ or s.
Now that the boom shape is solved for, the torque about the k* axis
due to solar radiation pressure can be foudn. Af ter reduction to first order
for'm, (the
.!?
torque component is somewhat simplér_'and for· 8, ~ constant theonly significant equation of motion, that about the k* axis becomes, for
+ where,
4
3 drp..
"/J ! silÎllil.! L dapJ
(-a 22x + a12y) COSV dxo
dz ds(Iv.8)
In the above, a12, a22,a~2 are components of the transformation matrix which expresses the space set 1n the body set. From Appendix A,
a
12 cos"/Jsin~ + cos8 C08~ sin"/J
a
22 sin?jJsi~ + cos8cos~cos"/J
a
32 - sin8cos~
M then is the torque about the ~* axis as effected by solar radiation strtking one boom at a given "/J. Therefore, knowing the boom shape for given 8, ~, "/J it is possible to find the torque about the k* axis from one
boom by performing the integration from 0 to L.
The average torque acting in ~he k* direction per revolution for an n-boomed satellite (all booms nominally in-the same plane) could then be found by integrating the k* torque from "/J = 0 to "/J
=
2~ for each boom, dividingthe result by 2v and sumffiing these average torques for each boom, ie •.
I
Mz
L
;1T
Î
(:
~
i
(1/J)
d1/J
i=l av i=l 0
4.2 getails of Torque Calculations
The actual solution procedure used is now described.
Since exact integration to find the boom shape, to find the:'k* torque for given 8,~,"/J and boom shape and to find the average torque per revo-lution is difficult, the integrations are done numerically.
To find the boom shape for a given e,~,~ and ~ the thermal torque
magnitude and direction at any point along the boom is found from KT and ~.
rhe thermal values of Tand TZ are used in Eqs. (IV.7) and since the Tyand
TZ
~alues,
if they are~hermal
contributions only, will not be functions of sEqs. (IV.7) can be integrated as
2 s Z
=
2" 1 2 -T S Y EIto find the boom shape from thermal torque only. From the thermal torque boom
shape a first approximation to the inertial torques on the boom can be found
from Eqs. (rv.6) where tbe integral terms are the inertial torque components.
The sum of these inertial torques and the thermal torques give a new boom shape
through integration of Eqs. (rV.7) using the inertial and thermal total torque
components as Tand T z . The new shape will give a better approximation to the
inertial
torque~,
hence an improved shape will re sult from using these torquesand the process is continued until there is no appreciable change in shape
through one iteration. The integration of Eqs. (IV.7) is done numerically by
dividing the boom into one-half foot segments. Once a boom shape has been
fOQ~d in this manner for a particular ~ the shape at any other ~, for the
same! e,~ can be found by using Eqs. (rv.4).
The value of ~ is determined by first finding.the value of ~ where
the thermal deflection
o~
the~*
direction is zero when~
=
O. This isaccom-plished, for the boom nominally along the ~* axis, by finding ~ such that
;i
x ~*-
~*The value of ~ to satisfy the above relation gives the ~ value where y is zero
changing from negative to positive if ~
=
O. An additional term is added tothis value of ~ to account for the time laî in boom bending which would be
needed when ~
f
O. The term added is tan- wT. The reason for adding this termis as follows. In the case where the radiation vector is in the boom plane the
differential equation for the curvature is (see Ref.l)
OKT 1
w
"è!iifJ
+ T KT=
K c os ~ which has, as its steady-state periodic solutionKT
=
KT cos (~- ~) 0 KT KT 0 .Jl + w2 T2P
tan -1 w'] w -if;In other words, the boom bending lags the radiation input which gives that
bend.ing by tan-lwT. In the more general case with which we are dealing, where
O,~
f
0, the radiation input leading to bending in the ~*,;i*
plane also variessinusoidally with ?/J and the thermal bending along the i* axis might be expected to lag the radiation input, for
~
f
0, by tan- l wT. ThusJ
x i*[ ('/1 f . * ) + tan- l ';1 T ]
- 'I' or -~
=
I~ x ~*I
'I'(rv.9)
wo~ld be a sensible proposal to find ?/J. This theoretical value of ?/J is . _
checked finding the thermal shape as aOfunction of ?/J for different 8,
8
and ~ values. Thermal sinusoidal y variation has a phase lag closely agreeing with that found from Eq. (IV.9) (see Figs.ll and 12). The thermal and actual shapes have the same frequency and phase (see Fig.9) therefore ?/Jo found from Eq. (IV.9)can e used for the lag of the actual shape.
M is found at a giveQ point in the satellite spin revolution by numericallyZintegrating Eq.
(rv.8).
~he integration to find the average torque per revolution is performed by dividing the 0 to 2~ range of ~ equivalent to one revolution into sixty equal sectors, finding M at these sixty different points in the revolution, summing these and dividiftg by sixty. This gives the average torque along the k* axis through one revolution as contributed by one boom. To account for the-satellite torque from a boom diametrically opposite to a boom whose average torque per revolution is known, the latter boorn's torque need only be multiplied by two. This is tmpossible because of the sinusoidal nature of boom shape with respect to ?/J.I~ the above described manner, the average torque per revolution for any boom can be found. Thus by summing the average torques per revolution for all the booms the total k* average torque per revolution can be fourrl.
•
?/J,8
of a
4.3
Once the average torque per revolut~n for a given boom configuration, and ~ values can be found, all that is required to calculàte the despin given satellite is the 8, ~ and percent sun history of the sateltite.
Calculations Applied to Alouette 1
The Alouette 1 is the first example chosen. Thee,~ history is found by transforming the given spin axis attitude history (Ref.2) from co-declination
and right ascension values into the co-ordinate values of a unit vector in
the spin axis direction in the ~, ~,
!
set. Appendix D depicts the co-declination and right ascension history of Alouette 1. The steps in the transformation are to write the spin axis unit vector in the ~', ~',!'
system from the right as-cension and co-declination, transform from this set to the ~o' ~o'!a
set and then to the ~, ~,!
set (see Fig.13). The co-declination is approximated by an absolute value sine function.codec
=
'2
TT s~n• ' ( TTxt40
(IV.10)~he right ascension values are approximately constant over a co-declination nodding from 0 to ~/2 to 0 so the constant right ascension value can easily be selected knowing which part of the approximately two hundred day cycle the
spin axis is in. Appendix C gives the orbit characteristics of Alouette 1. The sateèlite is not immersed in solar radiation for t~enty-four
hours per day and Fig.15 shows the percent sunlight history for the Alouette
1 which is used in computing the spin decay. The percent sun was approximated
by fitting the given data with straight lines for the one hundred percent sun
plateaux and with sine curves for the variations below the one hundred percent
levels. The change in ~ over a five day interval is computed by holding ~,
e
,cp
and percent sun constant over that interval and thus a spin rate versustime from 1aunch plot is made by dividing the flight time into five day
periods (sec Fig.l5).
4.4
Ca1culations Applied to ExplorerXX
The Explorer
XX
despin history is solved for in a manner exactlythe same as that used for the Alouette 1 except for the implementation of the
spin axis attitude and percent sun histories shown in Fig.l6. The spin axis
attitude history was approximated by fitting the curve given in Appendix D
with straight lines and the percent sun was fol~owed in the same manner as
for the Alouette 1. Five day intervals were again chosen as periods of
constant e,cp,~ and percent sun. Fig. 16 shows the despin plot for the
Ex-plorer
XX.
5. DISCUSSION
Ma~y assumptions and approximations are made in the foregoing
analysis. A summary and explanation of these practical measures which
have not been sufficiently discussed in previous sections is presented here.
The booms for Alouette 1 are actually Stored Tubular Extendible
Merribers rnanufactured by the SPAR Aerospace Company of Canada. This analysis
assumes the boom to be seamless in structure, however, it uses the flexural
stiffness (EI) for bending analysis and modulus of rigidity (G) for twmsting
analysis characteristic of the STEM booms. The error in not including the
overlap at the seam is introduced when the axis of bending is taken to be in
the direction of applied bending moment and the extent of this flaw remains
to be investigated. ~he assumptions made in using the first order
differen-tial equation for KT advanced in Ref.l are clearly stated in that reference.
'I'here is; however, a deviation from the intended application of thi.s equation
encountered in the way it is used in this work. The thermal bending equation'
proposed in Ref. l i s derived for the radiation vector lying in the boom
plane i.e., one diameter, always the same one, at any cross section always
has an extremum temperature difference across it compared to other diameters.
In our case, where the spin axis direct ion is arbitrary, the diameter with
the extremum 69 at a cross-section changes as the boom rotates. So the
relationship between 6 8max and KT in our case and in Ref.l is different
primarily because of the difference in the temperature distribution around
a
cross-section caused by a changing extremum 6 8 diameter. Equation (II.5)relates 68max and KT Dy a slightly higher constant than used in Ref.l. In sections 2 and
3
it was recognized that since there wilL be acomponent of inertial torque lying along the boom axis, the boom is subject
to a twisting torque. The angle of any diameter line relative to a
non-twisting reference line on a cross-section will therefore change as the axial component of torque changes. In computing the amount of twist,
n ,
due tothe axial torque, the time taken for the boom to twist tö its equilibrium
frequency of boom twisting is found to be forty-five to three hundred and
forty-five times larger than the greatest expected frequency of torque application
for the example satellites and therefore does not seriously jeopardize the
strength of the formulation in sections 2 and3 anilljnsection
4,
where thetwist-ing is neglected, has no influence.
The formula used to relate applied bending moment to curvature at
any point on the boom i .e.,
M 1
=
- KEI p
is yalid if the material being bent obeys Rooke's law and can be easily used if
the value of I, the bending moment of inertia of the cross-section does not
change during the bending. These conditions exist for small curvatures which
is the case for the satellite booms studied.
The most obvious approximations of the analysis are made in section
4
where the equations of shape and motion are solved. These approximations arediscussed below.
In order to make the equations formulated in 2 and 3 tractable, terms
which are second order or higher in deflection or slope values are dropped in
comparison to first order terms and boom twisting is not considered.
In section
4
,
when solving the equations for boom shape, a solutionform in which the z deflection is assumed to be_independent of ~ and the y
deflection sinusoidally depending on ~ is taken to be the boom shape solution.
The closeness between a pure dinusoid and the y deflection versus ~ is shown
in Fig.9. For purposes of comparison, the amplitude of the compared sine
wave is slightly greater than the amplitude of the deflection curve but has
exactly the same frequency and phase as the deflection curve.
A solution for a boom shape is found by converging to that shape
through a n~ber of iterations. Some criterion has to be established to
de-cide when the shape has converged closely enough to the solution shape. If
succeeding y and z deflection values at the end of a boom agree to wit~in
one-hundredth foot, th en the shape is accepted for an Alouette 1 boom. Succeeding
deflection values must agree to within two-hundredths for an Explorer XX beam.
The numerical solution for boom shape divides the boom into one half
foot segments but the numerical evalvation of satellite torque about the k*
axis divides the boom into one foot segments. The one foot intevvals
are-considered to be small enough to give an accurate value for the torque from a
boom of known shape since that shape has small curvature values everywhere on
the boom and a very slight change in deflection stope will occur over: a one
foot interval. Evaluation of the despin over a five day interval where the k*
torque is calculated with a one foot interval and with a one-half foot interval
shows agreement to the fourth decimal place in the value of ~ ~ in radians/
second.
In finding the attitude history of Alouette 1 in terms of 8 ,~ an
approximation is made of the co-declination and right ascension history shown
in Appendix~. The approximation is made by using Eq. (IV.10) to find the
co-declination, and the right ascension is chosen from 5.6, 1.71, 4.18, 0.384,
or 2.79 radians depending on which time region of the five co-declination
noddings cycle is wanted. The co-declination and right ascension found in this
WBnner give a reasonable approximation to the true history in Appendix D and
are certainly close enough for comparing the predicted spin decay with the
flight data. The errors in co-declination will not flignificantly change the
spin decay rate. The right ascension values listed earlier are mean values for
any given nod~ing from the earth's axis and therefore, on the average, are close
to the actual values. It should be noted that the spin axis attitude history
in Appendix D is :n..ot really accepted as an accurate account of attitude due to the lack of rneasur'ing devices on board Alouette 1 for spin axis attitude
measurements. The percent sun history of Alouette 1 given in Fig.15 is well
approximated by the sine waves and plateaux used.
The E-xplorer XX attitude history and percent sun history are well
approximated by the fittings made.
The earth's orbit about the sun is taken to be circular which develops
errors in rWo ways.
(i) the variation in radiation intensity as the earth~to-sun distance
varies is neglected
(ii) the rate at which the earth-to-sun line sweeps around the ecliptic
plane is taken to be constant which is not true for an ecliptic
path
Neither of these approximations seriously effect the validity of the match
between the predicted spin decay and flight data.
ll'he main results of this study are the boom shape and satellite
de-spin. Fligures 9, 10, I I and 12 depict the boom shape resul ts . In Fig. 9
it is clear that the z deflection remains constant throughout a revolution and
the y deflection varies sinusoidally with~. Both of these features are
ex-pected results. 'rlhe shape is determined, for the most part, by the temperature
gradients across the boom due to solar radiation heating, so the constant z
and the periodic y should be explainable from the way in which the radiation
ef~ects a shape. Referring to Fig. 14 a study of the radiation vector-boom
a~is orientation history will explain the boom shape variation with~. To
I
analyze hOVl the curvature, as effected by solar ;radiatio~, varies wi th ~ we
analyze how the component of € at right angles to the boom axis varies wit h
~ since it is the only componënt which causes heating. The first step is to
break the radiation vector into three components, one along" the boom axis, one
at ninety degrees to the boom axis and the boom plane and the third at ninety
dègrees to the boom axis but in the boom plane (see Fig .14) . The two components
which effect a ~
e
across a diameter of a cross-section and hence effect abending are the two at ninety degrees to the boom axis. Taking
-Ij;
= 0 we shalJ..now see how the two deflection values, y and z, should vary with ~ from thermal
effects. The radiation component normal to the boom axis multiplied by a
constant gives ~ 8 as stated in Eq. (11.3) and" mul~iplied by further constants
I'\:T and the thermal bending moment as stated in Eqs. (II.5) and
(n.6)respecti-vely. The thermal bending moment divided by EI and multiplied by the eosine of
the angle between the ~ 8max diameter line and the boom plane gives the y
z curvature i.e., referring to Fig.14.
and we see that, taking
t
=
0, the y thermal deflec~ion should be sinusoidalwith ~ and the z thermal deflection constant. For ~
f
0, the.y and z thermaldeflections will behave the same as ~ varies as they did for ~ =
°
except thatthe magnitude of the bending will be different and the bending will lag the
radiation input. Thus we have shown that the thermal boom deflection should
be sinusoidal in the y direction and constant in the z direction. This forms
the basis for the assumptioq of the solution forms given by Eq. (Iv.4).
The other major result, the satellite despin, is plotted in Figs.
15 and 16. It is clear that the despin rate changes through the plot. The
three influences on despin rate aside from the satellite configuration are
the fraction of ap orbit that the satellite is in sun-light, the spin axis
attitude and the spin rate.
Upon examining the despin plots for both satellites, it is clear
that the plateaux in the curve i.e., periods of small spin rate decay are
coincident with low percent sun and a near parallel condition between the
spin axis and solar vector.
6.
-CONCLUSIONSThe described work analyzes ~,I, ~.i how solar radiation interacts wi th
the long flexible booms of a long-boomed satel~ite to iqduce a torque which
effectively changes the spin rate of the satellites. The study is done for
an arbitrary orientation of the spin axis.
The results for first order boom deflections show that the boom
deflection parallel to the nominal boom plane is approximately sinusoidal with
the rotation angle aboMt the spin axis and that the boom deflection out of
the nominal boom plane is approximately constant with respect to the rotation
angle. The rate of despin has been shown to depend on the attitMde of the
spin axis, the fraction of the orbit that is sunlit, the spin rate and the
satellite configuration.
The predicted spin rate history matches well with the flight data
for Alouette 1 aRd Explorer XX which is reason to believe that the theory is
a good one.
With further verification, the program used to predict the spin
decay could be applied to other satellites of this class to foresee serious
spin~downs or spin-ups before launching. Spin-up behaviour analyzed for an
arbitrary spin axis attitude to the solar radiation vector could be
investi-gated by using higher initial spin rates.
1. Etkin, B. Hughes, P.C. 2. Graham, J. D. 3. Goldstein, H.
1
4.
Timoshenko,s.
Young,D. H.
5. Soko1nikoff,I.
S. Redheffer,:a.
M. 6. King-He1e, D.G. Ei1een, Quinn REF'ERENCES"Spin Decay of a C1ass of Satellites Caused by Solar Radiation". Ul'TAS Report 107, Ju1y 1965. Conàensed and extended in "Explanation e:f the Anomalous Spin Behaviour of Sate11ites with Long F1exib1e Antennae". JSR VoL
4 ,
No.9, Sept. 1967."Dynamics of Sate11ites Having Long F1exib1e Extend-ib 1e Member s" • De Havi11and Aircraft Co. Toronto. "C1assica1 Mechanics". Addison-Wes1ey, 1965. "E1ements of Steength of Materials" • D. Van Nostrand, 1962.
"Mathematics of Physics and Modern Engineering". McGraw-Hi11, 1958.
"Tab1e of Earth Sate11ites Launched in 1957-1966" R.A.E. Technica1 Report 67039, February 1967.
APPENDIX A: TIME RATE OF CHANGE OF w
To find the time rate of change of angular velocfuty,
w,
we shall needto describe the Euler angle rotations by three orthogonal transformations.
Referring to Fig.2 these transformations are
! '
,J.,
~
t
! l ' .J.l ,~l
about the k axis!l' .J.l ,
~l
ct
!2' ~,~2
ab out the !l axis!2 .J.2 ~2
t
i* .J.* k* ab out the ~2 axiswhere the matrices of rotation are
cos cp sin cp 0 1 0 0
-sin cos 0 D 0 cos 8 sin 8 C
0 0 1 0 -sin 8 cos 8
cos?jJ sin?jJ 0
-sin 7jJ cos 7jJ
o
Bo
o
1The matrix which transforms from
l,
.J., ~ to l*, 1*, k* is the product of thesethree matrices i.e.,
A
=
BCD ~s shown in Ref.3: c7jJccp -c8scps7jJ c7jJscp + c8ccps7jJ s7jJs8 A -s?jJccp -c8scpc?jJ -s7jJscp + c8ccpc?jJ c7jJs8 s8scp -s8ccp c8 all a12 aU = a 2l a22 a23 a 3l a32 a33 where,c7jJ = cos7jJ c8 cos8 ccp coscp
s7jJ simjJ s8 sin8 scp sincp Al
The angular veloei ty can be wri tten as
.
.
.
~=
ep"!.+
8~1+
?jJ"!.2 As shown in Ref,3: w=
Sc ?jJ+
cps ?jJs 8 x.
w=
és
'l/J
+
epc ?jJs 8 Y.
w=
epc 8+
?jJ z and since w=
(~)bOdY
+
~~
=
0 w=
w i* +W
J.*
+w
k* x - yz--
..
w=
[ -?jJ8s?jJ + 8c?jJ + ep8s?jJc8 + cp?fJs8c?jJ + eps?jJc8 ] i* ",
.,.
",
.
--[ -?jJ8c?fJ - 8s?jJ+
ep8e7fJe8 - cp?fJs8s?jJ+
epe?jJs8 ].J.*
•• I . [ - 8cps8+
epe8+
?jJ] ~S
i"*+S
i* +S
k* x- y""- z-A2APPENDIX B: MOMENTS OF INERT IA OF SATELLITE We wish to know I1' I
2 and I~, the rotationa1 moments of inertiaLbbout
i*, j*, k*, the three bodyaxes of thé satellite (Fig.2). Referring to the
sketëh bë1ow, the moment of inertia bout an axis of a body is in general given by
where the integration is over the entire body.
The moments of inertia of the sate11ite are thus found from n
L
i=l nL
i=l n I3=L
PiJ
i=l 0o
L. 1. L. 1. 2(Yi
+ z.) 2 ds + I 1. sl 2 2 (z. + x.) ds + I 1. 1. s2 2 2 (x. + y.) ds + I 1. 1. s3for a sate11ite with no booms. are the moments of inertia of the sate11ite body about the ~t ~*, and k* axes respective1y.
APPENDIX C: SATELLITE ORBIT CHARACTERISTICS
(i) Alouette 1 (from Ref.6)
orbital inclination nodal period semi-major axis perigee height apogee height eccentricity argument of perigee
(ii) Explorer XX (from Ref.6)
orbital inclination nodal period semi-major axis perigee height apogee height eccentricity argument of perigee 80.460 105.42 min. 7392 km. 1998 km. 1032 km. 0.002
8
0 79.870 103.97 min. 7323 km. 871 km. 1018 km. 0.01 2890APPENDIX D: SPIN AXIS AT.rITUDE HISTORY FOR ALOUErTE 1 2~ 191(
1jO
rr
160 210 1 150 \ / 220 1ÇKl 140"
/ 230 ll.O "- Assumed satellitte permant magnetic moment alon~he ~ spin axis =- !i microw~r metres -lO-A ouettä I - observe91 ---10--Al~l1ette ca. cLlla.ted I
-100 ~ -QJ -70 gGO -d '60 I c:: o _ .~ 80 '50 32b :§ 0 '40 / ~ 0 100 \ \ 30 ,~~~ 330 / u 340 I
L
' 20 ~~\.r:p 350 10 Ihg\'lt OSç:J,J::rj t-'
APPENDIX E
Satelli te Data
Alouette 1 Explorer XX Alouette 11
Boom lengths
2
booms @75
ft.2
booms @60
ft.2
booms @120
ft.2
boems @37.5
ft.4
booms @30
ft.2
booms @37.5
ft.Boom material steel Berylium-Copper Berylium-Copper
BoOf!1 diameter cl:
0.95
inches0.5
inches0.5
inchesAbsorptivity,
a
0·9
0.45
-
0.45
Reflectivit;y-, P!>
0.1
0.55
0.55
Mass/length, P
0.00212
slugs0.000441
slugs0.000441
slugsft. ft. ft.
Bending Stiffness EI
351
lbs - ft.2
15.49
lb _ ft.2
.
15. 9
4
lb. - ft.2
Time constant, T
17
sec.2.7
sec.2.7
sec.KT
o
.0064
ft.-l*
same as EKp~orer XXPolar moment of
2
2
2
inertia I m
681
slugs-ft.84.4
slugs. -ft.577
slugs-ft.**
'"
Solar pressure
.95
x
10-7
psf=-95
x
1
0-7
psf.
95
x
10-7
psfconstant, p
APPENDIX F: SOLUTION OF FIRST ORDER THERMAL BENDING EQ,UATION
;
It is neèessary to khow the steady-state solution of Eq. (IV.l) i.e.,
[-cos~sin~sin~ + cosecos~cos~cos~ - sin~sin8cos~]
Changing the independent variable from t to ~ we have
or
now by holding e,~ and ~ constant and co~sidering ~ to be constant the same
equation becomes
C
~ cos~sin~sin ~
C+
~ cosecos~cos~ cos~
(E.l)C
;:
i
.
sinecos~sir$
The steady-state solution of Eq. (E.l) is
(E.2)
+
C 3
t
_ _-
3r
...J_"--_ _ _ cos~sin~--=-_ _
+T
C~3
cosecos~cos~_ _ _ _ _ _
'
_
IJ
cos ~(T-q; )2 + 1
The above equation is solved to find the ~ value for a maximum 68
and'the value of 68
max itself by setting
~
(68)equal to zero andfinding which of the two~
values which satisfy this condition cause~2 ~~~
to be less than zero. , That ~ value (~) substituted into Eq. (E.2) ,gives 68 • The value .of KT can be found from Eq. (11.5). The product of C3 and C4~the
value 'K' given in Appendix E. A slight increase in'K' was necessary to fit the data; seeAPPENDIX G: TIME LAG IN BOOM TWISTING
There will be a time interval between the time when a twisting torque
is applied on the boom and when the boom reaches and stays at an angle of twist
determined by that torque. If the natural torsional vibration frequency of
the boom is much greater than the frequency at which the twisting torque is
applied then the time lag can be neglected. Analyzing the motion of a
canti-lever boom and using the following notation
P
d mass per unit volume
I = free rotational moment of inertia of boom I' fixed end rotational moment of inertia of boom
1'
-P polar moment of inertia of boom
GIp L
T twis~ing to~qae
n
angle of twistwe can say th at for a boom which is fixed at one end, straight, of constant
cross-section and has only its own structural rigidity giving a twisting torque
I' d 2
n
+l1t
n
=
0 dt2 pr d2n
+l1t
n
0=
dt2 I'I' ~! I however we shall use I for I' and the frequency calculated will be
3
lower than the true natural frequency. The natural frequency of torsional
vibration is Remembering that I P I W R
~~
~
radians!second!+
element of cross-sectional area
we can say for a hollow shaft of constant cr0ss-section and therefore, I so dm=
J
· 2 P d L r dA G1 P LPdLI P LdA Pd L1 P GComputing the value of ~ for Alouette 1, Alouette 2 and Explorer XX.
(i) Alouette 1 = (ii) Alouette 2 G 12 x 10
6
psi. L 75 ft. P d = 0.283 lb/irt 36
12 x 10 x 144 142.5 ·radians/second= 22.7
cycles/second.G
=
6
x 106
psi (for copper)L = 120 ft. P d = 0.322 lb/ in3. 6 x 10 x 144 (120)2 0.322 x 1728 x 32.174 58.9 radians/second
=
9.38 cycles/sec0nd (iii) Explorer XXG
=
6
x 106
psi (for copper)L = 60 ft.
P
6
x 10 x 144 x 0.322 x 172832.174
117.8 radians/second
=
18.7 cycles/second.The frequency ofaxial torque application is the same as the spin fre-quency of the satellite since the boom shape is periodic. Taking four cycles per minute or one-fifteenth of a cycle per second as the maximum frequency of torque application we can, by comparing this frequency to the ~ values for Alouette 1, Alouette 2 and Explorer XX, neglect the time lag in boom twisting for these
satel1ites.
x
FIG. 2.
k
Z
CO-ORDINATE SYSTEMS
y
1
1,
j,
k fixed in space
FIG.
3.
PRINCIPAL AND BODY AXES OF ALOUEl'TE 1o.
1
I
xp
o.
IIzp
FIG.
4.
PRINCIPAL AND BODY AXES OF EXPLORER XX
••
J
Ixp
••
1
,....
k
~--~--+-~---
-.
J
":"""
I