The rotational dynamics of space station Mir​
​By Michael Foale
NASA Johnson Space Center, Houston
Updated by Paul Abbott
Abstract
The Russian space station Mir is a solid body with unequal principal moments of inertia. Free body rotations initially begun about the middle principle axis, in general develop into rotations about the other pair of axes, in the course of free motion. This motion is problematic for a station that has lost attitude control, and that must maintain its solar arrays directed toward the sun, to maintain electrical power. This paper uses methods of Euler to calculate the resultant motion of the Mir in inertial space, with given total angular momentum and specified initial angular velocities about the solid body axes. The results are used to illustrate the difficulty of setting up an initial free rotation suitable to provide constant illumination of the Mir solar arrays.

Electric Power and Attitude Control on Station Mir

During the summer of 1997, the Russian space station Mir suffered a number of problems, resulting in the total loss of attitude control. A specific inertial attitude is essential for the Mir to keep its solar arrays illuminated by the sun. A rapid energy shortfall, and total loss of power to the station, can result if the arrays are not in a suitable orientation. The usual system of attitude control utilizes sets of momentum wheels, called gyrodynes, spun at high speeds, to provide reactive torques to keep the Mir inertially stable. When an attitude control failure occurs, the angular momentum stored in the gyrodynes is passed into the Mir body as a whole, as the gyrodynes slowly brake their rotation. So even if the Mir looses attitude control in a good orientation with no initial rotation rates, rates build up over the course of hours in all Mir axes, as the gyrodynes spin down.
The immediate problem presented to the crew of Mir, is the establishment of a free body rotation of the Mir to provide sufficient solar illumination of the arrays. With some electrical energy restored, contact with the ground controllers can be established, and workarounds developed to replace the attitude control system box that had failed. The solution developed using the Soyuz transport spacecraft was very crude, but effective. The Soyuz docked to the Mir has a power system independent of the Mir, enabling it to be powered even when the station is not. The Soyuz reaction control jets were manually fired to first arrest the resultant rotation caused by the gyrodynes, and then subsequently to reorient Mir, followed finally by a spin of Mir. However, because of the tiny roll moment arm of the Soyuz jets about the Mir long axis, rotation about this axis could not be influenced, complicating the resultant rotation dynamics of Mir.
Much time was spent by the crew considering the effects of initial rotations on the resultant rotation dynamics of Mir. This paper presents the equations and visualization model that permit the correct rotation impulse strategy to be applied.

Definition of the Euler Transformation Matrices

To transform motions about the Mir body axes into the inertial reference frame, define three Euler angles ϕ, θ, ψ as arguments of three rotation matrices. These matrices will be combined into one matrix to provide for a general multiaxis rotation. The Euler matrices
L
ϕ
,
L
θ
,
L
ψ
rotate from the inertial unprimed frame
x
to the primed frame
′
x
use the form
′
x
=
L
ϕ
x
where
x
is a vector in inertial space. First rotate about the inertial space axis
z
, to get to the first primed frame,
L
ϕ
=RotationMatrix[-ϕ,{0,0,1}]
cos(ϕ)
sin(ϕ)
0
-sin(ϕ)
cos(ϕ)
0
0
0
1
At this point
x
=
L
ϕ
x
. Now we rotate about the primed frame
y
axis, by
θ
,
L
θ
=RotationMatrix[-θ,{0,1,0}]
cos(θ)
0
-sin(θ)
0
1
0
sin(θ)
0
cos(θ)
At this point
_
x
=
L
θ
x
. Now rotate about the double primed frame axis about the primed frame
z
axis (note, not
x
, since this is the Euler system),
L
ψ
=RotationMatrix[-ψ,{0,0,1}]
cos(ψ)
sin(ψ)
0
-sin(ψ)
cos(ψ)
0
0
0
1
Finally
_
_
x
=
L
ψ
_
x
. Define
L
ψθϕ
to be the combined solid body rotation matrix
L=
L
ψ
L
θ
L
ϕ
:
L
ψθϕ
=
L
ψ
.
L
θ
.
L
ϕ
cos(θ)cos(ψ)cos(ϕ)-sin(ψ)sin(ϕ)
cos(θ)cos(ψ)sin(ϕ)+sin(ψ)cos(ϕ)
sin(θ)(-cos(ψ))
-cos(θ)sin(ψ)cos(ϕ)-cos(ψ)sin(ϕ)
cos(ψ)cos(ϕ)-cos(θ)sin(ψ)sin(ϕ)
sin(θ)sin(ψ)
sin(θ)cos(ϕ)
sin(θ)sin(ϕ)
cos(θ)
L
ψθϕ
EulerMatrix[{-ψ,-θ,-ϕ}]
True
and
T
L
be its inverse such that
T
L
ψθϕ
=
T
L
ϕ
T
L
θ
T
L
ψ
:
T
L
ψθϕ

-1
L
ψθϕ
EulerMatrix[{ϕ,θ,ψ}]//Simplify
True

Definition of Inertia Tensor and Euler's Equations

The angular momentum
J
of a collection of particles with momentum
α
P
for each particle α , with a position vector
α
(
r
)
relative to some origin in inertial space is
J

Σ
α
α
r
⋀
α
P
where
α
P
is given by
α
P
=
α
m
Ω
^
α
r
. Expanding the summation, and using Einstein summation indices, we can define the inertia tensor
I
ij
that combines with the body angular momentum vector ω to give the net angular momentum
J
i
.
J
i

Σ
α
α
m
(α)
2
r
δ
i,j
-
r
i
r
j
ω
j

I
ij
ω
j
I
ij

Σ
α
α
m
(α)
2
r
δ
i,j
-
r
i
r
j

I
ij
ω
j
In an arbitrary body coordinate system, for an irregular body like the Mir,
I
ij
is not diagonal. However a set of coordinate axes can be found in the solid body that correspond to the eigenvectors of
I
ij
, given by
I
ω
=λ
ω
=
J
. The eigenvalues λ of
I
are the three principal moments of inertia
A,B,C
about the three orthogonal eigenvectors defined as
ω
1
,
ω
2
,
ω
3
. If the body has axes of symmetry, then the
ω
's are in the same direction as the axes of symmetry. For arbitrary directions of
J
, the equation
J
=I
ω
implies that
ω
will not be in the same direction as
J
. The coupling of the components
ω
1
,
ω
2
,
ω
3
to each other through the principle moments of inertia
A,B,C
is given by the Euler equations, if no external forces are acting. The general Euler equations are
(C-B)
ω
2
ω
3
+A

ω
1
0​​(A-C)
ω
1
ω
3
+B

ω
2
=0​​(B-A)
ω
1
ω
2
+C

ω
3
=0
The ω's are evaluated in the body axes of the rotating frame. The total angular momentum is
J=
2
(A
ω
1
)
+
2
(B
ω
2
)
+
2
(C
ω
3
)
and
Ω
1
=
2
J
-
2
C
2
n
-
2
B
2
Ω
2
A
for initial values for
ω
1
,
ω
2
,
ω
3
to be
Ω
1
,
Ω
2
,n
respectively. Obtaining solutions for the ω's in the fixed body frame, these must be related to the Euler angles ϕ, θ, ψ of the body to inertial transformation matrices
L
ψ
,
L
θ
,
L
ϕ
.
L
ψ
.
L
θ
.
0
0
′
ϕ
+
0
′
θ
0
+
0
0
′
ψ

ω
1
(t)
ω
2
(t)
ω
3
(t)
′
θ
sin(ψ)-sin(θ)cos(ψ)
′
ϕ
′
θ
cos(ψ)+sin(θ)sin(ψ)
′
ϕ
cos(θ)
′
ϕ
+
′
ψ

ω
1
(t)
ω
2
(t)
ω
3
(t)
These three additional equations can be solved simultaneously with the Euler equations to give solutions for the six parameters
ω
1
,
ω
2
,
ω
3
, ϕ, θ, ψ as a function of time.

The Solution of Euler's Equations

Dynamic Equations

The Euler equations:
eqn1=
∂
ω
1
(t)
∂t
+(-ℬ)
ω
2
(t)
ω
3
(t)0;
eqn2=ℬ
∂
ω
2
(t)
∂t
+(-)
ω
1
(t)
ω
3
(t)0;
eqn3=
∂
ω
3
(t)
∂t
+(ℬ-)
ω
1
(t)
ω
2
(t)0;
The total angular momentum and initial conditions:
initialconditions={t0,
ω
1
(t)
Ω
1
,
ω
2
(t)
Ω
2
,
ω
3
(t)n};
rule=Total
2
(
ω
1
(t))
+
2
(ℬ
ω
2
(t))
+
2
(
ω
3
(t))
;
omega1rule=
Ω
1

-
2
ℬ
2
Ω
2
-
2

2
n
+
2


;
The Euler angles ϕ, θ, ψ using the transformation matrices
L
ψ
,
L
θ
,
L
ϕ
give the equations
L
ψ
.
L
θ
.
0
0
′
ϕ
+
0
′
θ
0
+
0
0
′
ψ

ω
1
(t)
ω
2
(t)
ω
3
(t)
′
θ
sin(ψ)-sin(θ)cos(ψ)
′
ϕ
′
θ
cos(ψ)+sin(θ)sin(ψ)
′
ϕ
cos(θ)
′
ϕ
+
′
ψ

ω
1
(t)
ω
2
(t)
ω
3
(t)
Explicitly:
eqn4=
′
θ
(t)sin(ψ(t))-sin(θ(t))cos(ψ(t))
′
ϕ
(t)
ω
1
(t);​​eqn5=
′
θ
(t)cos(ψ(t))+sin(θ(t))sin(ψ(t))
′
ϕ
(t)
ω
2
(t);​​eqn6=cos(θ(t))
′
ϕ
(t)+
′
ψ
(t)
ω
3
(t);
For display purposes, the initial Euler angles will be chosen so that output plots of the motion have the net angular momentum directed along the
z
-axis.
The Euler angles
ψ
0
,
θ
0
,
ϕ
0
at
t=0
are given by the transformation:
L
ψθϕ
.
0
0

==

Ω
1
ℬ
Ω
2
n
-sin(θ)cos(ψ)
sin(θ)sin(ψ)
cos(θ)


Ω
1
Ω
2
ℬ
n
Solve 2 equations for 2 initial angles, the angle ϕ being indeterminate:
Solve[{sin(
θ
0
)sin(
ψ
0
)
Ω
2
ℬ,cos(
θ
0
)n},{
θ
0
,
ψ
0
}]/.initialconditions

θ
0

-1
tan
n

,-
2

-
2
n
2


+2π
c
1
if
c
1
∈
,
ψ
0

-
-1
sin
ℬ
2

-
2
n
2

Ω
2
2
n
2

-
2

+2π
c
2
+π if
c
2
∈
,
θ
0

-1
tan
n

,-
2

-
2
n
2


+2π
c
1
if
c
1
∈
,
ψ
0

-1
sin
ℬ
2

-
2
n
2

Ω
2
2
n
2

-
2

+2π
c
2
if
c
2
∈
,
θ
0

-1
tan
n

,
2

-
2
n
2


+2π
c
1
if
c
1
∈
,
ψ
0

-
-1
sin
ℬ
Ω
2
2

-
2
n
2

+2π
c
2
+π if
c
2
∈
,
θ
0

-1
tan
n

,
2

-
2
n
2


+2π
c
1
if
c
1
∈
,
ψ
0

-1
sin
ℬ
Ω
2
2

-
2
n
2

+2π
c
2
if
c
2
∈

Simplify these solutions:

Initial Conditions

Define the total time of integration:
List our Mir physical properties as a single list:

Numerical Solutions

Three Dimensional Graphical Depiction of Mir Rotation

A Graphical Model of the Mir

Import a Mir model from NASA (just paste in the 3DS file):
The Mir model includes a depiction of the solar arrays, critical in their arrangement toward the sun.
Transform and rescale the model so that the center of mass of the Mir is roughly located at the central sphere, representing the node:

Animated 3D Rotation of Mir

Biography

The author is presently at the National Aeronautics and Space Administration (NASA) Johnson Space Center, Houston, Texas and is a NASA astronaut. The author was a crew member on board space station Mir from May until October of 1997.