AE4610
Rotary Gyro
Introduction
The objective of this experiment is to design a controller that maintains the direction of the gyroscope
module while the top base plate is rotated relative to the bottom base plate. While the disk spins, the SRV02
is used to apply the co
ect amount of counter torque and maintain the gyroscope heading in the event of
distu
ances (i.e., rotation of the bottom support plate).
Figure 0.1: Rotary Gyro Experiment
Gyroscopes are used in many different platforms, e.g., airplanes, large marine ships, submarines, and
satellites.
In the next sections, we will develop a model of the gyroscope device from first principles in order to obtain a
transfer function of the plant, which we will later use in a control scheme. To do so, however, we need some
preliminaries on angular momentum and the gyroscopic effect.
Part 0: Preliminaries
Angular Momentum
Let's consider an rigid body of mass , moving with any general motion in space. Assume that the axes
x-y-z are attached to the body with origin at its center of mass . Now, let be a mass element of the
ody situated by position vector with respect to the center of mass .
B m
G dm
ρ G
Figure 0.2: Angular Momentum
Given the body's linear velocity (i.e., the velocity of its center of mass) and its angular velocity , the total
angular momentum of the body with respect to the center of mass is given as
v ω
G
L = v × ρdm +
B
ρ ×
B
ω × ρ dm.( )
The first term is since the integral by definition of the center of mass. By using the triple
product rule for the cross product in the second term, it can be shown that
ρ dm =
B
L = ρ ×
B
ω × ρ dm =( ) (ρ ρI −
B
⊤
3 ρρ )dmω =⊤ Jω
where is the inertia tensor of the body about the center of mass .J 3 × 3 G
Time Rate of Change of Angular Momentum
Now, since the body's reference frame is rotating, the derivative of the total angular momentum about the
center of mass can be written asG
L =
dt
d
( ) +L̇ ω × L
where is the time derivative of the angular momentum as seen in the rotating x-y-z frame. We can view
the term as the part due to the change in magnitude of and the second term as the part due to
the change in direction of . By virtue of the Newton-Euler equation, we have
L̇
L̇ L ω × L
L
τ = L =
dt
d
( ) +L̇ ω × L
and we notice that either a change in magnitude of the angular moment or a change in direction of the
angular momentum generates a torque. Inversely, a torque can generate a change in either magnitude or
direction of the angular momentum, depending about which axis it is applied.
Gimbaling Gyroscope
Assume now that the total angular momentum of the gimbaling gyroscope is given by
L = (J +g J )ω +f g J ω f f
where is the angular velocity of the gyro module and is the angular velocity of the flywheel, is
the inertia tensor of the gyro module - without the flywheel - and is the inertia tensor of the flywheel.
ω g ω f J g
J f
Figure 0.3: Coordinate Frame and Angle Definitions
Assuming that the deflection of the flywheel gimbal axis is small and slow, we now have that, when
expressed in the rotating frame of the flywheel gimbal coordinates,
ω =g
0
0
ω + b θ̇l
and
ω =f .
ω f
0
0
Further assuming that the rotary base motor has achieved steady-state ( , ), and that the
flywheel's velocity is constant (i.e. ), we have that
=θ̇l 0 =θ̈l 0
=ω̇f 0
τ = (J +g J ) +f ω̇g ω ×g (J +g J )ω +f g ω ×g J ω f f
Assuming that the gimbal frame is aligned with the principal axes of he gyro module assembly and ignoring
the contribution of the wheel inertia to the principal moment of inertia about the vertical axis of the gyro
module, then the second term is 0, and we are left with the equations
τ =2 ω J ω b f f
τ =3 J g ω̇
co
esponds to the torque due to the gyroscopic effect and will be hereon denoted as . We will further
neglect the second equation pertaining to the vertical axis torque by assuming that is typically
small.
τ 2 τ g
≈ω̇b 0
Part 1: Modeling
1.1 Servo Model
The Servo Base Unit (SRV02) open-loop transfer function is given by
P(s) = =
V (s)m
Θ (s)l
s(τs + 1)
K
(1.1)
where is the load gear position and is the applied motor voltage. The
system steady-state gain and time constant are given by:
Θ (s) =l [θ (t)]l V (s) =m [v (t)]m
K = 1.53 rad/s/V
and
τ = XXXXXXXXXXs.
1.2 Gyroscope Gain
Consider the simplified model shown in Figure 1.1.
Figure 1.1: Simplified rotary gyroscope model.
The inertial disc, flywheel, spins at a relatively constant velocity, . When the base rotates at a speed of
, the resulting gyroscopic torque about the sensitive axis is
ω f
ω
τ =g ω L b f (1.2)
where is the angular momentum of the flywheel and is its moment of inertia. The springs
mounted on the gyroscope counteract the gyroscopic torque, , by the following amount
L =f J ω f f J f
τ g
τ =s K αr (1.3)
where is the rotational stiffness of the springs.K
Given that the spring torque equals the gyroscopic torque, , we can equate equations (1.2) and
(1.3) to obtain the expression
τ =s τ g
K α =r ω J ω .b f f (1.4)
Assume that the base speed is proportional to the deflection angle through the gain , thenGg
ω =b G α.g (1.5)
By examining equations (1.4) and (1.5), we find the gyroscopic sensitivity gain is given by
G =g =
α
ω
.
J ω f f
K r (1.6)
Thus, the deflection angle can be used to measure , the rotation rate of the platform relative to the
ase, without a direct measurement. NOTE: the dynamics in the sensitive axis (i.e., deflection axis) are
ignored. A more complete model would include these dynamics as the transfer function .
α ω
α(s)/ω (s)
1.3 Joint Stiffness
The two springs affecting the sensitive axis are shown in Figure 1.2. The stiffness at the axis of rotation is
derived in the following fashion. Assume the springs have a spring constant and an un-stretched length
. The length of the springs at the normal position, i.e. , is given by . If the axis is rotated by an
angle , then the two forces about the sensitive axis are given by (for a small )
K s
Lu α = 0 L
α α
F =1 K ΔL =s 1 K (L −s L −u αR)
and
F =2 K ΔL =s 2 K (L −s L +u αR)
Figure 1.2: Forces due to springs.
The spring torque about the pivot due to the two forces is
τ =s R(F −2 F ) =1 2R K α2 s
The rotational stiffness is given by
K =r =
α
τ s 2R K .2 s (1.7)
Part 2: Control design
2.1 Desired Position Control
The block diagram shown in Figure 2.1 is a general unity feedback compensator with compensator
(controller) and a transfer function representing the plant , . The measured output , is
supposed to track the reference and the tracking has to match certain desired specifications.
C(s) P(s) Y (s)
R(s)
Figure 2.1: Unity Feedback System.
The output of the system can be written as
Y (s) = C(s)P(s)(R(s) − Y (s)).
By solving for we get the closed-loop transfer functionY (s)
=
R(s)
Y (s)
.
1 + C(s)P(s)
C(s)P(s)
When a second-order system is placed in series with a proportional compensator in the feedback loop as in
Figure 1.2, the resulting closed-loop transfer function can be expressed as
=
C(s)
Y (s)
,
s + 2ζω s + ω 2 n n2
ω n
2
(2.1)
where is the natural frequency and is the damping ratio. This is called the standard second order
transfer function. Its response properties depend on the values of and .
ω n ζ
ω n ζ
2.2 Control Specifications
The desired time-domain specifications for stabilizing the gyroscope are:
ω =n 6π rad/s (2.2)
or 3Hz, and
ζ = XXXXXXXXXX)
2.3 Gyro PD Controlle
To stabilize the heading of the gyroscope, we will develop a Proportional-Derivative (PD) controller depicted
in Figure 2.2.
Figure 2.2: Gyroscope PD Control Block Diagram
Assuming that the support plate (and servo) rotates relative to the base by the angle (not measured) and
that the gyro module rotates relative to the servo module by the angle (measured) , the total rotation angle
of the gyro module relative to the base plate can be expressed by
γ
θ l
η = γ + θ l (2.4)
We want to design a controller that maintains the gyro heading, i.e. keeps , independent of and we
can only use the measurement from the gyro sensor, . In other terms, we want to stabilize the system such
that . Differentiating equation (2.4) gives
η = 0 γ
α
→η̇ 0
=η̇ +γ̇