Spacecraft Attitude Dynamics & Control - Lecture Notes, Dispense di Ingegneria Aerospaziale

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SPACECRAFT

ATTITUDE

DYNAMICS &

CONTROL

Prof. FRANCO BERNELLI ZAZZERA

A.Y. 2020/

Lecture Notes by Andrea Pizzetti

SADaC Pagina 1

The Attitude is the orientation of the spacecraft body axes relative to a reference frame. The Attitude Error will be the difference between the true and the desired attitude, in most of the cases is an angle, or to be more precise, 3 angles. We define the Attitude Determination when we use Sensors to estimate the attitude in real-time, while the Attitude Control is used to maintain specified attitude using Actuators, which usually are used to provide torques. In the first part of the course, Dynamics and Kinematics , we'll use Simulink to model a spacecraft in a space environment and understand how to exploit its dynamics for passive stabilization. The aim of the Determination module will be the implementation of attitude determination algorithms. The Ideal Control module will deal with the development of feedback controls to grant objectives such as the De- tumbling, Slew Motions, Three-axis stabilization. Finally, the Actuators part will allow to understand and implement algorithms to generate ideal torques using different types of actuators. Definitions mercoledì 23 settembre 2020 11: Introduction Pagina 2

Finally, in the real world we'll have this kind of model: Introduction Pagina 4

The assignment is an attitude control problem on a CubeSat of given size. The sensors are pretty much the same of a spacecraft of any size. Instead for the actuators we have a scaling problem. The inertia of a nano-spacecraft is very low, we need to miniaturize the actuators. Since we need to develop a virtual model, we'll first work on the dynamics and kinematic model and then add the control. The report should show the overall architecture of the model and its description (models used and assumptions). For control and determination algorithms, we need to justify the choices without copy and paste from the lecture notes. The algorithms should be compared and contrasted in different environment conditions or sets of parameters. Include all the references used, including theoretical and hardware data. In any case, do not copy and paste Simulink diagrams/plots because there is always a notation problem! An example of an AOCS architecture is: The sensors are 2 Sun sensors, 2 star trackers and an unit composed by 3 accelerometers and 3 gyroscopes. The actuators are 3 reaction wheels and 4 thrusters. There is obviously a redundancy, but in this way also the performances are higher. The project assignment mercoledì 23 settembre 2020 12: Introduction Pagina 5

Along the diagonal we always have positive terms, while in the off-diagonal terms we can have arbitrary signs. The inertia matrix is always symmetric. There are also some interesting properties, valid also if we cycle the indexes: The first two properties are related to distributions of mass in space. If there is no component in z direction, the sum of the first two inertial moment is equal to the third. If there is no component in both x and y direction, the difference of the first two inertial moments is equal to the third one. These properties work as a check for the physical reality of an object previously designed.

Rotational Kinetic Energy

We said that we neglect the linear velocities and so we can study only the Rotational Kinetic Energy. This can be written as: We can simplify the expression if the inertia matrix is diagonal, so if we select a Principal Axis frame. A symmetric matrix can be diagonalized always. In this case, we'll have:

It's common to distribute the masses inside the volume of the spacecraft in order to keep the principal axis as more equal as possible to the geometric ones.

Geometric Interpretation

To understand better, we can use a geometric interpretation using as variables the three components of angular velocity and imposing the conservation of energy and angular momentum respectively. In the 3D space, the equations result in ellipsoids: The kinetic energy is constant (no external torques), as well as the inertia moments. We have the surface of an ellipsoid, so specific triplets of angular velocity components. On the other hand, we can express the conservation of angular momentum with another ellipsoid: The intersections of the two ellipsoids give the locus of the angular velocity components which satisfy both balances, which are the triplets of angular velocity at each instant of time. These two intersection lines are called Polhodes. These triplets moves along the intersection, which represent the trajectory. If we assign initial conditions such that we have only one possible triplet combination in the intersection, we'll have a Minimum Energy Ellipsoid , the energy ellipsoid will be contained in the momentum sphere:

we need that the three terms in the bracket have different signs (not all negative nor all positive). If we assume those values for inertia moments, the x coefficient must be positive otherwise they would all be negative. Then, the z component cannot be positive otherwise all must be positive. The y component can be either positive or negative. In particular, h2/2T must be higher than the lowest inertial moment and lower than the highest inertial moment. In this way, we'll have a change of sign in any case. We can take the projections on the three planes and look at the shapes of the conic curves. For instance, considering the energy conservation: So, the first and second equations are ellipses (same signs), while the third one is an hyperbola (different signs). When looking from the x (z) direction, respectively projections on yz (xy) plane, we would see only ellipses, regardless of the solution. When looking from the y direction, projection on xz plane, we would see only hyperbolas, regardless of the solution.

would see only hyperbolas, regardless of the solution. Looking at xz plane, if the angular velocity is slightly shifted from the x-axis or from the z-axis it will remain confined in a region close to the axis, while a slight shift from the y-axis would cause a dramatic departure from the original condition. In this case the black ellipsoid is the kinetic energy and it's constant, and we have different values for the angular momentum such that we have different intersection curves. Through this figure it's possible to observe also the perturbations effect on the different axis: looking at yz plane, a perturbation on y or z will not cause a huge displacement (it could just change the position on the ellipse); instead, looking at xz plane, a perturbation on x or z will cause a huge displacement along the separatrix. This is why the rotation around the largest and smallest principal axis (inertia moments Ix and Iz) is stable while the rotation around the medium principal axis (inertia moment Iy) is unstable.

The effects of disturbances are divided by the inertia moments. So the same disturbance effect for a small spacecraft will be higher. The frequency content will also have a different behaviour, so it will affect the numerical integration. Using the chain rule, we can show the conservation of kinetic energy, in case of no torque applied:

Exact solution

In case of no torques applied, and considering a symmetric spacecraft (so two out of three inertia moments equal), we can have an exact solution. For instance, for a cylindrical shape. If that's the case, the last equation will be 0, so the z angular velocity will be constant and the first two equations are decoupled and can be solved separately.

Once we give the initial values, the solution is analytical because lambda can be computed separately. If the torque is present, we could still have an exact solution if we have a specific analytical expression for it.

Equilibrium Configuration

Equilibrium occurs when all the three derivatives are equal to 0. Other than all velocities equal to 0, we have other three equilibrium points, which can be actually combined in one: equilibrium occurs when only one component of angular velocity has a non-zero initial condition.

Stability

We can have Stable systems (or Lyapunov Stable systems) when a perturbation does not tend to zero as time grows. Instead, in Asymptotically Stable systems, any perturbation tends to be reduced to zero as time grows. To do that, we need damping mechanics. Otherwise, we'll have perturbations that keep oscillating in the solution. Looking at the equilibrium points, the case of only one component with initial conditions different from zero is a Lyapunov stable system. If we apply a perturbation:

The major axis is always stable, the intermediate axis is in any case unstable, while the minor axis even if it should be stable always, it can be unstable if we have energy dissipation. Energy dissipation is always present in real case, because there aren't objects completely rigid. In case of no external torques, kinetic energy is decreasing due to dissipations while momentum stays constant. This is the principle of planets (which can dissipate energy) or spinning objects in space. Considering the two cases of spinning around minimum and maximum inertia axis, we can see that to preserve angular momentum, the angular velocity around the major axis must be lower than the one around the minor axis. But this means that the kinetic energy in case of the minor axis must be higher than the one of the major axis, so the major inertia axis is the one with the lowest energy and the one where all the spinning objects will tend over an infinite time, because it's the least-energy state. So, in the Explorer I case, the flexible antennas were the elements which introduced dissipation and that's why the minor inertia axis was unstable. At a certain point it started to rotate and it found itself face up-down, spinning around the major inertia axis because the flexible antennas kept dissipating energy and the satellite had to pass to the lowest energy state.

Solutions in Phase Space

This kind of method provides the results in the phase space, so in the space of the angular velocity and angular acceleration. After taking the derivative of ωx and replacing the derivative of ωy and ωz, we

angular acceleration. After taking the derivative of ωx and replacing the derivative of ωy and ωz, we arrive to an equation which can be integrated: We have to understand if the coefficients P and Q are positive or negative. This gives an indication of which conic sections we have. Let's again assume values of inertia moments such as: For large angular velocities what matters is the value of Q, for small angular velocities instead P is predominant. From the explicit formula of the coefficients the following non-equalities can be retrieved in this specific case of inertia moments: In the x and z phase planes the conic sections are ellipses for large velocities, for small velocities the type of conic section is undefined and depends on the value of P. However, in any case for divergent phase plane trace (hyperbola) it is clear that as the angular velocity grows the trace will change into an ellipse, so an undefined section it's still ok. In the y phase plane instead, if the angular velocity is large, the conic section will be a hyperbola, if the angular velocity is small the conic section will be an ellipse. In this case, if the velocity grows, we would have a diverging motion. This is not possible due to the conservation of kinetic energy (it cannot grows infinitely), therefore the conclusion is that in the y phase plane the solution must be bounded such that the angular velocity is small enough to prevent the trace from being a hyperbola, leaving Py as predominant parameter.

We have always to check if the angular momentum and kinetic energy are constant in case of no torque applied: if that's not the case, it means that the applied numerical scheme is wrong.

We have analyzed the spacecraft as a single entity. Let's talk about another way of seeing the major axis rule, in case of energy loss. For a symmetric spacecraft with minor inertia axis z we have: Knowing that kinetic energy derivative is negative and angular momentum derivative is zero: So the z component of the angular acceleration must decrease in time, if the velocity is positive, and vice versa. So it will goes to zero in both cases. The component along the major axis n instead will goes to a maximum, as we can see from the second equation (the right hand term is always positive). Assuming to have a thruster that ejects mass downward, this means that the thrust can be guaranteed to be in the right direction. However, since the spacecraft is continuously rotating, it's hard to use an antenna and keep a stable connection. Dual-Spin We can have the single spin advantages while avoiding to spin one axis, and so keep the advantages of both cases. We speak of dual spin because usually one part spins far fastest than a second one (for instance, because the antenna should rotate very slow to always point towards the Earth). The conceptual model is a spacecraft with a Wheel which rotates relative to it, only around the z axis: Dual-Spin and Ring Damper mercoledì 7 ottobre 2020 11: