Geometric Insight into the Control Allocation Problem for OpenFrame ROVs and Visualisation of Solution
Abstract
:1. Introduction
2. Control Allocation Problem
2.1. Problem Definition
 Guidance: subsystem that continuously computes the reference (desired) position, velocity and acceleration of an ROV to be used by the motion control system.
 Navigation: subsystem to determine position/attitude, course, travelled distance and (optionally) velocity and acceleration of an ROV.
 Control: subsystem to determine necessary control forces and moments to be provided by the ROV to satisfy certain control objective (in conjunction with the guidance system).
 STEP 1 (Regulation Task): Design a control law, which specifies desired virtual control input (normalised vector of forces and moments ${\underset{\_}{\mathit{\tau}}}_{d}$) in the virtual control space;
 STEP 2 (Actuator Selection Task): Design control allocator, which finds the “best” feasible true control input $\underset{\_}{\mathit{u}}$ (normalised command vector to be applied to individual actuators) in the true control space.
 ${x}_{b}$—longitudinal axes (directed to front side),
 ${y}_{b}$—transversal axes (directed to starboard),
 ${z}_{b}$—normal axes (directed from top to bottom).
 $\mathfrak{J}$ is empty (i.e., no solution exists),
 $\mathfrak{J}$ has exactly one element (i.e., there is one unique solution),
 $\mathfrak{J}$ has more than one element (i.e., there are many solutions).
 If the intersection is a segment, there is an infinite number of solutions (each point that belongs to the segment is a solution),
 If the intersection is a point, there is only one solution,
 If the intersection is an empty set, no solution exists.
2.2. Nomenclature
2.3. Geometric Insight into Problem
2.4. Choice of Norm
 In the case when ${\mathit{W}}_{\mathit{u}}$ is the unity matrix, the ${l}_{2}$ norm distributes the virtual control demand among the control inputs in a uniform way, while the ${l}_{1}$ solution utilises as few control inputs as possible to satisfy the virtual control demand.
 The ${l}_{2}$ solution varies continuously with the parameters (elements) of $\mathit{B}$, while the ${l}_{1}$ solution does not. Change in a parameter (element) $b$ of $\mathit{B}$ will produce the change in the slope of $\ell $. The ${l}_{2}$ solution will vary continuously with $b$, while it can be shown that the ${l}_{1}$ solution will have discontinuity for some value of $b={b}^{*}$ and the solution in the breakpoint ${b}^{*}$ is not unique.
 If ${\mathit{W}}_{\mathit{u}}$ is a nonsingular, the problem $\underset{\mathit{u}}{\mathrm{min}}{\Vert {\mathit{W}}_{\mathit{u}}\mathit{u}\Vert}_{p}$ has a unique solution for $p=2$. For $p=1$, this is not always the case, as discussed above. The reason lies in the fact that the sphere ${S}_{{\mathit{W}}_{\mathit{u}}}{\left(\mathbf{0},r\right)}_{2}$ is a strictly convex set, while this is not the case for ${S}_{{\mathit{W}}_{\mathit{u}}}{\left(\mathbf{0},r\right)}_{1}$.
2.5. Choice of the Weighting Matrix ${\mathit{W}}_{u}$
3. Control Allocation Solution: Hybrid Method
3.1. Description
3.2. Weighted Pseudoinverse
3.2.1. Introduction
 Find ${\mathrm{\Phi}}_{p}\subset {\mathrm{\Phi}}_{v}$ such that ${B}_{{W}_{u}}^{\u2020}\left({\mathrm{\Phi}}_{p}\right)={\mathrm{\Omega}}_{P}$,
 Find ${B}_{{W}_{u}}^{\u2020}\left({\mathrm{\Phi}}_{v}\right)$.
 1.
 $S={S}_{1}\in {\Phi}_{p}$,
 2.
 $S={S}_{2}\in {\Phi}_{v}\backslash {\Phi}_{p}$,
 3.
 $S={S}_{3}\in \Phi \backslash {\Phi}_{v}$.
3.2.2. Approximation of Unfeasible Solution
 Tapproximation ${\mathit{u}}_{t}^{*}$ of the unfeasible pseudoinverse solution, $\mathit{u}$ introduces direction error ${\theta}_{t}\ne 0$, i.e., vectors $\mathit{v}$ and ${\mathit{v}}_{t}^{*}$ have not the same direction. At the same time, the direction error ${\theta}_{s}=0$ for Sapproximation ${\mathit{u}}_{s}^{*}$, i.e., vectors $\mathit{v}$ and ${\mathit{v}}_{s}^{*}$ always have the same direction, but the magnitude error ${\Vert {\mathit{e}}_{\mathit{s}}\Vert}_{2}=\Vert \mathit{v}{{\mathit{v}}_{s}^{*}\Vert}_{2}$ is greater than ${\Vert {\mathit{e}}_{\mathit{t}}\Vert}_{2}={\Vert \mathit{v}{\mathit{v}}_{t}^{*}\Vert}_{2}$.
 The fixedpoint method (Section 3.3) is able to improve the T or Sapproximation of the unfeasible weighted pseudoinverse solution $\mathit{u}$. Approximations ${\mathit{u}}_{t}^{*}$ or ${\mathit{u}}_{s}^{*}$ can be used as the initial iteration ${\mathit{u}}_{\mathbf{0}}$ and the algorithm will find the solution ${\mathit{u}}_{f}^{*}$ such that ${\mathit{v}}_{f}^{*}=\mathit{B}{\mathit{u}}_{f}^{*}$ is a better approximation of $\mathit{v}$ than ${\mathit{v}}_{t}^{*}$ or ${\mathit{v}}_{s}^{*}$. This feature is the main idea of the hybrid approach for control allocation.
3.3. FixedPoint Method
Introduction
3.4. Algorithm (Hybrid Method)
3.5. Extension of Concepts from “virtual” ROV to OpenFrame ROV
 If $\mathit{\tau}$ lies inside ${\mathsf{\Phi}}_{v}$, then $\Im $ has an infinite number of points, and the control allocation problem has an infinite number of solutions.
 If $\mathit{\tau}$ lies on the boundary of ${\mathsf{\Phi}}_{v}$, then $\Im $ is a single point, the unique solution for the control allocation problem.
 If $\mathit{\tau}$ lies outside ${\mathsf{\Phi}}_{v}$, then $\Im $ is an empty set, i.e., no exact solution exists, only approximation.
4. Testing and Validation
4.1. Evaluation of the FTC in Virtual Environment
4.1.1. Partial Fault in HT_{2}
4.1.2. Total Fault in HT_{2}
4.2. Evaluation of the FTC in RealWorld Environment
4.2.1. Path Following: Simulated Faults
4.2.2. Complex Tasks with Faulty Thruster
5. Conclusions
Author Contributions
Funding
Conflicts of Interest
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Virtual Control Input  $\mathbf{Horizontal}\text{}\mathbf{Plane}\text{}{\underset{\_}{\mathit{\tau}}}_{\mathit{H}{\mathit{T}}_{\mathit{d}}}$  $\mathbf{Vertical}\text{}\mathbf{Plane}\text{}{\underset{\_}{\mathit{\tau}}}_{\mathit{V}{\mathit{T}}_{\mathit{d}}}$  

${\underset{\_}{\mathit{\tau}}}_{d}=\left[\begin{array}{c}\begin{array}{c}{\underset{\_}{\tau}}_{{X}_{d}}\\ {\underset{\_}{\tau}}_{{Y}_{d}}\\ {\underset{\_}{\tau}}_{{Z}_{d}}\end{array}\\ \begin{array}{c}{\underset{\_}{\tau}}_{{K}_{d}}\\ {\underset{\_}{\tau}}_{{M}_{d}}\\ {\underset{\_}{\tau}}_{{N}_{d}}\end{array}\end{array}\right]$  ${\underset{\_}{\mathit{\tau}}}_{H{T}_{d}}=\left[\begin{array}{c}{\underset{\_}{\tau}}_{{X}_{d}}\\ {\underset{\_}{\tau}}_{{Y}_{d}}\\ {\underset{\_}{\tau}}_{{N}_{d}}\end{array}\right]$  ${\underset{\_}{\tau}}_{{X}_{d}}$—Surge Force ${\underset{\_}{\tau}}_{{Y}_{d}}$—Sway Force ${\underset{\_}{\tau}}_{{N}_{d}}$—Yaw Moment  ${\underset{\_}{\mathit{\tau}}}_{V{T}_{d}}=\left[\begin{array}{c}{\underset{\_}{\tau}}_{{Z}_{d}}\\ {\underset{\_}{\tau}}_{{K}_{d}}\\ {\underset{\_}{\tau}}_{{M}_{d}}\end{array}\right]$  ${\underset{\_}{\tau}}_{{Z}_{d}}$—Heave Force ${\underset{\_}{\tau}}_{{K}_{d}}$—Roll Moment ${\underset{\_}{\tau}}_{{M}_{d}}$—Pitch Moment 
Virtual ROV  OpenFrame ROV  

Virtual Control Input  $\mathit{v}=\left[\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right]\in {\mathrm{\mathbb{R}}}^{2}\left(k=2\right)$  $\mathit{\tau}=\left[\begin{array}{c}{\tau}_{X}\\ {\tau}_{Y}\\ {\tau}_{N}\end{array}\right]\in {\mathrm{\mathbb{R}}}^{3}\left(k=3\right)$ 
True Control Input  $\mathit{u}=\left[\begin{array}{c}{u}_{1}\\ {u}_{2}\\ {u}_{3}\end{array}\right]\in {\mathrm{\mathbb{R}}}^{3}\left(m=3\right)$  $\mathit{u}=\left[\begin{array}{c}\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\\ \begin{array}{c}{u}_{3}\\ {u}_{4}\end{array}\end{array}\right]\in {\mathrm{\mathbb{R}}}^{4}\left(m=4\right)$ 
Control Effectiveness Matrix  $\mathit{B}=\left[\begin{array}{ccc}\frac{1}{2}& \frac{1}{4}& \frac{1}{4}\\ 0& \frac{3}{5}& \frac{2}{5}\end{array}\right]$  $\mathit{B}=\left[\begin{array}{ccc}\frac{1}{4}& \frac{1}{4}& \begin{array}{cc}\frac{1}{4}& \frac{1}{4}\end{array}\\ \frac{1}{4}& \frac{1}{4}& \begin{array}{cc}\frac{1}{4}& \frac{1}{4}\end{array}\\ \frac{1}{4}& \frac{1}{4}& \begin{array}{cc}\frac{1}{4}& \frac{1}{4}\end{array}\end{array}\right]$ 
Actuator Position Constraints  $\underset{\_}{\mathit{u}}=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right],\overline{\mathit{u}}=\left[\begin{array}{c}+1\\ +1\\ +1\end{array}\right]$  $\underset{\_}{\mathit{u}}=\left[\begin{array}{c}\begin{array}{c}1\\ 1\end{array}\\ \begin{array}{c}1\\ 1\end{array}\end{array}\right],\overline{\mathit{u}}=\left[\begin{array}{c}\begin{array}{c}+1\\ +1\end{array}\\ \begin{array}{c}+1\\ +1\end{array}\end{array}\right]$ 
Control Allocation Problem (Inequalities (2) and (4) apply componentwise.)  For a given $\mathit{v}$, find $\mathit{u}$ such that
$$\mathit{B}\mathit{u}=\mathit{v}$$
$$\underset{\_}{\mathit{u}}\le \mathit{u}\le \overline{\mathit{u}}$$
 For a given $\mathit{\tau}$, find $\mathit{u}$ such that
$$\mathit{B}\mathit{u}=\mathit{\tau}$$
$$\underset{\_}{\mathit{u}}\le \mathit{u}\le \overline{\mathit{u}}$$

${\mathit{P}}_{23}$  ${\mathit{P}}_{45}$  ${\mathit{P}}_{13}$  ${\mathit{P}}_{46}$  ${\mathit{P}}_{15}$  ${\mathit{P}}_{26}$  

${\mathit{v}}_{\mathbf{1}}$  −0.7917  0.7917  −0.6875  0.6875  −0.2000  0.2000 
${\mathit{v}}_{\mathbf{2}}$  0.5333  −0.5333  −0.5500  0.5500  −1.0000  1.0000 
${\mathit{R}}_{23}$  ${\mathit{R}}_{45}$  ${\mathit{R}}_{13}$  ${\mathit{R}}_{46}$  ${\mathit{R}}_{15}$  ${\mathit{R}}_{26}$  

${\mathit{u}}_{\mathbf{1}}$  −1.0000  1.0000  −1.0000  1.0000  −0.4000  0.4000 
${\mathit{u}}_{\mathbf{2}}$  1.0000  −1.0000  −0.2500  0.2500  −1.0000  1.0000 
${\mathit{u}}_{\mathbf{3}}$  0.1667  −0.1667  1.0000  −1.0000  1.0000  −1.0000 
Virtual Control Space  True Control Space  

Partition  Polygon  Partition  Polygon 
${\mathsf{\Phi}}_{p}$  ${P}_{13}{P}_{15}{P}_{45}{P}_{46}{P}_{26}{P}_{23}$  ${\mathsf{\Omega}}_{p}$  ${R}_{13}{R}_{15}{R}_{45}{R}_{46}{R}_{26}{R}_{23}$ 
${\mathsf{\Phi}}_{v}\backslash {\mathsf{\Phi}}_{p}$  $\colorbox[rgb]{1,1,0}{$1$}$${P}_{15}{P}_{13}$  ${\mathsf{\Omega}}_{e}\backslash {\mathsf{\Omega}}_{p}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$1$}$${R}_{15}{R}_{13}$ 
$\colorbox[rgb]{1,1,0}{$3$}$${P}_{13}{P}_{23}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$3$}$${R}_{13}{R}_{23}$  
$\colorbox[rgb]{1,1,0}{$2$}$${P}_{23}{P}_{26}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$2$}$${R}_{23}{R}_{26}$  
$\colorbox[rgb]{1,1,0}{$6$}$${P}_{26}{P}_{46}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$6$}$${R}_{26}{R}_{46}$  
$\colorbox[rgb]{1,1,0}{$4$}$${P}_{46}{P}_{45}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$4$}$${R}_{46}{R}_{45}$  
$\colorbox[rgb]{1,1,0}{$5$}$${P}_{45}{P}_{15}$  $\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$5$}$${R}_{45}{R}_{15}$  
$\mathsf{\Phi}\backslash {\mathsf{\Phi}}_{v}$  ${\mathrm{V}}_{0}$$\colorbox[rgb]{1,1,0}{$1$}$$\colorbox[rgb]{1,1,0}{$3$}$  ${\mathsf{\Omega}}_{v}\backslash {\mathsf{\Omega}}_{e}$  ${\mathrm{U}}_{0}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$1$}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$3$}$ 
${\mathrm{V}}_{1}$$\colorbox[rgb]{1,1,0}{$3$}$$\colorbox[rgb]{1,1,0}{$2$}$  ${\mathrm{U}}_{1}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$3$}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$2$}$  
${\mathrm{V}}_{3}$$\colorbox[rgb]{1,1,0}{$6$}$$\colorbox[rgb]{1,1,0}{$4$}$  ${\mathrm{U}}_{3}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$6$}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$4$}$  
${\mathrm{V}}_{2}$$\colorbox[rgb]{1,1,0}{$4$}$$\colorbox[rgb]{1,1,0}{$5$}$  ${\mathrm{U}}_{2}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$4$}$$\colorbox[rgb]{1,0.627450980392157,0.47843137254902}{$5$}$ 
Initial Point  # Of Iterations  Last Iteration  Limit  Obtained Virtual Control Input  Desired Virtual Control Input  Direction Error  Magnitude Error 

${\mathit{u}}_{\mathbf{0}}$  $k$  ${\mathit{u}}_{{\mathit{f}}_{\mathit{k}}}$  ${\mathit{u}}_{f}$  ${\mathit{v}}_{{\mathit{f}}_{\mathit{k}}}$  $\mathit{v}$  ${\theta}_{k}$  ${\Vert {\mathit{e}}_{k}\Vert}_{2}$ 
${\mathit{u}}_{2t}^{*}$  19  $\left[\begin{array}{c}1.0000\\ 0.8585\\ 0.8874\end{array}\right]$  $\left[\begin{array}{c}1.0000\\ 0.8600\\ 0.8900\end{array}\right]$  $\left[\begin{array}{c}0.9365\\ 0.1601\end{array}\right]$  $\left[\begin{array}{c}0.9375\\ 0.1600\end{array}\right]$  $0.0181\xb0$  $0.0010$ 
${\mathit{u}}_{2s}^{*}$  20  $\left[\begin{array}{c}1.0000\\ 0.8582\\ 0.8870\end{array}\right]$  $\left[\begin{array}{c}1.0000\\ 0.8600\\ 0.8900\end{array}\right]$  $\left[\begin{array}{c}0.9363\\ 0.1601\end{array}\right]$  $\left[\begin{array}{c}0.9375\\ 0.1600\end{array}\right]$  $0.0208\xb0$  $0.0012$ 
$\mathbf{Vertices}\text{}\mathbf{of}\text{}\Omega $  $\mathbf{Vertices}\text{}\mathbf{of}\text{}{\Phi}_{\mathit{v}}$  

Label  Coordinates  Label  Coordinates  
0  −1  −1  −1  −1  0  −1.0  0.0  0.0 
1  −1  −1  −1  +1  1  −0.5  −0.5  0.5 
2  −1  −1  +1  −1  2  −0.5  0.5  −0.5 
3  −1  −1  +1  +1  3  0.0  0.0  0.0 
4  −1  +1  −1  −1  4  −0.5  −0.5  −0.5 
5  −1  +1  −1  +1  5  0.0  −1.0  0.0 
6  −1  +1  +1  −1  6  0.0  0.0  −1.0 
7  −1  +1  +1  +1  7  0.5  −0.5  −0.5 
8  +1  −1  −1  −1  8  −0.5  0.5  0.5 
9  +1  −1  −1  +1  9  0.0  0.0  1.0 
A  +1  −1  +1  −1  A  0.0  1.0  0.0 
B  +1  −1  +1  +1  B  0.5  0.5  0.5 
C  +1  +1  −1  −1  C  0.0  0.0  0.0 
D  +1  +1  −1  +1  D  0.5  −0.5  0.5 
E  +1  +1  +1  −1  E  0.5  0.5  −0.5 
F  +1  +1  +1  +1  F  1.0  0.0  0.0 
Segment  HT_{1}  HT_{2}  HT_{3}  HT_{4} 

01  ON  ON  ON  ON 
12  ON  ON  ON  ON 
23  ON  ON  ON  OFF 
34  ON  ON  ON  OFF 
45  ON  OFF  ON  ON 
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Omerdic, E.; Trslic, P.; Kaknjo, A.; Weir, A.; Rao, M.; Dooly, G.; Toal, D. Geometric Insight into the Control Allocation Problem for OpenFrame ROVs and Visualisation of Solution. Robotics 2020, 9, 7. https://doi.org/10.3390/robotics9010007
Omerdic E, Trslic P, Kaknjo A, Weir A, Rao M, Dooly G, Toal D. Geometric Insight into the Control Allocation Problem for OpenFrame ROVs and Visualisation of Solution. Robotics. 2020; 9(1):7. https://doi.org/10.3390/robotics9010007
Chicago/Turabian StyleOmerdic, Edin, Petar Trslic, Admir Kaknjo, Anthony Weir, Muzaffar Rao, Gerard Dooly, and Daniel Toal. 2020. "Geometric Insight into the Control Allocation Problem for OpenFrame ROVs and Visualisation of Solution" Robotics 9, no. 1: 7. https://doi.org/10.3390/robotics9010007