#+TITLE: Cubic configuration for the Stewart Platform
:DRAWER:
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#+PROPERTY: header-args:matlab  :session *MATLAB*
#+PROPERTY: header-args:matlab+ :tangle matlab/cubic_configuration.m
#+PROPERTY: header-args:matlab+ :comments org
#+PROPERTY: header-args:matlab+ :exports both
#+PROPERTY: header-args:matlab+ :results none
#+PROPERTY: header-args:matlab+ :eval no-export
#+PROPERTY: header-args:matlab+ :noweb yes
#+PROPERTY: header-args:matlab+ :mkdirp yes
#+PROPERTY: header-args:matlab+ :output-dir figs
:END:

* Introduction                                                       :ignore:
The discovery of the Cubic configuration is done in citenum:geng94_six_degree_of_freed_activ.
Further analysis is conducted in cite:jafari03_orthog_gough_stewar_platf_microm.

People using orthogonal/cubic configuration: cite:preumont07_six_axis_singl_stage_activ.


The specificity of the Cubic configuration is that each actuator is orthogonal with the others.

To generate and study the Cubic configuration, =initializeCubicConfiguration= is used (description in section [[sec:initializeCubicConfiguration]]).

According to cite:preumont07_six_axis_singl_stage_activ, the cubic configuration provides a uniform stiffness in all directions and *minimizes the crosscoupling* from actuator to sensor of different legs (being orthogonal to each other).

* Matlab Init                                               :noexport:ignore:
#+begin_src matlab :tangle no :exports none :results silent :noweb yes :var current_dir=(file-name-directory buffer-file-name)
  <<matlab-dir>>
#+end_src

#+begin_src matlab :exports none :results silent :noweb yes
  <<matlab-init>>
#+end_src

#+begin_src matlab :results none :exports none
  simulinkproject('./');
#+end_src

* Questions we wish to answer with this analysis
The goal is to study the benefits of using a cubic configuration:
- Equal stiffness in all the degrees of freedom?
- No coupling between the actuators?
- Is the center of the cube an important point?

* TODO Configuration Analysis - Stiffness Matrix
** Cubic Stewart platform centered with the cube center - Jacobian estimated at the cube center
We create a cubic Stewart platform (figure [[fig:3d-cubic-stewart-aligned]]) in such a way that the center of the cube (black dot) is located at the center of the Stewart platform (blue dot).
The Jacobian matrix is estimated at the location of the center of the cube.

#+name: fig:3d-cubic-stewart-aligned
#+caption: Centered cubic configuration
[[file:./figs/3d-cubic-stewart-aligned.png]]

#+begin_src matlab :results silent
  opts = struct(...
      'H_tot', 100, ... % Total height of the Hexapod [mm]
      'L',     200/sqrt(3), ... % Size of the Cube [mm]
      'H',     60, ... % Height between base joints and platform joints [mm]
      'H0',    200/2-60/2 ... % Height between the corner of the cube and the plane containing the base joints [mm]
      );
  stewart = initializeCubicConfiguration(opts);
  opts = struct(...
      'Jd_pos', [0, 0, -50], ... % Position of the Jacobian for displacement estimation from the top of the mobile platform [mm]
      'Jf_pos', [0, 0, -50]  ... % Position of the Jacobian for force location from the top of the mobile platform [mm]
      );
  stewart = computeGeometricalProperties(stewart, opts);

  save('./mat/stewart.mat', 'stewart');
#+end_src

#+begin_src matlab :results none :exports code
  K = stewart.Jf'*stewart.Jf;
#+end_src

#+begin_src matlab :results value table :exports results
  data = K;
  data2orgtable(data, {}, {}, ' %.2g ');
#+end_src

#+RESULTS:
|        2 |  1.9e-18 | -2.3e-17 |  1.8e-18 |  5.5e-17 | -1.5e-17 |
|  1.9e-18 |        2 |  6.8e-18 | -6.1e-17 | -1.6e-18 |  4.8e-18 |
| -2.3e-17 |  6.8e-18 |        2 | -6.7e-18 |  4.9e-18 |  5.3e-19 |
|  1.8e-18 | -6.1e-17 | -6.7e-18 |   0.0067 | -2.3e-20 | -6.1e-20 |
|  5.5e-17 | -1.6e-18 |  4.9e-18 | -2.3e-20 |   0.0067 |    1e-18 |
| -1.5e-17 |  4.8e-18 |  5.3e-19 | -6.1e-20 |    1e-18 |    0.027 |

** Cubic Stewart platform centered with the cube center - Jacobian not estimated at the cube center
We create a cubic Stewart platform with center of the cube located at the center of the Stewart platform (figure [[fig:3d-cubic-stewart-aligned]]).
The Jacobian matrix is not estimated at the location of the center of the cube.

#+begin_src matlab :results silent
  opts = struct(...
      'H_tot', 100, ... % Total height of the Hexapod [mm]
      'L',     200/sqrt(3), ... % Size of the Cube [mm]
      'H',     60, ... % Height between base joints and platform joints [mm]
      'H0',    200/2-60/2 ... % Height between the corner of the cube and the plane containing the base joints [mm]
      );
  stewart = initializeCubicConfiguration(opts);
  opts = struct(...
      'Jd_pos', [0, 0, 0], ... % Position of the Jacobian for displacement estimation from the top of the mobile platform [mm]
      'Jf_pos', [0, 0, 0]  ... % Position of the Jacobian for force location from the top of the mobile platform [mm]
      );
  stewart = computeGeometricalProperties(stewart, opts);
#+end_src

#+begin_src matlab :results none :exports code
  K = stewart.Jf'*stewart.Jf;
#+end_src

#+begin_src matlab :results value table :exports results
  data = K;
  data2orgtable(data', {}, {}, ' %.2g ');
#+end_src

#+RESULTS:
|        2 |  1.9e-18 | -2.3e-17 |  1.5e-18 |     -0.1 | -1.5e-17 |
|  1.9e-18 |        2 |  6.8e-18 |      0.1 | -1.6e-18 |  4.8e-18 |
| -2.3e-17 |  6.8e-18 |        2 | -5.1e-19 | -5.5e-18 |  5.3e-19 |
|  1.5e-18 |      0.1 | -5.1e-19 |    0.012 |   -3e-19 |  3.1e-19 |
|     -0.1 | -1.6e-18 | -5.5e-18 |   -3e-19 |    0.012 |  1.9e-18 |
| -1.5e-17 |  4.8e-18 |  5.3e-19 |  3.1e-19 |  1.9e-18 |    0.027 |

** Cubic Stewart platform not centered with the cube center - Jacobian estimated at the cube center
Here, the "center" of the Stewart platform is not at the cube center (figure [[fig:3d-cubic-stewart-misaligned]]).
The Jacobian is estimated at the cube center.

#+name: fig:3d-cubic-stewart-misaligned
#+caption: Not centered cubic configuration
[[file:./figs/3d-cubic-stewart-misaligned.png]]

The center of the cube is at $z = 110$.
The Stewart platform is from $z = H_0 = 75$ to $z = H_0 + H_{tot} = 175$.
The center height of the Stewart platform is then at $z = \frac{175-75}{2} = 50$.
The center of the cube from the top platform is at $z = 110 - 175 = -65$.

#+begin_src matlab :results silent
  opts = struct(...
      'H_tot', 100,         ... % Total height of the Hexapod [mm]
      'L',     220/sqrt(3), ... % Size of the Cube [mm]
      'H',     60,          ... % Height between base joints and platform joints [mm]
      'H0',    75           ... % Height between the corner of the cube and the plane containing the base joints [mm]
      );
  stewart = initializeCubicConfiguration(opts);
  opts = struct(...
      'Jd_pos', [0, 0, -65], ... % Position of the Jacobian for displacement estimation from the top of the mobile platform [mm]
      'Jf_pos', [0, 0, -65]  ... % Position of the Jacobian for force location from the top of the mobile platform [mm]
      );
  stewart = computeGeometricalProperties(stewart, opts);
#+end_src

#+begin_src matlab :results none :exports code
  K = stewart.Jf'*stewart.Jf;
#+end_src

#+begin_src matlab :results value table :exports results
  data = K;
  data2orgtable(data', {}, {}, ' %.2g ');
#+end_src

#+RESULTS:
|        2 | -1.8e-17 |  2.6e-17 |  3.3e-18 |     0.04 |  1.7e-19 |
| -1.8e-17 |        2 |  1.9e-16 |    -0.04 |  2.2e-19 | -5.3e-19 |
|  2.6e-17 |  1.9e-16 |        2 | -8.9e-18 |  6.5e-19 | -5.8e-19 |
|  3.3e-18 |    -0.04 | -8.9e-18 |   0.0089 | -9.3e-20 |  9.8e-20 |
|     0.04 |  2.2e-19 |  6.5e-19 | -9.3e-20 |   0.0089 | -2.4e-18 |
|  1.7e-19 | -5.3e-19 | -5.8e-19 |  9.8e-20 | -2.4e-18 |    0.032 |

We obtain $k_x = k_y = k_z$ and $k_{\theta_x} = k_{\theta_y}$, but the Stiffness matrix is not diagonal.

** Cubic Stewart platform not centered with the cube center - Jacobian estimated at the Stewart platform center
Here, the "center" of the Stewart platform is not at the cube center.
The Jacobian is estimated at the center of the Stewart platform.

The center of the cube is at $z = 110$.
The Stewart platform is from $z = H_0 = 75$ to $z = H_0 + H_{tot} = 175$.
The center height of the Stewart platform is then at $z = \frac{175-75}{2} = 50$.
The center of the cube from the top platform is at $z = 110 - 175 = -65$.

#+begin_src matlab :results silent
  opts = struct(...
      'H_tot', 100, ... % Total height of the Hexapod [mm]
      'L',     220/sqrt(3), ... % Size of the Cube [mm]
      'H',     60, ... % Height between base joints and platform joints [mm]
      'H0',    75 ... % Height between the corner of the cube and the plane containing the base joints [mm]
      );
  stewart = initializeCubicConfiguration(opts);
  opts = struct(...
      'Jd_pos', [0, 0, -60], ... % Position of the Jacobian for displacement estimation from the top of the mobile platform [mm]
      'Jf_pos', [0, 0, -60]  ... % Position of the Jacobian for force location from the top of the mobile platform [mm]
      );
  stewart = computeGeometricalProperties(stewart, opts);
#+end_src

#+begin_src matlab :results none :exports code
  K = stewart.Jf'*stewart.Jf;
#+end_src

#+begin_src matlab :results value table :exports results
  data = K;
  data2orgtable(data', {}, {}, ' %.2g ');
#+end_src

#+RESULTS:
|        2 | -1.8e-17 |  2.6e-17 | -5.7e-19 |     0.03 |  1.7e-19 |
| -1.8e-17 |        2 |  1.9e-16 |    -0.03 |  2.2e-19 | -5.3e-19 |
|  2.6e-17 |  1.9e-16 |        2 | -1.5e-17 |  6.5e-19 | -5.8e-19 |
| -5.7e-19 |    -0.03 | -1.5e-17 |   0.0085 |  4.9e-20 |  1.7e-19 |
|     0.03 |  2.2e-19 |  6.5e-19 |  4.9e-20 |   0.0085 | -1.1e-18 |
|  1.7e-19 | -5.3e-19 | -5.8e-19 |  1.7e-19 | -1.1e-18 |    0.032 |

We obtain $k_x = k_y = k_z$ and $k_{\theta_x} = k_{\theta_y}$, but the Stiffness matrix is not diagonal.

** Conclusion
#+begin_important
  - The cubic configuration permits to have $k_x = k_y = k_z$ and $k_{\theta\x} = k_{\theta_y}$
  - The stiffness matrix $K$ is diagonal for the cubic configuration if the Stewart platform and the cube are centered *and* the Jacobian is estimated at the cube center
#+end_important

* TODO Cubic size analysis
We here study the effect of the size of the cube used for the Stewart configuration.

We fix the height of the Stewart platform, the center of the cube is at the center of the Stewart platform.

We only vary the size of the cube.

#+begin_src matlab :results silent
  H_cubes = 250:20:350;
  stewarts = {zeros(length(H_cubes), 1)};
#+end_src

#+begin_src matlab :results silent
  for i = 1:length(H_cubes)
    H_cube = H_cubes(i);
    H_tot = 100;
    H = 80;

    opts = struct(...
        'H_tot', H_tot, ... % Total height of the Hexapod [mm]
        'L',     H_cube/sqrt(3), ... % Size of the Cube [mm]
        'H',     H, ... % Height between base joints and platform joints [mm]
        'H0',    H_cube/2-H/2 ... % Height between the corner of the cube and the plane containing the base joints [mm]
        );
    stewart = initializeCubicConfiguration(opts);

    opts = struct(...
        'Jd_pos', [0, 0, H_cube/2-opts.H0-opts.H_tot], ... % Position of the Jacobian for displacement estimation from the top of the mobile platform [mm]
        'Jf_pos', [0, 0, H_cube/2-opts.H0-opts.H_tot]  ... % Position of the Jacobian for force location from the top of the mobile platform [mm]
        );
    stewart = computeGeometricalProperties(stewart, opts);
    stewarts(i) = {stewart};
  end
#+end_src


The Stiffness matrix is computed for all generated Stewart platforms.
#+begin_src matlab :results none :exports code
  Ks = zeros(6, 6, length(H_cube));
  for i = 1:length(H_cubes)
    Ks(:, :, i) = stewarts{i}.Jd'*stewarts{i}.Jd;
  end
#+end_src

The only elements of $K$ that vary are $k_{\theta_x} = k_{\theta_y}$ and $k_{\theta_z}$.

Finally, we plot $k_{\theta_x} = k_{\theta_y}$ and $k_{\theta_z}$
#+begin_src matlab :results none :exports code
  figure;
  hold on;
  plot(H_cubes, squeeze(Ks(4, 4, :)), 'DisplayName', '$k_{\theta_x}$');
  plot(H_cubes, squeeze(Ks(6, 6, :)), 'DisplayName', '$k_{\theta_z}$');
  hold off;
  legend('location', 'northwest');
  xlabel('Cube Size [mm]'); ylabel('Rotational stiffnes [normalized]');
#+end_src

#+NAME: fig:stiffness_cube_size
#+HEADER: :tangle no :exports results :results raw :noweb yes
#+begin_src matlab :var filepath="figs/stiffness_cube_size.pdf" :var figsize="normal-normal" :post pdf2svg(file=*this*, ext="png")
  <<plt-matlab>>
#+end_src

#+NAME: fig:stiffness_cube_size
#+CAPTION: $k_{\theta_x} = k_{\theta_y}$ and $k_{\theta_z}$ function of the size of the cube
#+RESULTS: fig:stiffness_cube_size
[[file:figs/stiffness_cube_size.png]]


We observe that $k_{\theta_x} = k_{\theta_y}$ and $k_{\theta_z}$ increase linearly with the cube size.

#+begin_important
  In order to maximize the rotational stiffness of the Stewart platform, the size of the cube should be the highest possible.
  In that case, the legs will the further separated. Size of the cube is then limited by allowed space.
#+end_important

* TODO initializeCubicConfiguration
:PROPERTIES:
:HEADER-ARGS:matlab+: :exports code
:HEADER-ARGS:matlab+: :comments no
:HEADER-ARGS:matlab+: :eval no
:HEADER-ARGS:matlab+: :tangle src/initializeCubicConfiguration.m
:END:
<<sec:initializeCubicConfiguration>>

** Function description
#+begin_src matlab
  function [stewart] = initializeCubicConfiguration(opts_param)
#+end_src

** Optional Parameters
Default values for opts.
#+begin_src matlab
  opts = struct(...
      'H_tot', 90,  ... % Total height of the Hexapod [mm]
      'L',     110, ... % Size of the Cube [mm]
      'H',     40,  ... % Height between base joints and platform joints [mm]
      'H0',    75   ... % Height between the corner of the cube and the plane containing the base joints [mm]
      );
#+end_src

Populate opts with input parameters
#+begin_src matlab
  if exist('opts_param','var')
      for opt = fieldnames(opts_param)'
          opts.(opt{1}) = opts_param.(opt{1});
      end
  end
#+end_src

** Cube Creation
#+begin_src matlab :results none
  points = [0, 0, 0; ...
            0, 0, 1; ...
            0, 1, 0; ...
            0, 1, 1; ...
            1, 0, 0; ...
            1, 0, 1; ...
            1, 1, 0; ...
            1, 1, 1];
  points = opts.L*points;
#+end_src

We create the rotation matrix to rotate the cube
#+begin_src matlab :results none
  sx = cross([1, 1, 1], [1 0 0]);
  sx = sx/norm(sx);

  sy = -cross(sx, [1, 1, 1]);
  sy = sy/norm(sy);

  sz = [1, 1, 1];
  sz = sz/norm(sz);

  R = [sx', sy', sz']';
#+end_src

We use to rotation matrix to rotate the cube
#+begin_src matlab :results none
  cube = zeros(size(points));
  for i = 1:size(points, 1)
    cube(i, :) = R * points(i, :)';
  end
#+end_src

** Vectors of each leg
#+begin_src matlab :results none
  leg_indices = [3, 4; ...
                 2, 4; ...
                 2, 6; ...
                 5, 6; ...
                 5, 7; ...
                 3, 7];
#+end_src

Vectors are:
#+begin_src matlab :results none
  legs = zeros(6, 3);
  legs_start = zeros(6, 3);

  for i = 1:6
    legs(i, :) = cube(leg_indices(i, 2), :) - cube(leg_indices(i, 1), :);
    legs_start(i, :) = cube(leg_indices(i, 1), :);
  end
#+end_src

** Verification of Height of the Stewart Platform
If the Stewart platform is not contained in the cube, throw an error.

#+begin_src matlab :results none
  Hmax = cube(4, 3) - cube(2, 3);
  if opts.H0 < cube(2, 3)
    error(sprintf('H0 is not high enought. Minimum H0 = %.1f', cube(2, 3)));
  else if opts.H0 + opts.H > cube(4, 3)
    error(sprintf('H0+H is too high. Maximum H0+H = %.1f', cube(4, 3)));
    error('H0+H is too high');
  end
#+end_src

** Determinate the location of the joints
We now determine the location of the joints on the fixed platform w.r.t the fixed frame $\{A\}$.
$\{A\}$ is fixed to the bottom of the base.
#+begin_src matlab :results none
  Aa = zeros(6, 3);
  for i = 1:6
    t = (opts.H0-legs_start(i, 3))/(legs(i, 3));
    Aa(i, :) = legs_start(i, :) + t*legs(i, :);
  end
#+end_src

And the location of the joints on the mobile platform with respect to $\{A\}$.
#+begin_src matlab :results none
  Ab = zeros(6, 3);
  for i = 1:6
    t = (opts.H0+opts.H-legs_start(i, 3))/(legs(i, 3));
    Ab(i, :) = legs_start(i, :) + t*legs(i, :);
  end
#+end_src

And the location of the joints on the mobile platform with respect to $\{B\}$.
#+begin_src matlab :results none
  Bb = zeros(6, 3);
  Bb = Ab - (opts.H0 + opts.H_tot/2 + opts.H/2)*[0, 0, 1];
#+end_src

#+begin_src matlab :results none
  h = opts.H0 + opts.H/2 - opts.H_tot/2;
  Aa = Aa - h*[0, 0, 1];
  Ab = Ab - h*[0, 0, 1];
#+end_src

** Returns Stewart Structure
#+begin_src matlab :results none
  stewart = struct();
  stewart.Aa = Aa;
  stewart.Ab = Ab;
  stewart.Bb = Bb;
  stewart.H_tot = opts.H_tot;
end
#+end_src

* TODO Tests
** First attempt to parametrisation
#+name: fig:stewart_bottom_plate
#+caption: Schematic of the bottom plates with all the parameters
[[file:./figs/stewart_bottom_plate.png]]

The goal is to choose $\alpha$, $\beta$, $R_\text{leg, t}$ and $R_\text{leg, b}$ in such a way that the configuration is cubic.


The configuration is cubic if:
\[ \overrightarrow{a_i b_i} \cdot \overrightarrow{a_j b_j} = 0, \ \forall i, j = [1, \hdots, 6], i \ne j \]

Lets express $a_i$, $b_i$ and $a_j$:
\begin{equation*}
  a_1 = \begin{bmatrix}R_{\text{leg,b}} \cos(120 - \alpha) \\  R_{\text{leg,b}} \cos(120 - \alpha) \\ 0\end{bmatrix} ; \quad
  a_2 = \begin{bmatrix}R_{\text{leg,b}} \cos(120 + \alpha) \\  R_{\text{leg,b}} \cos(120 + \alpha) \\ 0\end{bmatrix} ; \quad
\end{equation*}

\begin{equation*}
  b_1 = \begin{bmatrix}R_{\text{leg,t}} \cos(120 - \beta) \\  R_{\text{leg,t}} \cos(120 - \beta\\ H\end{bmatrix} ; \quad
  b_2 = \begin{bmatrix}R_{\text{leg,t}} \cos(120 + \beta) \\  R_{\text{leg,t}} \cos(120 + \beta\\ H\end{bmatrix} ; \quad
\end{equation*}

\[ \overrightarrow{a_1 b_1} = b_1 - a_1 = \begin{bmatrix}R_{\text{leg}} \cos(120 - \alpha) \\  R_{\text{leg}} \cos(120 - \alpha) \\ 0\end{bmatrix}\]

** Second attempt
We start with the point of a cube in space:
\begin{align*}
  [0, 0, 0] ; \ [0, 0, 1]; \ ...
\end{align*}

We also want the cube to point upward:
\[ [1, 1, 1] \Rightarrow [0, 0, 1] \]

Then we have the direction of all the vectors expressed in the frame of the hexapod.

#+begin_src matlab :results none
  points = [0, 0, 0; ...
            0, 0, 1; ...
            0, 1, 0; ...
            0, 1, 1; ...
            1, 0, 0; ...
            1, 0, 1; ...
            1, 1, 0; ...
            1, 1, 1];
#+end_src

#+begin_src matlab :results none
  figure;
  plot3(points(:,1), points(:,2), points(:,3), 'ko')
#+end_src

#+begin_src matlab :results none
  sx = cross([1, 1, 1], [1 0 0]);
  sx = sx/norm(sx);

  sy = -cross(sx, [1, 1, 1]);
  sy = sy/norm(sy);

  sz = [1, 1, 1];
  sz = sz/norm(sz);

  R = [sx', sy', sz']';
#+end_src

#+begin_src matlab :results none
  cube = zeros(size(points));
  for i = 1:size(points, 1)
    cube(i, :) = R * points(i, :)';
  end
#+end_src

#+begin_src matlab :results none
  figure;
  hold on;
  plot3(points(:,1), points(:,2), points(:,3), 'ko');
  plot3(cube(:,1), cube(:,2), cube(:,3), 'ro');
  hold off;
#+end_src

Now we plot the legs of the hexapod.
#+begin_src matlab :results none
  leg_indices = [3, 4; ...
                 2, 4; ...
                 2, 6; ...
                 5, 6; ...
                 5, 7; ...
                 3, 7]

  figure;
  hold on;
  for i = 1:6
    plot3(cube(leg_indices(i, :),1), cube(leg_indices(i, :),2), cube(leg_indices(i, :),3), '-');
  end
  hold off;
#+end_src

Vectors are:
#+begin_src matlab :results none
  legs = zeros(6, 3);
  legs_start = zeros(6, 3);

  for i = 1:6
    legs(i, :) = cube(leg_indices(i, 2), :) - cube(leg_indices(i, 1), :);
    legs_start(i, :) = cube(leg_indices(i, 1), :)
  end
#+end_src

We now have the orientation of each leg.

We here want to see if the position of the "slice" changes something.

Let's first estimate the maximum height of the Stewart platform.
#+begin_src matlab :results none
  Hmax = cube(4, 3) - cube(2, 3);
#+end_src

Let's then estimate the middle position of the platform
#+begin_src matlab :results none
  Hmid = cube(8, 3)/2;
#+end_src

** Generate the Stewart platform for a Cubic configuration

First we defined the height of the Hexapod.
#+begin_src matlab :results none
  H = Hmax/2;
#+end_src

#+begin_src matlab :results none
  Zs = 1.2*cube(2, 3); % Height of the fixed platform
  Ze = Zs + H; % Height of the mobile platform
#+end_src

We now determine the location of the joints on the fixed platform.
#+begin_src matlab :results none
  Aa = zeros(6, 3);
  for i = 1:6
    t = (Zs-legs_start(i, 3))/(legs(i, 3));
    Aa(i, :) = legs_start(i, :) + t*legs(i, :);
  end
#+end_src

And the location of the joints on the mobile platform
#+begin_src matlab :results none
  Ab = zeros(6, 3);
  for i = 1:6
    t = (Ze-legs_start(i, 3))/(legs(i, 3));
    Ab(i, :) = legs_start(i, :) + t*legs(i, :);
  end
#+end_src

And we plot the legs.
#+begin_src matlab :results none
  figure;
  hold on;
  for i = 1:6
    plot3([Ab(i, 1),Aa(i, 1)], [Ab(i, 2),Aa(i, 2)], [Ab(i, 3),Aa(i, 3)], 'k-');
  end
  hold off;
  xlim([-1, 1]);
  ylim([-1, 1]);
  zlim([0, 2]);
#+end_src

* Bibliography                                                      :ignore:
bibliographystyle:unsrt
bibliography:ref.bib
