Tomography Experiment

Let's define the wanted position of each stage.

Ty = 1; % [m]
Ry = 3*pi/180; % [rad]
Rz = 180*pi/180; % [rad]

% Hexapod (first consider only translations)
Thx = 0; % [m]
Thy = 0; % [m]
Thz = 0; % [m]

Now, we compute the corresponding wanted translation and rotation of the sample with respect to the granite frame \(\{W\}\). This corresponds to \({}^WO_T\) and \(\theta_m {}^Ws_m\).

To do so, we have to define the homogeneous transformation for each stage.

% Translation Stage
Rty = [1 0 0 0;
       0 1 0 Ty;
       0 0 1 0;
       0 0 0 1];

% Tilt Stage - Pure rotating aligned with Ob
Rry = [ cos(Ry) 0 sin(Ry) 0;
        0       1 0       0;
       -sin(Ry) 0 cos(Ry) 0;
        0       0 0       1];

% Spindle - Rotation along the Z axis
Rrz = [cos(Rz) -sin(Rz) 0 0 ;
       sin(Rz)  cos(Rz) 0 0 ;
       0        0       1 0 ;
       0        0       0 1 ];

% Micro-Hexapod (only rotations first)
Rh = [1 0 0 Thx ;
      0 1 0 Thy ;
      0 0 1 Thz ;
      0 0 0 1 ];

We combine the individual homogeneous transformations into one homogeneous transformation for all the station.

Ttot = Rty*Rry*Rrz*Rh;

Using this homogeneous transformation, we can compute the wanted position and orientation of the sample with respect to the granite.

WOr = Ttot*[0;0;0;1];
WOr = WOr(1:3);

thetar = acos((trace(Ttot(1:3, 1:3))-1)/2)
if thetar == 0
  WSr = [0; 0; 0];
else
  [V, D] = eig(Ttot(1:3, 1:3));
  WSr = thetar*V(:, abs(diag(D) - 1) < eps(1));
end

WPr = [WOr ; WSr];

The measurement of the position of the sample using the metrology system gives the position and orientation of the sample with respect to the granite.

% Measurements: Xm, Ym, Zm, Rx, Ry, Rz
Dxm = 0; % [m]
Dym = 1; % [m]
Dzm = 0; % [m]

Rxm = 0*pi/180; % [rad]
Rym = 3*pi/180; % [rad]
Rzm = 0*pi/180; % [rad]

Let's compute the corresponding orientation using screw axis.

Trxm = [1 0         0;
        0 cos(Rxm) -sin(Rxm);
        0 sin(Rxm)  cos(Rxm)];
Trym = [ cos(Rym) 0 sin(Rym);
         0        1 0;
        -sin(Rym) 0 cos(Rym)];
Trzm = [cos(Rzm) -sin(Rzm) 0;
        sin(Rzm)  cos(Rzm) 0;
        0         0        1];

STw = [[ Trym*Trxm*Trzm , [Dxm; Dym; Dzm]]; 0 0 0 1];

We then obtain the orientation measurement in the form of screw coordinate \(\theta_m ({}^Ws_{x,m},\ {}^Ws_{y,m},\ {}^Ws_{z,m})^T\) where:

• \(\theta_m = \cos^{-1} \frac{\text{Tr}(R) - 1}{2}\)
• \({}^W\bm{s}_m\) is the eigen vector of the rotation matrix \(R\) corresponding to the eigen value \(\lambda = 1\)

thetam = acos((trace(STw(1:3, 1:3))-1)/2); % [rad]
if thetam == 0
  WSm = [0; 0; 0];
else
  [V, D] = eig(STw(1:3, 1:3));
  WSm = thetam*V(:, abs(diag(D) - 1) < eps(1));
end

WPm = [Dxm ; Dym ; Dzm ; WSm];

The wanted position expressed with respect to the granite is \({}^WO_T\) and the measured position with respect to the granite is \({}^WO_S\), thus the position error expressed in \(\{W\}\) is \[ {}^W E = {}^W O_T - {}^W O_S \] The same is true for rotations: \[ \theta_\epsilon {}^W\bm{s}_\epsilon = \theta_r {}^W\bm{s}_r - \theta_m {}^W\bm{s}_m \]

WPe = WPr - WPm;

Now we want to express this error in a frame attached to the base of the nano-hexapod with its origin at the same point where the Jacobian of the nano-hexapod is computed (175mm above the top platform + 90mm of total height of the nano-hexapod).

Or maybe should we want to express this error with respect to the top platform of the nano-hexapod? We are measuring the position of the top-platform, and we don't know exactly the position of the bottom platform. We could compute the position of the bottom platform in two ways:

• from the encoders of each stage
• from the measurement of the nano-hexapod top platform + the internal metrology in the nano-hexapod (capacitive sensors e.g)

A third option is to say that the maximum stroke of the nano-hexapod is so small that the error should no change to much by the change of base.

We now want the position error to be expressed in \(\{S\}\) (the frame attach to the sample): \[ {}^S E = {}^S T_W \cdot {}^W E \]

Thus we need to compute the homogeneous transformation \({}^ST_W\). Fortunately, this homogeneous transformation can be computed from the measurement of the sample position and orientation with respect to the granite.

Trxm = [1 0         0;
        0 cos(Rxm) -sin(Rxm);
        0 sin(Rxm)  cos(Rxm)];
Trym = [ cos(Rym) 0 sin(Rym);
         0        1 0;
        -sin(Rym) 0 cos(Rym)];
Trzm = [cos(Rzm) -sin(Rzm) 0;
        sin(Rzm)  cos(Rzm) 0;
        0         0        1];

STw = [[ Trym*Trxm*Trzm , [Dxm; Dym; Dzm]]; 0 0 0 1];

Translation Error.

SEm = STw * [WPe(1:3); 1];
SEm = SEm(1:3);

Rotation Error.

SEr = STw * [WPe(4:6); 0];
SEr = SEr(1:3);

Etot = [SEm ; SEr]