125 lines
5.0 KiB
Matlab
125 lines
5.0 KiB
Matlab
%% Clear Workspace and Close figures
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clear; close all; clc;
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%% Intialize Laplace variable
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s = zpk('s');
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simulinkproject('../');
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% Stewart architecture definition
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% Let's first define the Stewart platform architecture that we want to study.
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stewart = initializeStewartPlatform();
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stewart = initializeFramesPositions(stewart, 'H', 90e-3, 'MO_B', 45e-3);
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stewart = generateGeneralConfiguration(stewart);
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stewart = computeJointsPose(stewart);
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stewart = initializeStewartPose(stewart);
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stewart = initializeCylindricalPlatforms(stewart);
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stewart = initializeCylindricalStruts(stewart);
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stewart = initializeStrutDynamics(stewart);
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stewart = initializeJointDynamics(stewart);
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stewart = computeJacobian(stewart);
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% Wanted translations and rotations
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% Let's now define the wanted extreme translations and rotations.
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Tx_max = 50e-6; % Translation [m]
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Ty_max = 50e-6; % Translation [m]
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Tz_max = 50e-6; % Translation [m]
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Rx_max = 30e-6; % Rotation [rad]
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Ry_max = 30e-6; % Rotation [rad]
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Rz_max = 0; % Rotation [rad]
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% Needed stroke for "pure" rotations or translations
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% As a first estimation, we estimate the needed actuator stroke for "pure" rotations and translation.
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% We do that using either the Inverse Kinematic solution or the Jacobian matrix as an approximation.
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LTx = stewart.kinematics.J*[Tx_max 0 0 0 0 0]';
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LTy = stewart.kinematics.J*[0 Ty_max 0 0 0 0]';
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LTz = stewart.kinematics.J*[0 0 Tz_max 0 0 0]';
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LRx = stewart.kinematics.J*[0 0 0 Rx_max 0 0]';
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LRy = stewart.kinematics.J*[0 0 0 0 Ry_max 0]';
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LRz = stewart.kinematics.J*[0 0 0 0 0 Rz_max]';
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% The obtain required stroke is:
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ans = sprintf('From %.2g[m] to %.2g[m]: Total stroke = %.1f[um]', min(min([LTx,LTy,LTz,LRx,LRy])), max(max([LTx,LTy,LTz,LRx,LRy])), 1e6*(max(max([LTx,LTy,LTz,LRx,LRy]))-min(min([LTx,LTy,LTz,LRx,LRy]))))
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% Needed stroke for "combined" rotations or translations
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% We know would like to have a more precise estimation.
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% To do so, we may estimate the required actuator stroke for all possible combination of translation and rotation.
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% Let's first generate all the possible combination of maximum translation and rotations.
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Ps = [2*(dec2bin(0:5^2-1,5)-'0')-1, zeros(5^2, 1)].*[Tx_max Ty_max Tz_max Rx_max Ry_max Rz_max];
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% #+RESULTS:
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% | *Tx [m]* | *Ty [m]* | *Tz [m]* | *Rx [rad]* | *Ry [rad]* | *Rz [rad]* |
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% |----------+----------+----------+------------+------------+------------|
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% | -5.0e-05 | -5.0e-05 | -5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | -5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | -5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | -5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | 5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | 5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | 5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | -5.0e-05 | 5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | -5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | -5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | -5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | -5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | 5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | 5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | 5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | -5.0e-05 | 5.0e-05 | 5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | -5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | -5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | -5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | -5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | 5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | 5.0e-05 | -3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | 5.0e-05 | 3.0e-05 | -3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | -5.0e-05 | 5.0e-05 | 3.0e-05 | 3.0e-05 | 0.0e+00 |
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% | 5.0e-05 | 5.0e-05 | -5.0e-05 | -3.0e-05 | -3.0e-05 | 0.0e+00 |
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% For all possible combination, we compute the required actuator stroke using the inverse kinematic solution.
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L_min = 0;
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L_max = 0;
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for i = 1:size(Ps,1)
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Rx = [1 0 0;
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0 cos(Ps(i, 4)) -sin(Ps(i, 4));
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0 sin(Ps(i, 4)) cos(Ps(i, 4))];
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Ry = [ cos(Ps(i, 5)) 0 sin(Ps(i, 5));
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0 1 0;
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-sin(Ps(i, 5)) 0 cos(Ps(i, 5))];
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Rz = [cos(Ps(i, 6)) -sin(Ps(i, 6)) 0;
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sin(Ps(i, 6)) cos(Ps(i, 6)) 0;
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0 0 1];
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ARB = Rz*Ry*Rx;
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[~, Ls] = inverseKinematics(stewart, 'AP', Ps(i, 1:3)', 'ARB', ARB);
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if min(Ls) < L_min
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L_min = min(Ls)
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end
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if max(Ls) > L_max
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L_max = max(Ls)
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end
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end
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% We obtain the required actuator stroke:
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ans = sprintf('From %.2g[m] to %.2g[m]: Total stroke = %.1f[um]', L_min, L_max, 1e6*(L_max-L_min))
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