Parameters for Industrial Robot - Revision history

TobiasWeber at 14:51, 28 July 2016

2016-07-28T14:51:55Z

← Older revision		Revision as of 14:51, 28 July 2016
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	g_3 = 0.41398104470607;		g_3 = 0.41398104470607;

	%g(1) = 0;		g(1) = 0;
	%g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);		g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);
	%g(3) = g_2*sinq2q3;		g(3) = g_2*sinq2q3;

	f=zeros(3,1); % friction torque vector		f=zeros(3,1); % friction torque vector

FelixMueller at 08:37, 27 July 2016

2016-07-27T08:37:10Z

← Older revision		Revision as of 08:37, 27 July 2016
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	<source lang="Matlab">		<source lang="Matlab">
	% ~~Initialization~~		% Robot dynamics with 3 DOF, this function should be equal to zero at all
			% times. e.g. f(t,x=[q,dq],dx=[dq,ddq])=0! This model is only containing
			% the inerta Matrix and the torques of the joints. Friction, gravity and coriolis
			% effects are incorporated.

			function diff=f_full_effects(~,y,dy,tau)

	dq=zeros(3,1);		dq=zeros(3,1);
	dotq=zeros(3,1);		dotq=zeros(3,1);
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	ddq(1:3,1)=dy(4:6);		ddq(1:3,1)=dy(4:6);

	% ~~Intermediate~~ states		% intermediate states

	cosq=zeros(3,1);		cosq=zeros(3,1);
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	b_12= 0.422954084243519;		b_12= 0.422954084243519;

	b=zeros(3,3); % Inerta Matrix of Robot Modell (3DOF)		b=zeros(3,3); % Inerta Matrix of Robot Modell (3DOF)

	b(1,1) = b_1(sinq(2)^2) - b_2sin2q2 + b_3(cosq(2)^2) + b_4cosq(3) + b_5(cosq2q3^2) - b_6cos(2q(2) - 2q(3)) - b_4cos(2q(2) - q(3)) - b_7;		b(1,1) = b_1(sinq(2)^2) - b_2sin2q2 + b_3(cosq(2)^2) + b_4cosq(3) + b_5(cosq2q3^2) - b_6cos(2q(2) - 2q(3)) - b_4cos(2q(2) - q(3)) - b_7;
Line 72:		Line 77:
	g_3 = 0.41398104470607;		g_3 = 0.41398104470607;

	g(1) = 0;		%g(1) = 0;
	g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);		%g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);
	g(3) = g_2*sinq2q3;		%g(3) = g_2*sinq2q3;

	f=zeros(3,1); % friction torque vector		f=zeros(3,1); % friction torque vector
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	c(2,3) = c_13(-(-1)(-2)dq(2)/5 - (-1)2dq(3)/5)sinq(3);		c(2,3) = c_13(-(-1)(-2)dq(2)/5 - (-1)2dq(3)/5)sinq(3);
	c(3,3) = 0;		c(3,3) = 0;


			diff=[dq-dotq;...
			bddq+g+cdq+f-tau'];

	</source>		</source>

TobiasWeber at 08:28, 27 July 2016

2016-07-27T08:28:50Z

← Older revision		Revision as of 08:28, 27 July 2016
Line 2:		Line 2:

	<source lang="Matlab">		<source lang="Matlab">
	%~~initialization~~		% Initialization

	dq=zeros(3,1);		dq=zeros(3,1);
Line 13:		Line 13:
	ddq(1:3,1)=dy(4:6);		ddq(1:3,1)=dy(4:6);

	% ~~intermediate~~ states		% Intermediate states

	cosq=zeros(3,1);		cosq=zeros(3,1);

FelixMueller at 08:02, 27 July 2016

2016-07-27T08:02:51Z

← Older revision		Revision as of 08:02, 27 July 2016
Line 72:		Line 72:
	g_3 = 0.41398104470607;		g_3 = 0.41398104470607;

	%g(1) = 0;		g(1) = 0;
	%g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);		g(2) =-g_1sinq(2) - g_2sinq2q3 + g_3*cosq(2);
	%g(3) = g_2*sinq2q3;		g(3) = g_2*sinq2q3;

	f=zeros(3,1); % friction torque vector		f=zeros(3,1); % friction torque vector

FelixMueller at 15:54, 26 July 2016

2016-07-26T15:54:09Z

← Older revision		Revision as of 15:54, 26 July 2016
Line 2:		Line 2:

	<source lang="Matlab">		<source lang="Matlab">
	%initialization		%initialization

	dq=zeros(3,1);		dq=zeros(3,1);

FelixMueller: Created page with "This page contains a matlab script for the implicit formulation of the parameters for the Industrial robot problem. %initialization dq=zeros..."

2016-07-26T15:54:02Z

Created page with "This page contains a matlab script for the implicit formulation of the parameters for the Industrial robot problem. <source lang="Matlab"> %initialization dq=zeros..."

New page

This page contains a matlab script for the implicit formulation of the parameters for the [[Industrial robot]] problem.

<source lang="Matlab">
%initialization

dq=zeros(3,1);
dotq=zeros(3,1);
ddq=zeros(3,1);

q=y(1:3);
dq(1:3,1)=y(4:6);
dotq(1:3,1)=dy(1:3);
ddq(1:3,1)=dy(4:6);

% intermediate states

cosq=zeros(3,1);

cosq(1,1) = cos(q(1));
cosq(2,1) = cos(q(2));
cosq(3,1) = cos(q(3));

sinq=zeros(3,1);

sinq(1,1) = sin(q(1));
sinq(2,1) = sin(q(2));
sinq(3,1) = sin(q(3));

cosq2q3 = cos(q(2) - q(3));

sinq1q2= sin(q(1) - q(2));
sinq2q3= sin(q(2) - q(3));

sin2q2= sin(2*q(2));

sin2q22q3= sin(2*q(2) - 2*q(3));
sin2q2q3= sin(2*q(2) - q(3));

cos2q2= cos(2*q(2));

% Modell parameters

b_1 = 2.6491674286195;
b_2 = 0.0168799610481578;
b_3 = 1.43009916890323;
b_4 = 0.53524835732352;
b_5 = 0.791116620901176;
b_6 = 0.626240578068518;
b_7 = 1.57199673626875;
b_8 = 0.0175195488386143;
b_9 = 0.0204978721411787;
b_10= 1.07049671464704;
b_11= 1.81771647041588;
b_12= 0.422954084243519;

b=zeros(3,3); % Inerta Matrix of Robot Modell (3DOF)

b(1,1) = b_1*(sinq(2)^2) - b_2*sin2q2 + b_3*(cosq(2)^2) + b_4*cosq(3) + b_5*(cosq2q3^2) - b_6*cos(2*q(2) - 2*q(3)) - b_4*cos(2*q(2) - q(3)) - b_7;
b(2,1) = b_8*cosq(2) + b_9*cosq2q3;
b(3,1) = -b_9*cosq2q3;
b(1,2) = b_8*cosq(2) + b_9*cosq2q3;
b(2,2) = b_10*cosq(3) + b_11;
b(3,2) = -b_4*cosq(3) - b_12;
b(1,3) = -b_9*cosq2q3;
b(2,3) = -b_4*cosq(3) - b_12;
b(3,3) = b_12;

g=zeros(3,1); % gravity torque vector

g_1 = 32.4854155934467;
g_2 = 13.1269659633593;
g_3 = 0.41398104470607;

%g(1) = 0;
%g(2) =-g_1*sinq(2) - g_2*sinq2q3 + g_3*cosq(2);
%g(3) = g_2*sinq2q3;

f=zeros(3,1); % friction torque vector

f(1) = 0.186461043875677*atan(100*dq(1));
f(2) = 0.184489618216264*atan(100*dq(2));
f(3) = 0.156066608371045*atan(100*dq(3));

c=zeros(3,3); % coriolis matrix

c_1 = 0.609534129858137;
c_2 = 0.23068226761793;
c_3 = b_4;
c_4 = b_2;
c_5 = 0.26762417866176;
c_6 = 152.383532464534;
c_7 = 57.6705669044826;
c_8 = 133.81208933088;
c_9 = 4.21999026203945;
c_10= 66.90604466544;
c_11= b_8;
c_12= b_9;
c_13= 1.3381208933088;

c(1,1) = c_1*dq(2)*sin2q2 + c_2*dq(2)*sin2q22q3 + c_3*dq(2)*sin2q2q3 - c_4*dq(2)*cos2q2 - c_5*dq(3)*sinq(3) - c_2*dq(3)*sin2q22q3 - c_5*dq(3)*sin2q2q3;
c(2,1) = -(-1)*dq(1)*(-c_6*sin2q2 - c_7*sin2q22q3 - c_8*sin2q2q3 + c_9*cos2q2)/250;
c(3,1) = -(-1)*(-1)*dq(1)*(-c_10*sinq(3) - c_7*sin2q22q3 - c_10*sin2q2q3)/250;
c(1,2) = -(-1)*(-1)*dq(1)*(-c_6*sin2q2 - c_7*sin2q22q3 - c_8*sin2q2q3 + c_9*cos2q2)/250 + dq(2)*(-c_11*sinq(2) - c_12*sinq2q3) + c_12*dq(3)*sinq2q3;
c(2,2) = -c_3*dq(3)*sinq(3);
c(3,2) = c_3*dq(2)*sinq(3);
c(1,3) = -c_5*dq(1)*sinq(3) - c_2*dq(1)*sin2q22q3 - c_5*dq(1)*sin2q2q3 + c_12*dq(2)*sinq2q3 - c_12*dq(3)*sinq2q3;
c(2,3) = c_13*(-(-1)*(-2)*dq(2)/5 - (-1)*2*dq(3)/5)*sinq(3);
c(3,3) = 0;

</source>