Position Analysis 1Position AnalysisThe planar R-RTR-RTR mechanism considered is shown in Fig. 1. Thedriver link is the rigid link 1 (the link AB). The following numerical dataare given: AB = 0.140 m, AC = 0.060 m, AE = 0.250 m, CD = 0.150 m.The angle of the driver link 1 with the horizontal axis is φ = 30◦.Position analysis for an input anglePosition of joint AA Cartesian reference frame xOy is selected. The joint A is in the originof the reference frame, that is, A ≡ O,xA= 0, yA= 0. (1)Position of joint CThe coordinates of the joint C arexC= 0, yC= AC = 0.060 m. (2)Position of joint EThe coordinates of the joint E arexE= 0, yE= −AE = −0.250 m. (3)Position of joint BThe unknowns are the coordinates of the joint B, xBand yB. Becausethe joint A is fixed and the angle φ is known, the coordinates of the joint Bare computed from the following expressionsxB= AB cos φ = 0.140 cos 30◦= 0.121 m,yB= AB sin φ = 0.140 sin 30◦= 0.070 m. (4)Position of joint DThe unknowns are the coordinates of the joint D, xDand yD.The joint D is located on the line BC:yD− yCxD− xC=yB− yCxB− xCor(xB− xC)(yD− yC) = (xD− xC)(yB− yC) (5)Position Analysis 2Furthermore, the length of the segment CD is constant:(xC− xD)2+ (yC− yD)2= CD2. (6)The Eqs. (5) and (6) form a system from which the coordinates of thejoint D can be computed. To solve the system of equations, a specificMatlab/MathematicaT Mcommand will be used. Two sets of solutions arefound for the position of the joint D. These solutions are located at the inter-section of the line BC with the circle centered in C and radius CD (Fig. 2),and they have the following numerical values:xD1= −0.149 m, yD1= 0.047 m,xD2= 0.149 m, yD2= 0.072 m.To determine the correct position of the joint D for the mechanism, an ad-ditional condition is needed.For the first quadrant, 0 ≤ φ ≤ 90◦, the condition is xD≤ xC.Because xC= 0 m, the coordinates of the joint D arexD= xD1= −0.149 m,yD= yD1= 0.047 m.Angle φ2The angle of link 2 (or link 3) with the horizontal axis is calculated fromthe slope of the straight line BC:φ2= φ3= arctanyB− yCxB− xC.Angle φ4The angle of link 5 (or link 4) with the horizontal axis is obtained fromthe slope of the straight line ED:φ4= φ5= arctanyE− yDxE− xD.The Matlab /MathematicaT Mprogram for the input angle φ = 30◦isgiven in Program 1.Position Analysis 3Position analysis for a complete rotationFor a complete rotation of the driver link AB, 0 ≤ φ ≤ 360◦, a step angleof φ = 60◦is selected.Method IMethod I uses constraint conditions for the mechanism for each quadrant.For the mechanism, there are several conditions for the position of the jointD.For the angle φ located in the first quadrant 0◦≤ φ ≤ 90◦(Fig. 2), andthe fourth quadrant 270◦≤ φ ≤ 360◦(Fig. 5), the following relation existsbetween xDand xC:xD≤ xC.For the angle φ located in the second quadrant 90◦< φ ≤ 180◦(Fig. 3),and the third quadrant 180◦< φ < 270◦(Fig. 4), the following relation existsbetween xDand xC:xD≥ xC.The Matlab/MathematicaT Mprogram for a complete rotation of thedriver link using method I is given in Program 2. The graph of the mecha-nism for a complete rotation of the driver link is given in Fig. 6.Method IIAnother position analysis method for a complete rotation of the driverlink uses constraint conditions for the initial value of the angle φ. For themechanism, the correct position of the joint D is calculated using a simplefunction, the Euclidian distance between two points P and Q:d =q(xP− xQ)2+ (yP− yQ)2. (7)For the initial angle φ = 0◦, the constraint is xD≤ xC, so the first position ofthe joint D, that is, D0, is calculated for the first step D = D0= Dk, k = 0.For the next position of the joint, Dk+1, there are two solutions DIk+1andDIIk+1, k = 0, 1, 2, .... In order to choose the correct solution of the joint,Dk+1, it is compared the distances between the old position, Dk, and eachnew calculated positions DIk+1and DIIk+1. The distances between the knownsolution Dkand the new solutions DIk+1and DIIk+1are dIkand dIIk. If thedistance to the first solution is less than the distance to the second solution,Position Analysis 4dIk< dIIk, then the correct answer is Dk+1= DIk+1, or else Dk+1= DIIk+1(Fig. 7). With this algorithm the correct solution is selected using just oneconstraint relation for the initial step and then, automatically, the problemis solved. In this way it is not necessary to have different constraints fordifferent quadrants.The Matlab/MathematicaT Mprogram for a complete rotation of thedriver link using the second method is given in Program 3.% Program 1% R−RTR−RTR % Position analysis − input angle phi clear; clc; close all % Input data AB = 0.14 ; AC = 0.06 ; AE = 0.25 ; CD = 0.15 ; %(m) phi = pi/6 ; % (rad)% Select the dimensionsDF=0.4 ; EG=0.5 ; % (m) xA = 0 ; yA = 0 ; rA = [xA yA 0] ; % Position of A xC = 0 ; yC = AC ; rC = [xC yC 0] ; % Position of C xE = 0 ; yE = −AE ; rE = [xE yE 0] ; % Position of E xB = AB*cos(phi) ; yB = AB*sin(phi) ; rB = [xB yB 0] ; % Position of B % Position of joint D % Distance formula: CD=constanteqnD1 = ʼ’( xDsol − xC )^2 + ( yDsol − yC )^2 = CD^2 ʼ’;% Slope formula: B, C, and D are on the same straight line eqnD2 = ʼ’( yB − yC ) / ( xB − xC ) = ( yDsol − yC ) / ( xDsol − xC )ʼ’;% Simultaneously solve above equationssolD = solve(eqnD1, eqnD2, ʼ’xDsol, yDsolʼ’);% solve symbolic solution of algebraic equations% Two solutions for xD − vector formxDpositions = eval(solD.xDsol);% eval execute string as an expression or statement% Two solutions for yD − vector formyDpositions = eval(solD.yDsol);% Separate the solutions in scalar formxD1 = xDpositions(1); % first component of the vector xDpositionsxD2 = xDpositions(2); % second component of the vector xDpositionsyD1 = yDpositions(1); % first component of the vector yDpositionsyD2 = yDpositions(2); % second component of the vector yDpositions % Select the correct position for D for the given input angleif xD1 <= xC xD = xD1; yD=yD1;else xD = xD2; yD=yD2;endrD = [xD yD 0]; % Position of D % Angles of the links with the horizontalphi2 = atan((yB−yC)/(xB−xC));phi3 = phi2;phi4 = atan((yD−yE)/(xD−xE))+pi;phi5 = phi4; % Positions of the points F and G xF = xD + DF*cos(phi3) ; yF = yD + DF*sin(phi3) ; rF = [xF yF 0]; % Position vector of F xG = xE + EG*cos(phi5) ; yG = yE + EG*sin(phi5) ;rG = [xG yG 0]; %