Kinematic Analysis 1 Problem Chapter I 7 The planar R RTR RTR mechanism considered is shown in Fig 1 The driver link is the rigid link 1 the link AB The following numerical data are 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 30 The constant angular speed of the driver link 1 is 50 rpm Position analysis for an input angle Position of joint A A Cartesian reference frame xOy is selected The joint A is in the origin of the reference frame that is A O xA 0 yA 0 1 Position of joint C The coordinates of the joint C are xC 0 yC AC 0 060 m 2 Position of joint E The coordinates of the joint E are xE 0 yE AE 0 250 m 3 Position of joint B The unknowns are the coordinates of the joint B xB and yB Because the joint A is fixed and the angle is known the coordinates of the joint B are computed from the following expressions xB AB cos 0 140 cos 30 0 121 m yB AB sin 0 140 sin 30 0 070 m Position of joint D The unknowns are the coordinates of the joint D xD and yD The joint D is located on the line BC yB yC yD yC or xD xC xB xC xB xC yD yC xD xC yB yC or 0 121 0 yD 0 060 xD 0 0 070 0 060 4 5 Kinematic Analysis 2 Furthermore the length of the segment CD is constant xC xD 2 yC yD 2 CD2 or 0 xD 2 0 060 yD 2 0 1502 6 The Eqs 5 and 6 form a system from which the coordinates of the joint D can be computed Two sets of solutions are found for the position of the joint D These solutions are located at the intersection 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 additional condition is needed For the first quadrant 0 90 the condition is xD xC Because xC 0 m the coordinates of the joint D are xD xD1 0 149 m yD yD1 0 047 m Angle 2 The angle of link 2 or link 3 with the horizontal axis is calculated from the slope of the straight line BC 2 3 arctan yB yC 0 070 0 060 arctan 0 082 rad 4 715 xB xC 0 121 0 Angle 4 The angle of link 5 or link 4 with the horizontal axis is obtained from the slope of the straight line ED 4 5 arctan yE yD 0 250 0 047 arctan 1 105 rad 63 333 xE xD 0 0 149 Kinematic Analysis 3 Velocity and Acceleration Analysis Algebraic Method The velocity of the point B B1 on the link 1 is vB vB1 vA 1 rAB 1 rB where vA 0 is the velocity of the origin A O The angular velocity of link 1 is 1 1 k 50 n k k 5 235k rad s 30 30 the position vector of point B is rAB rB rA rB xB yB zB k 0 121 0 070 m The velocity of point B2 on the link 2 is vB2 vB1 because between the links 1 and 2 there is a rotational joint The velocity of B1 B2 is vB1 vB2 k k 0 0 0 0 5 235 0 366 0 634 m s xB yB 0 0 121 0 070 0 The acceleration of the point B B1 on the link 1 is aB aB1 aB2 aA 1 rB 1 1 rB 1 rB 21 rB 21 rB 5 2352 0 121 0 070 3 323 1 919 m s2 The angular acceleration of link 1 is 1 1 0 The velocity of the point B3 on the link 3 is calculated in terms of the velocity of the point B2 on the link 2 rel vB3 vB2 vB vB2 vB32 3 B2 7 rel where vB vB32 is the relative acceleration of B3 with respect to B2 on 3 B2 link 3 This relative velocity is parallel to the sliding direction BC vB32 BC or vB32 vB32 cos 2 vB32 sin 2 8 Kinematic Analysis 4 where 2 4 715 is known from position analysis The points B3 and C are on the link 3 and vB3 vC 3 rCB 3 rB rC 9 where vC 0 and the angular velocity of link 3 is 3 3 k Equations 7 8 and 9 give k 0 0 3 xB xC yB yC 0 vB2 vB32 cos 2 vB32 sin 2 10 Equation 10 represents a vectorial equations with two scalar components on x axis and y axis and with two unknowns 3 and vB32 3 yB yC vBx vB32 cos 2 3 xB xC vBy vB32 sin 2 or 3 0 070 0 060 0 366 vB32 cos 4 715 3 0 121 0 0 634 vB32 sin 4 715 It results 3 2 5 448 rad s and vB32 0 313 m s The acceleration of the point B3 on the link 3 is calculated in terms of the acceleration of the point B2 on the link 2 cor cor aB3 aB2 arel B3 B2 aB3 B2 aB2 aB32 aB32 11 where arel B3 B2 aB32 is the relative acceleration of B3 with respect to B2 on link 3 This relative acceleration is parallel to the sliding direction BC aB32 BC or aB32 aB32 cos 2 aB32 sin 2 12 Kinematic Analysis 5 The Coriolis acceleration of B3 realative to B2 is k 0 0 3 vB32 cos 2 vB32 sin 2 0 2 3 vB32 sin 2 3 vB32 cos 2 2 5 448 0 313 sin 4 715 5 448 0 313 cos 4 715 0 280 3 400 m s2 acor B32 2 3 vB32 2 2 vB32 2 13 The points B3 and C are on the link 3 and aB3 aC 3 rCB 32 rCB 14 where aC 0 and the angular acceleration of link 3 is 3 3 k Equations 11 12 13 and 14 give k 0 0 3 32 rB rC xB xC yB yC 0 aB2 aB32 cos 2 sin 2 2 3 vB32 15 Equation 15 represents a vectorial equations with two scalar components on x axis and y axis and with two unknowns 3 and aB32 3 yB yC 32 xB xC aBx aB32 cos 2 2 3 vB32 sin 2 3 xB xC 32 yB yC aBy aB32 sin 2 2 3 vB32 cos 2 or 3 0 070 0 060 5 4482 0 121 0 3 323 aB32 cos 4 715 2 5 448 0 313 sin 4 715 3 0 121 0 5 4482 0 070 0 060 1 919 …