WORKED SOLUTIONS (2019) Page 1 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 1a The reference frames, the first being fixed to the base while the other three are fixed to the links and are chosen according to the DH convention. The link parameters are summarized in the table. Table Link parameters for the first three links of the PUMA 560 manipulator Link no. ai, i di i Link 1 0 90° h 1 Link 2 a1 0 0 2 Link 3 a2 0 0 3 3 3 1b. The homogeneous transformation for the first link is, − − = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 cos sin 0 0 sin cos 11 11 1 ,0 h θθ θθ T , that is, − = − − = 1 0 0 0 0 1 0 0 cos 0 sin 0 sin 0 cos 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 cos sin 0 0 sin cos 11 11 11 11 1 ,0 hh θθ θθ θθ θθ T . The homogeneous transformation for the second link is, − = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 cos sin 0 0 sin cos 1 22 22 2 ,1 a θθ θθ T , − = 1 0 0 0 0 1 0 0 sin 0 cos sin cos 0 sin cos 2122 2122 2 ,1 θθθ θθθ a a T . The homogeneous transformation for the third link is, − = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 0 0 cos sin 0 0 sin cos 1 33 33 3 ,2 a θθ θθ T , − = 1 0 0 0 0 1 0 0 sin 0 cos sin cos 0 sin cos 3233 3233 3 ,2 θθθ θθθ a a T . 4 4 4 1c. The co-ordinates of a point in a frame located at the end effector may be expressed in terms of the base coordinates as, 3 3 0,1 1,2 1,3 0,31 1 1 = = 0p p pT T T T where pi is the position vector of the point in the ith frame, i= 0, 1, 2, 3. Hence, 0,3 0,1 1,2 1,3=T T T T . 4 WORKED SOLUTIONS (2019) Page 2 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 1c WORKED SOLUTIONS (2019) Page 3 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 2a) 5 2a) Projecting the end effector position on the x, y, and z axes, where the x axis is aligned with the x2 axis, y is aligned with the y2 axis and z with the z2 axis, the position of joint 2 relative to the end-effector axis is, ( )[ ]Te slslclcl 02332223322,22 ++−== rr , Projecting the position of joint 3 in the x, y, and z axes, the position of joint 2 relative to the joint 3 is, ( )[ ]Tslcl 022223,21 −== rr . Hence, the vector positions of joint 2, 2r relative to the end-effector and 1r relative to joint 3, ( )[ ]Tslslclcl 023322233222 ++−=r , ( )[ ]Tslcl 022221 −=r , [ ]T1000 =z , where the quantities ic , is , ikc and iks are defined as iic θcos = , iis θsin = , ( iikc θ= cos the ( )000 ,, zyx frame. The top view of the first two links in the robot manipulator with the local joint coordinates is shown below. The first two links are equivalent to a two-link manipulator. 2+1 WORKED SOLUTIONS (2019) Page 4 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 2b) As far as first joint, it is prismatic with displacement d1, assumed positive in the positive z direction. The linear velocity of the prismatic joint is: 10 1 d z y x z= . Assembling the preceding results, for the joints 1 it follows that, 11 0 dd J z 0 v = = ω , = 0z 0J . The unit vector in the positive z direction is: [ ]T1 0 00 =z . 5 1 1 2c) Now consider the second and third joints. To determine the relevant components of the screw vectors, consider the second joint by itself. The second and third joints are both revolute joints rotating about parallel axes with axes passing through the joint axes and are in the plane of the paper but in the positive z direction. The first joint is prismatic while the next two joints are revolute joints rotating about parallel axes with axes passing through the joint axes and are in the plane of the paper but in the positive z direction. Hence the angular velocity vectors of each of the revolute joints are, i i θ ω ω ω 0 3 2 1 z= , i =2 and 3, where = 1 0 0 0z . The velocity vector of the end effector relative to centre of rotation of joint 2 while assuming that joint 3 is fixed, is given in the ( )000 ,, zyx frame by: 220 2 rz ×= θ z y x where 2r is the vector distance of the location of joint 2 relative to the end effector location in the ( )000 ,, zyx frame. 2 3 WORKED SOLUTIONS (2019) Page 5 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 2c) Considering the third joint only, the velocity vector of a fixed point at joint 3 relative to the end effector, while assuming that the joint 2 is fixed is given in the ( )000 ,, zyx frame by:, ( )1220 2 rrz −×= θ z y x where 1r is the vector distance of joint 2 relative to the joint 3 in the ( )000 ,, zyx frame. Assembling the preceding results, it follows that, ( ) = −×× = 3 2 1 3 2 1 120200 003 2 1 0 θ θ θ θω ω ω dd z y x J rrzrzz zz and that, ( ) −×× = 120200 00 0 rrzrzz zzJ . 3 2c) Thus the 46 × Jacobian matrix for the transformation from the end-effector to the base co- ordinates is of the form, ( ) −×× = 120200 00 0 rrzrzz zzJ , = 1 0 0 0z , − −= 0 22 22 1 sl cl r and + −− −= 0 23322 23322 2 slsl clcl r . 2 WORKED SOLUTIONS (2019) Page 6 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 3a Let q1= 1θ , angle of rotation of the first link w.r.t. the local horizontal, positive counter clockwise and 2q = 2θ , angle of rotation of the second link w.r.t. the first, positive counter clockwise. The height of the c.g. of the first link from the axis of the first revolute joint is, 111 sinθcgLY = . For the second link it is, ( )122112 sinsin θθθ ++= cgLLY and for the tip mass it is, ( )12211 sinsin θθθ ++= LLYtip . Increase in the potential energy of the body is, [ ] ( )[ ] [ ] ( )[ ]122112111 122112111 sinsinsin sinsinsin θθθθ θθθθ ++++ +++= LLgMLgM LLgmLgmV cgcg . Hence, ( ) ( ) ( )212222112111211 sinsin θθθ ++++++= LMLmgLMLMLmLmgV cgcg which may be written as, ( )2122111 sinsin θθθ +Γ+Γ= ggV where, ( )1211121111 LMLMLmLm cg +++=Γ , ( )222222 LMLm cg +=Γ 3 3a The horizontal position of the c.g. of the first & second link and the tip mass, positive east, are, 111 cosθcgLX = , ( )122112 coscos θθθ ++= cgLLX , ( )12211 coscos θθθ ++= LLX tip , 111 cosθLX j = Horizontal velocities of the c.g.’s of the masses, 1111 sinθθ cgLX −= , 1111 sinθθ LX j −= , ( ) ( )121221112 sinsin θθθθθθ ++−−= cgLLX , ( ) ( )12122111 sinsin θθθθθθ ++−−= LLX tip . Vertical velocity of the c.g.’s of the masses, 1111 cosθθ cgLY = , 1111 cosθθ LY j = , ( ) ( )121221112 coscos θθθθθθ +++= cgLLY , ( ) ( )12122111 coscos θθθθθθ +++= LLYtip ( ) ( ) ( ) ( )2222222221211212111 21212121 tiptipjj YXMYXmYXMYXmT +++++++= . 3 3a ( ) ( )( ) ( ) ( )( )21212211121212211122 coscossinsin θθθθθθθθθθθθ +++++++=+ LLLLYX tiptip ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )12112112121 12 2 12 22 12 2 21 2 1 22 1 2 1 22 coscossinsin2 cossincossin θθθθθθθθθ θθθθθθθθθ +++++ ++++++=+ LL LLYX tiptip ( ) ( ) 21212121222212122 cos2 θθθθθθθ ++++=+ LLLLYX tiptip ; 21212121 θ LYX jj =+ ( ) ( ) 2121212122221212222 cos2 θθθθθθθ ++++=+ cgcg LLLLYX ; 21212121 θ cgLYX =+ . Substituting & Simplifying, ( ) ( ) +++ ++++= 2 21 2 2 2 1 2 12 2 21 2 2 2 1 2 12 2 1 2 11 2 1 2 111 2 1 2 1 2 1 2 1 θθθθθθθθ LLMLLmLMLmT cgcg ( ) ( )( )211212222 cos θθθθ +++ LLMLm cg 3 WORKED SOLUTIONS (2019) Page 7 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 3a The kinetic energy of rotation of the rods is, ( )221222212112 2 1 2 1 θθθ ++= kmkmT . The total kinetic energy is, =+= 21 TTT ( ) ( )( ) ( )( )( )22122222222212121221211 2121 θθθ ++++++++ LMkLmLMMmkLm cgcg( ) ( )( )211212222 cos θθθθ +++ LLMLm cg . 1 3a Hence the total kinetic energy may be expressed as, ( ) ( )211221221222111 cos2121 θθθθθθθ ++++= IIIT , where ( ) ( ) 21222112121111 LMmLMkLmI cgcg ++++= , ( ) 1221222221 LLLMLmI cg Γ=+= , ( ) 2222222222 LMkLmI cgcg ++= , 3 3b Hence the Lagrangian may be defined as, L T V= − . The Euler Lagrange equations are, d dt L q L q Q i i i ∂ ∂ ∂ ∂ − = where, iiq θ= ; Qi are the generalised forces other than those accounted for by the potential energy function and are equal to the torques applied by the joint servo motors, iT . ( ) ( ) 221212122111 1 cos2 θθθθθθ θ∂ ∂ ++++= IIIT ( )21222121 2 cos θθθθ θ∂ ∂ ++= IIT ; 0 1 = θ∂ ∂ T ; ( )( )211221 2 sin θθθθ θ∂ ∂ +−= IT ; ( ) ( )2122111 1 coscos θθθ∂ ∂ +Γ+Γ= gg q V ; ( )2122 2 cos θθ∂ ∂ +Γ= g q V . 3 ( ) ( ) 221212122111 1 cos2 θθθθθθ θ∂ ∂ ++++= IIIL , ( )21222121 2 cos θθθθ θ∂ ∂ ++= IIL , 1 1 Γ−= g q L ∂ ∂ , ( )( )2112212 2 sin θθθθ∂ ∂ +−Γ−= Ig q L , With ( ) ( )21221111 coscos θθθ +Γ+Γ=Γ , ( )21222 cos θθ +Γ=Γ . 3 WORKED SOLUTIONS (2019) Page 8 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 3c Hence the two Euler-Lagrange equations of motion are, ( )2 1 2111 21 2 22 21 2 21 2 221 2 22 1 2 1 1 1 2 2 2 cos cos sin cos I I I I I I I T g T θ θ θθθ θ θ θ θ θ θ − + + + + + Γ + = Γ . 3 3c Let + = 21 1 2 1 θθ θ ω ω Hence the two Euler-Lagrange equations of motion are, 2 2 1 1 111 21 2 22 21 2 1 2 21 2 221 2 22 2 2 21 cos cos sin cos TI I I I I g I I T ωθ θ ω ωθ θ ω ω Γ+ + − + + = Γ . 2 3c The equations may also be written in the state space and in matrix form in terms of 1ω and 2ω , and in terms of 1θ and 2θ , where, 11 ωθ = ; 122 ωωθ −= . The matrix equations of motion in terms of 1ω and 2ω , are: 2 2 1 1 111 21 2 22 21 2 1 2 21 2 221 2 22 2 2 21 cos cos sin cos TI I I I I g I I T ωθ θ ω ωθ θ ω ω Γ+ + − + + = Γ , where ( ) ( )21221111 coscos θθθ +Γ+Γ=Γ ; ( )21222 cos θθ +Γ=Γ . 1 WORKED SOLUTIONS (2019) Page 9 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks Q4a The strategy of Computed torque control involves feeding back, to each of the joint servos, a signal that cancels all of the effects of gravity, friction, the manipulator inertia torques as well as the Coriolis and centrifugal torques. All of these forces are computed on the basis of Euler-Lagrange dynamic model. All these effects are treated as disturbances that must be cancelled at each of the joints. Additional feedbacks are then provided to put in place the desired inertia torques, viscous friction torques and stiffness torques which are selected so the error dynamics behaves in a desirable and prescribed manner. 4+3 4b) Thus the feedback law may be assumed to be a computed torque controller. In the general case the computed torque controller inputs are, ( )qqhD ,ˆˆ 2 1 2 1 2 1 + ≡ = v v T T T T c c . In the case of the inverted pendulum on a cart, 2 2 M m mLc x FmL s mLc mL mgLs θ θ τ + − = . Hence traction force and joint torque are, 2 1 2 2 M m mLc vF mL s vmLc mL mgLs θ τ + = − , where the auxiliary control inputs are defined as, 1 2 2 2 2 2des des desn n n n desdes des x x xxv x v ω ω ω ω θ θθ θ θ = + + − − . Thus the equations of motion reduce to, 2 1 2 2 vxd vdt θ = . 3 3 4 WORKED SOLUTIONS (2019) Page 10 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 4c Let [ ] TTdes des desx θ=q be the demanded angular position commands to the joint servo motors . The tracking error is given by, 1 2 des des act des xe x e θθ = − = − q q and satisfies the equation, = + + 0 0 2 2 12 2 1 2 1 e e e e e e nn ωω and is exponentially stable, i.e. poles are strictly negative real parts. Robot arms are designed so they don’t oscillate and consequently an optimum choice of the damping ratio ζ is unity or very slightly greater than unity. Hence the poles (eigenvalues) are chosen to be strictly negative. 2 3 1 2 WORKED SOLUTIONS (2019) Page 11 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 5a For the measurement of the acceleration a sensor based on the piezo-electric effect is often employed. A piezoelectric crystal or ceramic is employed and it can be modelled as a mass supported on a spring and damper. The inertial force acting on the piezoelectric material generates a stress across two faces on the material which in turn generates an output charge. 3 5a Consider a schematic diagram of an accelerometer as shown in the figure above. The direction of sensitivity is shown on the right of the figure The ground motion is given as ( )ty . The equation of motion in terms of ( )tx , ( ) ( ) mgyxcyxkxm −−−−−= . Let the relative displacement of the mass, relative to the base be, ( ) ( ) ( )tytxtxr −= , Then it follows that, mgymkxxcxm rrr −−=++ . As the stiffness constant is usually very high, neglecting the relative inertial term in comparison with the spring force, ( ) ( )GymgymmgymxckxF rr −−=+−=−−=+= , Where F is the force acting on the piezoelectric material surface, gG −= is the component of the acceleration due to gravity in the direction of measurement. Thus, GAvpiezo −∝ where A is the ground acceleration given by, yA = . This is the principle of acceleration measurement. 2 2 1 WORKED SOLUTIONS (2019) Page 12 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 5b A DC servo motor includes a feedback loop around a conventional DC motor, which is driven by a voltage. The motor output shaft of a conventional DC motor, rotates at constant speed depending on the applied input voltage. Typically for the position control of a DC motor it is essential that the measured position must be the primary feedback as it is compared with the demanded position. A voltage proportional to the error between the demanded and actual positions is used to drive the motor. In addition a signal proportional to the rate of position is also added to the feedback. The closed loop system is of the form shown below. + K G(D)+yi _ e H(D) y In the figure, ( ) 2 21 1H D K d dt K D= + ≡ + . Unlike servo motor, stepper motors are versatile, long-lasting, simple motors that can be used in many applications. Stepper motors are used without feedback in many applications. This is because unless a step is missed, a stepper motor steps a known angle each time it is moved. Thus, its angular position is always known and no feedback is nec- essary. 3 3 1 3 5c One popular method of implement a control system is to use pulses are illustrated in figure 5c and is known as Pulse Width Modulation (PWM) control. The cycle time, Tc , is maintained a constant and the pulse on-time is T+ . 1 The ratio of the pulse on-time of the pulse cycle time defines the duty cycle and the average D. C. value of the drive signal is: d cV V T T+= . Referring to the figure, the error (e) input is defined by the voltage V. The gain K is effectively varied by changing the pulse width or on-time. The control input to the motor is dV . t O V Tc T+ T- Figure 5c Pulse Width Modulation 1 2 WORKED SOLUTIONS (2019) Page 13 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks The principal advantage of PWM control over the traditional method described earlier is the simplicity of the drive electronics and the ease of interfacing to a micro-controller circuit. PWM is easy to implement with microprocessors. Commonly, to control the speed of a DC motor, the applied voltage is changed; as the voltage increases or decreases, the speed of the motor is changed accordingly. However, when the speed is to be controlled by a microprocessor, changing the voltages to servomotors requires many bits of information. Alternatively, using PWM, many voltage levels can be achieved with one single-level input voltage, namely, the high voltage (say, 5 volts) and one output bit. To do this, the voltage on the output port of the processor is turned on and off repeatedly, many times a second, so that by varying the length of time that the voltage is on or off, the average effective voltage will vary. A major disadvantage of the PWM method is the high levels of noise generated by the drive electronics which makes it unreliable for precision applications. 3 WORKED SOLUTIONS (2019) Page 14 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks 6a The primary factors that must be considered in selecting a sensor for a robotic application apart from its weight and size are, Resolution and dynamic range: The resolution may be defined as the smallest increment of the measurand that can be detected with certainty by the instrument. The dynamic range is the minimum reading of the instrument output corresponding to the minimum input which is necessary to cause a measurable change in the instrument from zero input to the maximum reading or output corresponding to the maximum allowable input to the instrument. Thus the range represents to which the sensor may be employed in practice. It is also stated as a ratio of the maximum to the minimum reading that may be obtained from a sensor while the difference between the maximum and minimum is specified as the span. Consistency or Repeatability: When a measurement of a quantity made by an observer is identical or sufficiently close to another measurement of the same quantity, repeated under identical experimental conditions, methods and apparatus, it is said to be repeatable or reproducible or consistent. Accuracy: Accuracy is a term used to qualify the error as being within certain limits with a specified probability. Accuracy is the degree of correctness with which a method of measurement yields the true value of a quantity, usually expressed in terms of the relative magnitude of the error. Predictability: Predictability is the property of being able to estimate the static and dynamic errors in the mesurand, to the same level of accuracy as in a measurement of it, with a specified sensor, without actually performing the measurement. Sensitivity: Typically, the sensitivity is the ratio of the sensor output to the corresponding incremental change in the magnitude of the measurand. 4 2 2 2 2 6b Consider the equivalent circuit of a linear potentiometer. x vin vout com Resistance = R/unit length Moving wiper i i1 iload Rload L Equivalent circiuit of a linear potentiometer The total current is related to the load current by the relation: loadiii += 1 . The voltage drop aross the load is equal to the voltage drop across the wiper and the com port. Thus, ( ) loadload RixLRi ×=−×1 . Hence, 3 WORKED SOLUTIONS (2019) Page 15 of 15 Module: DEN408+DENM011 Title: Robotics First examiner: Dr Ranjan Vepa Second examiner: Dr M. Hasan Shaheed Qu. No. Worked Solution Marks ( )xLR R ii loadload − ×=1 and ( ) − +×= xLR R ii loadload 1 . It follows that, ( ) ( ) +− − ×= load load RxLR xLRii and ( )( ) +− ×− ×= load load load RxLR RxLR iv . The total circuit rsistance with the wiper in position is, ( ) ( ) ( ) ( )( ) ( ) load load load load load Total RxLR LRxLxRR RxLR RxLR xR RRxL xRR +− +−× = +− ×− += + − += 111 and ( )( )( ) load load Total RxLR LRxLxRR iRiv +− +−× ×=×= . Thus, ( )( )( ) ( ) ( )( ) +− ×− = +−× ×− = LRxLxR RxL LRxLxRR RxLR v v load load load loadload . When loadR is very large relative to R, i.e. ∞→loadR , ( ) ( )( ) ( ) ( ) L xL LR RxL LRxLxR RxL v v load load load loadload − = ×− ≈ +− ×− = Hence, the output voltage is a linear function of the position of the wiper from one end. 4 6c The need for tactile sensing: The sense of touch, one of 5 human senses (vision, hearing, smell, taste and touch), provides particular information that cannot be perceived by any of the other senses. Touch may interpret object texture, hardness and temperature as well as the mechanical state of an object such as vibration or movement. Tactile sensors are devices that measure the parameters of contact between the sensory surface and an object. Thus tactile sensing involves detection and measurement of the spatial distribution of forces perpendicular to a predetermined sensory area, and the subsequent interpretation of the spatial force distribution. The application in Robotics: A tactile sensors are used to control a manipulator for grasping an unknown object based on tactile feedback. The tactile sensor placed on each jaw of the manipulator's gripper and serves as force distribution sensor, which is then used to control the applied force/pressure applied by the gripper on the object by force-feedback control. 3 3
欢迎咨询51作业君
