Robotics I Midterm classroom test – November 24, 2017 Exercise 1 [8 points] Consider the 3-dof (RPR) planar robot in Fig. 1, where the joint coordinates r = have been defined in a free, arbitrary way, with reference to a base frame RF b .
r
1
r2 r3
T
Figure 1: A RPR planar robot.
• Determine position and orientation of the end-effector frame RF e in terms of the coordinates r .
T
• Assign the Denavit-Hartenberg (DH) frames and define the joint coordinates q = q 1 q 2 q 3 according to the standard DH convention, starting from the reference frame RF 0 . Provide Provide the the DH table, the expression of the homogeneous transformation matrices between the successive frames frames that have been assigned. assigned. Determine Determine position and orientatio orientation n of the end-effect end-effector or frame RF e in terms of the coordinates q. • Find the transformation q = f (r) between the two sets of coordinates so as to associate the same kinematic kinematic configurati configurations ons of the robot. Does this mapping mapping have have singularit singularities? ies? Exercise 2 [5 points]
Figure 2: A DC servomotor that drives a robot link through transmissions. With reference to the servo-drive sketched in Fig. 2 and the data therein, we need to measure the position of the tip point P ee 0 .01 [mm]. A suitable suitable incrementa incrementall ee of the link with a resolution of 0. 1
encoder with quadrature detection is chosen to measure the angular position θ m of the DC motor, which is capable of delivering a maximum torque τ m . How many pulses per turn should generated by the optical disk on each of the channels A and B? How many bits should be used by the digital counter of the encoder? Neglecting dissipative effects, what should be the value of the motor inertia J m in order to maximize the angular acceleration of the link for a given motor torque? With this choice, what is the maximum linear acceleration achievable by the tip of the link? Exercise 3 [12 points] Consider the 6-dof robot St¨aubli RX 160 in Fig. 3. In the extra sheet provided separately, the Denavit-Hartenberg (DH) table of parameters is specified, in part numerically and in part symbolically. The two DH frames 0 and 6 are already drawn on the manipulator (in two views). In the shown ‘straight upward’ robot configuration, the first and last joint variables take the values q 1 = q 6 = 0. Draw directly on the extra sheet the remaining DH frames, according to the DH table. Provide all parameters labeled in red in the table, i.e., the missing numerical values of the constant parameters and of the joint variables q 2 to q 5 when the robot is in the ‘straight upward’ configuration. [Please, make clean drawings and return the sheet with your name written on it.] ..
Staubli RX 160 Assignment of frames according to the given DH table Name: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! z6
D 3
z6 y6
E 4
x6
C F
5 6 B
R
2
1
A
R
ai
di
" i
/2
a1 > 0
d1 > 0
q 1 = 0
0
a2 > 0
0
q 2
! i
i
1
!
2 3
!
4
!
z0 z0 x0
/2
0
0
q 3
/2
0
d4 > 0
q 4
5
!/2
0
0
q 5
6
0
0
d6 > 0
q 6 = 0
y0
view from the right side
view from the back side
Robotics 1, Midterm classroom test, November 24, 2017
Figure 3: The St¨aubli RX 160 robot manipulator and the extra sheet (provided separately). Exercise 4 [5 points] The orientations of two right-handed frames RF A and RF B with respect to a third right-handed frame RF 0 (all having the same origin) are specified, respectively, by the rotation matrices
− = − − 3 4
0
R A
3 8
3 8
,
− 14
1 4
1 2
1 2
3 8
3 8
and
0
R B
√ 0 = √
1 2
3 4
1 0 − √ 2
1
0
0
1 √
2
.
Determine, if possible, a unit vector r and an angle θ < 0 such that the axis-angle rotation matrix R (r , θ) provides the orientation of the frame RF B with respect to the frame RF A . [180 minutes, open books but no computer or smartphone] 2
Solution of Midterm Test November 24, 2017 Exercise 1 The direct kinematics in terms of the coordinates r =
(r + N )cos r + L cos r ( ) = (r + N )sin r + L sin r , 2
b
pe r
1
2
r
1
T
is easily computed as
cos r ( ) = sin r
3
1
r2 r3
3
b
3
R e r
0
3
0
− sin r3 cos r3 0
0 0 1
.
Indeed, if these quantities have to be expressed with respect to the (DH) frame 0 indicated in Fig. 1, we need to introduce a constant rotation. Since the two frames have the same origin, we obtain
(r + N )sin r + L sin r ( ) = −(r + N )cos r − L cos r , sin r cos0r 0 ( ) = − cos r sin r 0 . 2
0
R b
0 = −1 0
1 0 0 0 0 1
0 p
e (r )
=
0 R b p b
⇒
e
r
1
2
1
3
0 R
b 0 e (r ) = R b R e r
3
3
3
0
3
3
0
1
Keeping into account the already specified assignment of frame 0 in Fig. 1, a feasible assignment of DH frames for the RPR robot is shown in Fig. 4, with associated parameters given in Tab. 1.
Figure 4: Assignment of DH frames for the RPR planar robot. i
αi
ai
di
θi
1
π/2
0
0
q 1
2
−π/2
0
q 2
0
3
0
L
0
q 3
Table 1: Parameters associated to the DH frames in Fig. 4. 3
The DH homogenous transformations take the following expressions:
cos q sin q (q ) = 0
1 0 0 0 0 1 (q ) = 0 −1 0 0 0 0 0 cos q − sin q 0 L cos q sin q cos q 0 L sin q . (q ) = 0 0 1 0 0 0 0 1 is then The direct kinematics in terms of the coordinates = q q q ( ) ( ) 0 sin q 1 0 − cos q 1 1 0 0 0
1
0
A1
1
1
0 0 0 1
,
1
3
2
A3
,
3
3
3
T
q
0
2
3
3
3
A2
0 0 q 2 1
1
T e (q ) = 0A1 (q 1 ) 1A2 (q 2 ) 2A3 (q 3 ) =
2
3
0 R
e
q
0 p
0 0 0
e
q
,
1
with
q sin q + L cos(q + q ) ( ) = −q cos q + L sin(q + q ) , 2
0
pe q
1
2
1
1
cos(q + q ) ( ) = sin(q + q )
3
1
3
0
R e q
0
1
3
1
3
− sin(q 1 + q 3 ) 0 cos(q 1 + q 3 ) 0 0 1
0
.
The transformation that matches the robot configurations using the different sets of coordinates is
r r + N ( )= r − r − π
q − ( )= q − N π , q + q +
1
q = f r
1
3
1
r = f
⇐⇒
2
1
q
2
1
2
3
2
which is invertible and without singularities. It is easy to check that, e.g., 0
pe (q )|q=f (r) = 0 pe (r),
0
R e (q )|q=f (r) = 0 R e (r ).
Exercise 2 The requested resolution ∆ = 10−5 [m] on the linear motion at the tip of the link of length L = 0.55 [m] needs to be transformed into an angular one δ = ∆/L = 1.8182 · 10−5 [rad] at the base of the link and then, via the two transmission gears with reduction ratios N bevel = r 4 /r3 = 1 and N spur = r 2 /r1 = 0.35/0.08 = 4.375 respectively, into an angular resolution on the motor axis δ m = N spur · N bevel ·
∆ = 7.9545 · 10−5 [rad] = (4.56 · 10−3 )◦ . L
Taking into account the factor 4 introduced by the quadrature electronics, the number of pulses per turn N ppt of the optical disk should be at least N ppt =
2π 4 · δ m
= 19748.
Accordingly, the digital counter should have at least a number of bits N bit = log2 19748 = 15. 4
Being the link inertia (around its axis of rotation) J L = 0.3025 [kg·m2 ] and the reduction ratio of the transmissions N = N spur · N bevel = 4.375, the optimal value of the motor inertia according to the requested criterion is J L J m = 2 = 0.0158 [kg·m2 ]. N Accordingly, with a maximum available motor torque τ m = 0.8 [Nm] on the motor axis, the maximum torque at the link base is τ L,max = N · τ m = 3.5 [Nm]. Since N 2 = J L /J m , the balance of torques on the link side provides τ L = J L θ¨L + N (J m θ¨m ) = (J L + J m N 2 )θ¨L = 2J L θ¨L . The maximum tip acceleration a max of the link tip will then be amax = L ¨ θL,max = L
τ L,max = 0.55 [m] · 5.7851 [rad·s−2 ] = 3.1818 [m·s−2 ]. 2 J L
Exercise 3 view from the right side
view from the back side
z6
z6
D 3
x6
z5
E 4
C
y6
z5 z4
x4, x5
F
z3
5 6
x3
B
z3
x2 x2
R
z2
2
1
A
y1
z1
ai
di
" i
/2
150
550
q 1 = 0
0
825
0
q 2 = !/2
/2
0
0
q 3 = !/2
/2
0
625
q 4 =
!
5
!/2
0
0
q 5 =
0
6
0
0
110
i
1
! i !
2 3
!
4
!
y1
x1
z0
z0 x0
y0
q 6 = 0
Figure 5: The DH frames and the complete table of parameters for the St¨aubli RX 160 robot manipulator. The numerical values of q = θ refer to the shown ‘straight upward’ configuration. The assignment of Denavit-Hartenberg frames for the St¨aubli robot RX 160 according to the given table is shown in Fig. 5. The numerical values of all symbolic parameters are reported, with the values of the joint variables q = θ when the robot is in the ‘straight upward’ configuration. 5
Exercise 4 The relative orientation of frame RF B with respect to frame RF A is expressed by the rotation matrix
√ − √ =− √ − 1
3 8
2 2
A
R B =
0
0 R T A · R B
3 2
1 2
1 2 2
3 8
1 − √ 2
0 1 √
2
0.3536 = 0.8660
−0.6124 −0.7071 0.5
0.3536 −0.6124
0 0.7071
,
whose elements will be denoted by Rij . Therefore, the equation R (r, θ) = A R B should be solved for r and θ, using the inverse mapping of the axis-angle representation. Since sin θ = ±
1 2
(R
2 12 − R21 )
+ (R13 − R31)2 + (R23 − R32 )2 = ± 0.9599 = 0,
(1)
the problem at hand is regular, and two distinct solutions can be found depending on the choice of the + or − sign in the expression of sin θ. From 1 cos θ = (R11 + R22 + R33 − 1) = 0.2803, 2 taking the − sign in (1) will yield a solution angle θ < 0, as requested. Thus θ = ATAN2 {−0.9599, 0.2803} = − 1.2867 [rad] = − 73.72◦ and r
R 1 = R 2sin θ
32 − R23 13 − R31 R21 − R12
=
∗∗∗∗∗
6
0.3190 0.5525 −0.7701
.
