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Introduction to Robotics Analysis, Control, Applications Solution Manual Saeed B. Niku © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system and the Internet without prior expressed consent of the copyright owner. 2 CHAPTER ONE Problem 1.1 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 15-25 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 3 Problem 1.2 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 4 Problem 1.3 Draw the approximate workspace for the following robot. Assume the dimensions of the base and other parts of the structure of the robot are as shown. Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) 1.14 Solution: The workspace shown is approximate. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 5 CHAPTER TWO Problem 2.1 Write a unit vector in matrix form that describes the direction of the cross product of p = 5i + 3k and q = 3i + 4 j + 5k . Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: ⎡i j k⎤ r = p × q = ⎢⎢5 0 3 ⎥⎥ = i ( 0 − 12 ) − j ( 25 − 9 ) + k ( 20 − 0 ) = −12i − 16 j + 20k ⎢⎣3 4 5 ⎥⎦ λ = rx2 + ry2 + rz2 = 144 + 256 + 400 = 28.28 ⎡ −12 ⎤ ⎢ 28.28 ⎥ ⎢ ⎥ ⎡ −0.424 ⎤ −16 ⎥ ⎢ ⎢ = −0.566 ⎥⎥ r= ⎢ 28.28 ⎥ ⎢ ⎢ ⎥ ⎢⎣ 0.707 ⎥⎦ ⎢ 20 ⎥ ⎣⎢ 28.28 ⎦⎥ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 6 Problem 2.2 A vector p is 8 units long and is perpendicular to vectors q and r described below. Express the vector in matrix form. ⎡ 0.3⎤ ⎡ rx ⎤ ⎢q ⎥ ⎢ 0.5⎥ qunit = ⎢ y ⎥ runit = ⎢ ⎥ ⎢ 0.4 ⎥ ⎢0.4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 ⎦ ⎣ 0 ⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: The two vectors given are unit vectors. Therefore, each missing component can be found as: q y = 1 − 0.09 − 0.16 = 0.866 rx = 1 − 0.25 − 0.16 = 0.768 Since p is perpendicular to the other two vectors, it is in the direction of the cross product of the two. Therefore: j k ⎤ ⎡ i ⎢ λ p = ⎢ 0.3 0.866 0.4 ⎥⎥ = i ( 0.346 − 0.2 ) − j ( 0.12 − 0.307 ) + k ( 0.15 − 0.665 ) ⎢⎣0.768 0.5 0.4 ⎥⎦ = i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) Since q and r are not perpendicular to each other, the resulting p is not a unit vector. Vector p can be found as: λ p = i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) λp = ( 0.146 ) + ( 0.187 ) + ( 0.515) 2 2 2 = 0.567 8 = 14.1 0.567 p = w ( i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) ) w= p = i ( 2.06 ) + j ( 2.64 ) − k ( 7.27 ) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 7 Problem 2.3 Will the three vectors p, q, and r in Problem 2.2 form a traditional frame? If not, find the necessary unit vector s to form a frame between p, q, and s. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: As we saw in Problem 2.2, since q × r is not a unit vector, it means that q and r and not perpendicular to each other, and therefore, they cannot form a frame. However, p and q are perpendicular to each other, and we can select s to be perpendicular to those two. Of course, p is not a unit length, therefore we use the unit vector representing it. λp = ( 0.146 ) + ( 0.187 ) + ( 0.515) 2 2 2 = 0.567 1 = 1.764 0.567 p = w ( i ( 0.146 ) + j ( 0.187 ) − k ( 0.515 ) ) w= p = i ( 0.257 ) + j ( 0.33) − k ( 0.908 ) p = i ( 0.257 ) + j ( 0.33) − k ( 0.908 ) j k ⎤ ⎡ i ⎢ s = ⎢ 0.257 0.33 −0.908⎥⎥ = i ( 0.918 ) − j ( 0.375 ) + k ( 0.124 ) ⎢⎣ 0.3 0.866 0.4 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 8 Problem 2.4 T Suppose that instead of a frame, a point P = ( 3,5, 7 ) in space was translated a distance of d = ( 2,3, 4 ) . Find the new location of the point relative to the reference frame. T Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) 2.6 Solution: As for a frame, Pnew ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 2⎤ ⎡ 3⎤ ⎡ 5 ⎤ 3 ⎥⎥ ⎢⎢ 5 ⎥⎥ ⎢⎢ 8 ⎥⎥ = 4 ⎥ ⎢ 7 ⎥ ⎢11⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 9 Problem 2.5 T The following frame B was moved a distance of d = ( 5, 2, 6 ) . Find the new location of the frame relative to the reference frame. ⎡0 ⎢1 B=⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 2⎤ 4 ⎥⎥ 6⎥ ⎥ 1⎦ Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 2.6 Solution: The transformation matrix representing the translation is used to find the new location as: Bnew ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 5 ⎤ ⎡0 2 ⎥⎥ ⎢⎢1 6 ⎥ ⎢0 ⎥⎢ 1 ⎦ ⎣0 1 0 0 0 0 −1 0 0 2⎤ ⎡0 4 ⎥⎥ ⎢⎢1 = 6 ⎥ ⎢0 ⎥ ⎢ 1 ⎦ ⎣0 1 0 7⎤ 0 0 6 ⎥⎥ 0 −1 12⎥ ⎥ 0 0 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 10 Problem 2.6 For frame F, find the values of the missing elements and complete the matrix representation of the frame. ⎡ ? 0 −1 5 ⎤ ⎢ ? 0 0 3⎥ ⎥ F =⎢ ⎢ ? −1 0 2 ⎥ ⎢ ⎥ ⎣0 0 0 1 ⎦ Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: ⎡ nx ⎢n F =⎢ y ⎢ nz ⎢ ⎣0 0 −1 0 0 −1 0 0 0 5⎤ 3 ⎥⎥ 2⎥ ⎥ 1⎦ j k⎤ ⎡i ⎢ ⎥ From n × o = a ⎢ nx n y nz ⎥ = − i ⎢⎣ 0 0 −1⎥⎦ Or: i ( −n y ) − j ( −nx ) + k ( 0 ) = −i , and therefore: n y = 1, nx = 0, nz = 0 ⎡ 0 0 −1 ⎢1 0 0 F =⎢ ⎢ 0 −1 0 ⎢ ⎣0 0 0 5⎤ 3 ⎥⎥ 2⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 11 Problem 2.7 Find the values of the missing elements of frame B and complete the matrix representation of the frame. ? ⎡0.707 ⎢ ? 0 B=⎢ ⎢ ? −0.707 ⎢ 0 ⎣ 0 0 1 0 0 2⎤ 4 ⎥⎥ 5⎥ ⎥ 1⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.4 Solution: ox ⎡0.707 ⎢ n 0 B=⎢ y ⎢ nz −0.707 ⎢ 0 ⎣ 0 From n × o = a 0 1 0 0 2⎤ 4 ⎥⎥ 5⎥ ⎥ 1⎦ j ⎡ i ⎢ 0.707 n y ⎢ ⎢⎣ ox 0 k ⎤ nz ⎥⎥ = j 0.707 ⎥⎦ j k ⎤ ⎡ i ⎢ Therefore: ⎢ 0.707 n y nz ⎥⎥ = j ⎢⎣ ox 0 0.707 ⎥⎦ i ( 0.707n y ) − j ( 0.5 − nz ox ) + k ( − ny ox ) = j → ny = 0 And From length equations: n = 1 or 0.707 2 + n y2 + nz2 = 1 → nz = ±0.707 ox2 + 0.5 = 1 → ox = ±0.707 Therefore, there are two possible acceptable solutions: ⎡0.707 0.707 ⎢ 0 0 B=⎢ ⎢0.707 −0.707 ⎢ 0 ⎣ 0 0 1 0 0 2⎤ 4 ⎥⎥ 5⎥ ⎥ 1⎦ and ⎡ 0.707 −0.707 ⎢ 0 0 B=⎢ ⎢ −0.707 −0.707 ⎢ 0 ⎣ 0 0 1 0 0 2⎤ 4 ⎥⎥ 5⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 12 Problem 2.8 Derive the matrix that represents a pure rotation about the y-axis of the reference frame. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.2. Solution: From the figure: px = pn cos θ + pa sin θ p y = po pz = − pn sin θ + pa cos θ and ⎡ px ⎤ ⎡ C ⎢p ⎥ = ⎢ 0 ⎢ y⎥ ⎢ ⎢⎣ pz ⎥⎦ ⎢⎣ − S S ⎤ ⎡ pn ⎤ 1 0 ⎥⎥ ⎢⎢ po ⎥⎥ 0 C ⎥⎦ ⎢⎣ pa ⎥⎦ 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 13 Problem 2.9 Derive the matrix that represents a pure rotation about the z-axis of the reference frame. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.2. Solution: From the Figure: px = pn cos θ − po sin θ p y = pn sin θ + po cos θ pz = pa and ⎡ p x ⎤ ⎡C ⎢ p ⎥ = ⎢S ⎢ y⎥ ⎢ ⎢⎣ pz ⎥⎦ ⎢⎣ 0 −S C 0 0 ⎤ ⎡ pn ⎤ 0 ⎥⎥ ⎢⎢ po ⎥⎥ 1 ⎥⎦ ⎢⎣ pa ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 14 Problem 2.10 Verify that the rotation matrices about the reference frame axes follow the required constraint equations set by orthogonality and length requirements of directional unit vectors. Estimated student time to complete: 10 minutes. Prerequisite knowledge required: Text Section(s) 2.4.5 and 2.6.2 Solution: ⎡1 0 For Rot ( x, θ ) = ⎢⎢0 C ⎢⎣0 S 0⎤ − S ⎥⎥ we have: C ⎥⎦ nio = 0 or nx ox + ny oy + nz oz = 0 n ia = 0 ai o = 0 n =1 o = S 2 + C2 = 1 a= ( −S ) 2 + C2 = 1 Rot ( y, θ ) and Rot ( z,θ ) will be the same. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 15 Problem 2.11 Find the coordinates of point P (2,3, 4)T relative to the reference frame after a rotation of 45 D about the x-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution: U 0 0 ⎤ ⎡ 2⎤ ⎡ 2 ⎤ ⎡ 2 ⎤ ⎡1 ⎢ ⎥ ⎢ P = Rot ( x, 45) ⎢ 3 ⎥ = ⎢0 0.707 −0.707 ⎥⎥ ⎢⎢ 3 ⎥⎥ = ⎢⎢ −0.707 ⎥⎥ ⎢⎣ 4 ⎥⎦ ⎢⎣0 0.707 0.707 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ 4.95 ⎥⎦ Note that the rotation is written in a 3 × 3 , not homogeneous form, because we are only concerned about the rotation part. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 16 Problem 2.12 Find the coordinates of point P (3,5, 7)T relative to the reference frame after a rotation of 30 D about the z-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution: U ⎡ 3 ⎤ ⎡ 0.866 −0.5 0 ⎤ ⎡ 3 ⎤ ⎡ 0.1 ⎤ P = Rot ( z ,30) ⎢⎢ 5 ⎥⎥ = ⎢⎢ 0.5 0.866 0 ⎥⎥ ⎢⎢ 5 ⎥⎥ = ⎢⎢5.83⎥⎥ ⎢⎣7 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 7 ⎥⎦ ⎢⎣ 7 ⎥⎦ Note that the rotation is written in a 3 × 3 form not homogeneous form, because we are only concerned about the rotation part. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 17 Problem 2.13 Find the new location of point P(1, 2,3)T relative to the reference frame after a rotation of 30D about the z-axis followed by a rotation of 60D about the y-axis. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution: U P = Rot ( y, 60) Rot ( z ,30) P ⎡ 0.5 ⎢ 0 U P=⎢ ⎢ −0.866 ⎢ ⎣ 0 0 0.866 0 ⎤ ⎡0.866 −0.5 0 0 ⎤ ⎡ 1 ⎤ ⎡ 2.531⎤ 1 0 0 ⎥⎥ ⎢⎢ 0.5 0.866 0 0 ⎥⎥ ⎢⎢ 2 ⎥⎥ ⎢⎢ 2.232 ⎥⎥ = 0 0.5 0 ⎥ ⎢ 0 0 1 0 ⎥ ⎢ 3 ⎥ ⎢1.616 ⎥ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 0 0 1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 18 Problem 2.14 A point P in space is defined as B P = (5,3, 4)T relative to frame B which is attached to the origin of the reference frame A and is parallel to it. Apply the following transformations to frame B and find AP. Using the 3-D grid, plot the transformations and the result and verify it. Also verify graphically that you would not get the same results if you apply the transformations relative to the current frame: • Rotate 90 D about x-axis, then • Translate 3 units about y-axis, 6 units about z-axis, and 5 units about x-axis. Then, • Rotate 90 D about z-axis. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.3. Solution: A P = Rot ( z ,90)Trans (5,3, 6) Rot ( x,90) B P ⎡ 0 −1 ⎢1 0 A P=⎢ ⎢0 0 ⎢ ⎣0 0 0 0 ⎤ ⎡1 0 0 ⎥⎥ ⎢⎢0 1 0⎥ ⎢0 ⎥⎢ 0 1⎦ ⎣0 0 0 5 ⎤ ⎡1 1 0 3 ⎥⎥ ⎢⎢0 0 1 6⎥ ⎢0 ⎥⎢ 0 0 1 ⎦ ⎣0 0⎤ ⎡5⎤ ⎡ 1 ⎤ 0 −1 0 ⎥⎥ ⎢⎢ 3⎥⎥ ⎢⎢10 ⎥⎥ = 1 0 0⎥ ⎢ 4⎥ ⎢ 9 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣1 ⎦ ⎣ 1 ⎦ 0 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 19 Problem 2.15 A point P in space is defined as B P = (2,3,5)T relative to frame B which is attached to the origin of the reference frame A and is parallel to it. Apply the following transformations to frame B and find AP. Using the 3-D grid, plot the transformations and the result and verify it: D • Rotate 90 about x-axis, then • Rotate 90 D about local a-axis, then • Translate 3 units about y-, 6 units about z-, and 5 units about x-axes. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution: A P = Trans (5,3, 6) Rot ( x,90) Rot (a,90) B P ⎡1 ⎢0 A P=⎢ ⎢0 ⎢ ⎣0 0 0 5⎤ ⎡1 1 0 3⎥⎥ ⎢⎢0 0 1 6⎥ ⎢0 ⎥⎢ 0 0 1 ⎦ ⎣0 0 ⎤ ⎡ 0 −1 0 −1 0 ⎥⎥ ⎢⎢1 0 1 0 0⎥ ⎢0 0 ⎥⎢ 0 0 1⎦ ⎣0 0 0 0 0 0⎤ ⎡ 2⎤ ⎡ 2 ⎤ 0 0 ⎥⎥ ⎢⎢ 3 ⎥⎥ ⎢⎢ −2 ⎥⎥ = 1 0⎥ ⎢5⎥ ⎢ 8 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 0 1⎦ ⎣1⎦ ⎣ 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 20 Problem 2.16 A frame B is rotated 90D about the z-axis, then translated 3 and 5 units relative to the nand o-axes respectively, then rotated another 90D about the n-axis, and finally, 90D about the y-axis. Find the new location and orientation of the frame. ⎡0 ⎢1 B=⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 1⎤ 1⎥⎥ 1⎥ ⎥ 1⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution: Bnew = Rot ( y,90) Rot ( z,90) BTrans(3,5, 0) Rot (n,90) Bnew ⎡0 ⎢0 =⎢ ⎢ −1 ⎢ ⎣0 0 1 0 ⎤ ⎡0 −1 1 0 0 ⎥⎥ ⎢⎢1 0 0 0 0⎥ ⎢0 0 ⎥⎢ 0 0 1⎦ ⎣0 0 0 0⎤ ⎡0 0 0 ⎥⎥ ⎢⎢1 1 0⎥ ⎢0 ⎥⎢ 0 1 ⎦ ⎣0 1⎤ ⎡1 0 0 1⎥⎥ ⎢⎢ 0 0 −1 1⎥ ⎢0 ⎥⎢ 0 0 1⎦ ⎣ 0 1 0 0 0 3⎤ ⎡1 1 0 5⎥⎥ ⎢⎢ 0 0 1 0⎥ ⎢0 ⎥⎢ 0 0 1⎦ ⎣0 0 ⎤ ⎡ 0 −1 0 0 −1 0⎥⎥ ⎢⎢0 0 −1 = 1 0 0 ⎥ ⎢1 0 0 ⎥ ⎢ 0 0 1 ⎦ ⎣0 0 0 0 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 1⎤ 6 ⎥⎥ 4⎥ ⎥ 1⎦ 21 Problem 2.17 The frame B of problem 2.16 is rotated 90D about the a-axis, 90D about the y-axis, then translated 2 and 4 units relative to the x- and y-axes respectively, then rotated another 90D about the n-axis. Find the new location and orientation of the frame. ⎡0 ⎢1 B=⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 1⎤ 1⎥⎥ 1⎥ ⎥ 1⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 2.6.4. Solution: Bnew = Trans (2, 4, 0) Rot ( y,90) B Rot ( a,90) Rot ( n,90) Bnew ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 2⎤ ⎡ 0 4 ⎥⎥ ⎢⎢ 0 0 ⎥ ⎢ −1 ⎥⎢ 1⎦ ⎣ 0 0 1 0 0 1 0 0 0 0⎤ ⎡0 0 ⎥⎥ ⎢⎢1 0⎥ ⎢0 ⎥⎢ 1 ⎦ ⎣0 1 0 0 0 0 −1 0 0 1⎤ ⎡ 0 −1 1⎥⎥ ⎢⎢1 0 1⎥ ⎢ 0 0 ⎥⎢ 1⎦ ⎣ 0 0 0 0 1 0 0⎤ ⎡1 0 ⎥⎥ ⎢⎢ 0 0⎥ ⎢0 ⎥⎢ 1⎦ ⎣0 0 0 0 −1 1 0 0 0 0 ⎤ ⎡ 0 −1 0⎥⎥ ⎢⎢ 0 0 = 0⎥ ⎢ −1 0 ⎥ ⎢ 1⎦ ⎣ 0 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 0 3⎤ 1 5 ⎥⎥ 0 −1⎥ ⎥ 0 1⎦ 22 Problem 2.18 Show that rotation matrices about the y- and the z-axes are unitary. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 2.7. Solution: ⎡C A = Rot ( y, θ ) = ⎢⎢ 0 ⎢⎣ − S S⎤ 1 0 ⎥⎥ 0 C ⎥⎦ 0 det A = 1(C 2θ + S 2θ ) + 0 = 1 0 0 ⎤ ⎡1 ⎢ T A = ⎢ 0 Cθ Sθ ⎥⎥ ⎢⎣ 0 − Sθ Cθ ⎥⎦ adjA = A T minor ⎡C 0 − S ⎤ = ⎢⎢ 0 1 0 ⎥⎥ /1 = A−1 ⎢⎣ S 0 C ⎥⎦ A−1 = AT Exact same is true for rotation about the z-axis. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 23 Problem 2.19 Calculate the inverse of the following transformation matrices: ⎡ 0.527 −0.574 0.628 2 ⎤ ⎢ 0.369 0.819 0.439 5 ⎥ ⎥ T1 = ⎢ ⎢ −0.766 0 0.643 3 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 and ⎡ 0.92 ⎢ 0 T2 = ⎢ ⎢ −0.39 ⎢ ⎣ 0 0 0.39 5 ⎤ 1 0 6 ⎥⎥ 0 0.92 2 ⎥ ⎥ 0 0 1⎦ Estimated student time to complete: 15-20 minutes. Prerequisite knowledge required: Text Section(s) 2.7. Solution: Since transformation matrices are unitary, we calculate the inverses simply by transposing the rotation part and calculating the position part by: 0.527 × 2 + 0.369 × 5 − 0.766 × 3 = −0.601 −0.574 × 2 + 0.819 × 5 − 0 × 3 = −2.947 0.628 × 2 + 0.439 × 5 + 0.643 × 3 = −5.38 Therefore: ⎡ 0.527 0.369 −0.766 −0.601⎤ ⎢ −0.574 0.819 0 −2.947 ⎥⎥ T1−1 = ⎢ ⎢ 0.628 0.439 0.643 −5.38 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 And similarly, ⎡ 0.92 ⎢ 0 −1 T2 = ⎢ ⎢ 0.39 ⎢ ⎣ 0 0 −0.39 −3.82 ⎤ 1 0 −6 ⎥⎥ 0 0.92 −3.79 ⎥ ⎥ 0 0 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 24 Problem 2.20 Calculate the inverse of the matrix B of Problem 2.17. Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) 2.7. Solution: ⎡0 ⎢1 B=⎢ ⎢0 ⎢ ⎣0 ⎡0 ⎢1 B −1 = ⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 1⎤ 1⎥⎥ 1⎥ ⎥ 1⎦ 1 0 −1⎤ 0 0 −1⎥⎥ 0 −1 1 ⎥ ⎥ 0 0 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 25 Problem 2.21 Write the correct sequence of movements that must be made in order to restore the original orientation of the spherical coordinates and make it parallel to the reference frame. About what axes are these rotations supposed to be? Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 2.9.3. Solution: ⎡1 ⎢0 Tsph (r , β , γ ) Rot (o, − β ) Rot (a, −γ ) = ⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 rS β Cγ ⎤ 0 rS β Sγ ⎥⎥ 1 rC β ⎥ ⎥ 0 1 ⎦ As shown, the rotations are relative to o- and a-axes in the sequence shown. These rotations must be relative to the current moving frame in order to prevent changing the position of the frame. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 26 Problem 2.22 A spherical coordinate system is used to position the hand of a robot. In a certain situation, the hand orientation of the frame is later restored in order to be parallel to the reference frame, and the matrix representing it is described as: ⎡1 0 0 3.1375⎤ ⎢ 0 1 0 2.195 ⎥ ⎥ Tsph = ⎢ ⎢ 0 0 1 3.214 ⎥ ⎢ ⎥ 1 ⎦ ⎣0 0 0 • Find the necessary values of r , β , γ to achieve this location. • Find the components of the original matrix n, o, a vectors for the hand before the orientation was restored. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.9.3. Solution: a) The equations representing position in a spherical coordinates may be set equal to the given values as: i. rCγ S β = 3.1375 ii. rSγ S β = 2.195 iii. rC β = 3.214 I. Assuming S β is positive, γ = 35D from ii and iii from iii II. If S β were negative, then from i and ii β = 50D r=5 units. γ = 215D β = −50D r=5 units. Since orientation is not specified, no more information is available to check the results. b) For case I, substitute corresponding values of S β , C β , Sγ , Cγ and r in spherical coordinates to get: ⎡ 0.5265 −0.5735 0.6275 3.1375⎤ ⎢ 0.3687 0.819 0.439 2.195 ⎥⎥ Tsph (r , β , γ ) = Tsph (35,50,5) = ⎢ ⎢ −0.766 0 0.6428 3.214 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 27 Problem 2.23 Suppose that a robot is made of a Cartesian and RPY combination of joints. Find the necessary RPY angles to achieve the following: ⎡ 0.527 −0.574 0.628 4 ⎤ ⎢ 0.369 0.819 0.439 6 ⎥ ⎥ T =⎢ ⎢ −0.766 0 0.643 9 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.10.1 Solution: The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units along the x-, y-, and z-axes. The orientation is achieved by RPY rotations, therefore: φa = ATAN 2(n y , nx ) = ATAN 2(0.369, 0.527) = 35D or φa = ATAN 2(−n y , −nx ) = ATAN 2(−0.369, −0.527) = 215D For φa = 35D φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.766, ( 0.527 × 0.819 + 0.369 × 0.574 ) ) = 50D For φa = 215D φo = ATAN 2 ( 0.766, −0.643) = 130D For φa = 35D φn = ATAN 2(( − a y Cφa + ax Sφa ), (o y Cφa − ox Sφa )) = ATAN 2 ⎡⎣( −0.439 × 0.819 + 0.628 × 0.574 ) , ( 0.819 × 0.819 + 0.574 × 0.574 ) ⎤⎦ = ATAN 2 ( 0,1) = 0D For φa = 215D φn = ATAN 2 ( 0, −1) = 180D ⎧φa = 35D ⎪ Either solution is acceptable: ⎨φo = 50D ⎪ φ = 0D ⎩ n ⎧φa = 215D ⎪ D ⎨φo = 130 ⎪φ = 180D ⎩ n © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 28 Problem 2.24 Suppose that a robot is made of a Cartesian and Euler combination of joints. Find the necessary Euler angles to achieve the following: ⎡ 0.527 −0.574 0.628 4 ⎤ ⎢ 0.369 0.819 0.439 6 ⎥ ⎥ T =⎢ ⎢ −0.766 0 0.643 9 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.10.2 Solution: The position is achieved by Cartesian joints. Therefore, the robot moves 4, 6, and 9 units along the x-, y-, and z-axes. The orientation is achieved by Euler rotations, therefore: φ = ATAN 2(a y , ax ) or φ = ATAN 2(−a y , − ax ) ⎧⎪φ = ATAN 2(0.439, 0.628) = 35D or ⎨ D ⎪⎩φ = ATAN 2(−0.439, −0.628) = 215 ψ = ATAN 2[(− nx Sφ + n y Cφ ), (−ox Sφ + o y Cφ )] ⎧⎪ψ = ATAN 2[0,1] = 0 or ⎨ ⎪⎩ψ = ATAN 2[0, −1] = 180 θ = ATAN 2[(ax Cφ + a y Sφ ), az )] ⎧⎪θ = ATAN 2[(0.628 × 0.819 + 0.439 × 0.573), 0.643)] = 50D or ⎨ D ⎪⎩θ = ATAN 2[(0.628 × ( −0.819 ) + 0.439 × ( −0.573)), 0.643)] = −50 ⎧φ = 35D ⎪ Then ⎨ ψ = 0D ⎪θ = 50D ⎩ ⎧φ = 215D ⎪ D ⎨ψ = 180 ⎪θ = −50D ⎩ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 29 Problem 2.25 Assume that the three Euler angles used with a robot are 30D , 40D ,50D respectively. Determine what angles should be used to achieve the same result if RPY is used instead. Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.10 Solution: For Euler angles: Euler(φ ,θ ,ψ ) = Rot (a, φ ) Rot (o,θ ), Rot (a,ψ ) = Rot (a,30) Rot (o, 40), Rot (a,50) ⎡ 0.0435 −0.8296 0.5568 0 ⎤ ⎢ 0.9096 0.2635 0.3215 0 ⎥ ⎥ =⎢ ⎢ −0.4134 0.4925 0.766 0 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 φa = ATAN 2(n y , nx ) = ATAN 2(0.9096, 0.0435) = 87.26D or φa = ATAN 2(−n y , −nx ) = ATAN 2(−0.9096, −0.0435) = 267.26D φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.4134, 0.9106 ) = 24.5D or φo = ATAN 2 ( 0.4134, −0.9106 ) = 155.5D φn = ATAN 2((− a y Cφa + ax Sφa ), (o y Cφa − ox Sφa )) = ATAN 2 ( 0.5408, 0.8412 ) = 32.7D or φn = ATAN 2(−0.5408, −0.8412) = 212.7D © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 30 Problem 2.26 A frame UB was moved along its own o-axis a distance of 6 units, then rotated about its n-axis an angle of 60D , then translated about the z-axis for 3 units, followed by a rotation of 60D about the z-axis, and finally rotated about x-axis for 45D . • Calculate the total transformation performed. • What angles and movements would we have to make if we were to create the same location and orientation using Cartesian and Euler configurations? Estimated student time to complete: 25-30 minutes Prerequisite knowledge required: Text Section(s) 2.9- 2.10 Solution: U TB = Rot ( x, 45) Rot ( z , 60)Trans(0, 0,3)Trans (0, 6, 0) Rot (n, 60) Bnew −0.433 0.75 −5.196 ⎤ ⎡ 0.5 ⎢0.6123 −0.4355 −0.6596 0 ⎥⎥ =⎢ ⎢0.6123 0.789 0.0474 4.242 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 The Cartesian translations are: px = −5.196, px = 0, px = 4.242 . Euler angles are: φ = ATAN 2(a y , ax ) or φ = ATAN 2(−a y , − ax ) D ⎪⎧φ = ATAN 2(−0.6596, 0.75) = −41.33 or ⎨ D ⎪⎩φ = ATAN 2(0.6596, −0.75) = 138.67 ψ = ATAN 2[(− nx Sφ + n y Cφ ), (−ox Sφ + o y Cφ )] ⎪⎧ψ = ATAN 2[0.79, −0.613] = 127.8 or ⎨ ⎪⎩ψ = ATAN 2[−0.79, 0.613] = −52.2 θ = ATAN 2[(ax Cφ + a y Sφ ), az )] ⎧⎪θ = ATAN 2[0.9988, 0.0474] = 87.28D or ⎨ D ⎪⎩θ = ATAN 2[−0.9988, 0.0474] = −87.28 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 31 Problem 2.27 A frame UF was moved along its own n-axis a distance of 5 units, then rotated about its o-axis an angle of 60D , followed by a rotation of 60D about the z-axis, then translated about its a-axis for 3 units, and finally rotated 45D about the x-axis. • Calculate the total transformation performed. • What angles and movements would we have to make if we were to create the same location and orientation using Cartesian and RPY configurations? Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.11 Solution: U TB = Rot ( x, 45) Rot ( z, 60)Trans (5, 0, 0) Rot (o, 60)Trans (0, 0,3) Bnew ⎡ 0.25 −0.866 0.433 3.8 ⎤ ⎢ 0.918 0.354 0.177 3.59 ⎥ ⎥ =⎢ ⎢ −0.306 0.354 0.884 5.71⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 For Cartesian coordinates: px = 3.8, p y = 3.59, pz = 5.71 For RPY angles: φa = ATAN 2(n y , nx ) = ATAN 2(0.918, 0.25) = 74.8D or φa = ATAN 2(− n y , − nx ) = 254.8D φo = ATAN 2(− nz , (nx Cφa + n y Sφa )) = ATAN 2 ( 0.306, 0.951) = 17.8D φo = ATAN 2 ( 0.306, −0.951) = 162.2D φn = ATAN 2((− a y Cφa + ax Sφa ), (o y Cφa − ox Sφa )) = ATAN 2 [ 0.372, 0.928] = 21.8D φn = ATAN 2 ( −0.372, −0.928 ) = 201.8D © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 32 Problem 2.28 Frames describing the base of a robot and an object are given relative to the Universe frame. • Find a transformation RTH of the robot configuration if the hand of the robot is to be placed on the object. • By inspection, show whether this robot can be a 3-axis spherical robot, and if so, find α, β , r . • Assuming that the robot is a 6-axis robot with Cartesian and Euler coordinates, find px , p y , pz , φ , θ ,ψ . U Tobj ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 0 0 0 −1 1 0 0 0 ⎡ 0 −1 ⎢1 0 U TR = ⎢ ⎢0 0 ⎢ ⎣0 0 1⎤ 4 ⎥⎥ 0⎥ ⎥ 1⎦ 0 2⎤ 0 −1⎥⎥ 1 0⎥ ⎥ 0 1⎦ Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 2.11 Solution: a. We want to place the hand on the part. Therefore, RTH = U TH−1 U Tobj U Tobj = U TR RTH = U TR RTobj ⎡0 ⎢ −1 R Substitute: TH = ⎢ ⎢0 ⎢ ⎣0 1 0 0 0 → 0 0 1 0 1 ⎤ ⎡1 2 ⎥⎥ ⎢⎢ 0 0⎥ ⎢0 ⎥⎢ 1 ⎦ ⎣0 TH = U TR−1 U Tobj R 0 0 0 −1 1 0 0 0 1⎤ ⎡ 0 4 ⎥⎥ ⎢⎢ −1 = 0⎥ ⎢ 0 ⎥ ⎢ 1⎦ ⎣ 0 0 −1 0 0 1 0 0 0 5⎤ 1 ⎥⎥ 0⎥ ⎥ 1⎦ b. No. The 3,2 element of spherical coordinate transformation matrix should be zero. This is not. c. px = 5, p y = 1, pz = 0 φ = ATAN 2[0, −1] = 180 or φ = ATAN 2[0,1] = 0 ψ = ATAN 2[−1(−1), 0] = 90 or ψ = ATAN 2[−1(1), 0] = 270 θ = ATAN 2[−1(−1), 0] = 90D or θ = ATAN 2[−1(1), 0] = 270D ⎧φ = 180D ⎧ φ = 0D ⎪ ⎪ D Then ⎨ψ = 90D ⎨ψ = 270 ⎪ θ = 90D ⎪θ = 270D ⎩ ⎩ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 33 Problem 2.29 A 3-DOF robot arm has been designed for applying paint on flat walls, as shown. • Assign coordinate frames as necessary based on the D-H representation. • Fill out the parameters table. • Find the UTH matrix. Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. # 0-1 1-2 2-H U θ θ1 0 90 d l4 a 0 0 l6 l5 0 α 0 -90 0 TH = U TR RTH = U TR A1 A2 A3 ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 ⎡ 0 −C1 ⎢0 S1 =⎢ ⎢ −1 0 ⎢ 0 ⎣0 l1 ⎤ ⎡C1 l2 ⎥⎥ ⎢⎢ S1 l3 ⎥ ⎢ 0 ⎥⎢ 1⎦ ⎣ 0 − S1 C1 0 0 − S1 C1 0 0 0 0 ⎤ ⎡1 0 0 0 ⎥⎥ ⎢⎢ 0 0 1 l4 ⎥ ⎢ 0 −1 ⎥⎢ 0 1 ⎦ ⎣0 0 − S1l6 + C1l5 + l1 ⎤ C1l6 + S1l5 + l2 ⎥⎥ ⎥ l4 + l3 ⎥ 1 ⎦ 0 1 0 0 l5 ⎤ ⎡ 0 −1 0 ⎥⎥ ⎢⎢ 1 0 0 ⎥ ⎢0 0 ⎥⎢ 1 ⎦ ⎣0 0 0 0 1 0 0⎤ 0 ⎥⎥ l6 ⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 34 Problem 2.30 In the 2-DOF robot shown, the transformation matrix 0TH is given in symbolic form, as well as in numerical form for a specific location. The length of each link l1 and l2 is 1 ft. Calculate the values of θ1 and θ 2 for the given location. ⎡C12 ⎢S 0 TH = ⎢ 12 ⎢ 0 ⎢ ⎣ 0 − S12 C12 0 0 0 l2C12 + l1C1 ⎤ ⎡ −0.2924 −0.9563 0 l2 S12 + l1S1 ⎥⎥ ⎢⎢ 0.9563 −0.2924 = ⎥ ⎢ 0 1 0 0 ⎥ ⎢ 0 1 0 ⎦ ⎣ 0 0 0.6978⎤ 0 0.8172 ⎥⎥ 1 0 ⎥ ⎥ 0 1 ⎦ y x x1 θ2 z θ1 xH zH z1 x0 z0 Figure P.2.30 Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 2.13 Solution: This example is very simple, with only two degrees of freedom. Therefore, there is really no need to use the traditional method of inverse kinematic mentioned in Section 2.13. The joint angles can be determined as follows: θ1 + θ 2 = ATAN 2( S12 , C12 ) = ATAN 2(0.9563, −0.2924) = 107D ⎧l2C12 + l1C1 = px ⎨ ⎩ l2 S12 + l1S1 = p y Substitute: ⎧⎪C1 = ( px − l2C12 ) / l1 → ⎨ → θ1 = ATAN 2 S = p − l S l / ( ) y 2 12 1 ⎪⎩ 1 ( θ1 = ATAN 2 ( p y − l2 S12 ) / l1 , ( px − l2C12 ) / l1 (( p y − l2 S12 ) / l1 , ( px − l2C12 ) / l1 ) θ1 = ATAN 2 ( ( 0.8172 − 0.9563) , ( 0.6978 + 0.2924 ) ) = ATAN 2 ( −0.1391, 0.9902 ) = −8D θ 2 = 107 − (−8) = 115D © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. ) 35 Problem 2.31 For the following SCARA-type robot: • Assign the coordinate frames based on the D-H representation. • Fill out the parameters table. • Write all the A matrices. • Write the UTH matrix in terms of the A matrices. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. # 0-1 1-2 2-H U θ θ1 θ2 θ3 α d 0 a d3 0 d4 180 d5 0 0 0 TH = U T0 0TH = U T0 A1 A2 A3 ⎡1 ⎢0 =⎢ ⎢0 ⎢ ⎣0 ⎤ ⎡C1 1 0 0 ⎥⎥ ⎢⎢ S1 0 1 d1 + d 2 ⎥ ⎢ 0 ⎥⎢ 0 0 1 ⎦⎣ 0 0 0 0 − S1 0 d3C1 ⎤ ⎡C2 C1 0 d 3 S1 ⎥⎥ ⎢⎢ S 2 0 1 0 ⎥⎢ 0 ⎥⎢ 0 0 1 ⎦⎣ 0 S2 −C 2 0 0 d 4C2 ⎤ ⎡C3 0 d 4 S 2 ⎥⎥ ⎢⎢ S3 −1 0 ⎥⎢ 0 ⎥⎢ 0 1 ⎦⎣ 0 0 − S3 C3 0 0 0⎤ 0 0 ⎥⎥ 1 d5 ⎥ ⎥ 0 1⎦ 0 Multiplying these matrices will yield the total transformation. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 36 Problem 2.32 A special 3-DOF spraying robot has been designed as shown: • Assign the coordinate frames based on the D-H representation. • Fill out the parameters table. • Write all the A matrices. • Write the UTH matrix in terms of the A matrices. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: Please note that no particular reset position is specified for this robot. # 0-1 1-2 2-3 3-H θ α d 0 a 0 90 θ3 l2 0 0 0 -90 90 0 l3 0 0 90 + θ1 0 0 0 l1 ⎤ ⎡ − S1 0 C1 0 ⎤ ⎥ ⎢ C 0 S 0⎥ 1 0 0⎥ 1 ⎥ A1 = ⎢ 1 ⎢ 0 1 0 0⎥ 0 1 0⎥ ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 0 1 1⎦ ⎡1 0 0 0 ⎤ ⎡C3 0 S3 0 ⎤ ⎡1 0 0 0 ⎤ ⎢0 0 1 0 ⎥ ⎢ S 0 −C 0 ⎥ ⎢0 1 0 0 ⎥ 3 ⎥ ⎥ ⎥ A2 = ⎢ A3 = ⎢ 3 A4 = ⎢ ⎢ 0 −1 0 l2 ⎥ ⎢ 0 1 0 0⎥ ⎢ 0 0 1 l3 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 0 0 1 ⎦ ⎣ 0 0 0 1⎦ ⎣0 0 0 1 ⎦ U TH = U T0 0TH = U T0 A1 A2 A3 A4 The total transformation can be summarized by multiplying these matrices. ⎡1 ⎢0 U T0 = ⎢ ⎢0 ⎢ ⎣0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 37 Problem 2.33 For the Unimation Puma 562, 6-axis robot shown, • Assign the coordinate frames based on the D-H representation. • Fill out the parameters table. • Write all the A matrices. • Find the RTH matrix for the following values: Base height=27 in, d2=6 in, a2=15 in, a3=1 in, d4=18 in θ1 = 0D , θ 2 = 45D ,θ3 = 0D ,θ 4 = 0D ,θ5 = −45D ,θ 6 = 0D Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. To express the height from the table, the 27-in base height must be added to the pz value. Please also note that no particular reset position is specified. Therefore, the results relate to a position and orientation that would correspond to the given transformations from a reset position. We assume at reset, there is 90 degrees between x0 and x1 . Please note how frames 2, 3, and 4 relate to each other by referring to the side view of the arm. (Reprinted with permission from Staubli Robotics.) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 38 ⎡0 0 −1 ⎢1 0 0 A1 = ⎢ ⎢0 −1 0 ⎢ ⎣0 0 0 ⎡1 ⎢0 A4 = ⎢ ⎢0 ⎢ ⎣0 0 0 0 −1 1 0 0 0 ⎡0 ⎢1 0 T6 = ⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 0⎤ ⎡0.707 −0.707 0 10.61⎤ ⎡1 ⎢0.707 0.707 0 10.61⎥ ⎢0 0 ⎥⎥ ⎥ A2 = ⎢ A3 = ⎢ ⎢ 0 ⎢0 0⎥ 0 1 6 ⎥ ⎥ ⎢ ⎥ ⎢ 1⎦ 0 0 1 ⎦ ⎣ 0 ⎣0 0⎤ ⎡ 0.707 0 0.707 0 ⎤ ⎡1 0 ⎥ ⎢ ⎥ ⎢0 1 0⎥ −0.707 0 0.707 0 ⎥ ⎢ ⎢ A6 = A5 = ⎢0 0 ⎢ 0 18⎥ 0 0⎥ −1 ⎥ ⎢ ⎥ ⎢ 1⎦ 0 0 1⎦ ⎣0 0 ⎣ 0 −6 ⎤ −1.41⎥⎥ −24 ⎥ ⎥ 1 ⎦ 0 0 −1 0 0 1 0 0 0 0 1 0 0⎤ 0 ⎥⎥ 0⎥ ⎥ 1⎦ 0⎤ 0 ⎥⎥ 0⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 39 Problem 2.34 For the given 4-DOF robot: • Assign appropriate frames for the Denavit-Hartenberg representation. • Fill out the parameters table. • Write an equation in terms of A-matrices that shows how U TH can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. ⎡1 ⎢0 0 U U U TH = T0 TH = T0 A1 A2 A3 A4 = ⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 l1 ⎤ l2 ⎥⎥ A1 A2 A3 A4 l3 ⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 40 Problem 2.35 For the given 4-DOF robot designed for a specific operation: • • • Assign appropriate frames for the Denavit-Hartenberg representation. Fill out the parameters table. Write an equation in terms of A-matrices that shows how U TH can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. * L = ( l4 + l5 ) / sin β The total transformation for the robot can be determined from: ⎡1 0 0 l1 ⎤ ⎢0 1 0 0 ⎥ U U U 0 ⎥AA A A TH = T0 TH = T0 A1 A2 A3 A4 = ⎢ ⎢0 0 1 0 ⎥ 1 2 3 4 ⎢ ⎥ ⎣0 0 0 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 41 Problem 2.36 For the given specialty-designed 4-DOF robot: • Assign appropriate frames for the Denavit-Hartenberg representation. • Fill out the parameters table. • Write an equation in terms of A-matrices that shows how U TH can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. ⎡1 ⎢0 0 U U U TH = T0 TH = T0 A1 A2 A3 = ⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 0 1 0 l1 ⎤ l2 ⎥⎥ A1 A2 A3 0⎥ ⎥ 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 42 Problem 2.37 For the given 3-DOF robot: • Assign appropriate frames for the Denavit-Hartenberg representation. • Fill out the parameters table. • Write an equation in terms of A-matrices that shows how U TH can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. Please also note that there can be an additional frame between frames 1 and H. In that case, the two joint variables will be represented by two separate matrices. # 0-1 θ −90 + θ1 d −l3 − l4 a 0 α -90 1-H θ2 l5 + l6 + l7 0 0 ⎡1 ⎢0 U TH = U T0 0TH = U T0 A1 A2 = ⎢ ⎢0 ⎢ ⎣0 0 1 0 0 0 l1 ⎤ 0 −l2 ⎥⎥ A1 A2 1 0 ⎥ ⎥ 0 1 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 43 Problem 2.38 For the given 4-DOF robot: • Assign appropriate frames for the Denavit-Hartenberg representation. • Fill out the parameters table. • Write an equation in terms of A matrices that shows how U TH can be calculated. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 2.12 Solution: In this solution, the locations of the origins of some of the frames are arbitrary. Therefore, intermediate matrices might be different for each case. However, the final answer should be the same. Please also note that no particular reset position is specified. # 0-1 θ θ1 d l3 1-2 90 + θ 2 2-H θ3 α 90 −l5 a l4 0 l6 0 0 ⎡1 ⎢0 0 U U U TH = T0 TH = T0 A1 A2 A3 = ⎢ ⎢0 ⎢ ⎣0 0 1 0 0 90 0 l1 ⎤ 0 l2 ⎥⎥ A1 A2 A3 1 0⎥ ⎥ 0 1⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 44 Problem 2.39 Derive the inverse kinematic equations for the robot of Problem 2.36. Estimated student time to complete: 20-30 minutes if Problem 2.36 is already solved Prerequisite knowledge required: Text Section(s) 2.13 Solution: We assume all positions are made relative to the base of the robot, and therefore, U T0 is not included in the solution. Using the Denavit-Hartenberg representation and the joint parameters shown, we get: ⎡C1 ⎢S A1 = ⎢ 1 ⎢0 ⎢ ⎣0 0 1 0 TH = A1 A2 A3 S1 0 −C1 0 → A1−1 0TH = A2 A3 0 −C1 0 ⎡C1 ⎢0 ⎢ ⎢ S1 ⎢ ⎣0 0 0 0 ⎤ 0 ⎥⎥ l3 + l4 ⎥ ⎥ 1 ⎦ S1 ⎤ ⎡ nx 1 −l3 − l4 ⎥⎥ ⎢⎢ n y 0 0 ⎥ ⎢ nz ⎥⎢ 0 1 ⎦⎣ 0 0 0 ⎡ C2 ⎢S A2 = ⎢ 2 ⎢0 ⎢ ⎣0 ox ax oy ay oz az 0 0 0 − S2 0 C2 −1 0 0 0 p x ⎤ ⎡ C2 p y ⎥⎥ ⎢⎢ S 2 = pz ⎥ ⎢ 0 ⎥ ⎢ 1⎦ ⎣0 ⎡ −C3 ⎢ −S A3 = ⎢ 3 ⎢ 0 ⎢ ⎣ 0 l5C2 ⎤ l5 S 2 ⎥⎥ 0 ⎥ ⎥ 1 ⎦ 0 − S2 0 C2 −1 0 0 0 S3 −C3 0 0 l5C2 ⎤ ⎡ −C3 l5 S 2 ⎥⎥ ⎢⎢ − S3 0 ⎥⎢ 0 ⎥⎢ 1 ⎦⎣ 0 0 ⎤ 0 0 ⎥⎥ 1 l6 + l7 ⎥ ⎥ 0 1 ⎦ 0 S3 −C3 0 0 0 ⎤ 0 0 ⎥⎥ 1 l6 + l7 ⎥ ⎥ 0 1 ⎦ 0 Multiplying these matrices we get: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 45 ⎡C1nx + S1n y ⎢ nz ⎢ LHS = ⎢ S1nx − C1n y ⎢ 0 ⎣ ⎡ −C2C3 ⎢ −S C RHS = ⎢ 2 3 ⎢ S3 ⎢ ⎣ 0 C1ox + S1o y oz S1ox − C1o y 0 C2 S 3 S 2 S3 C3 0 − S2 C2 0 0 C1ax + S1a y az S1ax − C1a y 0 C1 px + S1 p y ⎤ pz − l3 − l4 ⎥⎥ S1 px − C1 p y ⎥ ⎥ 1 ⎦ − S 2 (l6 + l7 ) + l5C2 ⎤ C2 (l6 + l7 ) + l5 S2 ⎥⎥ ⎥ 0 ⎥ 1 ⎦ ⎛a ⎞ S1ax − C1a y = 0 → θ1 = tan −1 ⎜ y ⎟ ⎝ ax ⎠ From 3,3 or 3,4 elements we get: ⎛p ⎞ S1 px − C1 p y = 0 → θ1 = tan −1 ⎜ y ⎟ ⎝ px ⎠ From 1,3 and 2,3 elements we ⎧ S 2 = −(C1ax + S1a y ) get: ⎨ → θ 2 = ATAN 2 ( −C1ax − S1a y , az ) ⎩C2 = az From 2,1 and 2,2 elements we get: ⎧⎪ S3 = S1nx − C1n y ⎨ ⎪⎩C3 = S1ox − C1o y → θ3 = ATAN 2 ( S1nx − C1n y , S1ox − C1o y ) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 46 Problem 2.40 Derive the inverse kinematic equations for the robot of Problem 2.37. Estimated student time to complete: 20-30 minutes if Problem 2.37 is already solved Prerequisite knowledge required: Text Section(s) 2.13 Solution: We assume all positions are made relative to the base of the robot, and therefore, U T0 is not included in the solution. Using the Denavit-Hartenberg representation and the joint parameters shown, we get: ⎡ S1 ⎢ −C A1 = ⎢ 1 ⎢ 0 ⎢ ⎣ 0 0 C1 0 S1 −1 0 0 0 # 0-1 θ −90 + θ1 d −l3 − l4 a 0 α -90 1-H θ2 l5 + l6 + l7 0 0 0 ⎤ 0 ⎥⎥ −l3 − l4 ⎥ ⎥ 1 ⎦ ⎡ C2 ⎢S A2 = ⎢ 2 ⎢0 ⎢ ⎣0 −S2 C2 0 0 0 0 ⎤ ⎥ 0 0 ⎥ 1 l6 + l6 + l7 ⎥ ⎥ 0 1 ⎦ TH = A1 A2 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 47 ⎡ nx ⎢n ⎢ y ⎢ nz ⎢ ⎣0 ox ax oy ay oz az 0 0 px ⎤ ⎡ S1 0 C1 0 ⎤ ⎡ C2 − S 2 0 0 ⎤ ⎥ ⎢ ⎥ ⎢ ⎥ p y ⎥ ⎢ −C1 0 S1 0 ⎥ ⎢ S 2 C2 0 0 ⎥ = pz ⎥ ⎢ 0 −1 0 −l3 − l4 ⎥ ⎢ 0 0 1 l5 + l6 + l7 ⎥ ⎥ ⎢ ⎥⎢ ⎥ 1⎦ ⎣ 0 0 0 1 ⎦⎣ 0 0 0 1 ⎦ ⎡ S1C2 − S1S 2 C1 C1 (l5 + l6 + l7 ) ⎤ ⎢ −C C S 2C1 S1 S1 (l5 + l6 + l7 ) ⎥⎥ 1 2 ⎢ = ⎢ − S2 −C2 0 −l3 − l4 ⎥ ⎢ ⎥ 0 0 1 ⎣ 0 ⎦ ⎧⎪ S1 = a y From 1,3 and 2,3 elements we get: ⎨ ⎪⎩C1 = ax → θ1 = ATAN 2 ( a y , ax ) ⎧ S = − nz From 3,1 and 3,2 elements we get: ⎨ 2 ⎩C2 = −oz → θ 2 = ATAN 2 ( − nz , −oz ) From 1,4 elements we get: C1 (l5 + l6 + l7 ) = px → l5 + l6 + l7 = px C1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 48 CHAPTER THREE Problem 3.1 Suppose the location and orientation of a hand frame is expressed by the following matrix. What is the effect of a differential rotation of 0.15 radians about the z-axis, followed by a differential translation of [0.1, 0.1, 0.3]? Find the new location of the hand. ⎡0 ⎢1 R TH = ⎢ ⎢0 ⎢ ⎣0 0 0 1 0 1 0 0 0 2⎤ 7 ⎥⎥ 5⎥ ⎥ 1⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6, 3.7. Solution: For δ z = 0.15, dx = 0.1, dy = 0.1, dz = 0.3, we get: −0.15 ⎡ 0 ⎢0.15 0 Δ=⎢ ⎢ 0 0 ⎢ 0 ⎣ 0 0 0.1⎤ 0 0.1⎥⎥ 0 0.3⎥ ⎥ 0 0⎦ −0.15 ⎡ 0 ⎢0.15 0 d ⎡⎣ RTH ⎤⎦ = [ Δ ] ⎡⎣ RTH ⎤⎦ = ⎢ ⎢ 0 0 ⎢ 0 ⎣ 0 ⎡⎣ RTH ⎤⎦ = ⎡⎣ RTH ⎤⎦ new old 0 0.1⎤ ⎡0 0 0.1⎥⎥ ⎢⎢1 0 0.3⎥ ⎢0 ⎥⎢ 0 0 ⎦ ⎣0 ⎡ −0.15 ⎢ 1 + d ⎡⎣ RTH ⎤⎦ = ⎢ ⎢ 0 ⎢ ⎣ 0 0 1 2 ⎤ ⎡ −0.15 0 0 7 ⎥⎥ ⎢⎢ 0 = 1 0 5⎥ ⎢ 0 ⎥ ⎢ 0 0 1⎦ ⎣ 0 −0.95⎤ 0 0.15 0.4 ⎥⎥ 0 0 0.3 ⎥ ⎥ 0 0 0 ⎦ 0 0 1.05⎤ 0 0.15 7.4 ⎥⎥ 1 0 5.3 ⎥ ⎥ 0 0 1 ⎦ 0 1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 49 Problem 3.2 As a result of applying a set of differential motions to frame T shown, it has changed an amount dT as shown. Find the magnitude of the differential changes made ( dx, dy, dz , δ x, δ y, δ z ) and the differential operator with respect to frame T. ⎡1 0 ⎢0 0 T =⎢ ⎢ 0 −1 ⎢ ⎣0 0 0 1 0 0 −0.1 −0.1 0.6 ⎤ ⎡ 0 ⎢ 0.1 0 0 0.5 ⎥⎥ ⎢ dT = ⎢ −0.1 0 0 −0.5⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 5⎤ 3⎥⎥ 8⎥ ⎥ 1⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8. Solution: [ dT ] = [ Δ ][T ] → [ Δ ] = [ dT ][T ] −1 −0.1 −0.1 0.6 ⎤ ⎡1 ⎡ 0 ⎢ 0.1 0 0 0.5 ⎥⎥ ⎢⎢ 0 ⎢ [Δ] = ⎢ 0 −0.1 0 −0.5⎥ ⎢ 0 ⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣ 0 −0.1 0.1 0.1⎤ 0 0 −5⎤ ⎡ 0 ⎥ ⎢ 0 −1 8 ⎥ ⎢ 0.1 0 0 0 ⎥⎥ = 1 0 −3⎥ ⎢ −0.1 0 0 0⎥ ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 0 0 0⎦ By inspection: d = [0.1,0,0], δ = [0,0.1,0.1]. [ dT ] = [T ] ⎡⎣ T Δ ⎤⎦ ⎡1 ⎢0 ⎡⎣ T Δ ⎤⎦ = ⎢ ⎢0 ⎢ ⎣0 0 0 1 0 → ⎡⎣ T Δ ⎤⎦ = [T ] [ dT ] −0.1 −0.1 0.6 ⎤ ⎡ 0 −0.1 −0.1 0.6 ⎤ 0 −5⎤ ⎡ 0 ⎥ ⎢ −1 8 ⎥ ⎢ 0.1 0 0 0.5 ⎥⎥ ⎢⎢ 0.1 0 0 0.5⎥⎥ = −0.5⎥ ⎢ 0.1 0 0 −3⎥ ⎢ −0.1 0 0 0 0.5⎥ ⎥⎢ ⎥ ⎢ ⎥ 0 1 ⎦⎣ 0 0 0 0 ⎦ ⎣0 0 0 0 ⎦ −1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 50 Problem 3.3 Suppose the following frame was subjected to a differential translation of d = [1 0 0.5] units and a differential rotation of δ = [0 0.1 0] . a. What is the differential operator relative to the reference frame? b. What is the differential operator relative to the frame A? ⎡0 ⎢1 A=⎢ ⎢0 ⎢ ⎣0 0 0 1 0 1 10 ⎤ 0 5 ⎥⎥ 0 0⎥ ⎥ 0 1⎦ Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8 Solution: a. ⎡ 0 ⎢ 0 Δ=⎢ ⎢ −0.1 ⎢ ⎣ 0 0 0.1 1 ⎤ 0 0 0 ⎥⎥ 0 0 0.5⎥ ⎥ 0 0 0⎦ b. ⎡0 ⎢0 −1 A ⎡⎣ Δ ⎤⎦ = [ A] [ Δ ][ A] = ⎢ ⎢1 ⎢ ⎣0 −5 ⎤ ⎡ 0 0 1 0 ⎥⎥ ⎢⎢ 0 0 0 −10 ⎥ ⎢ −0.1 ⎥⎢ 0 0 1 ⎦⎣ 0 1 0 0 0.1 0 0 0 0 0 0 1 ⎤ ⎡0 0 ⎥⎥ ⎢⎢1 0.5⎥ ⎢ 0 ⎥⎢ 0 ⎦ ⎣0 0 1 10 ⎤ 0 0 5 ⎥⎥ 1 0 0⎥ ⎥ 0 0 1⎦ 0 0 ⎤ ⎡0 0 ⎢0 0 −0.1 −0.5⎥ ⎥ =⎢ ⎢0 0.1 0 1 ⎥ ⎢ ⎥ 0 0 ⎦ ⎣0 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 51 Problem 3.4 The initial location and orientation of a robot’s hand are given by T1, and its new location and orientation after a change are given by T2. a. Find a transformation matrix Q that will accomplish this transform (in the Universe frame). b. Assuming the change is small, find a differential operator Δ that will do the same. c. By inspection, find a differential translation and a differential rotation that constitute this operator. ⎡1 ⎢0 T1 = ⎢ ⎢0 ⎢ ⎣0 0 0 0 −1 1 0 0 0 5⎤ 3⎥⎥ 6⎥ ⎥ 1⎦ ⎡1 ⎢0.1 T2 = ⎢ ⎢0 ⎢ ⎣0 0 0.1 4.8⎤ 0 −1 3.5 ⎥⎥ 1 0 6.2 ⎥ ⎥ 0 0 1 ⎦ Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.8 Solution: a. T2 = QT1 ⎡1 ⎢0.1 → Q = T2T1−1 = ⎢ ⎢0 ⎢ ⎣0 0 0.1 4.8⎤ ⎡1 0 0 −1 3.5 ⎥⎥ ⎢⎢0 0 1 0 6.2 ⎥ ⎢0 −1 ⎥⎢ 0 0 1 ⎦ ⎣0 0 0 −5⎤ ⎡ 1 −0.1 1 −6 ⎥⎥ ⎢⎢ 0.1 1 = 0 3⎥ ⎢0 0 ⎥ ⎢ 0 1⎦ ⎣0 0 0 0.1⎤ 0 0 ⎥⎥ 1 0.2 ⎥ ⎥ 0 1 ⎦ b. T2 = T1 + dT = T1 + ΔT1 = ( Δ + I ) T1 = QT1 Therefore: Δ = Q − I ⎡ 0 −0.1 0 0.1⎤ ⎢ 0.1 0 0 0 ⎥⎥ ⎢ or Δ = ⎢0 0 0 0.2 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣0 c. By inspection: d = [0.1, 0, 0.2] and δ = [0, 0, 0.1]. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 52 Problem 3.5 The hand frame of a robot and the corresponding Jacobian are given. For the given differential changes of the joints, compute the change in the hand frame, its new location, and corresponding Δ . ⎡0 ⎢1 T6 = ⎢ ⎢0 ⎢ ⎣0 1 0 10 ⎤ 0 0 5 ⎥⎥ 0 −1 0 ⎥ ⎥ 0 0 1⎦ ⎡8 0 0 0 0 ⎢ −3 0 1 0 0 ⎢ ⎢ 0 10 0 0 0 T6 J =⎢ ⎢0 1 0 0 1 ⎢0 0 0 1 0 ⎢ ⎣ −1 0 0 0 0 0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎦ ⎡ 0 ⎤ ⎢ 0.1 ⎥ ⎢ ⎥ ⎢ −0.1⎥ Dθ = ⎢ ⎥ ⎢ 0.2 ⎥ ⎢ 0.2 ⎥ ⎢ ⎥ ⎣ 0 ⎦ Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: ⎡8 0 0 0 0 ⎢ −3 0 1 0 0 ⎢ ⎢ 0 10 0 0 0 ⎡⎣ T6 D ⎤⎦ = ⎡⎣ T6 J ⎤⎦ [ Dθ ] = ⎢ ⎢0 1 0 0 1 ⎢0 0 0 1 0 ⎢ ⎣ −1 0 0 0 0 0 ⎤ ⎡ 0 ⎤ ⎡ 0 ⎤ ⎡ T6 dx ⎤ ⎢ ⎥ 0 ⎥⎥ ⎢⎢ 0.1 ⎥⎥ ⎢⎢ −0.1⎥⎥ ⎢ T6 dy ⎥ 0 ⎥ ⎢ −0.1⎥ ⎢ 1 ⎥ ⎢ T6 dz ⎥ ⎥ ⎥⎢ ⎥=⎢ ⎥=⎢ 0 ⎥ ⎢ 0.2 ⎥ ⎢ 0.3 ⎥ ⎢ T6δ x ⎥ 0 ⎥ ⎢ 0.2 ⎥ ⎢ 0.2 ⎥ ⎢ T6δ y ⎥ ⎥ ⎥⎢ ⎥ ⎢ ⎥ ⎢ 1 ⎦ ⎣ 0 ⎦ ⎣ 0 ⎦ ⎢⎣ T6δ z ⎥⎦ 0 0.2 0 ⎤ ⎡ 0 ⎢ 0 0 −0.3 −0.1⎥⎥ T6 T6 ⎢ Substitute in Δ to get: Δ = ⎢ −0.2 0.3 0 1 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 Then 0 0.2 0 ⎤ ⎡ 0 0 −0.3 −0.1⎤ ⎡ 0 1 0 10 ⎤ ⎡ 0 ⎢1 0 0 5 ⎥ ⎢ 0 0 −0.3 −0.1⎥⎥ ⎢⎢ 0 0 0.2 0 ⎥⎥ ⎥⎢ = dT6 = [T6 ] ⎡⎣ T6 Δ ⎤⎦ = ⎢ ⎢ 0 0 −1 0 ⎥ ⎢ −0.2 0.3 −1 ⎥ 0 1 ⎥ ⎢0.2 −0.3 0 ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 0 0 0 ⎦ ⎣0 0 0 1 ⎦ ⎣ 0 1 −0.3 9.9 ⎤ ⎡ 0 ⎢1 0 0.2 5 ⎥⎥ [T6 ]new = [T6 ]old + [ dT6 ] = ⎢⎢ 0.2 −0.3 −1 −1 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 53 0 0.2 0 ⎤ ⎡0 ⎡0 1 0 10 ⎤ ⎡ 0 ⎢1 0 0 5 ⎥ ⎢ 0 0 −0.3 −0.1⎥⎥ ⎢⎢1 −1 ⎥⎢ Δ = [T6 ] ⎡⎣ T6 Δ ⎤⎦ [T6 ] = ⎢ ⎢0 0 −1 0 ⎥ ⎢ −0.2 0.3 0 1 ⎥ ⎢0 ⎢ ⎥⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣0 0 0 1 ⎦ ⎣ 0 0 0.3 −0.1⎤ ⎡ 0 ⎢ 0 0 −0.2 0 ⎥⎥ ⎢ = ⎢ −0.3 0.2 0 1 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 −5 ⎤ 0 0 −10 ⎥⎥ 0 −1 0 ⎥ ⎥ 0 0 1 ⎦ 1 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 54 Problem 3.6 Two consecutive frames describe the old (T1) and new (T2) positions and orientations of the end of a 3-DOF robot. The corresponding Jacobian relative to T1, relating to T1 dz , T1δ x, T1δ z is also given. Find values of joint differential motions ds1 , dθ 2 , dθ 3 of the robot that caused the given frame change. 0.01 1 8.1⎤ ⎡0 0 1 8 ⎤ ⎡ 0 ⎡5 10 0 ⎤ ⎢1 0 0 5 ⎥ ⎢ 1 ⎥ − 0.05 0 5 T 1 ⎥ ⎥ T1 = ⎢ T2 = ⎢ J = ⎢⎢ 3 0 0 ⎥⎥ ⎢0 1 0 2 ⎥ ⎢0.05 −0.01 2 ⎥ 1 ⎢⎣0 1 1 ⎥⎦ ⎢ ⎥ ⎢ ⎥ 0 0 1⎦ ⎣0 0 0 1 ⎦ ⎣ 0 Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: 0.01 ⎡ 0 ⎢ 0 −0.05 T2 − T1 = dT = ⎢ ⎢0.05 0 ⎢ 0 ⎣ 0 ⎡0 1 0 ⎢0 0 1 −1 T1 ⎡⎣ Δ ⎤⎦ = [T1 ] dT = ⎢ ⎢1 0 0 ⎢ ⎣0 0 0 ⎧ T1δ z = 0.05 ⎪ Therefore, ⎨ T1δ x = 0.01 ⎪ T1 ⎩ d z = 0.1 → And: ⎡⎣ T1 D ⎤⎦ = ⎡⎣ T1 J ⎤⎦ [ Dθ ] → 0.1⎤ 0 0 ⎥⎥ = [ Δ ][T1 ] = [T1 ] ⎡⎣ T1 Δ ⎤⎦ −0.01 0 ⎥ ⎥ 0 0⎦ 0.01 0 0.1⎤ ⎡ 0 0 0⎤ −5⎤ ⎡ 0 −0.05 ⎥ ⎢ ⎥ ⎢ 0 0 ⎥ ⎢ 0.05 0 −2 ⎥ ⎢ 0 −0.05 −0.01 0 ⎥⎥ = 0 0.01 0 0.1⎥ −8⎥ ⎢ 0.05 −0.01 0 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ 1 ⎦⎣ 0 0 0 0⎦ ⎣ 0 0 0 0⎦ 0 T1 ⎡ 0.1 ⎤ D = ⎢⎢ 0.01⎥⎥ ⎣⎢ 0.05⎦⎥ [ Dθ ] = ⎡⎣ T J ⎤⎦ 1 −1 ⎡⎣ T1 D ⎤⎦ Please see Appendix A for methods to calculate the inverse of a non-unitary matrix. Substitute to get: −1 ⎡ 5 10 0 ⎤ ⎡ 0.1 ⎤ ⎡ 0.0033 ⎤ −1 T T1 [ Dθ ] = ⎡⎣ J ⎤⎦ ⎡⎣ 1 D ⎤⎦ = ⎢⎢3 0 0⎥⎥ ⎢⎢ 0.01⎥⎥ = ⎢⎢0.00833⎥⎥ ⎢⎣ 0 1 1 ⎥⎦ ⎢⎣ 0.05⎥⎦ ⎢⎣ 0.0417 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 55 Problem 3.7 Two consecutive frames describe the old (T1) and new (T2) positions and orientations of the end of a 3-DOF robot. The corresponding Jacobian, relating to dz , δ x, δ z is also given. Find values of joint differential motions ds1 , dθ 2 , dθ 3 of the robot that caused the given frame change. 0 1 9.75⎤ ⎡0 0 1 10 ⎤ ⎡ −0.05 ⎢1 0 0 5 ⎥ ⎢ 1 −0.1 0.05 5.2 ⎥⎥ ⎢ ⎥ ⎢ T1 = T2 = ⎢0 1 0 3 ⎥ ⎢ 0.1 1 0 3.7 ⎥ ⎢ ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣0 0 0 1 ⎦ ⎣ 0 Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 ⎡5 10 0 ⎤ J = ⎢⎢ 3 0 0 ⎥⎥ ⎢⎣0 1 1 ⎥⎦ Solution: 0 0 −0.25⎤ ⎡ −0.05 ⎢ 0 −0.1 0.05 0.2 ⎥⎥ ⎢ T2 − T1 = dT = ⎢ 0.1 0 0 0.7 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 [ dT ] = [ Δ ][T1 ] 0 0 −0.25⎤ ⎡0 ⎡ −0.05 ⎢ 0 −0.1 0.05 0.2 ⎥⎥ ⎢⎢0 −1 ⎢ → [ Δ ] = [ dT ][T1 ] = ⎢ 0.1 0 0 0.7 ⎥ ⎢1 ⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣ 0 −5 ⎤ 0 1 −3 ⎥⎥ 0 0 −10 ⎥ ⎥ 0 0 1 ⎦ 1 0 0 0 ⎤ −0.05 ⎡ 0 ⎢0.05 0 −0.1 0 ⎥⎥ =⎢ ⎢ 0 0.1 0 0.2 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 ⎧d z = 0.2 ⎪ Therefore, ⎨δ x = 0.1 ⎪δ = 0.05 ⎩ z ⎡ 0.2 ⎤ → D = ⎢⎢ 0.1 ⎥⎥ ⎢⎣ 0.05⎥⎦ −1 ⎡ 5 10 0 ⎤ ⎡ 0.2 ⎤ ⎡ 0.0334 ⎤ −1 Substitute to get: [ Dθ ] = [ J ] [ D ] = ⎢⎢ 3 0 0 ⎥⎥ ⎢⎢ 0.1 ⎥⎥ = ⎢⎢0.00334 ⎥⎥ ⎢⎣ 0 1 1 ⎥⎦ ⎢⎣0.05⎥⎦ ⎢⎣ 0.0467 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 56 Problem 3.8 A camera is attached to the hand frame T of a robot as given. The corresponding inverse Jacobian of the robot at this location is also given. The robot makes a differential motion, as a result of which, the change in the frame dT is recorded as given. a. Find the new location of the camera after the differential motion. b. Find the differential operator. c. Find the joint differential motion values associated with this move. d. Find how much the differential motions of the hand-frame ( T D ) should have been instead, if measured relative to frame T, to move the robot to the same new location as in part a. 0 0 0 0 0⎤ ⎡1 ⎢ −1 0 0 0 ⎥⎥ 0 0 −0.1 0.79 ⎤ ⎡ −0.03 ⎡0 1 0 3⎤ ⎢2 ⎢ ⎢1 0 0 2 ⎥ ⎢ 0 −0.2 0 0 0 0 ⎥ 0 0.03 0 0.09 ⎥⎥ −1 ⎢ ⎢ ⎥ T= J =⎢ ⎥ dT = ⎢ 0 ⎢ 0 0 −1 8 ⎥ −0.1 0 −0.4 ⎥ ⎢ 0 −1 0 0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 0 0 1 0 0⎥ 0 0 0 ⎦ ⎣0 0 0 1 ⎦ ⎣ 0 ⎢ ⎥ 0 0 0 0 1⎦ ⎣1 Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: a. Tnew = Told −0.1 3.79 ⎤ ⎡ −0.03 1 ⎢ 1 0.03 0 2.09 ⎥⎥ + dT = ⎢ ⎢ 0 −0.1 −1 7.6 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 b. −0.1 3.79 ⎤ ⎡ 0 1 0 −2 ⎤ ⎡ 0 −0.03 0.1 0.05⎤ ⎡ −0.03 1 ⎢ 1 ⎥ ⎢ ⎥ ⎢ 0.03 0 2.09 ⎥ ⎢1 0 0 −3⎥ ⎢ 0.03 0 0 0 ⎥⎥ Δ = dT ⋅ T −1 = ⎢ = ⎢ 0 −0.1 −1 7.6 ⎥ ⎢ 0 0 −1 8 ⎥ ⎢ −0.1 0 0 −0.1⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣0 0 0 1 ⎦ ⎣ 0 0 0 0 ⎦ ⎣ 0 c. D = [ 0.05 0 −0.1 0 0.1 0.03] 0 0 ⎡1 ⎢2 −1 0 ⎢ ⎢ 0 −0.2 0 Therefore: Dθ = J −1 ⋅ D = ⎢ ⎢ 0 −1 0 ⎢0 0 0 ⎢ 0 0 ⎣1 T 0 0 0 0 1 0 0 0 0 1 0 0 0 ⎤ ⎡ 0.05⎤ ⎡ 0.05⎤ 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥ 0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥ ⎥⎢ ⎥=⎢ ⎥ 0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥ 0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎦ ⎣ 0.03⎦ ⎣ 0.08⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 57 d. ⎡0 ⎢1 T T −1 Δ = T ⋅ dT = ⎢ dT = T ⋅ Δ → ⎢0 ⎢ ⎣0 0.03 0 0.09 ⎤ ⎡ 0 ⎢ −0.03 0 −0.1 0.79 ⎥⎥ ⎢ = ⎢ 0 0.1 0 0.4 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 T −2 ⎤ ⎡ −0.03 0 −0.1 0.79 ⎤ ⎥ ⎢ 0 0 −3⎥ ⎢ 0 0.03 0 0.09 ⎥⎥ −0.1 0 −0.4 ⎥ 0 −1 8 ⎥ ⎢ 0 ⎥⎢ ⎥ 0 0 1 ⎦⎣ 0 0 0 0 ⎦ 1 0 D = [ 0.09 0.79 0.4 0.1 0 −0.03] T © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 58 Problem 3.9 A camera is attached to the hand frame T of a robot as given. The corresponding inverse Jacobian of the robot relative to the frame at this location is also given. The robot makes a differential motion, as a result of which, the change dT in the frame is recorded as given. a. Find the new location of the camera after the differential motion. b. Find the differential operator. Find the joint differential motion values Dθ associated with this move. c. ⎡0 ⎢1 T =⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 3⎤ 2 ⎥⎥ 8⎥ ⎥ 1⎦ 0 ⎡ −0.02 ⎢ 0 0.02 dT = ⎢ ⎢ 0 −0.1 ⎢ 0 ⎣ 0 ⎡1 ⎢ ⎢2 ⎢0 T −1 J =⎢ ⎢0 ⎢0 ⎢ ⎣1 −0.1 0.7 ⎤ 0 0.08 ⎥⎥ −0.3⎥ 0 ⎥ 0 0 ⎦ 0 0 −1 0 −0.1 0 −1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎦ Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: a. Tnew = Told 1 −0.1 3.7 ⎤ ⎡ −0.02 ⎢ 1 0.02 0 2.08⎥⎥ + dT = ⎢ ⎢ 0 −0.1 −1 7.7 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 b. 0 −0.1 0.7 ⎤ ⎡0 ⎡ −0.02 ⎢ 0 0.02 0 0.08 ⎥⎥ ⎢⎢1 −1 ⎢ Δ = dT ⋅ T = ⎢ 0 −0.1 0 −0.3⎥ ⎢0 ⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣ 0 1 0 −2 ⎤ ⎡ 0 −0.02 0.1 −0.06 ⎤ ⎥ ⎢ 0 0 −3⎥ ⎢0.02 0 0 0.02 ⎥⎥ = 0 −1 8 ⎥ ⎢ −0.1 0 0 0 ⎥ ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 0 0 0 ⎦ c. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 59 T Δ = ⎡⎣T −1 ⎤⎦ [ Δ ][T ] = ⎡⎣T −1 ⎤⎦ [ dT ] = ⎡0 ⎢1 =⎢ ⎢0 ⎢ ⎣0 −0.1 0.7 ⎤ ⎡ 0 1 0 −2 ⎤ ⎡ −0.02 0 0.02 0 0.08⎤ −0.1 0.7 ⎥⎥ 0 0 −3⎥⎥ ⎢⎢ 0 0.02 0 0.08 ⎥⎥ ⎢⎢ −0.02 0 = −0.1 0 −0.3⎥ ⎢ 0 0 −1 8 ⎥ ⎢ 0 0.1 0 0.3 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 0 0 1 ⎦⎣ 0 0 0 0 ⎦ ⎣ 0 0 0 0 ⎦ Therefore: T D = [ 0.08 0.7 0.3 0.1 0 −0.02] T T D = T J ⋅ Dθ 0 0 ⎡1 ⎢2 0 −1 ⎢ ⎢ 0 −0.1 0 → Dθ = T J −1 ⋅ T D = ⎢ ⎢ 0 −1 0 ⎢0 0 0 ⎢ 0 0 ⎣1 0 0 0 0 1 0 0 0 0 1 0 0 0 ⎤ ⎡ 0.08 ⎤ ⎡ 0.08 ⎤ 0 ⎥⎥ ⎢⎢ 0.7 ⎥⎥ ⎢⎢ −0.14 ⎥⎥ 0 ⎥ ⎢ 0.3 ⎥ ⎢ −0.07 ⎥ ⎥⎢ ⎥=⎢ ⎥ 0 ⎥ ⎢ 0.1 ⎥ ⎢ −0.7 ⎥ 0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎦ ⎣ −0.02 ⎦ ⎣ 0.06 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 60 Problem 3.10 The hand frame TH of a robot is given. The corresponding inverse Jacobian of the robot at this location relative to this frame is also shown. The robot makes a differential motion relative to this frame described as T TH D = [ 0.05 0 −0.1 0 0.1 0.1] . a. Find which joints must make a differential motion, and by how much, in order to create the indicated differential motions. b. Find the change in the frame. c. Find the new location of the frame after the differential motion. d. Find how much the differential motions (given above) should have been if measured relative to the Universe, to move the robot to the same new location as in Part c. 0 0 0 0 0⎤ ⎡5 ⎢ −1 0 0 0 ⎥⎥ 0 ⎡ 0 1 0 3⎤ ⎢2 ⎢ 1 0 0 3⎥ ⎢ 0 −0.2 0 0 0 0 ⎥ TH −1 ⎥ TH = ⎢ J =⎢ ⎥ ⎢ 0 0 −1 8 ⎥ ⎢ 0 −1 0 0 1 0 ⎥ ⎢ ⎥ ⎢0 0 0 1 0 0⎥ ⎣0 0 0 1⎦ ⎢ ⎥ 0 0 0 0 1⎦ ⎣1 Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: a. 0 0 0 0 0 ⎤ ⎡ 0.05⎤ ⎡0.25⎤ ⎡5 ⎢2 −1 0 0 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥ 0 ⎢ ⎢ 0 −0.2 0 0 0 0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥ −1 [ Dθ ] = ⎡⎣ TH J ⎤⎦ ⎡⎣ TH D ⎤⎦ = ⎢ ⎥⎢ ⎥=⎢ ⎥ − 0 1 0 0 1 0 0 ⎢ ⎥⎢ ⎥ ⎢ 0.1 ⎥ ⎢0 0 0 1 0 0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ 0 0 0 0 1 ⎦ ⎣ 0.1 ⎦ ⎣0.15⎦ ⎣1 Therefore, joints 1, 2, 4, and 6 must move as shown. b. Substituting values from ⎡0 ⎢1 TH dT = TH ⋅ Δ = ⎢ ⎢0 ⎢ ⎣0 c. 1 0 0 0 0 −1 0 0 TH D to form TH Δ , we get: 3⎤ ⎡ 0 0 0 ⎤ −0.1 0.1 0.05⎤ ⎡0.1 0 ⎥ ⎢ ⎥ ⎢ 3⎥ ⎢ 0.1 0 0 0 ⎥ ⎢ 0 −0.1 0.1 0.05⎥⎥ = 8⎥ ⎢ −0.1 0 0 −0.1⎥ ⎢0.1 0 0 0.1 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 1⎦ ⎣ 0 0 0 0 ⎦ ⎣0 0 0 0 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 61 TH new = TH old ⎡0 ⎢1 + dT = ⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 3⎤ ⎡0.1 0 0 0 ⎤ ⎡0.1 1 0 3 ⎤ ⎥ ⎢ ⎥ ⎢ 3⎥ ⎢ 0 −0.1 0.1 0.05⎥ ⎢ 1 −0.1 0.1 3.05⎥⎥ + = −1 8.1 ⎥ 8⎥ ⎢0.1 0 0 0.1 ⎥ ⎢0.1 0 ⎥ ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣ 0 0 0 0 ⎦ ⎣ 01 0 0 1 ⎦ d. dT = TH ⋅ TH Δ = Δ ⋅ TH 0 0 ⎤ ⎡0 ⎡0.1 0 ⎢ 0 −0.1 0.1 0.05⎥ ⎢1 ⎥⎢ Δ = dT ⋅ TH −1 = ⎢ ⎢0.1 0 0 0.1 ⎥ ⎢ 0 ⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣0 −3⎤ ⎡ 0 −0.3⎤ 0.1 0 ⎥ ⎢ 0 0 −3⎥ ⎢ −0.1 0 −0.1 1.15 ⎥⎥ = −0.2 ⎥ 0 −1 8 ⎥ ⎢ 0 0.1 0 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 0 0 0 ⎦ 1 0 Therefore, D = [ −0.3 1.15 −0.2 0.1 0 −0.1] T © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 62 Problem 3.11 The hand frame T of a robot is given. The corresponding inverse Jacobian of the robot at this location is also shown. The robot makes a differential motion described as T D = [ 0.05 0 −0.1 0 0.1 0.1] . Find which joints must make a differential motion, and by how much, in order to a. create the indicated differential motions. b. Find the change in the frame. c. Find the new location of the frame after the differential motion. d. Find how much the differential motions (given above) should have been, if measured relative to Frame T, to move the robot to the same new location as in Part c. ⎡0 ⎢1 = T ⎢ ⎢0 ⎢ ⎣0 1 0 0 0 0 −1 0 0 0 0 ⎡5 ⎢2 −1 0 ⎢ ⎢ 0 −0.2 0 J −1 = ⎢ ⎢ 0 −1 0 ⎢0 0 0 ⎢ 0 0 ⎣1 3⎤ 3⎥⎥ 8⎥ ⎥ 1⎦ 0 0 0 0 1 0 0 0 0 1 0 0 0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 1⎦ Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 3.6-3.11 Solution: a. 0 0 ⎡5 ⎢2 −1 0 ⎢ ⎢ 0 −0.2 0 Dθ = J −1 ⋅ D = ⎢ ⎢ 0 −1 0 ⎢0 0 0 ⎢ 0 0 ⎣1 0 0 0 0 1 0 0 0 0 1 0 0 0 ⎤ ⎡ 0.05⎤ ⎡ 0.25⎤ 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ 0.2 ⎥⎥ 0 ⎥ ⎢ −0.1⎥ ⎢ 0 ⎥ ⎥⎢ ⎥=⎢ ⎥ 0 ⎥ ⎢ 0 ⎥ ⎢ 0.1 ⎥ 0 ⎥ ⎢ 0.1 ⎥ ⎢ 0 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎦ ⎣ 0.1 ⎦ ⎣ 0.15⎦ Therefore, joints 1, 2, 4, and 6 must move as shown. b. Using D given above to form Δ , −0.1 0.1 0.05⎤ ⎡ 0 ⎡ 0 ⎢ 0.1 0 0 0 ⎥⎥ ⎢⎢1 ⎢ dT = Δ ⋅ T = ⎢ −0.1 0 0 −0.1⎥ ⎢ 0 ⎢ ⎥⎢ 0 0 0 ⎦ ⎣0 ⎣ 0 1 0 0 0 0 −1 0 0 3⎤ ⎡ −0.1 0 −0.1 0.55 ⎤ ⎥ ⎢ 3⎥ ⎢ 0 0.1 0 0.3 ⎥⎥ = −0.1 0 −0.4 ⎥ 8⎥ ⎢ 0 ⎥ ⎢ ⎥ 1⎦ ⎣ 0 0 0 0 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 63 c. Tnew = Told −0.1 3.55⎤ ⎡ −0.1 1 ⎢ 1 0.1 0 3.3 ⎥⎥ ⎢ + dT = ⎢ 0 −0.1 −1 7.6 ⎥ ⎢ ⎥ 0 0 1 ⎦ ⎣ 0 d. ⎡0 ⎢1 T −1 −1 Δ = ⎡⎣T ⎤⎦ [ Δ ][T ] = ⎡⎣T ⎤⎦ [ dT ] = ⎢ ⎢0 ⎢ ⎣0 0.1 0 0.3 ⎤ ⎡ 0 ⎢ −0.1 0 −0.1 0.55⎥ ⎥ =⎢ ⎢ 0 0.1 0 0.4 ⎥ ⎢ ⎥ 0 0 0 ⎦ ⎣ 0 −3⎤ ⎡ −0.1 0 −0.1 0.55 ⎤ ⎥ ⎢ 0 0 −3 ⎥ ⎢ 0 0.1 0 0.3 ⎥⎥ −0.1 0 −0.4 ⎥ 0 −1 8 ⎥ ⎢ 0 ⎥⎢ ⎥ 0 0 1 ⎦⎣ 0 0 0 0 ⎦ 1 0 The differential motions should have been: T D = [ 0.3 0.55 0.4 0.1 0 −0.1] © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 64 Problem 3.12 Calculate the T6 J 21 element of the Jacobian for the revolute robot of Example 2.25. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 3.10 Solution: The robot has 6 revolute joints. From Equation 3.26: T6 J 2i = ( −ox p y + o y px ) where for i=1 we use oT6 = A1 A2 A3 A4 A5 A6 given in Equation 2.59. Substitute to get: T6 J 21 = − [C1 (−C234C5C6 − S234C6 ) + S1S5 S6 ] × [ S1 (C234 a4 + C23 a3 + C2 a2 ) ] + [ S1 (−C234C5C6 − S234C6 ) − C1S5 S6 ] × [C1 (C234 a4 + C23 a3 + C2 a2 ) ] © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 65 Problem 3.13 Calculate the T6 J16 element of the Jacobian for the revolute robot of Example 2.25. Estimated student time to complete: 20 minutes Prerequisite knowledge required: Text Section(s) 3.10 Solution: The robot has 6 revolute joints. From Equation 3.26: T6 J1i = ( −nx p y + n y px ) where for i=6 we use 5T6 = A6 given in Equation 2.57. Substitute to get: T6 J 21 = − ⎡⎣( C6 )( 0 ) ⎤⎦ + ⎡⎣( S6 )( 0 ) ⎤⎦ = 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 66 Problem 3.14 Using Equation (2.34), differentiate proper elements of the matrix to develop a set of symbolic equations for joint differential motions of a cylindrical robot and write the corresponding Jacobian. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: For a cylindrical coordinate (without additional rotation) there can only be 3 variables. The Jacobian will be a 3 × 3 matrix. ⎡Cα ⎢ Sα Tcyl (r , α , l ) = ⎢ ⎢ 0 ⎢ ⎣ 0 − Sα Cα 0 0 0 rCα ⎤ 0 rSα ⎥⎥ 1 l ⎥ ⎥ 0 1 ⎦ ⎧ px = rCα ⎧dpx = drCα − rSα dα ⎪ ⎪ Then ⎨ p y = rSα and ⎨dp y = drSα + rCα dα ⎪ ⎪ ⎩ pz = l ⎩dpz = dl ⎡ dpx ⎤ ⎡Cα Therefore, ⎢⎢ dp y ⎥⎥ = ⎢⎢ Sα ⎣⎢ dpz ⎦⎥ ⎣⎢ 0 − rSα rCα 0 0 ⎤ ⎡ dr ⎤ 0 ⎥⎥ ⎢⎢ dα ⎥⎥ 1 ⎦⎥ ⎣⎢ dl ⎦⎥ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 67 Problem 3.15 Using Equation (2.36), differentiate proper elements of the matrix to develop a set of symbolic equations for joint differential motions of a spherical robot and write the corresponding Jacobian. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: For a spherical coordinate (without additional rotation) there can only be 3 variables. The Jacobian will be a 3 × 3 matrix. ⎡ C β Cγ ⎢ C β Sγ Tsph (r , β , γ ) = ⎢ ⎢ −S β ⎢ ⎣ 0 − Sγ Cγ 0 0 S β Cγ S β Sγ Cβ 0 rS β Cγ ⎤ rS β Sγ ⎥⎥ rC β ⎥ ⎥ 1 ⎦ ⎧ px = rS β .Cγ ⎧dpx = drS β .Cγ + rC β .Cγ d β − rS β .Sγ d γ ⎪ ⎪ Then ⎨ p y = rS β .Sγ and ⎨dp y = drS β .S γ + rC β .Sγ d β + rS β .Cγ d γ ⎪ ⎪ ⎩ pz = rC β ⎩dpz = drC β − rS β d β ⎡ dpx ⎤ ⎡ S β .Cγ Therefore, ⎢⎢ dp y ⎥⎥ = ⎢⎢ S β .S γ ⎢⎣ dpz ⎥⎦ ⎢⎣ C β rC β .Cγ rC β .Sγ −rS β − rS β .S γ ⎤ ⎡ dr ⎤ rS β .Cγ ⎥⎥ ⎢⎢ d β ⎥⎥ ⎥⎦ ⎢⎣ d γ ⎥⎦ 0 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 68 Problem 3.16 For a cylindrical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r =0.1 in/sec, α =0.05 rad/sec, l =0.2 in/sec, r=15 in, α = 30D , l=10 in. Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.14, and dividing both sides by dt, substitute the given values into the equation to get: ⎡ p x ⎤ ⎡Cα ⎢ p ⎥ = ⎢ Sα ⎢ y⎥ ⎢ ⎢⎣ p z ⎥⎦ ⎢⎣ 0 − rSα rCα 0 0 ⎤ ⎡ r ⎤ ⎡ 0.866 −7.5 0⎤ ⎡ 0.1 ⎤ ⎡ −0.288⎤ 0 ⎥⎥ ⎢⎢α ⎥⎥ = ⎢⎢ 0.5 13 0⎥⎥ ⎢⎢ 0.05⎥⎥ = ⎢⎢ 0.7 ⎥⎥ in/sec 1 ⎥⎦ ⎢⎣ l ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.2 ⎥⎦ ⎢⎣ 0.2 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 69 Problem 3.17 For a spherical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r =2 in/sec, β =0.05 rad/sec, γ =0.1 rad/sec, r=20 in, β = 60D , γ = 30D . Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the given values into the equation to get: ⎡ p x ⎤ ⎡ S β Cγ ⎢ p ⎥ = ⎢ S β S γ ⎢ y⎥ ⎢ ⎣⎢ p z ⎦⎥ ⎣⎢ C β rC β Cγ rC β Sγ − rS β −rS β Sγ ⎤ ⎡ r ⎤ ⎡ 0.75 8.66 −8.66⎤ ⎡ 2 ⎤ ⎡1.067 ⎤ ⎥ ⎢ ⎥ ⎢ rS β Cγ ⎥ ⎢ β ⎥ = ⎢ 0.433 5 15 ⎥⎥ ⎢⎢ 0.05⎥⎥ = ⎢⎢ 2.616⎥⎥ 0 ⎦⎥ ⎣⎢ γ ⎦⎥ ⎣⎢ 0.5 −17.32 0 ⎦⎥ ⎢⎣ 0.1 ⎦⎥ ⎣⎢ 0.134 ⎦⎥ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 70 Problem 3.18 For a spherical robot, the three joint velocities are given for a corresponding location. Find the three components of the velocity of the hand frame. r =1 unit/sec, β =1 rad/sec, γ =1 rad/sec, r=5 units, β = 45D , γ = 45D . Estimated student time to complete: 10 or 20 minutes Prerequisite knowledge required: Text Section(s) 3.10-3.11 Solution: Referring to the solution of Problem 3.15, and dividing both sides by dt, substitute the given values into the equation to get: ⎡ p x ⎤ ⎡ S β Cγ ⎢ p ⎥ = ⎢ S β S γ ⎢ y⎥ ⎢ ⎣⎢ p z ⎦⎥ ⎣⎢ C β rC β Cγ rC β Sγ − rS β − rS β Sγ ⎤ ⎡ r ⎤ ⎡ 0.5 2.5 −3.535⎤ ⎡1⎤ ⎡ −0.535⎤ ⎥ ⎢ ⎥ ⎢ rS β Cγ ⎥ ⎢ β ⎥ = ⎢ 0.5 2.5 3.535 ⎥⎥ ⎢⎢1⎥⎥ = ⎢⎢ 6.535 ⎥⎥ 0 ⎦⎥ ⎣⎢ γ ⎦⎥ ⎣⎢ 0.707 −3.535 0 ⎦⎥ ⎣⎢1⎦⎥ ⎣⎢ −2.828⎦⎥ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 71 Problem 3.19 For a cylindrical robot, the three components of the velocity of the hand frame are given for a corresponding location. Find the required three joint velocities that will generate the given hand frame velocity. x = 1 in/sec, y = 3 in/sec, z = 5 in/sec, α = 45D , r = 20 in, l = 25 in Estimated student time to complete: 10 or 25 minutes Prerequisite knowledge required: Text Section(s) 3.12 Solution: Referring to the solution of Problem 3.16, substitute to get: ⎡ p x ⎤ ⎡Cα ⎢ p ⎥ = ⎢ Sα ⎢ y⎥ ⎢ ⎢⎣ p z ⎥⎦ ⎢⎣ 0 − rSα rCα 0 0 ⎤ ⎡ r ⎤ ⎡ 1 ⎤ ⎡ 0.707 −14.14 0 ⎤ ⎡ r ⎤ ⎥ ⎢ ⎥ 0 ⎥ ⎢α ⎥ → ⎢⎢ 2⎥⎥ = ⎢⎢ 0.707 14.14 0 ⎥⎥ ⎢⎢α ⎥⎥ ⎢⎣ 3⎥⎦ ⎢⎣ 0 1 ⎥⎦ ⎢⎣ l ⎥⎦ 0 1 ⎥⎦ ⎢⎣ l ⎥⎦ ⎧0.707 r − 14.14α = 1 ⎧r = 2.83 in/sec ⎪ ⎪ D ⎨0.707 r + 14.14α = 3 → ⎨α =0.071 /sec ⎪ ⎪ in/sec ⎩l = 5 ⎩l = 5 Please note that this problem could have been solved by calculating the inverse of the Jacobian as well. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 72 Problem 3.20 For a spherical robot, the three components of the velocity of the hand frame are given for a corresponding location. Find the required three joint velocities that will generate the given hand frame velocity. x = 5 in/sec, y = 9 in/sec, z = 6 in/sec, β = 60D , r = 20 in, γ =30D Estimated student time to complete: 15 or 30 minutes Prerequisite knowledge required: Text Section(s) 3.12 Solution: Referring to the solution of Problem 3.17, substitute to get: ⎡ p x ⎤ ⎡ S β Cγ ⎢ p ⎥ = ⎢ S β S γ ⎢ y⎥ ⎢ ⎢⎣ p z ⎥⎦ ⎢⎣ C β rC β Cγ rC β Sγ − rS β −rS β Sγ ⎤ ⎡ r ⎤ 8.66 −8.66⎤ ⎡ r ⎤ ⎡ 5 ⎤ ⎡ 0.75 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ rS β Cγ ⎥ ⎢ β ⎥ → ⎢9 ⎥ = ⎢ 0.433 5 15 ⎥⎥ ⎢⎢α ⎥⎥ ⎢⎣ 6 ⎥⎦ ⎢⎣ 0.5 −17.32 0 ⎥⎦ ⎢⎣ γ ⎥⎦ 0 ⎥⎦ ⎢⎣ l ⎥⎦ ⎧0.75r + 8.66β − 8.66γ = 5 ⎧r = 10.647 in/sec ⎪ ⎪ → ⎨ β = − 0.039 D /sec ⎨0.433r + 5β + 15γ = 9 ⎪ ⎪γ = 0.306 D /sec ⎩ ⎩0.5r − 17.32 β = 6 Please note that this problem could have been solved by calculating the inverse of the Jacobian as well. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 73 CHAPTER FOUR Problem 4.1 Using Lagrangian mechanics derive the equations of motion of a cart with two tires under the cart as shown: k F m r, I r, I x Figure P.4.1 Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 4.2 Solution: I. Kinematics: Since the rollers are in contact with the block at the top, the contact-point velocity is x . Therefore: x = 2rω → ω = x 2r x vc II. Dynamics: m I 1 2 1 ⎛1 ⎞ 1 mx + 2 ⎜ mwvc2 + I cω 2 ⎟ = mx 2 + w x 2 + c2 x 2 = meff x 2 2 2 4 4r ⎝2 ⎠ 2 m I ⎞ ⎛1 where ⎜ m + w + c2 ⎟ = meff 4 4r ⎠ ⎝2 1 P = kx 2 2 1 Therefore, L = K − P = meff x 2 − kx 2 2 ∂L d ⎛ ∂L ⎞ = 2meff x → x ⎜ ⎟ = 2meff ∂x dt ⎝ ∂x ⎠ ∂L = −kx ∂x F = 2meff x + kx K= © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 74 Problem 4.2 Calculate the total kinetic energy of the link AB, attached to a roller with negligible mass, as shown. B l, I y C x θ m A Vo Figure P.4.2 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 4.2 Solution: First, find the velocity of point C, assuming that the roller has negligible mass too: l l ⎛l ⎞ v C = v A + v C / A = v o + θ × = (vo )i + θk × ⎜ cos θ i + sin θ j ⎟ 2 2 ⎝2 ⎠ l sin θ ⎞ l cos θ ⎛l ⎞ ⎛l ⎞ ⎛ θ ⎟i + θj = (vo )i + ⎜ cos θ ⎟ θ j − ⎜ sin θ ⎟ θi = ⎜ vo − 2 2 ⎝2 ⎠ ⎝2 ⎠ ⎝ ⎠ l 2 sin 2 θ 2 l 2 cos 2 θ 2 l2 θ − vol sin θθ + θ = vo 2 + θ 2 − vol sin θθ 4 4 4 2 ⎞ 1 1 1 1 ⎛ l K = mvC2 + Iθ 2 = m ⎜ vo 2 + θ 2 − vo l sin θθ ⎟ + Iθ 2 2 2 2 ⎝ 4 ⎠ 2 vC2 = vo 2 + © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 75 Problem 4.3 Derive the equations of motion for the 2-link mechanism with distributed mass, as shown. y0 O l 1, I1 x0 θ1 C y1 m1 A l2 , θ2 D I2 x1 B m2 Figure P.4.3 Estimated student time to complete: 1.5-2 hours Prerequisite knowledge required: Text Section(s) 4.2 Solution: I. Kinematic: First we write equations describing the kinematic relationships. l2 ⎧ x l C C12 = + D 1 1 ⎪⎪ 2 ⎨ ⎪ y = l S + l2 S ⎪⎩ D 1 1 2 12 ⎧ S − l2 θ + θ S x l θ = − D 1 1 1 1 2 12 ⎪⎪ 2 → ⎨ ⎪ y = l θ C + l2 θ + θ C ⎪⎩ D 1 1 1 2 1 2 12 ( ) ( ) The velocity of point D will be: l2 2 2 2 2 2 2 2 2 2 θ1 + θ 2 + 2θ1θ2 S12 2 + l1l2θ1 θ1 + θ2 S1S12 vD = xD + y D = l1 θ1 S1 + 4 2 l + l12θ12C12 + 2 θ12 + θ2 2 + 2θ1θ2 C12 2 + l1l2θ1 θ1 + θ2 C1C12 4 2 l = l12θ12 + 2 θ12 + θ2 2 + 2θ1θ2 + l1l2θ1 θ1 + θ2 C2 4 Rearrange: ⎛ ⎞ ⎛l 2 ⎞ ⎛l 2 ⎞ l2 vD2 = θ12 ⎜ l12 + 2 + l1l2C2 ⎟ + θ2 2 ⎜ 2 ⎟ + θ1θ2 ⎜ 2 + l1l2C2 ⎟ 4 ⎝ ⎠ ⎝ 4 ⎠ ⎝ 2 ⎠ ( ) ( ( ) ) ( ( ( ) ) ) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 76 II: Kinetics: Kinetic Energy: ( ) 2 1 2 1 1 I Oθ1 + I D θ1 + θ2 + m2 vD2 2 2 2 2 1⎛1 1⎛ 1 1 ⎞ ⎞ = ⎜ m1l12 ⎟ θ12 + ⎜ m2l22 ⎟ θ1 + θ2 + m2 vD2 2⎝3 2 ⎝ 12 2 ⎠ ⎠ K = K1 + K 2 = ( ) 1 1 1 1 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ = θ12 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 2 ⎜ m2l22 ⎟ + θ1θ2 ⎜ m2l22 + m2l1l2C2 ⎟ 6 2 2 2 ⎝6 ⎠ ⎝6 ⎠ ⎝3 ⎠ Potential Energy: P = P1 + P2 = m1 g l1 l (1 − C1 ) + m2 gl1 (1 − C1 ) + m2 g 2 (1 − C12 ) 2 2 Lagrangian: 1 1 1 1 ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ L = K − P = θ12 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 2 ⎜ m2l22 ⎟ + θ1θ2 ⎜ m2l22 + m2l1l2C2 ⎟ 6 2 2 2 ⎝6 ⎠ ⎝6 ⎠ ⎝3 ⎠ l l l l ⎛ ⎞ + ⎜ m1 g 1 + m2 gl1 ⎟ C1 + m2 g 2 C12 − m1 g 1 − m2 gl1 − m2 g 2 2 2 2 2 ⎝ ⎠ Derivatives and equations of motion: ∂L 1 1 1 1 ⎛1 ⎞ ⎛1 ⎞ = 2θ1 ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ + θ2 ⎜ m2l22 + m2l1l2C2 ⎟ ∂θ 6 2 2 2 ⎝6 ⎠ ⎝3 ⎠ 1 1 1 1 d ⎛ ∂L ⎞ ⎛1 2 1 ⎞ ⎛1 ⎞ 2 2 2 ⎜ ⎟ = 2θ1 ⎜ m1l1 + m2l2 + m2l1 + m2l1l2C2 ⎟ + θ2 ⎜ m2l2 + m2l1l2C2 ⎟ 6 2 2 2 dt ⎝ ∂θ1 ⎠ ⎝6 ⎠ ⎝3 ⎠ 1 − θ1θ2 m2l1l2 S2 − θ2 2 m2l1l2 S 2 2 ∂L l l ⎛ ⎞ = − ⎜ m1 g 1 + m2 gl1 ⎟ S1 − m2 g 2 S12 ∂θ1 2 2 ⎝ ⎠ Substitute to get: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 77 1 1 ⎛1 ⎞ ⎛1 ⎞ T1 = ⎜ m1l12 + m2l22 + m2l12 + m2l1l2C2 ⎟ θ1 + ⎜ m2l22 + m2l1l2C2 ⎟ θ2 3 2 ⎝3 ⎠ ⎝3 ⎠ 1 1 ⎛1 ⎞ − m2l1l2 S2θ1θ2 − m2l1l2 S 2θ2 2 + ⎜ m1 + m2 ⎟ gl1S1 + m2 gl2 S12 2 2 ⎝2 ⎠ For the second variable: ∂L 1 1 ⎛1 ⎞ = m2l22θ2 + θ1 ⎜ m2l22 + m2l1l2C2 ⎟ ∂θ 2 3 2 ⎝3 ⎠ d ⎛ ∂L ⎞ 1 1 ⎞ 1 2 2 ⎛ 1 ⎜ ⎟ = m2l2 θ 2 + θ1 ⎜ m2l2 + m2l1l2C2 ⎟ − m2l1l2 S2θ1θ 2 dt ⎝ ∂θ 2 ⎠ 3 3 2 2 ⎝ ⎠ ∂L 1 1 1 = − m2l1l2 S2θ12 − m2l1l2 S2θ1θ2 − m2 gl2 S12 ∂θ 2 2 2 2 Substitute to get: 1 1 1 1 ⎛1 ⎞ T2 = ⎜ m2l22 + m2l1l2C2 ⎟ θ1 + m2l22θ2 + m2l1l2 S2θ12 + m2 gl2 S12 2 3 2 2 ⎝3 ⎠ The result can be written in a matrix form too. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 78 Problem 4.4 Write the equations that express U62, U35, U53, U623, and U534 for a 6-axis cylindrical-RPY robot in terms of the A and Q matrices. Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 4.4.1. Solution: For joints 1 and 3: For joints 2,4,5,6: U ij = ⎡0 ⎢0 Q ( prismatic ) = ⎢ ⎢0 ⎢ ⎣0 0 0 0 0⎥ 0 0 1⎥ 0 0 0⎦ ⎡0 −1 ⎢1 0 Q ( revolute ) = ⎢ ⎢0 0 ⎢ ⎣0 0 ∂ 0Ti = A1 A2 ..Q j Aj .. Ai ∂q j 0⎤ 0 ⎥ = QP ⎥ 0 0 0 0 0⎤ 0 ⎥⎥ = QR 0⎥ ⎥ 0⎦ j≤i U ijk = ∂U ij ∂qk ∂ 0T6 ∂ ( A1 A2 A3 A4 A5 A6 ) = = A1QR A2 A3 A4 A5 A6 ∂α ∂α ∂ 0T3 ∂ ( A1 A2 A3 ) U 35 = = = 0 (i ≥ j ) ∂φ0 ∂φ0 U 62 = ∂ 0T5 ∂ ( A1 A2 A3 A4 A5 ) = = A1 A2QP A3 A4 A5 ∂l ∂l ∂U 62 ∂ ( A1QR A2 A3 A4 A5 A6 ) = = = A1QR A2QP A3 A4 A5 A6 ∂l ∂l ∂U 53 ∂ ( A1 A2QP A3 A4 A5 ) = = = A1 A2QP A3QR A4 A5 ∂qa ∂qa U 53 = U 623 U 534 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 79 Problem 4.5 Using Equations 4.49 to 4.54, write the equations of motion for a 3-DOF revolute robot and describe each term. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 4.4.4. Solution: We have: T1 = D11θ1 + D12θ2 + D13θ3 + I1( act )θ1 + D111θ12 + D122θ22 + D133θ32 + 2 D112θ1θ2 + 2 D113θ1θ3 + 2 D123θ2θ3 + D1 T2 = D21θ1 + D22θ2 + D23θ3 + I 2( act )θ2 + D211θ12 + D222θ22 + D233θ32 + 2 D212θ1θ2 + 2 D213θ1θ3 + 2 D223θ2θ3 + D2 T3 = D31θ1 + D32θ2 + D33θ3 + I 3( act )θ3 + D311θ12 + D322θ22 + D333θ32 + 2 D312θ1θ2 + 2 D313θ1θ3 + 2 D323θ2θ3 + D3 ( D )θi terms are caused by angular accelerations. ( D )θi 2 terms represent centripetal accelerations. ( D )θiθj terms are related to the Coriolis accelerations. ( Di ) terms are gravity terms. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 80 Problem 4.6 Expand the D134 and D15 terms of Equation 4.49 in terms of their constituent matrices. Estimated student time to complete: 1 hour Prerequisite knowledge required: Text Section(s) 4.4 Solution: a. From Equation 4.44: Dijk = n ∑ p = max( i , j , k ) Trace(U pjk J pU Tpi ) For D134 we have i =1, j =3, k =4, p =4, n =6. Then: T D134 = ∑ Trace(U p 34 J pU Tp1 ) = Trace (U 434 J 4U 41 ) + Trace (U 534 J 5U 51T ) + Trace (U 634 J 6U 61T ) 6 4 ∂ 0T4 ∂ ( A1 A2 A3 A4 ) = = A1 A2Q3 A3 A4 ∂q3 ∂q3 U 43 = U 434 = ∂U 43 ∂ ( A1 A2Q3 A3 A4 ) = = A1 A2Q3 A3Q4 A4 ∂q4 ∂q4 ∂ 0T4 ∂ ( A1 A2 A3 A4 ) = = Q1 A1 A2 A3 A4 U 41 = ∂q1 ∂q1 T = A4T A3T A2T A1T Q1T From Equation A.11 of Appendix A: U 41 U 53 = ∂ ( A1 A2 A3 A4 A5 ) = A1 A2Q3 A3 A4 A5 ∂q3 U 534 = U 51 = U 63 = ∂ ( A1 A2Q3 A3 A4 A5 ) = A1 A2Q3 A3Q4 A4 A5 ∂q4 ∂ ( A1 A2 A3 A4 A5 ) = Q1 A1 A2 A3 A4 A5 ∂q1 T → U 51 = A5T A4T A3T A2T A1T Q1T ∂ ( A1 A2 A3 A4 A5 A6 ) = A1 A2Q3 A3 A4 A5 A6 ∂q3 U 634 = ∂U 63 = A1 A2Q3 A3Q4 A4 A5 A6 ∂q4 T = A6T A5T A4T A3T A2T A1T Q1T U 61 Substitute to get: D134 = Trace ( A1 A2Q3 A3Q4 A4 J 4 A4T A3T A2T A1T Q1T ) + Trace ( A1 A2Q3 A3Q4 A4 A5 J 5 A5T A4T A3T A2T A1T Q1T ) + Trace ( A1 A2Q3 A3Q4 A4 A5 A6 J 6 A6T A5T A4T A3T A2T A1T Q1T ) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 81 b. From Equation 4.43: Dij = n ∑ p = max( i , j ) Trace(U pj J pU Tpi ) For D15 we have i =1, j =5, p =5, n =6. Then: T D15 = ∑ Trace(U p 5 J pU Tp1 ) = Trace (U 55 J 5U 51 ) + Trace (U 65 J 6U 61T ) 6 5 U 55 = A1 A2 A3 A4Q5 A5 U 51 = Q1 A1 A2 A3 A4 A5 T → U 51 = A5T A4T A3T A2T A1T Q1T U 65 = A1 A2 A3 A4Q5 A5 A6 U 61 = Q1 A1 A2 A3 A4 A5 A6 T → U 61 = A6T A5T A4T A3T A2T A1T Q1T Substitute to get: D15 = Trace ( A1 A2 A3 A4Q5 A5 J 5 A5T A4T A3T A2T A1T Q1T ) + Trace ( A1 A2 A3 A4Q5 A5 A6 J 6 A6T A5T A4T A3T A2T A1T Q1T ) Although a large number of matrices are involved, they are all known. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 82 Problem 4.7 An object is subjected to the following forces and moments relative to the reference frame. Attached to the object is a frame, which describes the orientation and the location of the object. Find the equivalent forces and torques acting on the object relative to the current frame. ⎡0.707 0.707 0 2 ⎤ ⎢ 0 0 1 5 ⎥⎥ ⎢ = B F T = [10, 0,5,12, 20, 0] N, N.m ⎢0.707 −0.707 0 3 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣ 0 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 4.6. Solution: Substitute: f = [10, 0,5]T m = [12, 20, 0]T n = [0.707, 0, 0.707]T i p = [2,5,3]T o = [0.707, 0, −0.707]T a = [0,1, 0]T j k f × p = 10 0 5 = i (−25) − j(20) + k (50) 2 5 3 (f × p) + m = −13i + 50k B f x = n ⋅ f = 7.07 + 3.54 = 10.61 N B f y = o ⋅ f = 7.07 − 3.54 = 3.53 N B fz = a ⋅ f = 0 B mx = n ⋅ [(f × p) + m] = −9.19 + 35.35 = 26.16 Nm B my = o ⋅ [(f × p) + m] = −9.19 − 35.35 = −44.54 Nm B mz = a ⋅ [(f × p) + m] = 0 and ⎡ 10.61 ⎤ ⎢ 3.53 ⎥ ⎢ ⎥ ⎢ 0 ⎥ F =⎢ ⎥ ⎢ 26.16 ⎥ ⎢ −44.54 ⎥ ⎢ ⎥ ⎣ 0 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 83 Problem 4.8 In order to assemble two parts together, one part must be pushed into the other with a force of 10 lb in the x-axis direction, 5 lb in the y-direction, and be turned with a torque of 5 lb ⋅ in along the x-axis direction. The object’s location relative to the base frame of a robot is described by RT0. Assuming that the two parts must be aligned together for this purpose, find the necessary forces and moments that the robot must apply to the part relative to its hand frame. ⎡ 0 −0.707 0.707 4 ⎤ ⎢1 0 0 6 ⎥⎥ R ⎢ T0 = ⎢ 0 0.707 0.707 3 ⎥ ⎢ ⎥ 0 0 1⎦ ⎣0 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 4.6. Solution: Since the robot must be aligned with the object for this operation, then Therefore we have: f = [10,5, 0]T m = [5, 0, 0]T n = [0,1, 0]T i R TH = R TO . p = [4, 6,3]T o = [−0.707, 0, 0.707]T a = [0.707, 0, 0.707]T j k f × p = 10 5 0 = i (15) − j(30) + k (40) 4 6 3 (f × p) + m = 20i − 30 j + 40k H f x = n ⋅ f = 5 units H f y = o ⋅ f = −7.07 units H f z = a ⋅ f = 7.07 units H mx = n ⋅ [(f × p) + m] = −30 units H m y = o ⋅ [(f × p) + m] = 14.14 units H mz = a ⋅ [(f × p) + m] = 42.42 units and ⎡ 5 ⎤ ⎢ −7.07 ⎥ ⎢ ⎥ ⎢ 7.07 ⎥ F =⎢ ⎥ ⎢ −30 ⎥ ⎢14.14 ⎥ ⎢ ⎥ ⎣ 42.42 ⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 84 CHAPTER FIVE Problem 5.1 It is desired to have the first joint of a 6-axis robot go from an initial angle of 50D to a final angle of 80D in 3 seconds. Calculate the coefficients for a third-order polynomial joint-space trajectory. Determine the joint angles, velocities, and accelerations at 1, 2, and 3 seconds. It is assumed that the robot starts from rest, and stops at its destination. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 5.5.1. Solution: For a 3rd-order polynomial: θ (t ) = c0 + c1t + c2t 2 + c3t 3 θ(t ) = c1 + 2c2t + 3c3t 2 Substitute boundary conditions to get: θi = 50 = c0 + 0 θi = 0 = c1 + 0 θ f = 80 = 50 + 9c2 + 27c3 θ f = 0 = 6c2 + 27c3 Solve to get: c0 = 5 c1 = 0 c2 = 10 θ (t ) = 50 + 10t − 2.222t 2 c3 = −2.222 3 θ(t ) = 20t − 6.666t 2 θ(t ) = 20 − 13.332t ⎧ θ = 57.78 D ⎪ @t=1 sec ⎨θ = 13.334 D sec ⎪ θ = 6.668 D sec ⎩ ⎧ θ = 72.22 D ⎪ @t=2 sec ⎨ θ = 13.34 D sec ⎪θ = −6.664 D sec ⎩ @t=3 sec ⎧ θ = 80 D ⎪ D sec ⎨ θ =0 ⎪θ = −20 D sec ⎩ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 85 Problem 5.2 It is desired to have the third joint of a 6-axis robot go from an initial angle of 20D to a final angle of 80D in 4 seconds. Calculate the coefficients for a third-order polynomial joint-space trajectory and plot the joint angles, velocities, and accelerations. The robot starts from rest, but should have a final velocity of 5D / sec . Estimated student time to complete: 20-30 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.5.1. Solution: For a 3rd-order polynomial: θ (t ) = c0 + c1t + c2t 2 + c3t 3 θ(t ) = c1 + 2c2t + 3c3t 2 Substitute boundary conditions to get: θi = 20 = c0 + 0 θi = 0 = c1 + 0 θ f = 80 = 20 + 16c2 + 64c3 θ f = 5 = 8c2 + 48c3 Solve to get: c0 = 20 c1 = 0 c2 = 10 θ (t ) = 20 + 10t − 1.5625t 2 c3 = −1.5625 3 θ(t ) = 20t − 4.6875t 2 θ(t ) = 20 − 9.375t The plot is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 86 Problem 5.3 The second joint of a 6-axis robot is to go from initial angle of 20D to an intermediate angle of 80D in 5 seconds and continue to its destination of 25D in another 5 seconds. Calculate the coefficients for third-order polynomials in joint-space. Plot the joint angles, velocities, and accelerations. Assume the joint stops at intermediate points. Estimated student time to complete: 30-40 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.5.1 Solution: There are two segments to this motion. Therefore: Segment 1 ⎧⎪θ (t ) = c0 + c1t + c2t 2 + c3t 3 ⎨ 2 ⎪⎩θ (t ) = c1 + 2c2t + 3c3t Segment 2 ⎧⎪θ (t ) = b0 + b1t + b2t 2 + b3t 3 ⎨ 2 ⎪⎩θ (t ) = b1 + 2b2t + 3b3t There are eight unknowns; eight equations are needed. Known initial and final conditions are: θ1i , θ1 f , θ 2i , θ 2 f , θ1i , θ2 f , and θ1 f = θ2i . This results in seven equations. The eighth equation can be generated by making assumptions such as a maximum allowable acceleration or an intermediate velocity. For this problem we will assume that the joint will come to a stop at the intermediate point. Therefore, ⎧20 = c0 + 0 ⎪0 = c + 0 ⎪ 1 1 ⎨ ⎪80 = 20 + 25c2 + 125c3 ⎩⎪0 = 10c2 + 75c3 ⎧80 = b0 + 0 ⎪0 = b + 0 ⎪ 1 2 ⎨ ⎪25 = 80 + 25b2 + 125b3 ⎩⎪0 = 10b2 + 75b3 → → ⎧c0 = 20 ⎪ ⎪c1 = 0 ⎨ ⎪c2 = 7.2 ⎪⎩c3 = −0.96 ⎧b0 = 80 ⎪ ⎪b1 = 0 ⎨ ⎪b2 = −6.6 ⎪⎩b3 = 0.88 Substitute to get: 1 ⎧θ (t ) = 20 + 7.2t 2 − 0.96t 3 ⎪ 2 ⎨θ (t ) = 14.4t − 2.88t ⎪ ⎩θ (t ) = 14.4 − 5.76t and 2 ⎧θ (t ) = 80 − 6.6t 2 + 0.88t 3 ⎪ 2 ⎨θ (t ) = −13.2t + 2.64t ⎪ ⎩θ (t ) = −13.2 + 5.28t The position, velocities, and accelerations are shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 87 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 88 Problem 5.4 A fifth-order polynomial is to be used to control the motions of the joints of a robot in joint-space. Find the coefficients of a fifth-order polynomial that will allow a joint to go from an initial angle of 0D to a final joint angle of 75D in 3 seconds, while the initial and final velocities are zero and initial acceleration and final decelerations are 10 D / sec 2 . Estimated student time to complete: 20-30 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.5.2 Solution: For a 5th-order polynomial: ⎧θ (t ) = c0 + c1t + c2t 2 + c3t 3 + c4t 4 + c5t 5 ⎪ 2 3 4 ⎨θ (t ) = c1 + 2c2t + 3c3t + 4c4t + 5c5t ⎪ 2 3 ⎩θ (t ) = 2c2 + 6c3t + 12c4t + 20c5t Substitute the given initial and final conditions and get: ⎧0 = c0 ⎪75 = 3c + 9c + 27c + 81c + 243c 1 2 3 4 5 ⎪ ⎪⎪0 = c1 ⎨ ⎪0 = 6c2 + 27c3 + 108c4 + 405c5 ⎪10 = 2c2 ⎪ ⎪⎩10 = 2c2 + 18c3 + 108c4 + 540c5 Solve to get: c0 = 0, c1 = 0, c2 = 5, c3 = 24.24, c4 = −13.33, c5 = 1.85 . Substitute to get: ⎧θ (t ) = 5t 2 + 24.44t 3 − 13.33t 4 + 1.85t 5 ⎪ 2 3 4 ⎨θ (t ) = 10t + 73.33t − 53.33t + 9.26t ⎪ 2 3 ⎩θ (t ) = 10 + 146.66t − 160t + 37.04t The position, velocity, and acceleration plots are shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 89 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 90 Problem 5.5 Joint 1 of a 6-axis robot is to go from an initial angle of θi = 30D to the final angle of θ f = 120D in 4 seconds with a cruising velocity of ω1 = 30D sec . Find the necessary blending time for a trajectory with linear segments and parabolic blends and plot the joint positions, velocities, and accelerations. Estimated student time to complete: 15-30 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.5.3 Solution: tb = θi − θ f + ωt f 30 − 120 + 30 ( 4 ) = = 1 sec 30 ω For θ = θi to θ A For θ = θ A to θ B For θ = θ B to θ f 1 ⎧ 2 ⎪θ = 30 + 2 ( 30 ) t ⎪⎪ ⎨θ = 30t ⎪ ⎪θ = 30 ⎪⎩ ⎧θ = θ A + 30 ( t − 1) ⎪ ⎨θ = 30 ⎪ ⎩θ = 0 1 ⎧ 2 ⎪θ = 120 − 2 ( 30 ) (4 − t ) ⎪⎪ ⎨θ = 30(4 − t ) ⎪ ⎪θ = −30 ⎪⎩ 140 120 100 Position 80 60 Velocity 40 20 0 Acceleration -20 -40 1 21 2 41 4 Seconds © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 91 Problem 5.6 A robot is to be driven from an initial position through two via points before it reaches its final destination using a 4-3-4 trajectory. The positions, velocities, and time duration for the three segments for one of the joints are given below. Determine the trajectory equations and plot the position, velocity, and acceleration curves for the joint. θ1 = 20D θ1 = 0 θ1 = 0 θ 2 = 60D τ 2i = 0 τ2 f = 2 θ 3 = 100D τ 3i = 0 τ3 f = 1 θ 4 = 40D θ4 = 0 θ4 = 0 τ 1i = 0 τ1 f = 1 Estimated student time to complete: 50-60 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.5.5 Solution: Using Equations 5.19 to 5.32 we get the following equations. The same can be found using the matrix equation of Equation 5.33: ⎧a0 = 20 ⎪a = 0 ⎪ 1 ⎪a2 = 0 ⎪ ⎪b0 = 60 ⎪c = 100 ⎪ 0 ⎪60 = 20 + a3 + a4 ⎪3a + 4a = b ⎪ 3 4 1 ⎨ ⎪6a3 + 12a4 = 2b2 ⎪100 = 60 + 2b1 + 4b2 + 8b3 ⎪ ⎪b1 + 4b2 + 12b3 = c1 ⎪2b + 12b = 2c 3 2 ⎪ 2 ⎪40 = 100 + c1 + c2 + c3 + c4 ⎪ ⎪0 = c1 + 2c2 + 3c3 + 4c4 ⎪⎩0 = 2c2 + 6c3 + 12c4 → ⎧a3 = 80.95 ⎪a = −40.95 ⎪ 4 ⎪b1 = 79.05 ⎪ ⎪b2 = −2.85 ⎪ ⎨b3 = −13.34 ⎪c = −92.4 ⎪1 ⎪c2 = −82.875 ⎪ ⎪c3 = 202.9 ⎪⎩c4 = −87.62 These equations can be solved by elimination or an equation solver. Substituting the coefficients into each segment we get: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 92 θ1 (t ) = 20 + 80.95t 3 − 40.95t 4 θ 2 (t ) = 60 + 79.05t − 2.85t 2 − 13.34t 3 θ3 (t ) = 100 − 92.4t − 82.875t 2 + 202.9t 3 − 87.62t 4 The results are plotted below. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 93 Problem 5.7 A 2-DOF planar robot is to follow a straight line in Cartesian-space between the start (2,6) and the end (12,3) points of the motion segment. Find the joint variables for the robot if the path is divided into 10 sections. Each link is 9 inches long. Estimated student time to complete: 60-75 minutes (with plotting) Prerequisite knowledge required: Text Section(s) 5.6 Solution: The inverse kinematic equations can be found in different ways. We will use a simple Denavit-Hartenberg representation as follows: y0 z2 z1 θ2 x2 1-2 x1 θ1 z0 ⎡C1 ⎢S A1 = ⎢ 1 ⎢0 ⎢ ⎣0 − S1 0 9C1 ⎤ C1 0 9 S1 ⎥⎥ 0 1 0 ⎥ ⎥ 0 0 1 ⎦ # 0-1 θ θ1 θ2 d 0 a 9 0 9 α 0 0 x0 ⎡ C2 ⎢S A2 = ⎢ 2 ⎢0 ⎢ ⎣0 − S2 C2 0 0 0 9C2 ⎤ 0 9 S 2 ⎥⎥ 1 0 ⎥ ⎥ 0 1 ⎦ T2 = A1 A2 0 Multiplying A1 by A2 and setting px and py equal to the components of position for 0T2, we get: ⎧⎪ px = 9C1C2 − 9 S1S 2 + 9C1 = 9C12 + 9C1 ⎨ ⎪⎩ p y = 9S1S2 + 9C1C2 + 9S1 = 9S12 + 9S1 (1) (2) Squaring these equations and adding them together, and then regrouping, we get: ⎛ px2 + p y2 ⎞ − 1 → θ 2 = cos ⎜ − 1⎟ C2 = ⎜ 162 ⎟ 162 ⎝ ⎠ Substitute into (1) to get: px2 + p y2 −1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 94 px + S1S 2 9 (3) C2 + 1 Substitute (3) into (2) to get: C1 = py S 2 px + S1S 22 = ( C2 + 1) 9 9 ⎛ p y ( C2 + 1) − S 2 px p y ( C2 + 1) − S 2 px → θ1 = sin −1 ⎜⎜ S1 = 18 ( C2 + 1) 18 ( C2 + 1) ⎝ S1 ( C2 + 1) + 2 ⎞ ⎟⎟ ⎠ Next, calculate the slope of the straight line motion: y −3 3−6 = = −0.3 x − 12 12 − 2 The following table shows the coordinates of the intermediate points and the joint angles from the above equations. The joint angles are shown on the graph as well: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 95 Problem 5.8 The 3-DOF robot of Example 5.7, as shown in Figure 5.18, is to move from point (3, 5, 5) to point (3, -5, 5) along a straight line, divided into 10 sections. Find the angles of the three joints for each intermediate point and plot the results. Estimated student time to complete: 30 minutes, depending on programming expertise Prerequisite knowledge required: Text Section(s) 5.6 Solution: Using the inverse kinematic equations of the robot from Example 5.7, the joint angles can be found as shown. The values are also plotted: # 1 2 3 4 5 6 7 8 9 10 11 Px 3 3 3 3 3 3 3 3 3 3 3 Py 5 4 3 2 1 0 -1 -2 -3 -4 -5 Pz 5 5 5 5 5 5 5 5 5 5 5 Theta 1 30.96 36.87 45 56.31 71.57 90 108.4 123.7 135 143.1 149 Theta 2 48.59 49.87 52.04 54.87 57.51 99.6 122.5 125.1 128 130.1 131.4 Theta 3 137.27 142.2 146.45 149.8 151.98 160.82 151.98 149.8 146.45 142.2 137.27 180 160 Joint 3 140 120 100 80 60 Joint 2 40 Joint 1 20 0 1 2 3 4 5 6 7 8 9 10 11 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 96 CHAPTER SIX Problem 6.1 Derive the inverse Laplace transform of the following equation: Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 6.5 and 6.6 Solution: F (s) = 3 3 a a = = 1 + 2 ( s + 5s + 4) ( s + 1)( s + 4 ) s + 1 s + 4 2 ⎡ ⎤ 3 3 = =1 a1 = ⎢( s + 1) ⎥ ( s + 1)( s + 4 ) ⎦ s =−1 3 ⎣ ⎡ ⎤ 3 3 = = −1 a2 = ⎢( s + 4 ) ⎥ ( s + 1)( s + 4 ) ⎦ s =−4 −3 ⎣ 1 1 F (s) = − s +1 s + 4 1 −1 f (t ) = L−1 + L−1 = e −t − e−4t s +1 s+4 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 97 Problem 6.2 Derive the inverse Laplace transform of the following equation: ( s + 6) F (s) = 2 s ( s + 5s + 6) Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 6.5, 6.6 Solution: F (s) = = ( s + 6) ( s + 6) = s ( s + 5s + 6) s ( s + 5) ( s + 3) 2 a a1 a + 2 + 3 s s+2 s+3 ⎡ ⎤ s+6 6 a1 = ⎢( s ) ⎥ = =1 ⎣ s ( s + 2 )( s + 3) ⎦ s =0 6 ⎡ ⎤ s+6 4 a2 = ⎢ ( s + 2 ) = = −2 ⎥ s ( s + 2 )( s + 3) ⎦ s =−2 −2 ⎣ ⎡ ⎤ s+6 3 = =1 a3 = ⎢( s + 3) ⎥ s ( s + 2 )( s + 3) ⎦ s =−3 ( −3)( −1) ⎣ 1 −2 1 + F (s) = + s s+2 s+3 1 ⎞ ⎛ 1 −2 −3t −2 t + f (t ) = L−1 ⎜ + ⎟ = u (t ) − e + e ⎝ s s+2 s+3⎠ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 98 Problem 6.3 Derive the inverse Laplace transform of the following equation: 1 F (s) = 2 ( s + 1) ( s + 2) Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 6.5, 6.6 Solution: F (s) = 1 b2 b a1 = + 1 + 2 ( s + 1) ( s + 2) ( s + 1) ( s + 1) ( s + 2) 2 ⎡ ⎤ 1 2 =1 b2 = ⎢( s + 1) 2 ( s + 1) ( s + 2) ⎥⎦ s =−1 ⎣ b1 = −1 d ⎡ 1 ⎤ = = −1 ⎢ ⎥ ds ⎣ ( s + 2) ⎦ s =−1 ( s + 2) 2 ⎡ ⎤ 1 =1 a1 = ⎢( s + 2 ) 2 ( s + 1) ( s + 2) ⎥⎦ s =−2 ⎣ 1 1 1 − + F (s) = 2 ( s + 1) ( s + 1) ( s + 2) ⎛ 1 1 1 ⎞ −t −2 t − + f (t ) = L−1 ⎜ ⎟ = ( t − 1) e + e 2 + + + s s s ( 1) ( 1) ( 2) ⎝ ⎠ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 99 Problem 6.4 Derive the inverse Laplace transform of the following equation: 10 F (s) = ( s + 4)( s + 2)3 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.5-6.6 Solution: F (s) = b3 b2 b1 a1 10 = + + + 3 3 2 ( s + 4)( s + 2) ( s + 2) ( s + 2) ( s + 2) ( s + 4) ⎡ ⎤ 10 3 =5 b3 = ⎢( s + 2 ) 3⎥ ( s + 4)( s + 2) ⎦ s =−2 ⎣ b2 = b1 = ⎤ −10 d ⎡ d ⎛ 10 ⎞ 10 3 = ⎜ = ( s + 2) ⎟ ⎢ 3⎥ ds ⎣ ( s + 4)( s + 2) ⎦ s =−2 ds ⎝ s + 4 ⎠ s =−2 ( s + 4 )2 1 d 2 ⎛ 10 ⎞ 1 d −10 = ⎟ 2 ⎜ 2 ds ⎝ s + 4 ⎠ s =−2 2 ds ( s + 4 )2 = s =−2 1 20 2 ( s + 4)3 = s =−2 = s =−2 −10 = −2.5 4 5 4 ⎡ ⎤ 10 −5 a1 = ⎢( s + 4 ) = 3⎥ ( s + 4)( s + 2) ⎦ s =−4 4 ⎣ 5 5 5 5 F (s) = − + − 3 2 ( s + 2) 2( s + 2) 4( s + 2) 4( s + 4) ⎛ 5 5 5 5 ⎞ ⎛ 5 2 5 5 ⎞ −2t 5 −4t f (t ) = L−1 ⎜ − + − ⎟ = ⎜ t − t + ⎟e − e 3 2 2 4⎠ 4 ⎝ ( s + 2) 2( s + 2) 4( s + 2) 4( s + 4) ⎠ ⎝ 2 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 100 Problem 6.5 Simplify the block diagram of Figure P.6.5. H2 A +- +- B +- G1 G2 H1 H3 Figure P.6.5 Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 6.8. Solution: The following sequence is used to simplify the block diagram: H2 /G1 B A +- +- +- G2 G1 H1 H3 H2 /G1 A +- +- B G1 1 + G1 H1 G2 H3 H2 /G1 A +- +- B G1G2 1 + G1 H1 H3 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 101 A +- G1G2 1 + G1 H1 + G2 H 2 B H3 A G1G2 1 + G1 H1 + G2 H 2 + G1G2 H 3 B © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 102 Problem 6.6 Simplify the block diagram of Figure P.6.6. A +- +- G1 +- H1 G2 B H2 H3 Figure P.6.6 Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 6.8. Solution: The following sequence is used to simplify the block diagram: A +- G2 1 + G2 H 2 G1 1 + G1 H1 B H3 A +- G1G2 (1 + G1H1 )(1 + G2 H 2 ) B H3 A +- K B where K is calculated as follows: G1G2 (1 + G1H1 )(1 + G2 H 2 ) = G1G2 K= G1G2 H 3 (1 + G1H1 )(1 + G2 H 2 ) + G1G2 H 3 1+ (1 + G1H1 )(1 + G2 H 2 ) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 103 Problem 6.7 Simplify the block diagram of Figure P.6.7. G4 A +- +- G1 G3 G2 +- B H1 H3 Figure P.6.7 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.8. Solution: The following sequence is used to simplify the block diagram: A +- G1G2 1 + G1G2 H1 G3 + G4 B G3 + G4 B H3 A +- G1G2 1 + G1G2 H1 H3 A +- K B G1G2 ( G3 + G4 ) G1G2 ( G3 + G4 ) 1 + G1G2 H1 = where K = G G ( G + G4 ) H 3 1 + G1G2 H1 + G1G2 ( G3 + G4 ) H 3 1+ 1 2 3 1 + G1G2 H1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 104 Problem 6.8 Write an equation that describes the output of the system of Figure P.6.8. U1 U2 +- G1 ++ Y G2 H1 Figure P.6.8 Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) 6.8. Solution: ( (U 1 − YH1 ) G1 + U 2 ) G2 = Y (U1G1 − YG1H1 + U 2 ) G2 = Y U1G1G2 − YG1G2 H1 + U 2G2 = Y U1G1G2 + U 2G2 = Y (1 + G1G2 H1 ) Y= U1G1G2 + U 2G2 1 + G1G2 H1 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 105 Problem 6.9 Write the equations that describe the input-output relationships for Figure P.6.9. U2 U1 ++ + G1 H1 U3 ++ Y1 H2 Y2 G2 Figure P.6.9 Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 6.8. Solution: ⎧⎪U1G1 + U 2 + Y2 H 2 = Y1 ⎨ ⎪⎩(U 3 + U1G1 H1 ) G2 = Y2 ⎧U1G1 + U 2 + Y2 H 2 = Y1 ⎨ ⎩U 3G2 + U1G1G2 H1 = Y2 Substitute Y2 into Y1 to get: ⎧Y1 = U1G1 + U 2 + U 3G2 H 2 + U1G1G2 H1 H 2 ⎨ ⎩Y2 = U 3G2 + U1G1G2 H1 ⎡ Y1 ⎤ ⎡G1 + G1G2 H1 H 2 ⎢Y ⎥ = ⎢ G1G2 H1 ⎣ 2⎦ ⎣ ⎡U ⎤ 1 G2 H 2 ⎤ ⎢ 1 ⎥ U 0 G2 ⎥⎦ ⎢ 2 ⎥ ⎢⎣U 3 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 106 Problem 6.10 Sketch the root locus for the following: K GH = s ( s + 1)( s + 3)( s + 4) Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.13. Solution: Poles are at p = 0, p = −1, p = −3, p = −4 Number of asymptotes: α = 4 − 0 = 4 , angles of asymptotes: 45, 135, 225, 315. ∑ σ p − ∑ σ Z = ( 0 − 1 − 3 − 4 ) − 0 = −2 Asymptotic center: σ A = 4 α The root locus, as drawn by MATLAB, is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 107 Problem 6.11 Sketch the root locus for the following: K ( s + 6) GH = s ( s + 4) Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 6.13. Solution: Roots are at z = −6, p = 0, p = −4 Number of asymptotes: α = 2 − 1 = 1 , angles of asymptote: 180. The root locus, as drawn by MATLAB, is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 108 Problem 6.12 Sketch the root locus for the following: GH = K ( s + 6) s ( s + 10 − j10)( s + 10 + j10)( s + 12) Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.13. Solution: Roots are at z = −6, p = 0, p = −10 ± j10, p = −12 Number of asymptotes: α = 4 − 1 = 3 , angles of asymptotes: 60, 180, 300. ∑ σ p − ∑ σ Z = ( 0 − 10 − 10 − 12 ) − (−6) = −6.5 Asymptotic center: σ A = 4 α The root locus, as drawn by MATLAB, is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 109 Problem 6.13 For the system of Problem 6.10, assume that two of the roots are chosen at s = −5 ± 2.55 j . Find the system’s gain, damping ratio, and natural frequency. Show that the angle criterion is met. Can you determine from the root locus whether or not the system is stable? Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 6.13, 6.14. Solution: ∏ M pi = 52 + 2.552 42 + 2.552 22 + 2.552 12 + 2.552 = 236 K= ∏ M zi ⎛ 2.55 ⎞ ⎟ = 180 − 27 = 153 ⎝ 5 ⎠ ⎛ 2.55 ⎞ θ 2 = 180 − tan −1 ⎜ ⎟ = 180 − 32.52 = 147.48 ⎝ 4 ⎠ ⎛ 2.55 ⎞ θ3 = 180 − tan −1 ⎜ ⎟ = 180 − 51.9 = 128.1 ⎝ 2 ⎠ ⎛ 2.55 ⎞ θ 4 = 180 − tan −1 ⎜ ⎟ = 180 − 68.59 = 111.4 ⎝ 1 ⎠ ∑θtotal = 153 + 147.48 + 128.1 + 111.4 = 540 =3 ×180 It meets the criterion. θ1 = 180 − tan −1 ⎜ ⎛ ⎝ ζ = cos θ = cos ⎜ tan −1 2.55 ⎞ ⎟ = 0.891 5 ⎠ s = −ζωn ± ωn 1 − ζ 2 j = −0.891ωn ± ωn 1 − 0.8912 j = −5 ± 2.55 j ωn = 5.61 rad / sec The root have to potential to fall on the right side, and therefore, may become unstable.. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 110 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 111 Problem 6.14 For the system of Problem 6.10, assume that two of the roots are chosen at s = −4 ± 1.24 j . Find the system’s gain, damping ratio, and natural frequency. Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 6.13,6.14 Solution: ∏ M pi = 42 + 1.242 32 + 1.242 12 + 1.242 1.242 = 26.85 K= ∏ M zi ⎛ ⎝ ζ = cos θ = cos ⎜ tan −1 1.24 ⎞ ⎟ = 0.955 4 ⎠ s = −ζωn ± ωn 1 − ζ 2 j = −0.891ωn ± ωn 1 − 0.9552 j = −4 ± 1.24 j ωn = 4.19 rad / sec © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 112 Problem 6.15 For the system of Problem 6.11, assume that the roots are chosen at s = −3 ± 1.73 j . Find the system’s gain, damping ratio, and natural frequency. Show that the angle criterion is met. Can you determine whether or not the system may become unstable as the gain changes? Estimated student time to complete: 30 minutes Prerequisite knowledge required: Text Section(s) 6.13, 6.14. Solution: K= ∏Mp ∏Mz i = 32 + 1.732 12 + 1.732 32 + 1.732 i 1.73 ⎞ ⎛ ζ = cos θ = cos ⎜ tan −1 ⎟ = 0.866 3 ⎠ ⎝ =2 s = −ζωn ± ωn 1 − ζ 2 j = −0.866ωn ± ωn 1 − 0.8662 j = −3 ± 1.73 j ωn = 3.46 rad / sec ⎛ 1.73 ⎞ ⎟ = 180 − 30 = 150 ⎝ 3 ⎠ ⎛ 1.73 ⎞ θ p−4 = 180 − tan −1 ⎜ ⎟ = 60 ⎝ 1 ⎠ ⎛ 1.73 ⎞ θ z6 = tan −1 ⎜ ⎟ = 30 ⎝ 3 ⎠ ∑θtotal = 150 + 60 − 30 = 180 θ p = 180 − tan −1 ⎜ 0 It meets the criterion. The roots are always on the left side, and therefore, the system is always stable. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 113 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 114 Problem 6.16 For the system of Problem 6.11, find the roots, the gain, and the steady-state error for a settling time of less that 1 second and overshoot of 4% or less. Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 6.14 Solution: GH = K ( s + 6) s ( s + 4) By trial and error, %OS = e ( −ζπ / 1−ζ 2 ) ×100% = 4% → ζ ≥ 0.715, θ = cos −1 ζ = 44.4D The chosen roots must be below this value, therefore choose s = −4 ± 2.83 j 4 ζωn (approximate) = = 4 1 System type: 1 → Ess = 0 ∏Mp K= ∏Mz i i = −42 + 2.832 2.832 −22 + 2.832 = ( 4.9 )( 2.83) = 4 3.465 2.83 = 35.28D 4 = cos θ = 0.816 θ = tan −1 ζ actual The root locus, as drawn by MATLAB is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 115 Problem 6.17 For the following system, find the roots, the gain, and steady-state error for the fastest response and a settling time of less than 2 seconds and an overshoot of less than 4%. K GH = ( s + 1)( s + 3) Estimated student time to complete: 30-40 minutes Prerequisite knowledge required: Text Section(s) 6.14 Solution: For root locus: σ A = −1 − 3 = −2 and θ A = ±90D 2 ( −ζπ / 1−ζ 2 ) ×100% = 4% → ζ ≅ 0.72 4 4 θ = cos −1 ζ = 44D and Ts = = =2 ζωn 2 The root locus and the limitations are shown: For overshoot of 4%: %OS = e Choose poles for fastest response: tan 44D = Therefore: s = −2 ± 1.93 j K= ∏Mp ∏Mz i = i System type: 0 l → l = 1.93 2 (3 − 2) 2 + 1.932 (1 − 2) 2 + 1.932 = 4.72 1 4.72 = 1.57 s → 0 ( s + 1)( s + 3 ) → K P = lim G ( s ) = lim s →0 Ess = 1 = 0.39 1 + 1.57 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 116 Problem 6.18 For the system of problem 6.17, select the locations and the proportional and integral gains to change it to a proportional-plus-integral system with zero steady-state error. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.15 Solution: We add s = 0 at the origin and a zero near it (location to be found). Therefore, K (s + z) K ( s + 0.1) = s ( s + 1)( s + 3) s ( s + 1)( s + 3) Assume ζ = 0.73 for 4% OS even though the system is not second-order (this is close enough). Therefore, GH = θ = 43D approximately at s = −2 ± 1.87 j . K= ∏Mp ∏Mz i i = 22 + 1.87 2 12 + 1.87 2 12 + 1.87 2 1.92 + 1.87 2 = 4.6 The gain can be calculated as: ⎛ K KP ⎜ s + I KP K s + KI ⎝ = K= P s s K P = 4.6 and K I = 0.46 ⎞ ⎟ ⎠ = 4.6 ( s + 0.1) s Since this is Type-1, Ess = 0. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 117 Problem 6.19 For the system of Problem 6.17, we would like to improve the settling time to 1 second by adding a zero to the system, therefore a proportional-plus-derivative system. Find a proper location for the zero and the loop gain. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.16 Solution: Although not accurate, assume that σ = to be at s = −4 ± 3.85 j . 4 = 4 . Therefore, we arbitrarily choose the poles 1 Angle deficiency: 3.85 = 127.9D 3 3.85 θ −3 = 180 − tan −1 = 104.6D 1 θ z = 127.9 + 104.6 − 180 = 52.5D θ −1 = 180 − tan −1 3.85 → z = 6.95 z−4 The loop gain is: tan 52.5 = K= ∏Mp ∏Mz i = 32 + 3.852 12 + 3.852 i GH = 2.952 + 3.852 =4 4 ( z + 6.95 ) ( s + 1)( s + 3) © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 118 Problem 6.20 For the system of problem 6.19, add an integrator to the system to make it into a PID system in order to achieve a zero steady-state error. Find the location of an additional zero and proportional, derivative, and integral gains. Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 6.17 Solution: K ( s + 6.95 )( s + z ) s ( s + 1)( s + 3) Magnitude criterion: GH = Assume: K= ∏Mp ∏Mz i i = 42 + 3.852 32 + 3.852 12 + 3.852 2.952 + 3.852 3.92 + 3.852 = 4.06 ⎛ K K ⎞ KD ⎜ s2 + P s + I ⎟ KD KD ⎠ V K ⎝ G = a = KP + I + KDs = E s s 2 4.06( s + 6.95)( s + 0.1) 4.06( s + 7.05s + 0.695) = = s s ⎧ ⎪ ⎪ K D = 4.06 ⎧ K D = 4.06 ⎪ KP ⎪ = 7.05 → ⎨ K P = 28.6 ⎨ ⎪ KD ⎪ K = 2.82 ⎩ I ⎪ KI = 0.695 ⎪ ⎩ KD © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 119 CHAPTER SEVEN Problem 7.1 A motor with a rotor inertia of 0.030 Kgm2 and maximum torque of 12 Nm is connected to a uniformly distributed arm with a concentrated mass at its end, as shown in Figure P.7.1. Ignoring the inertia of a pair of reduction gears and viscous friction in the system, calculate the total inertia felt by the motor and the maximum angular acceleration it can develop if the gear ratio is a) 5, b) 50, c) 100. Compare the results. z y 0.5 m Tl N x Im m=3Kg Tm m=3Kg Figure P.7.1. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 7.2 Solution: 1 1 I load = marml 2 + mmass l 2 = (3)(0.5) 2 + (3)(0.5) 2 = 1 Kgm 2 3 3 a) For N = 5, b) For N = 50, c) For N = 100, ⎧ 1 ⎛ 1 ⎞ 2 ⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 25 =0.07 Kgm ⎪ ⎝ ⎠ ⎨ ⎪θ = Tmax = 12 = 171 r / s 2 ⎪⎩ max I total 0.07 ⎧ 1 ⎛ 1 ⎞ 2 ⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 2500 =0.0304 Kgm ⎪ ⎝ ⎠ ⎨ ⎪θ = Tmax = 12 = 395 r / s 2 ⎪⎩ max I total 0.0304 ⎧ 1 ⎛ 1 ⎞ 2 ⎪ I total = I motor + ⎜ N 2 ⎟ I load = 0.03 + 10000 =0.0301 Kgm ⎪ ⎝ ⎠ ⎨ ⎪θ = Tmax = 12 = 399 r / s 2 ⎪⎩ max I total 0.0301 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 120 Problem 7.2 Repeat Problem 1, but assume that the two gears have 0.002 Kgm2 and 0.005 Kgm2 inertias respectively. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 7.2 Solution: 1 1 I load = marml 2 + mmass l 2 = (3)(0.5) 2 + (3)(0.5) 2 = 1 Kgm 2 3 3 0.005 I gears = 0.002 + N2 0.005 1 1.005 ⎛ 1 ⎞ I total = I motor + I gears + ⎜ 2 ⎟ I load = 0.03 + 0.002 + + 2 = 0.032 + 2 N N N2 ⎝N ⎠ a) For N = 5, 1.005 ⎧ 2 ⎪⎪ I total = 0.032 + 25 =0.0722 Kgm ⎨ ⎪θmax = Tmax = 12 = 166 r / s 2 I total 0.0722 ⎪⎩ b) For N = 50, 1.005 ⎧ 2 ⎪⎪ I total = 0.032 + 2500 = 0.0324 Kgm ⎨ ⎪θmax = Tmax = 12 = 370 r / s 2 I total 0.0324 ⎪⎩ c) For N = 100, 1.005 ⎧ 2 ⎪ I total = 0.032 + 10, 000 =0.0321 Kgm ⎪ ⎨ ⎪θmax = Tmax = 12 = 374 r / s 2 ⎪⎩ I total 0.0321 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 121 Problem 7.3 The three-axis robot shown in Figure P.7.3. is powered by geared servomotors attached to the joints by worm gears. Each link is 22 cm long, made of hollow aluminum bars, each weighing 0.5 Kg. The center of mass of the second motor is 20 cm from the center of rotation. The gear ratio is 1/3 in the servomotor and 1/5 in the worm gear set. The worst case scenario for the elbow joint is when the arm is fully extended, as shown. Calculate the torque needed to accelerate both arms together, fully extended, at a rate of 90 rad/s2. Assume the inertias of the worm gears are negligible. Worm gear N=5 m4=0.5 Kg m2=0.5 Kg I1=0.01 Kgm2 N=3 m3=0.5 Kg Motors Figure P.7.3. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 7.2 Solution: Total gear ratio N = 5 × 3 = 15 ( I total = I rotor1 + I arm1+2 + I motor2 ) N1 2 2 2⎞ ⎛1 = 0.01 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟ ⎝3 ⎠ 2 2⎤ 1 ⎡1 = 0.01 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.01038 ⎣3 ⎦ 15 T = Iθ = 0.01038 × 90 = 0.934 Nm © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 122 Problem 7.4 Repeat Problem 3, but suppose the maximum torque this motor can provide is 0.9 Nm. Therefore, a new motor must be picked. Two other motors are available, one with the inertia of 0.009 Kgm2 and torque of 0.85 Nm, one with inertia of 0.012 Kgm2 and torque of 1 Nm. Which one would you use? Estimated student time to complete:15-20 minutes Prerequisite knowledge required: Text Section(s) 7.2. Solution: For N = 5 × 3 = 15 , For case 1: ( I total = I rotor1 + I arm1+2 + I motor2 ) N1 2 2 2⎞ ⎛1 = 0.009 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟ ⎝3 ⎠ 2 2⎤ 1 ⎡1 = 0.009 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.00938 ⎣3 ⎦ 15 T = Iθ = 0.00938 × 90 = 0.844 Nm For case 2: ( I total = I rotor1 + I arm1+2 + I motor2 ) N1 2 2 2⎞ ⎛1 = 0.012 + ⎜ ( 2marm )( 2l ) + mmotor ( lmotor ) ⎟ ⎝3 ⎠ 2 2⎤ 1 ⎡1 = 0.009 + ⎢ (1)( 0.44 ) + ( 0.5 )( 0.2 ) ⎥ 2 = 0.01238 ⎣3 ⎦ 15 T = Iθ = 0.01238 × 90 = 1.114 Nm The smaller motor is better. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 123 Problem 7.5 Estimate how much the torque/inertia ratio of a disk motor might be if it can go from zero to 2000 rpm in one millisecond, and compare it to the motor of Problem 1. Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 7.2 Solution: 2000 × 2π = 209.44 rad / sec 60 θ − θ 209.44 − 0 = 209440 rad / sec 2 θ = 2 1 = 0.001 t2 − t1 The torque/inertia ratio is: θ2 = T T = Iθ → = θ = 209440 I For the motor of Problem 1: T 12 = = 400 rad / sec 2 T = Iθ → I 0.03 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 124 Problem 7.6 Using a timer circuit, design a pulse generating circuit that will deliver a range of 5-500 pulses per second to a stepper motor driver. Estimated student time to complete:15-20 minutes Prerequisite knowledge required: Text Section(s) 7.6.7. Solution: We will pick a 50% duty cycle for each signal. Therefore: 1 tlow = thigh = ttotal 2 ttotal ranges between 0.002 and 0.2 seconds. Therefore tlow = 0.001 sec or tlow = 0.1 sec. Pick C2 = 0.1 μ F . Therefore: tlow = 0.693C2 ( R2 ) 0.001 = 0.693 ( 0.1× 10−6 ) R2 0.1 = 0.693 ( 0.1× 10−6 ) R2 → R2 = 14, 430 Ω → R2 = 1, 443, 000 Ω Therefore, a potentiometer with R2 = 1.5 M Ω enables us to achieve the desired values. For 50% duty cycle, as shown above, tlow = thigh 0.693C2 ( R2 ) = 0.693C2 ( RH + R2 ) → RH = R1 + R3 + R4 = 0 Pick R1 = R3 = R4 = 1 K Ω . Therefore: ⎧C2 = 0.1 μ F ⎪R = 1 K Ω ⎪⎪ 1 ⎨ R2 = 1.5 M Ω ⎪R = 1 K Ω ⎪ 3 ⎪⎩ R4 = 1 K Ω © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 125 Problem 7.7 Calculate the gear ratio for a Harmonic drive if NL=100, NF=95, N2=90, N3=95. Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) 7.11 Solution: From Equation 7.25: e= ωL N 2 N L − N F N 3 ( 90 )(100 ) − ( 95 )( 95 ) 1 = = =− ωA N2 NL 360 ( 90 )(100 ) The gear ratio between the Arm and the Last gear (with Arm as input) is: ωA = −360 ωL The minus sign indicates that the input (Arm) and the output (Last) gears rotate in opposite directions. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 126 Problem 7.8 Write a program to generate a variable pulse stream to drive a motor with pulse-widthmodulated voltages of 1, 2, 3, 4, and 5 volts for a 5 volt input. Estimated student time to complete: Variable depending on expertise. Prerequisite knowledge required: Text Section(s) 7.7. Solution: The accuracy with which this program will work depends on the processing speed of the microprocessor. It also depends on the level of programming language and how efficiently it is written. We assume that the processor can accept and run a compiled “C” program, and that it is fast enough such that the execution of commands will not negatively affect the accuracy of the output timing. Since only five distinct output voltages of 1, 2, 3, 4, and 5 volts are desired, we only need to generate 5 variations in the timing loops. We assume the output level will be indicated by an input port. Therefore, if input port 1 is high, the desired output level is 1 volts, etc. The following program is written in “C” language for execution by a Mini-Board™ microprocessor. It is the essentials of the program that are important here, not the actual program and the way it is written. This program must be modified for use for any other specific microprocessor with its own unique programming syntax and requirements. In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at n VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 15 “msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 1 is turned off. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 127 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 128 Problem 7.9 Write a program to generate a sinusoidal pulse-width-modulated output for a constant input voltage. Estimated student time to complete: Variable depending on expertise. Prerequisite knowledge required: Text Section(s) 7.7. Solution: The accuracy with which this program will work depends on the processing speed of the microprocessor. It also depends on the level of programming language and how efficiently it is written. We assume that the processor can accept and run a compiled “C” program, and that it is fast enough such that the execution of commands will not negatively affect the accuracy of the output timing. Since a sinusoidal out put is desired for a 5 volt input voltage, we need to generate a variable control timing for the pulse width modulation. We calculate the sine value for each degree between zero and 90 degrees. This allows for approximately 10 pulses per second. The following program is written in “C” language for execution by a Mini-Board™ microprocessor. It is the essentials of the program that are important here, not the actual program and the way it is written. This program must be modified for use for any other specific microprocessor with its own unique programming syntax and requirements. Since Mini-Board can generate voltages in both polarities, the program can generate a full sine wave. The program must be modified accordingly if the microprocessor cannot do this. In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at n VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 15 “msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 1 is turned off. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 129 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 130 Problem 7.10 If you have access to a microprocessor and electronic components such as transistors, make an H-bridge and write a control program to drive a motor in either direction or to brake it. Be mindful of the problems associated with an H-bridge’s transistors turning on and off at inappropriate times. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) Solution: The following program is written in “C” language for execution by a Mini-Board™ microprocessor. It is the essentials of the program that are important here, not the actual program and the way it is written. This program must be modified for use for any other specific microprocessor with its own unique programming syntax and requirements. In this program, the command “motor(1,n)” indicates that the output port 1 is turned on at n VCC . If the input voltage is 5 volts, for n=15, the output is 5 volts, etc. The command 15 “msleep(m)” indicates a stop in execution for m milliseconds. “Off(1)” means output port 1 is turned off. It is assumed that an input to the microprocessor equal to 1 indicates rotation in the CCW direction, 2 indicates CW rotation, and a 0 means the motor should be braked. /************************************************************/ /* H-Bridge Controller */ /************************************************************/ . . main( ) { int direction while(1) {direction = digital(0); if (direction = = 1) {msleep (1); motor (1,15); motor (4,15);} off (2); off (3); if (direction = = 2) {msleep (1); motor (2,15); motor (3,15);} off (1); off (4); /* motor rotation direction input */ /* CCW rotation */ /* CW rotation */ if (direction = = 0) {msleep (1); /* brake the motor */ motor (3,15); motor (4,15);} off (1); off (2); }} © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 131 CHAPTER EIGHT There are no problems in Chapter 8. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 132 CHAPTER NINE Problem 9.1 Calculate the necessary memory requirement for a still color image from a camera with 10 megapixels at: 8-bits per pixel (256 levels) • 16-bits per pixel (65536 levels) • Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) 9.5 Solution: For an 8-bit per pixel system, each pixel requires 1 byte. The total memory needed is 10 megabytes. For a 16-bit per channel system, each pixel requires 2 bytes. The total memory needed is 20 megabytes. The assumption here is that the pixel count given is the total number for all three colors. The actual memory is greatly influenced by the format in which the image is saved. A bitmap image needs much more memory than JPEG. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 133 Problem 9.2 Consider the pixels of an image, with values as shown in Figure P.9.2, as well as a convolution mask with the given values. Calculate the new values for the given pixels. 4 3 5 8 7 2 3 6 1 6 2 5 8 6 3 9 1 1 1 0 1 0 1 1 1 Figure P.9.2 Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 9.10 Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 4 3 5 8 7 3.3 4.6 6 1 5 4.4 5 8 6 3 9 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 134 Problem 9.3 Consider the pixels of an image, with values as shown in Figure P.9.3, as well as a convolution mask with the given values. Calculate the new values for the given pixels. Substitute 0 for negative grey levels. What conclusion can you make from the result? 5 5 10 5 5 5 10 5 5 5 10 5 7 8 10 9 8 8 8 9 7 8 10 9 9 8 9 10 8 10 1 0 1 0 -4 0 1 0 1 Figure P.9.3 Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 9.10. Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 5 5 10 5 5 10 0 16 5 12 0 17 7 0 0 1 8 8 8 9 7 3 0 3 9 8 9 10 8 10 This can be an indication of an edge. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 135 Problem 9.4 Repeat Problem 9.3, but substitute the absolute value of negative grey levels. What conclusion can you make from the result? Estimated student time to complete: 5 -10 minutes Prerequisite knowledge required: Text Section(s) 9.10 Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 5 5 10 5 5 10 20 16 5 12 13 17 7 0 13 1 8 8 8 9 7 3 6 3 9 8 9 10 8 10 When absolute values of negative pixels are shown, edges are no longer discernable. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 136 Problem 9.5 Repeat Problem 9.3, but apply the mask of Figure P.9.5 and compare your results with Problem 9.3. Which one is better? -1 0 -1 0 4 0 -1 0 -1 Figure P.9.5 Estimated student time to complete: 5 -10 minutes Prerequisite knowledge required: Text Section(s) 9.10 Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 5 5 10 5 5 0 20 0 5 0 13 0 7 0 13 0 8 8 8 9 7 0 6 0 9 8 9 10 8 10 This mask also shows the edge, but unlike the one in problem 9.3., the pixels are brighter than the neighboring pixels. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 137 Problem 9.6 Repeat Problem 9.3, but apply the mask of Figure P.9.6 and compare your results with Problem 9.3. Which one is better? 0 1 0 1 -4 1 0 1 0 Figure P.9.6 Estimated student time to complete: 5 -10 minutes Prerequisite knowledge required: Text Section(s)9.10 Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 5 5 10 5 5 5 0 8 5 8 0 12 7 0 0 0 8 8 8 9 7 2 0 0 9 8 9 10 8 10 At least for this image, the effect is similar to the mask in problem 9.3. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 138 Problem 9.7 Repeat Problem 9.3, but apply the mask of Figure P.9.7 and compare your results with Problem 9.3. Which one is better? 1 1 1 1 -8 1 1 1 1 Figure P.9.7 Estimated student time to complete: 5 -10 minutes Prerequisite knowledge required: Text Section(s) 9.10 Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: 5 5 10 5 5 15 0 24 5 20 0 29 7 0 0 0 8 8 8 9 7 5 0 3 9 8 9 10 8 10 At least for this image, the effect is similar to Problem 9.3, but darker. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 139 Problem 9.8 An image is represented by values shown below. Find the value of pixel 2c when mask 1 is applied. a) b) Find the value of pixel 3b when mask 2 is applied. Find the values of pixels 2b and 3c when a 3 × 3 median filter is applied. c) Find the area of the major object that results when a threshold of 4.5 is applied d) based on a +4-connectivity (start at the first on-pixel). 1 2 3 4 5 a b c d e 4 4 6 1 2 9 3 2 6 8 6 7 6 5 4 2 4 5 9 4 6 6 3 2 7 1 1 1 1 1 1 1 1 1 Mask 1 0 1 0 1 -4 1 0 1 0 Mask 2 Figure P.9.8. Estimated student time to complete: 10 -15 minutes Prerequisite knowledge required: Text Section(s) 9.10 and 9.17. Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: a) For 2c: R = (9× 1 + 6× 1 +2× 1 +3× 1 +7× 1 +4× 1 +2× 1 +6× 1 +5× 1) / 9 = 4.9 (or 5) b) For 3b: R = (4× 0 + 3× 1 +7× 0 +6× 1 +2× -4 +6× 1 +1× 0 +6× 1 +5× 0) = 13 c) For 2b: 2,3,4,4,6,6,6,7,9. The median is 6. For 3c: 2,3,4,5,5,6,6,7,9. The median is 5. d) The resulting image after the application of the threshold is: a 1 b c X X X X X X X X X X 2 3 4 5 X X X d e © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 140 Applying the +4 connectivity rule starting at the first on-pixel results in: 1b, 1c, 2c, 3c, 3d, 4c, 4d, 4b, 5b, for a total of 9 pixels. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 141 Problem 9.9 An image is represented by values shown below. Find the value of pixel 3b when mask 1 is applied. a. b. Find the values of pixels 2b, 2c, 2d when mask 2 is applied. Find the value of pixel 3c when a 5 × 5 median filter is applied. c. Find the area of the major object that results when a threshold of 4.5 is applied d. based on a × 4-connectivity (start at the first on-pixel). 1 2 3 4 5 a b c d e 8 4 6 1 2 9 6 7 6 2 5 5 4 2 4 6 9 3 5 6 5 4 2 10 8 0 1 0 1 4 1 0 1 0 Mask 1 1 1 1 -2 -2 -2 1 1 1 Mask 2 Figure P.9.9. Estimated student time to complete: 10 -15 minutes Prerequisite knowledge required: Text Section(s) 9.10 and 9.17. Solution: The mask will only affect the middle pixels. The remaining pixels on the perimeter will not be affected. Depending on how a new image is formatted, the pixels on the perimeter may be set to zero or the current values. The new values for the image, assuming that the old file is copied into a new file, are: a) For 3b: R = (4× 0 + 6× 1 +2× 0 +6× 1 +7× 4 +5× 1 +1× 0 +10× 1 +5× 0)/8 = 6.875 (or 7) b) For 2b: R = (8× 1 + 9× 1 +6× 1 +4× -2 +6× -2 +2× -2 +6× 1 +7× 1 +5× 1) = 17 For 2c: R = (9× 1 + 6× 1 +2× 1 +6× -2 +2× -2 +4× -2 +7× 1 +5× 1 +6× 1) = 11 For 2d: R = (6× 1 + 2× 1 +5× 1 +2× -2 +4× -2 +6× -2 +5× 1 +6× 1 +5× 1) = 5 c) For 3c: 1,2,2,2,2,3,4,4,4,4,5,5,5,5,6,6,6,6,6,7,8,8,9,9,10. The median is 5. d) The resulting image after the application of the threshold is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 142 1 a b c X X X X X X X 2 3 4 5 X d e X X X X X X X Applying the × 4 connectivity rule starting at the first on-pixel results in: 1a, 2b, 3c, 1c, 3a, 4d, 4b, 3e, for a total of 8 pixels. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 143 Problem 9.10 Write a computer program for the application of a 3× 3 averaging convolution mask unto a 15× 15 image. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on the expertise Prerequisite knowledge required: Text Section(s) 9.10 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 144 Problem 9.11 Write a computer program for the application of a 5× 5 averaging convolution mask unto a 15× 15 image. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on the expertise Prerequisite knowledge required: Text Section(s) 9.10 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 145 Problem 9.12 Write a computer program for the application of a 3× 3 high pass convolution mask unto a 15× 15 image for edge detection. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on expertise Prerequisite knowledge required: Text Section(s) 9.10 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 146 Problem 9.13 Write a computer program for the application of an n× n convolution mask unto a k× k image. Refer to the note on page 415 for more information. You should write the routine such that the user can choose the size of the mask and the values of each mask cell individually. Estimated student time to complete: Variable depending on expertise Prerequisite knowledge required: Text Section(s) 9.10 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 147 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 148 Problem 9.14 Write a computer program that will perform the Left-Right search routine for a 15× 15 image. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on expertise Prerequisite knowledge required: Text Section(s) 9.13 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 149 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 150 Problem 9.15 Using the Left-Right search technique, find the outer edge of the object in Figure P.9.15: a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 Figure P.9.15 Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) 9.13 Solution: a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 The edge consists of: 2c, 2d, 2e, 2f, 2g, 2h, 2i, 2j, 2k, 3k, 4k, 5j, 6j, 7k, 7j, 8i, 9h, 10h, 11h, 12h, 12g, 12f, 11f, 10g, 9g, 8f, 8e, 7e, 7d, 7c, 6b, 5b, 4b, 4c, 3c. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 151 Problem 9.16 The x and y coordinates of 5 points are given as (2.5, 0), (4,2), (5,4), (7,6), and (8.5,8). Using the Hough transform, determine which of these points form a line and find its slope and intercept. Estimated student time to complete: 20-25 minutes Prerequisite knowledge required: Text Section(s) 9.15. Solution: y x x,y m,c 0 2.5 0=2.5m+c c= -2.5m+0 2 4 2=4m+c c= -4m+2 4 5 4=5m+c c= -5m+4 6 7 6=7m+c c= -7m+6 8 8.5 8=8.5m+c c= -8.5m+8 The following graph shows the lines and where they intersect. Line 3 (for point 5,4) does not intersect the remaining lines. The approximate slope and intercept of the line on which the other four points lay are 1.4 and -3.5. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 152 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 153 Problem 9.17 Write a computer program that will perform a region growing operation based on +4 connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a region with a chosen index number, and after finishing that region, must continue searching for another nuclei until all object pixels have been checked. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on expertise. Prerequisite knowledge required: Text Section(s) 9.17 Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. In this program, when a nucleus is found, all the pixels connected to it based on a +4connectivity that are “on” (have a value in them and therefore, are part of the image not the background) are placed in a stack by changing the value of the cell form zero to 1 in an array of 256x256. The cell’s value is later checked until all are found and the cell’s are all zero. Each corresponding pixel’s value is changed from 1 to the region number. At the end of execution, the original image is changed to as many regions as necessary. Since there will be no cells (pixels) left with a value of 1, execution ends. All segments will be 2 and larger in value. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 154 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 155 Problem 9.18 Write a computer program that will perform a region growing operation based on × 4 connectivity. The routine should start at the 1,1 corner pixel, search for a nucleus, grow a region with a chosen index number, and after finishing that region, must continue searching for another nuclei until all object pixels have been checked. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on expertise. Prerequisite knowledge required: Text Section(s) 9.17 Solution: Please see the solution for Problem 9.17. The program for growing a region based on × 4 connectivity is very similar to that program, except that the search will follow the × 4 sequence given in Equation 9.7. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 156 Problem 9.19 Using +4 connectivity logic and starting from 1a pixel, write the sequence of pixels in correct order that will be detected by a region growing routine for the object in Figure P.9.19: a b c d e f 1 2 3 4 5 6 7 8 9 g h 4 1 2 3 +4-connectivity Figure P.9.19 Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 9.17 Solution: The sequence of pixels detected is: 2d, 3d, 3e, 3f, 4e, 3g, 4f, 5e, 5f, 5d, 5g, 6d, 6c, 7c, 7b, 8b. The remaining pixels are not connected to this region and will not be detected. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 157 Problem 9.20 Using × 4 connectivity logic and starting from 1a pixel, write the sequence of pixels in correct order that will be detected by a region growing routine for the object in Figure P.9.20: a b c d e f g h 1 2 3 4 5 6 7 8 9 1 4 3 2 x4-connectivity Figure P.9.20 Estimated student time to complete: 10 minutes Prerequisite knowledge required: Text Section(s) 9.17 Solution: The sequence of pixels detected is: 2c, 3d, 4e, 4c, 5f, 5d, 3f, 4b, 6g, 6c, 7b. The remaining pixels are not connected to this region and will not be detected. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 158 Problem 9.21 Find the union between the two objects in Figure P.9.21. 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Figure P.9.21 Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) 9.18 Solution: The union is shown: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 159 Problem 9.22 Apply a single-pixel erosion based on 8-connectivity on the image of Figure P.9.22. a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 Figure P.9.22 Estimated student time to complete: 10-20 minutes Prerequisite knowledge required: Text Section(s) 9.18 Solution: The result is shown: a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 160 Problem 9.23 Apply a one-pixel dilation to the result of Problem 9.22 and compare your result to Figure P.9.22. Estimated student time to complete: 15-25 minutes Prerequisite knowledge required: Text Section(s)9.18 Solution: The result is shown. As you notice, the image is different from the original. a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 161 Problem 9.24 Apply an open operation to Figure P.9.24. a b c d e f g h i j k l m f g h 1 2 3 4 5 6 7 8 9 10 11 12 13 Figure P.9.24 Estimated student time to complete: 10-20 minutes Prerequisite knowledge required: Text Section(s) 9.18 Solution: The result is shown: a 1 2 3 4 5 6 7 8 9 10 11 12 13 b c d e f g h i j k l m a b c d e i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 162 Problem 9.25 Apply a close operation to Figure P.9.24. Estimated student time to complete: 10-20 minutes Prerequisite knowledge required: Text Section(s) 9.18 Solution: The result is shown: a b c 1 2 3 4 5 6 7 8 9 10 11 12 13 d e f g h i j k l m a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 163 Problem 9.26 Apply a skeletonization operation to Figure P.9.26. a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 Figure P.9.26 Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 9.18 Solution: The result is shown: a b c d e f g h i j k l m 1 2 3 4 5 6 7 8 9 10 11 12 13 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 164 Problem 9.27 Write a computer program in which different moments of an object in an image can be calculated. The program should ask you for moment indices. The results may be reported to you in a new file, or may be stored in memory. Refer to the note on page 415 for more information. Estimated student time to complete: Variable depending on expertise Prerequisite knowledge required: Text Section(s) 9.21. Solution: The following program is written in “C” language. It is the essentials of the programming that are important here, not the syntax. You may use the proper syntax for the program you are using. This program may be written in countless other languages, both graphical and scientific. In this program, the assumption is that the image input file is read from a data file or it is entered after initializing the array. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 165 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 166 Problem 9.28 Calculate the M 0,2 moment for the result of Problem 9.8d based on +4-connectivity. Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 9.21. Solution: The resulting image after the application of the threshold is: a 1 b c X X X X X X X X X X 2 3 X X X 4 5 d e Applying the +4 connectivity rule starting at the first on-pixel results in the following area: a 1 b c X X X X X X X 2 3 X X 4 5 d e x y The moment, calculated relative to the first row and column of pixels with whole-number distances is: M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 1(1) 2 + 2(2) 2 + 3(3) 2 + 1(4) 2 = 52 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 167 Problem 9.29 For the binary image of a key in Figure P.9.29, calculate the following: • Perimeter, based on the Left-Right search technique. 2 . • Thinness, based on P Area • Center of gravity • Moment M 0,1 about the origin (pixel 1,1) and about the lowest pixel of a rectangular box around the key (2,2). 1 2 3 4 5 6 7 8 9 10 x 1 2 3 4 5 6 7 8 9 10 y Figure P.9.29 Estimated student time to complete: 15-20 minutes Prerequisite knowledge required: Text Section(s) 9.13, 9.21. Solution: 1 2 3 4 5 6 7 8 9 10 x 1 2 3 4 5 6 7 8 9 10 y The left-right search result is shown. The perimeter based on the pixels on the search line is P = 23. The area may be calculated by M 0,0 , by region growing or by counting. P2 232 Thinness: = = 17 Area 31 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 168 The 1st moments are: M 1,0 = ∑ x1 y 0 = ∑ x = 2(1) + 3(2) + 4(3) + 4(4) + 5(5) + 5(6) + 5(7) + 3(8) = 150 M 0,1 = ∑ x 0 y1 = ∑ y = 2(1) + 4(2) + 4(3) + 5(4) + 5(5) + 4(6) + 4(7) + 3(8) =143 Center of gravity is: x = M 1,0 M 0,0 1 = 150 = 4.84 and 31 2 3 4 5 6 7 8 9 y= M 0,1 M 0,0 = 143 = 4.61 31 10 x 1 2 3 4 5 6 7 8 9 10 x' y' y M 0,1 @(1,1) = ∑ x 0 y1 =143 M 0,1 @(2, 2) = ∑ x 0 y1 = 4(1) + 4(2) + 5(3) + 5(4) + 4(5) + 4(6) + 3(7) = 112 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 169 Problem 9.30 Using moment equations, calculate M 0,2 and M 2,0 about the centroidal axes of the part in Figure P.9.30. 1 2 3 4 5 6 7 8 9 10 x 1 2 3 4 5 6 7 8 9 10 y Figure P.9.30 Estimated student time to complete: 20-30 minutes Prerequisite knowledge required: Text Section(s) 9.21 Solution: The moment values for the image are: M 0,0 = ∑ x 0 y 0 = 16(1) =16 M 1,0 = ∑ x1 y 0 = ∑ x = 1(2) + 3(3) + 3(4) + 4(5) + 2(6) + 3(7) =76 M 0,1 = ∑ x 0 y1 = ∑ y = 3(2) + 2(3) + 4(4) + 3(5) + 3(6) + 1(7) =68 x= M 1,0 M 0,0 = 76 = 4.75 16 and y= M 0,1 M 0,0 = 68 = 4.25 16 M 2,0 = ∑ x 2 y 0 = ∑ x = 1(2) 2 + 3(3) 2 + 3(4) 2 + 4(5) 2 + 2(6) 2 + 3(7) 2 = 398 2 M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 3(2) 2 + 2(3) 2 + 4(4) 2 + 3(5) 2 + 3(6) 2 + 1(7) 2 =326 The second moments about the centroids can be calculates as follows: M 2,0 @ x = M 2,0 − M 0,0 ( x ) = 398 − 16(4.75) 2 = 37 2 M 0,2 @ y = M 0,2 − M 0,0 ( y ) 2 = 326 − 16(4.25) 2 = 37 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 170 Problem 9.31 Using moment equations, calculate M 0,2 and M 2,0 about the centroidal axes of the part in Figure P.9.31. 1 2 3 4 5 6 7 8 9 10 x 1 2 3 4 5 6 7 8 9 10 y Figure P.9.31 Estimated student time to complete:20 -30 minutes Prerequisite knowledge required: Text Section(s) 9.21 Solution: The moment values for the image are: M 0,0 = ∑ x 0 y 0 = 18(1) =18 M 1,0 = ∑ x1 y 0 = ∑ x = 3(2) + 2(3) + 4(4) + 4(5) + 2(6) + 3(7) =81 M 0,1 = ∑ x 0 y1 = ∑ y = 4(2) + 5(3) + 5(4) + 4(5) =63 x= M 1,0 M 0,0 = 81 = 4.5 18 and y= M 0,1 M 0,0 = 63 = 3.5 18 M 2,0 = ∑ x 2 y 0 = ∑ x = 3(2) 2 + 2(3) 2 + 4(4) 2 + 4(5) 2 + 2(6) 2 + 3(7) 2 = 413 2 M 0,2 = ∑ x 0 y 2 = ∑ y 2 = 4(2) 2 + 5(3) 2 + 5(4) 2 + 4(5) 2 = 241 The second moments about the centroids can be calculates as follows: M 2,0 @ x = M 2,0 − M 0,0 ( x ) = 413 − 18(4.5) 2 = 48.5 2 M 0,2 @ y = M 0,2 − M 0,0 ( y ) 2 = 241 − 18(3.5) 2 = 20.5 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 171 CHAPTER TEN Problem 10.1 Develop a fuzzy inference system for a robot, where the force exerted at the hand and the velocity of the hand are the inputs and the % power to the actuators is the output. Estimated student time to complete:-------Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Force; Small, Average, Large Velocity; Slow, Fast, Fast, Very-Fast Outputs: Power; Low, Medium, High, Very-High The fuzzy sets are (arbitrary numbers): © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 172 The rules are: The resulting output is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 173 Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 174 Problem 10.2 Develop a fuzzy inference system for a washing machine. The inputs are how dirty the fabrics are and how much clothes are being washed, and the output is the wash time. Estimated student time to complete:-------Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Dirty; Not-Much, Some, Very. Clothes; Small, Med, Much, A-Lot. Outputs: Wash-Time; Five, Ten, Twenty, Thirty The fuzzy sets are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 175 The rules are: The resulting output is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 176 Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 177 Problem 10.3 Develop a fuzzy inference system for a barbecue. The inputs may be the thickness of the steak and how cooked or rare it is desired to be. The output may be the temperature of the flame and/or the time of cooking. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Thickness; See-Through, Average, Hefty How Cooked; Rare, Medium-Rare, Medium, Well-Done Outputs: Temp; Low, Medium, High Time; Three, Five, Eight, Ten The fuzzy sets are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 178 The rules are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 179 The resulting out puts are: And Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 180 Problem 10.4 Develop a fuzzy inference system for an automatic gearbox. The inputs are the speed of the car and the load on the engine, and the output is the gear ratio of the transmission. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Speed; Slow, Medium, Cruise, Fast, Maximum Load; Low, Average, High Outputs: Gear-Ratio; Fourth, Third, Second, First The fuzzy sets are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 181 The rules are: The resulting out put is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 182 Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 183 Problem 10.5 Develop a fuzzy logic system for a vision system in which the inputs are the intensities of the three colors of red, green, and blue (RGB) in a color image and the output is the relationship of the combination to the colors of the rainbow. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Color-Red; Red-Off, Red-Medium, Red-High Color-Green; Green-Off, Green-Medium, Green-High Color-Blue; Blue-Off, Blue-Medium, Blue-High Outputs: Color; Black, Red, Orange, Yellow, Green, Cyan, Blue, Violet, White The fuzzy sets are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 184 The rules are: The resulting out put is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 185 Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 186 Problem 10.6 Develop a fuzzy inference system for grading a robotics course. The inputs are your effort level in the course and your exam grade, and the output is your letter grade. Estimated student time to complete: Prerequisite knowledge required: Text Section(s) 10.3-10.9 Solution: As in any other design problem, the solution is not unique. For example, different inputs and outputs may be selected, different fuzzy sets may be chosen, the ranges selected may be different, and alternate rules may be assigned. Therefore, not only the solution is not unique, the result will also differ. This suggested solution may only be used as a guide. This solution is implemented on the MATLAB Fuzzy Toolbox. You may use this or any other available system. In this solution, the inputs, outputs, the sets, and the rules are as follows: Inputs: Effort; Low, Medium, High Exam-Grade; eA, eB, eC, eD, eF Outputs: Course-Grade; A, B, C, D, F The fuzzy sets are: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 187 The rules are: The resulting out put is: © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 188 Changing the rules or the membership functions will change this result. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 189 APPENDICES Problem A.1 Show that the determinant of a matrix can be calculated by picking any row or column. Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) A.1 Solution: For an arbitrary matrix: ⎡a A = ⎢⎢ d ⎢⎣ g b e h c⎤ f ⎥⎥ i ⎥⎦ det A = a ( ei − fh ) − b ( di − fg ) + c ( dh − eg ) = aei − afh − bdi + bfg + cdh − ceg det A = −d ( bi − ch ) + e ( ai − cg ) − f ( ah − bg ) = −dbi + dch + eai − ecg − fah + fbg det A = −b ( di − fg ) + e ( ai − cg ) − h ( af − cd ) = −bdi + bfg + eai − ecg − haf + hcd They are all the same. © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 190 Problem A.2 Calculate the determinant of the following (4 × 4) matrix. ⎡1 ⎢0 A=⎢ ⎢3 ⎢ ⎣1 1 1 0 0 0 2 1 0 0⎤ 0 ⎥⎥ 1⎥ ⎥ 1⎦ Estimated student time to complete: 5 minutes Prerequisite knowledge required: Text Section(s) A.1. Solution: det A = 1(1(1 − 0 ) − 0 + 0 ) − 1( 0 − 2 ( 3 − 1) − 0 ) + 0 − 0 = 1 + 4 = 5 © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 191 Problem A.3 Calculate the inverse of the following matrix using method 1: ⎡1 1 2⎤ B = ⎢0 1 0⎥ ⎢ ⎥ ⎢⎣2 0 3⎥⎦ Estimated student time to complete: 5-10 minutes Prerequisite knowledge required: Text Section(s) A.1. Solution: det B = 1(3 − 0) − 0 + 2(−2) = 3 − 4 = −1 ⎡1 0 B = ⎢⎢1 1 ⎢⎣ 2 0 ⎡3 adjB = ⎢⎢ 0 ⎢⎣ −2 T 2⎤ 0 ⎥⎥ 3 ⎥⎦ −3 −2 ⎤ −1 0 ⎥⎥ 2 1 ⎥⎦ ⎡ −3 3 2 ⎤ −1 B = ⎢⎢ 0 1 0 ⎥⎥ ⎢⎣ 2 −2 −1⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner. 192 Problem A.4 Calculate the inverse of the following matrix using method 2: ⎡1 0 1 ⎤ C = ⎢0 2 1⎥ ⎢ ⎥ ⎢⎣ 3 1 0⎥⎦ Estimated student time to complete: 10-15 minutes Prerequisite knowledge required: Text Section(s) A.1. Solution: ⎡1 0 1 ⎤ ⎡ x11 ⎢0 2 1 ⎥ ⎢ x ⎢ ⎥ ⎢ 21 ⎣⎢ 3 1 0 ⎦⎥ ⎢⎣ x31 ⎧ x11 + x31 = 1 ⎪ ⎨2 x21 + x31 = 0 ⎪3 x + x = 0 ⎩ 11 21 ⎧ x12 + x32 = 0 ⎪ ⎨2 x22 + x32 = 1 ⎪3 x + x = 0 ⎩ 12 22 ⎧ x31 + x33 = 0 ⎪ ⎨2 x23 + x33 = 0 ⎪3 x + x = 1 ⎩ 13 23 1 ⎡1 − ⎢7 7 ⎢ −3 3 C −1 = ⎢ ⎢7 7 ⎢ 1 ⎢6 7 ⎣⎢ 7 x13 ⎤ ⎡1 0 0 ⎤ x22 x23 ⎥⎥ = ⎢⎢0 1 0 ⎥⎥ x32 x33 ⎦⎥ ⎣⎢0 0 1 ⎥⎦ ⎧ x11 = 1/ 7 ⎪ → ⎨ x31 = 6 / 7 ⎪ x = −3 / 7 ⎩ 21 x12 ⎧ x12 = −1/ 7 ⎪ → ⎨ x22 = 3 / 7 ⎪ x = 1/ 7 ⎩ 32 ⎧ x13 = 2 / 7 ⎪ → ⎨ x23 = 1/ 7 ⎪ x = −2 / 7 ⎩ 33 2⎤ 7⎥ ⎥ 1⎥ 7⎥ ⎥ −2 ⎥ 7 ⎥⎦ © Copyrighted 2010. This solution manual may not be copied, posted, made available to students, placed on BlackBoard or any other electronic media system or the Internet without prior expressed consent of the copyright owner.