ME5286 Robotics Spring 2014 Quiz 1 Solution. Total Points: 30

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1 Page 1 of 7 ME5286 Robotics Spring 2014 Quiz 1 Solution Total Points: 30 (Note images from original quiz are not included to save paper/ space. Please see the original quiz for additional information and images)

2 Page 2 of 7 A. (2 points) You are given the rotational axis for each joint in one leg of a robot similar the S-one robot (Figures 1.2 and 1.3).The world coordinate frame X w, Y w, Z w is inertial with the robot s forward motion in the +X w direction. Determine the number of degrees of freedom. Then decide on a base coordinate frame for the leg and state why. Draw the base coordinate frame X 0, Y 0, Z 0 for the leg on Figure 1.6. The number of joints is 7, therefore the degrees of freedom is 7. The axis is drawn on the figure. The chosen base coordinate frame is at the foot of leg. Note that this is one of several options. Either the foot or hip could be chosen as the base coordinate frame with any orientation at that point. The foot was chosen assuming that the foot is currently in contact with the ground and the hip is thus acting as the end effector. B. (3 points) Write the unit vectors B i 0 for each of the joint axes J i for the shown figures (Figures 1.2 and 1.3), as expressed in the base coordinate frame X 0, Y 0, Z 0. The direction of these vectors should be determined from the joints rotational direction as shown (Figure 1.5), based on the right-hand rule convention. Assume that joint axes B and F are parallel and have no offset in the Y axis. B 1 0 = 0 0 1, B 2 0 = 1 0 0, B 3 0 = 0 1 0, B 4 0 = 0 1 0, B 5 0 = 0 1 0, B 6 0 = 1 0 0, B 7 0 = (Note that this depends on the chosen base coordinate frame. Each answer in transposed to make a column vector) C. (4 points) Given the base coordinate frame you chose in Part A, clearly draw and label the remaining link coordinate frames on Figure 1.6. Note that the order in which you label the remaining coordinate frames will depend on your selection of the base coordinate frame. The coordinate frames are all drawn on the following figure. Again this is one of many possible solutions. Furthermore since the choice of joint numbering was arbitrary we could have switched the ordering of joints 1 and 2 as well as joints 4 and 5 without changing the leg s kinematics. The labeling of the coordinate frames was done simultaneously with the filling out of the D-H table

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4 Page 4 of 7 D. (6 points) The elements of the A matrices from the i 1 th to the i th joint can be calculated using the Denavit-Hartenburg convention. Fill in the table below with the Denavit-Hartenburg variables θ, d, a, α for each of the A matrices. Note that the joints will start from your choice of base coordinate frame and work up (or down) the leg. Use the appropriate geometric dimensions from Figures 1.4 and 1.5. Joint i θ (degrees) d (mm) a (mm) α (degrees) 1 θ θ θ θ θ θ E. (5 points) At one instant in time, the joints move through their zero positions (as shown in Figures 1.4 and 1.5). Compute the matrices A 0 1, A 1 2, A 2 3, and A 3 4 using the Denavit-Hartenburg variables from part C for this instant. Using the standard form and the D-H table: cθ n sθ n cα n sθ n sα n a n cθ n n sθ = [ n cθ n cα n cθ n sα n a n sθ n A sα n cα n d 0 = [, n A n A = [, A = [ A = [ A quick check of the 4 th column in each A matrix gives us a sanity check for correctness. (units in mm)

5 Page 5 of 7 F. (4 points) Find the homogeneous transformation matrix T 0 4 from your chosen base coordinate frame X 0, Y 0, Z 0 to the coordinate frame at the knee joint (Joint D). The standard formula provides: T 4 0 = A 1 0 A 2 1 A 3 2 A = [ The matrix multiplication was done using a TI calculator. A quick check of the 3x3 rotation matrix gives us a sanity check since the 4 th coordinate frame is rotated in the XZ plane of the base coordinate frame. The translation column look correct since we moved 0.1 m in the X and 0.75 m in the Z base coordinate frame. (Note rounding errors may occur) G. (1 point) Explain in words what you think the advantage is of using the rear bending knee in the S-one robot. This question can have multiple answers. Correct answers include improved forward thrust, ability to approach objects without interference from knees, improved ability to climb ladders, etc.

6 Page 6 of 7 A. (1 point) Determine the homogenous transformation matrix to go from the camera coordinate frame to the GPS coordinate frame. In other words, find T C G. According the diagram given and offset between the camera and the GPS unit we can write: T G Note that this makes sense since the two coordinate frame s axis are aligned. The translations are obvious. B. (1 point) At a given instant in time a general offset from the object to the camera unit is given by (J,K,L) (Figures 2.3 and 2.4). Determine the homogenous transformation matrix to go from the camera coordinate frame to the object coordinate frame. In other words, find T C O. Notice that these offsets are given as variables and therefore do not have signs (that will come later). The offsets on the diagram should not be used to infer the signs J T O K L This makes sense since the coordinate frames are also aligned with each other and the offsets are obvious. C. (3 points) At a given moment in time the GPS receiver is found to be at [ m, m relative to a state plane coordinate system with a heading angle θ G = 45 (Figure 2.5). At the same moment in time, the object on the road is found by the camera to be at [5.0 m, 0.0 m, -2.5 m. Determine the position and orientation of the object on the road relative to the state plane coordinate system. We can assume the Z height of the GPS in the state coordinate frame to be any value. We will try setting it to zero and observing the outcome. Next we note that in general the form for the overall transfer function will be: T S 0 = T S G T G C T C O Notice that we already have the form for T C O so we can substitute in the given offsets for the variables as:

7 Page 7 of T O Notice that the signs on this matrix would have been incorrect had we used signs on our variables in step B. We also have T C G. So we can use the identity that T G C = T C G 1 : T C G G = T Next we can use the given offset of the GPS [ m, m, 0 m and the fact that the GPS is rotated by -45 degrees in the XY plane to compute T S G. It is important to note the -45 degrees given the sketch of the car in the state coordinate frame T G S = [ Notice that the 3x3 rotation matrix component is easily found using: cos θ sin θ R = [ and substituting θ = 45 sin θ cos θ Finally we compute the complete transformation matrix as: T 0 S = T G S T C G T O As a sanity check we note that it makes sense that our x coordinate is a slightly larger since the object was off to the right of the car. We also note that the Y coordinate of the object should be slightly less than that of the car given the -45 degree angle relative to the state plane coordinate system. Also notice that because of our choice of GPS Z = 0, the Z coordinate of the object is 1.5 meters into the ground.