Question

Assume the coordinate frame assignments shown in the figure for the Rhino robot used in the lab and the transformation matrix from the base frame to the end effector frame Ts is as given below Suppose we want to use the Rhino for a painting application where the end-effector should point at 4j degrees from the horizontal. In this case, we would need to align the end-effector frame Z axis (Z4) at 135 degrees to Z axis Given the desired end effector position coordinates, x, y, and z, write four equations relating joint angles, 0,, θ2.93 and θ4 to end effector coordinates and robot link parameters. (20 points) Solve the four equations for the four joint variables θι, θ-,83 and θ4 in terms of end effector coordinates x, y, and z and link parameters. (20 points) a. Z X X) b. a3 23

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Answer #1

Solution:

.......................................................................................................................................................

The equations are:

Take the last column of the matrix. It represents position vector of the end effector.

x = a_3 C_1 C_{23}+a_2 C_1 C_2

y = a_3 S_1 C_{23}+a_2 S_1 C_2

3 523

If Z-axis of End Effector makes 1350 with Base X-axis

Angle with Base Z axis is 360 - 135 - 90 = 1350

-Cos( heta_2+ heta_3+ heta_4) = Cos(135)

Square and add x, y, z:

x^2+y^2 = a_3 ^2 C_{23} ^2(C_1 ^2 + S_ 1 ^2)+a_2 ^2 C_{2} ^2(C_1 ^2 + S_ 1 ^2)+2a_3a_2(C_1^2C_2C_{23}+S_1^2C_2C_{23})

x^2+y^2 = a_3 ^2 C_{23} ^2+a_2 ^2 C_{2} ^2+2a_3a_2 C_2C_{23}

3 523

z^2 = (a_3 S_{23})^2+(a_2 S_2)^2+2*a_3 S_{23}*a_2 S_2

x^2+y^2+z^2 = a_3 ^2 C_{23} ^2+a_2 ^2 C_{2} ^2+2a_3a_2 C_2C_{23}+ (a_3 S_{23})^2+(a_2 S_2)^2+2 a_2 a_3 S_{23} S_2

x^2+y^2+z^2 = a_3 ^2 +a_2 ^2 +2 a_2 a_3(C_{23} C_2+ S_{23} S_2)

x^2+y^2+z^2 = a_3 ^2 +a_2 ^2 +2 a_2 a_3 C_3

2a2a3

з = cos 2a2a3

Substituting heta_3 in

3 523 You get heta_2

Substituting heta_2 and heta_3   in

x = a_3 C_1 C_{23}+a_2 C_1 C_2

You get heta_1

You get heta_4 by using the equation:

-Cos( heta_2+ heta_3+ heta_4) = Cos(135)

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