DH Parameter for OpenManipulator X from Robotis

Question

I try to solve forward kinematics problem for OpenManipulator X from Robotis by using standard DH convention. The figure and dimension of the robot like this pic

I try to find DH parameter and got result like this

To match frame assignment requirement, I move frame 2 to the left 0.024m. But when I verify the forward kinematics result, the position of end effector is wrong when I move joint 3. For example if I move joint 3 to 90 degree, the position of end effector should be X = 0.024m, Y = 0, and Z = 0.455M. But I got from the software X = 0, Y = 0, and Z = 0.455. It is because of I move the centre of frame 3 to the left. But if I dont move it, the DH frame assignment rule doesn't match.

How to figure out this problem?

Here is my code to find forward kinematics

def std_DH(theta, alpha, a, d):
    DH = np.array([[np.cos(theta),  -np.sin(theta)*np.cos(alpha),   np.sin(theta)*np.sin(alpha),    a*np.cos(theta)],
                   [np.sin(theta),  np.cos(theta)*np.cos(alpha),    -np.cos(theta)*np.sin(alpha),   a*np.sin(theta)],
                   [0,              np.sin(alpha),                  np.cos(alpha),                  d],
                   [0,              0,                              0,                              1]])
    return DH

def forward_kinematics(q1, q2, q3, q4):
    _0T1 = std_DH(q1, np.pi/2, 0, 0.077)
    _1T2 = std_DH(q2+np.pi/2, 0, 0.128, 0)
    _2T3 = std_DH(q3-np.pi/2, 0, 0.124, 0)
    _3T4 = std_DH(q4, 0, 0.126, 0)
    _0T4 = _0T1.dot(_1T2).dot(_2T3).dot(_3T4)
    return _0T4

Is there any method to find forward kinematics without using DH convention?

Answer 1

DH parameterisation is always hard. Using Robotics Toolbox for MATLAB we can easily do the forward kinematics without DH parameters. First create a string of simple transforms which I took directly from your diagrams

>> s = 'Rz(q1)Tz(0.077)Ry(q2)Tz(0.128)Tx(0.024)Ry(q3)Tx(0.124)Ry(q4)Tx(0.126)';

which describes the kinematic chain with length constants and variable angles represented by qj. Now the FK is simply

>> trchain(s, [0 0 0 0])
ans =
    1.0000         0         0    0.2740
         0    1.0000         0         0
         0         0    1.0000    0.2050
         0         0         0    1.0000

where the values in the second argument are substituted for the joint angles, in this case all zero. The resulting SE(3) matrix shows a translation in the x- and z-directions which is correct.

You could use this to check your DH version.

My Toolbox also has an automated symbolic manipulation tool, not perfect, to convert a similar string format (all length parameters must be symbolic not numeric) to DH parameters

>> s = 'Rz(q1)Tz(L1)Ry(q2)Tz(L2)Tx(L3)Ry(q3)Tx(L4)Ry(q4)Tx(L5)'
>> DHFactor(s)
DHFactor: error: Incomplete factorization, no DH equivalent found

ans =

DH(q1, L1, 0, -90).DH(q2, 0, L3, 0).Ty(-L2).DH(q3, 0, L4, 0).DH(q4, 0, L5, 90)

but it fails for your example. The translation L2 hasn't been assigned to any DH term. This is more likely a problem with my code than your mechanism.

Answer 2

Not sure if it still helps, but for both traditional and modified D-H convention, you need your Z-axis to be at the direction of the joint axis. Your frame 2 did not comply to this requirement. This probably explains the missing link parameter 0.024m.

Answer 3

I struggled with the same thing. I solved it rotating the second frame to a fixed angle, and using the 0.130 m distance. I think this satisfies the DH convention.

Answer 4

Actually, your forward kinematic solution is correct. You made a silly mistake in the code. While definining _2T3 you used a as 0.124 instead of 0.148 (as you mentioned in the DH parameters).

Corrected Code would be

def forward_kinematics(q1, q2, q3, q4):
    _0T1 = std_DH(q1, np.pi/2, 0, 0.077)
    _1T2 = std_DH(q2+np.pi/2, 0, 0.128, 0)
    _2T3 = std_DH(q3-np.pi/2, 0, 0.148, 0)
    _3T4 = std_DH(q4, 0, 0.126, 0)
    _0T4 = _0T1.dot(_1T2).dot(_2T3).dot(_3T4)
    return _0T4

I checked this and everything is working fine.