Kinematic and Singularity Analysis of 10 DOF Lower Body of Humanoid Robot

Now, the individual transformation matrices that specify the relationship between adjacent links as below.

$T_{1}^{0}=\operatorname{Rot}\left(z, \theta_{1}\right) \operatorname{Trans}\left(0,0, l_{1}\right) \operatorname{Trans}(0,0,0) \operatorname{Rot}(x, 90)$     (1)

$T_{2}^{1}=\operatorname{Rot}\left(z, \theta_{2}-90\right) \operatorname{Trans}\left(0,0, l_{2}\right) \operatorname{Trans}(0,0,0) \operatorname{Rot}(x,-90)$    （2）

$T_{3}^{2}=\operatorname{Rot}\left(z, \theta_{3}-90\right) \operatorname{Trans}(0,0,0) \operatorname{Trans}\left(0,0, l_{3}\right) \operatorname{Rot}(x, 90)$     (3)

$T_{4}^{3}=\operatorname{Rot}\left(z, \theta_{4}\right) \operatorname{Trans}(0,0,0) \operatorname{Trans}\left(0,0, l_{4}\right) \operatorname{Rot}(x, 0)$    (4)

$T_{5}^{4}=\operatorname{Rot}\left(z, \theta_{5}\right) \operatorname{Trans}(0,0,0) \operatorname{Trans}\left(0,0, l_{5}\right) \operatorname{Rot}(x, 0)$     (5)

where, $i-1_{T_{i}}$ is a general link transformation matrix relating the i-th coordinate frame to the (i-1)th coordinate frame. The workspace created by the right foot is obtained as follows:

$T_{5}^{0}=\left[\begin{array}{llll}r_{11} & r_{12} & r_{13} & r_{14} \\ r_{21} & r_{22} & r_{23} & r_{24} \\ r_{31} & r_{32} & r_{33} & r_{34} \\ r_{41} & r_{42} & r_{43} & r_{44}\end{array}\right]$    (6)

3.1 Inverse kinematics

A humanoid robot control requires knowledge of the toe position and orientation for the instantaneous location of each joint as well as knowledge of the joint displacement required to place the foot at a new location. Closed-form solution, joint displacement is determined as explicit functions of the position and orientation of the toe. The closed-form in the present context means a solution method based on an analytical algebraic or kinematic approach. In the present model, five unknown joint parameters are to be determined [18].

The overall transformation matrix $T_{5}^{0}$ and end effector foot matrix $T_{\text {foot }}$ represent the same transformations.

$T_{f \circ o t}=\left[\begin{array}{cccc}n_{x} & o_{x} & a_{x} & p_{x} \\ n_{y} & o_{y} & a_{y} & p_{y} \\ n_{z} & o_{z} & a_{z} & p_{z} \\ 0 & 0 & 0 & 1\end{array}\right]=T_{5}^{0}$    (7)

where n, o, a is the unit vector along the normal, sliding, and approach. The inverse kinematics o the system helps to take the static gait and dynamic gait during walking and the running of the humanoid robot. The step size of the humanoid robot depends on the distance between the object and the robot.

Applying the closed-form solution and analytical algebraic approach resulted in no solution. To obtain the solution for the joint displacement.

It is possible to view the description of tool frame {5}, for frame {1}, the foot endpoint frame, along two different paths.

Path1: frame{0)------frame{4}--------frame{5}

Along this path, the transformation can be obtained as $T_{5}^{0}$.

$T_{5}^{0}=T_{1}^{0} T_{4}^{3} T_{5}^{4}=\left[\begin{array}{cccc}n_{x} & o_{x} & a_{x} & p_{x} \\ n_{y} & o_{y} & a_{y} & p_{y} \\ n_{z} & o_{z} & a_{z} & p_{z} \\ 0 & 0 & 0 & 1\end{array}\right]$     (8)

Since $\theta_{1}, \theta_{4}$ and $\theta_{5}$ are unknown. The inverse of a matrix $T_{5}^{4}$ can be computed. Substituting these values and post multiplying computed.

$T_{5}^{0} T_{5}^{4-1}=T_{1}^{0} T_{4}^{3}$    (9)

$\left[\begin{array}{ccccc}n_{x} \cos \theta_{5}-o_{x} \sin \theta_{5} & o_{x} \cos \theta_{5}+n_{x} \sin \theta_{5} & a_{x} & p_{x}-l_{5} n_{x} \\ n_{y} \cos \theta_{5}-o_{y} \sin \theta_{5} & o_{y} \cos \theta_{5}+n_{y} \sin \theta_{5} & a_{y} & p_{y}-l_{5} n_{y} \\ \sin \theta_{4} & \cos \theta_{4} & 0 & P_{z}-l_{5} n_{z} \\ 0 & 0 & 0 & 1\end{array}\right]=\left[\begin{array}{cccc}\cos \theta_{1} \cos \theta_{4} & -\cos \theta_{1} \sin \theta_{4} & \sin \theta_{1} & l_{4} \cos \theta_{1} \cos \theta_{4} \\ \cos \theta_{1} \sin \theta_{4} & -\sin \theta_{1} \sin \theta_{4} & -\cos \theta_{1} & l_{4} \cos \theta_{4} \sin \theta_{1} \\ \sin \theta_{4} & \cos \theta_{4} & 0 & l_{1}+l_{4} \sin \theta_{4} \\ 0 & 0 & 0 & 1\end{array}\right]$

$\theta_{1}=\arctan \left(\frac{p_{y}-l_{5} n_{y}}{p_{x}-l_{5} n_{x}}\right)$

$\theta_{4}=\arctan \left(\frac{p_{z}-l_{5} n_{z}-l_{1}}{\left.\left.\pm \sqrt{(}\left(p_{x}-l_{5} n_{x}\right)^{2}\right)+\left(p_{y}-l_{5} n_{y}\right)^{2}\right)}\right)$

Since $\theta_{5}$ is unknown. The inverse of a matrix $T_{4}^{3}$ can be computed. Substituting these values and post multiplying with the matrix $T_{5}^{0} T_{5}^{4-1}$ can be computed.

$T_{5}^{0} T_{5}^{4-1} T_{4}^{3^{-1}}=T_{1}^{0}$     (10)

$\left[\begin{array}{lllll}n_{x} \cos \theta_{45}-o_{x} \cos \theta_{45} & o_{x} \cos \theta_{45}+n_{x} \sin \theta_{45} & a_{x} & p_{x}-l_{5} n_{x}-l_{4} n_{x} \cos \theta_{5}+l_{4} o_{x} \sin \theta_{5} \\ n_{y} \cos \theta_{45}-o_{y} \sin \theta_{45} & o_{y} \cos \theta_{45}+n_{y} \sin \theta_{45} & a_{y} & p_{y}-l_{5} n_{y}-l_{4} n_{y} \cos \theta_{5}+l_{4} o_{y} \sin \theta_{5} \\ n_{z} \cos \theta_{45}-o_{z} \sin \theta_{45} & o_{z} \cos \theta_{45}+n_{z} \sin \theta_{45} & a_{z} & p_{z}-l_{5} n_{z}-l_{4} n_{z} \cos \theta_{5}+l_{4} o_{z} \sin \theta_{5}\end{array}\right]=\left[\begin{array}{cccc}\cos \theta_{1} & 0 & \sin \theta_{1} & 0 \\ \sin \theta_{1} & 0 & -\cos \theta_{1} & 0 \\ 0 & 0 & 0 & l_{1} \\ 0 & 0 & 0 & 1\end{array}\right]$

$\theta_{5}=\arctan \left(\frac{\left(p_{y}+l_{5} n_{y}\right) l_{4} n_{x}-\left(p_{x}+l_{5} n_{x}\right) l_{4} n_{y}}{\left(p-x+l_{5} n_{x}\right) l_{4} o_{y}+\left(p_{y}+l_{5} n_{y}\right) l_{4} o_{x}}\right)$

Path2: frame{0)------frame{1}--------frame{2}-------frame{3}-----Frame{4}------ frame{5}

$\theta_{2}$ and $\theta_{3}$ were obtained by post multiplying of $T_{4}^{3^{-1}}$ and $T_{5}^{4^{-1}}$ with the matrix $T_{5}^{0-1}$.

$$\begin{gathered}\theta_{2}=\arctan \left(\frac{-\left(o_{z} \cos \theta_{45}+n_{z} \sin \theta_{45}\right)}{\left.\pm \sqrt{(} o_{x} \cos \theta_{45}+n_{x} \sin \theta_{45}\right)^{2}+\left(o_{y} \cos \theta_{45}+n_{y} \sin \theta_{45}\right)^{2}}\right) \\\theta_{3}=\arctan \left(\frac{n_{z} \cos \theta_{45}-o_{z} \sin \theta-45}{-a_{z}}\right)\end{gathered}$$

3.2 Jacobian

The transformation from joint velocities to the end effector velocity is described by a matrix, called Jacobian. The Jacobian matrix, which is dependent on manipulator configuration is a linear mapping from velocities in joint space to velocities in Cartesian space. The Jacobian is one of the most important tools for the characterization of differential motions of the manipulator. The mapping between differential changes is linear and can be expressed as:

$V_{e}(t)=J(q) \dot{q}$    (11)

where, $V_{e}(t)$ is the cartesian velocity vector and $J(q)=6 \times 1$ manipulator Jacobian matrix and $\dot{q}=n \times 1$ vector of n joints.

Eq. (11) can be written in column vectors of the Jacobian, that is,

$V_{e}(t)=\left[\begin{array}{llllll}J_{1}(q) & J_{2}(q) & J_{3}(q) & J_{4}(q) & \mathrm{J}(5) & J_{6}(q)\end{array}\right]$     (12)

The velocity can be described in the terms of linear velocity as well as angular velocity.

$V_{e}(t)=\left[\begin{array}{c}v \\ w\end{array}\right]=\left[\begin{array}{l}\dot{d} \\ \dot{\theta}\end{array}\right]=J(q) \dot{q}$   (13)

The manipulator Jacobian can be described as follows:

$J_{i}(q)=\left[\begin{array}{c}P_{i-1} \times\left(P_{n}^{i-1}\right) \\ P_{i-1}\end{array}\right]$    (14)

The origin of frame {n} at the end effector i.e. toe is:

$O_{n}=\left[\begin{array}{l}0 \\ 0 \\ 0 \\ 1\end{array}\right]$

This applies to the origin of any frame i.e. for any value of n. The above vector equation can be written as:

$P_{n}^{i-1}=T_{n}^{0}-T_{i-1}^{0} O_{n}$     (15)

Each column of the Jacobian matrix is computed separately and all the columns are combined from the total Jacobian matrix.

$J_{1}(q)=\left[\begin{array}{c}P_{0} \times\left(P_{5}^{0}\right) \\ P_{0}\end{array}\right], J_{2}(q)=\left[\begin{array}{c}P_{1} \times\left(P_{5}^{0}\right) \\ P_{1}\end{array}\right]$

$J_{3}(q)=\left[\begin{array}{c}P_{2} \times\left(P_{5}^{0}\right) \\ P_{2}\end{array}\right]$

$J_{4}(q)=\left[\begin{array}{c}P_{3} \times\left(P_{5}^{0}\right) \\ P_{3}\end{array}\right], J_{5}(q)=\left[\begin{array}{c}P_{4} \times\left(P_{5}^{0}\right) \\ P_{4}\end{array}\right]$

The Jacobian for the manipulator for the right leg is as follows:

$J=\left[\begin{array}{lllll}J_{1} & J_{2} & J_{3} & J_{4} & J_{5}\end{array}\right]$$=\left[\begin{array}{lllll}J_{11} & J_{21} & J_{31} & J_{41} & J_{51} \\ J_{12} & J_{22} & J_{32} & J_{42} & J_{52} \\ J_{13} & J_{23} & J_{33} & J_{43} & J_{53} \\ J_{14} & J_{24} & J_{34} & J_{44} & J_{54} \\ J_{15} & J_{25} & J_{35} & J_{45} & J_{55}\end{array}\right]$     (16)

3.3 Singularity posture

At a singular configuration [19], the humanoid robot loses one or more degrees of freedom. The singular configuration is classified into two categories based on the location of the leg in the workspace. The boundary singularities occur when the leg is on the boundary of the workspace i.e. the leg is either fully stretched out or fully retracted. The interior singularities occur when the leg is inside the reachable workspace of the humanoid robot. They can occur anywhere in the reachable workspace. Other than the boundary singularities and configurations singularities, the actuation singularities affect the motion of the humanoid robot. The actuation singularities do not allow the humanoid robot to take the higher step so that the balancing of the body can be maintained throughout the movement. The computation of internal singularities can be carried out by analyzing the rank of the Jacobian matrix. The Jacobian matrix loses its rank and becomes ill-conditioned at values of joint variables, that is,

$\operatorname{det}(J)=0$   (17)

Since the matrix consists of a $6 \times 5$ matrix array, it is not possible to obtain the determinant of the matrix. In this work, some assumptions had been made for finding the determinant of the Jacobin matrix for finding the singularity posture of the lower body of a humanoid robot. Considering the last three rows and first three columns of the Jacobean matrix. The determinant was obtained $\operatorname{det}(J)=\cos \theta_{2}$.

Considering the first three rows and last three-column of the Jacobin matrix $\operatorname{det}(J)=0.008 \sin \theta_{4}-0.004 \sin \theta_{5}-$ $0.008 \cos \theta_{4} \sin \theta_{5}-0.004 \cos \theta_{5} \cos \theta_{5} \sin \theta_{4}-$ $0.008 \cos \theta_{4} \cos \theta_{5} \sin \theta_{5}$.

Considering the last two rows and last two-column of the Jacobean matrix $\operatorname{det}(J)=-0.004\left(\cos \theta_{2} \cos \theta_{3} \sin \theta_{5}\right)$.

The advantage of the singularity posture of the joint provides safety to the system. If any complicated case arises, in that condition the structures of the system come to the locking state and provide safety to the system. The singularity posture is avoided by the controller and does not aloe the link to achieve a particular point. The singularity point can be computed by taking the determinant of the equation and substituting it to zero.