Forward Kinematic and Jacobian Matrix for the Prosthetic Human Finger Actuated by Links

Prosthetics play an important part in compensating amputees in terms of form and appearance once and in terms of their ability to perform tasks in acceptable proportions again, the manufacture of prosthetics began around the sixteenth century, and they were metal sheets linked to gears in the palm of the hand, Research and technological development of the prosthetic hand coincided during and after World War I, as a number of companies in London began supplying mechanical hands [1], After the World War II, work began with the manufacture of prosthetic limbs, which were characterized by being limbs that worked with the power-body of the amputee and needed effort, or were cosmetic limbs more than they had functional characteristics. so many prosthetics companies have gone on to improve the prosthetics' ability to perform in a large technique, with different capabilities and at several cost e.g. Be Bionic hand and Michelangelo hand [2]. The Revolutionizing Prosthetics 2009 program is a Defence Advanced Research Projects Agency task aimed to develop a neutrally controlled upper prosthetic limb [3], Researchers have made significant progress in this area, and many of these species are designed to reach the desired level at acceptable cost, It differs according to the operating mechanisms by which these prosthetics hand operate. A hand called Khefa, which used artificial muscle wires instead of the traditional motor, with a small size and easy to operate without noise [4]. A robotic hand Sandy was designed, equipped with touch and temperature sensors, optical sensors, and fingers connected to the palm of the hand by a magnet [5]. Design a prosthetic hand that is structurally similar to a natural hand made of artificial bone and rubber joints [6]. A continuum differential mechanism used in prosthetic hand design with only one actuator [7]. A prosthetic hand was designed with a biologically-inspired parallel operating system that includes two types of actuators, the (DC) and (SMA) actuators [8]. Others have developed prosthetics available on the market, development of the bionic robotic finger by computer simulation of movement [9]. Improving the finger mechanism of the Larm prosthetic hand [10].

In this paper, a prosthetic hand with operative fingers was designed by embedding links in their structures; the finger structural properties were analysed and the kinematic model was created for the finger. And studied the forward kinematic for the prosthetic finger and the Jacobin matrix was calculated. The results were validated using MATLAB, and the finger movement was simulated using the Solidwork program 2018 to provide theoretical support to the designed hand.

3.1 Forward kinematic for the index finger

As per Table 1. And concluded parameters of (D.H) method, it is possible to find (Homogenous Transformation Matrix) between each two adjacent links to reach to the (H.T.M) which connects the finger base links to its tip and as shown in Eq. (1).

$H_{n}^{n-1}=\left[\begin{array}{cccc}c \theta_{n} & -s \theta_{n} c \alpha_{n} & s \theta_{n} s \alpha_{n} & r_{n} c \theta_{n} \\ s \theta_{n} & c \theta_{n} c \alpha_{n} & -c \theta_{n} s \alpha_{n} & r_{n} s \theta_{n} \\ 0 & s \alpha_{n} & c \alpha_{n} & d_{n} \\ 0 & 0 & 0 & 1\end{array}\right]$      (1)

When (s) refer to (sin) and (c) refer to $(\cos )$, so to find (H.T.M) between the base of the finger and the tip of it, we must find $H_{1}^{0}, H_{2}^{1}, H_{3}^{2}, H_{4}^{3}, H_{5}^{4}$.

When:

$H_{5}^{0}=H_{1}^{0} \cdot H_{2}^{1} \cdot H_{3}^{2} \cdot H_{4}^{3} \cdot H_{5}^{4}$     (2)

From Eq. (1). When (n=1),

$H_{1}^{0}=\left[\begin{array}{cccc}c \theta_{1} & -s \theta_{1} & 0 & a_{2} c \theta_{1} \\ s \theta_{1} & c \theta_{1} & 0 & a_{2} s \theta_{1} \\ 0 & 0 & 1 & a_{1} \\ 0 & 0 & 0 & 1\end{array}\right]$      (3)

For (n=2),

$H_{2}^{1}=\left[\begin{array}{cccc}c \theta_{2} & -s \theta_{2} & 0 & a_{7} c \theta_{2} \\ s \theta_{2} & c \theta_{2} & 0 & a_{7} s \theta_{2} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$     (4)

For (n=3),

$H_{3}^{2}=\left[\begin{array}{cccc}c \theta_{3} & -s \theta_{3} & 0 & a_{8} c \theta_{3} \\ s \theta_{3} & c \theta_{3} & 0 & a_{8} s \theta_{3} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$     (5)

For (n=4),

$H_{4}^{3}=\left[\begin{array}{cccc}c \theta_{4} & -s \theta_{4} & 0 & a_{5} c \theta_{4} \\ s \theta_{4} & c \theta_{4} & 0 & a_{5} s \theta_{4} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$      (6)

For (n=5),

$H_{5}^{4}=\left[\begin{array}{cccc}c \theta_{5} & -s \theta_{5} & 0 & a_{6} c \theta_{5} \\ s \theta_{5} & c \theta_{5} & 0 & a_{6} s \theta_{5} \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]$     (7)

when: $\alpha_{1,2,3,4,5}=0, c \alpha_{1,2,3,4,5}=1, s \alpha_{1,2,3,4,5}=0$,

From Eq. (2). We can get $H_{5}^{0}$ :

$H_{5}^{0}=\left[\begin{array}{cccc}A_{11} & -A_{12} & 0 & A_{14} \\ A_{21} & -A_{22} & 0 & A_{24} \\ 0 & 0 & 1 & a_{1} \\ 0 & 0 & 0 & 1\end{array}\right]$     (8)

$A_{11}=\left(m c \theta_{4}+n s \theta_{4}\right) c \theta_{5}+\left(n c \theta_{4}-m s \theta_{4}\right) s \theta_{5}$

$A_{21}=\left(l c \theta_{4}+f s \theta_{4}\right) c \theta_{5}+\left(f c \theta_{4}-l s \theta_{4}\right) s \theta_{5}$

$A_{12}=\left(m c \theta_{4}+n s \theta_{4}\right) s \theta_{5}-\left(n c \theta_{4}-m s \theta_{4}\right) c \theta_{5}$

$A_{22}=\left(l c \theta_{4}+f s \theta_{4}\right) s \theta_{5}+\left(f c \theta_{4}-l s \theta_{4}\right) c \theta_{5}$

$A_{14}=r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g$

$A_{24}=w a_{6} c \theta_{5}+y a_{6} s \theta_{5}+h$

When:

$m=\left(c \theta_{1} c \theta_{2}-s \theta_{1} s \theta_{2}\right) c \theta_{3}-\left(c \theta_{1} s \theta_{2}\right.$

$\left.+s \theta_{1} c \theta_{2}\right) s \theta_{3}$

$n=\left(s \theta_{1} s \theta_{2}-c \theta_{1} c \theta_{2}\right) s \theta_{3}-\left(c \theta_{1} s \theta_{2}\right.$

$\left.+s \theta_{1} c \theta_{2}\right) c \theta_{3}$

$\begin{aligned} l=\left(s \theta_{1} c \theta_{2}+c \theta_{1} s \theta_{2}\right) c \theta_{3} & \\ &+\left(c \theta_{1} c \theta_{2}-s \theta_{1} s \theta_{2}\right) s \theta_{3} \end{aligned}$

$f=\left(c \theta_{1} c \theta_{2}-s \theta_{1} s \theta_{2}\right) c \theta_{3}-\left(s \theta_{1} c \theta_{2}\right.$

$\left.+c \theta_{1} s \theta_{2}\right) s \theta_{3}$

$r=\left(m c \theta_{4}+n s \theta_{4}\right), t=\left(n c \theta_{4}-m s \theta_{4}\right)$,

$w=\left(l c \theta_{4}+f s \theta_{4}\right) \quad y=\left(f c \theta_{4}-l s \theta_{4}\right)$,

$g=\left(m a_{5} c \theta_{4}+n a_{5} s \theta_{4}+A\right)$

$h=\left(l a_{5} c \theta_{4}+f a_{5} s \theta_{4}+B\right)$

$A=\left(c \theta_{1} c \theta_{2}-s \theta_{1} s \theta_{2}\right) a_{8} c \theta_{3}-\left(c \theta_{1} s \theta_{2}\right.$

$\left.+s \theta_{1} c \theta_{2}\right) a_{8} s \theta_{3}+\left(a_{7} c \theta_{2} c \theta_{1}\right.$

$\left.-a_{7} s \theta_{1} s \theta_{2}+a_{2} c \theta_{1}\right)$

$B=\left(s \theta_{1} c \theta_{2}+c \theta_{1} s \theta_{2}\right) a_{8} c \theta_{3}+\left(c \theta_{1} c \theta_{2}\right.$

$\left.-s \theta_{1} s \theta_{2}\right) a_{8} s \theta_{3}+\left(a_{7} c \theta_{2} s \theta_{1}\right.$

$\left.+a_{7} c \theta_{1} c \theta_{2}+a_{2} s \theta_{1}\right)$

3.2 The Jacobian matrix for the index finger

The [J] matrix was established for the determining the speed of the tip of the finger, since this matrix represents the relationship between the speed of the links and the speed of the top of the finger [10]; when the [J] matrix is:

$\left[\begin{array}{c}\dot{x} \\ \dot{y} \\ \dot{z} \\ \dot{\omega}_{x} \\ \dot{\omega}_{y} \\ \dot{\omega}_{z}\end{array}\right]=J\left[\begin{array}{c}\dot{\theta_{1}} \\ \vdots \\ \vdots \\ \vdots \\ \dot{\theta_{n}}\end{array}\right]$     (9)

When (θ) refer to the joint variable this is mean that the Jacobin matrix has (number of columns equal to number of joints in the manipulator), also the number of rows of the Jacobin matrix always =6; in this paper the design finger has (5) joints, so the [J] matrix will consist of (6) rows and (5) columns, and equal to the total number of joints when i= the joint no.

The [J] matrix consists of two parts, one for the linear velocity of the fingertip and another part of the rotational velocity of the fingertip as shown in the Eq. (10).

$[J]=\left[\begin{array}{l}J_{v} \\ J_{\omega}\end{array}\right]$    (10)

$\left[J_{v}\right]$ Refer to the linear part; $\left[J_{\omega}\right]$ Refer to the rotating part of the Jacobin matrix, we can get the linear and rotational relation of the revolute joint that used in the design finger from the Table 2 .

Table 2. Joint model formula

Thus, the Jacobian matrix will be in the following form:

$[J]=\left[\begin{array}{ccccc}J_{v 1} & J_{v 2} & J_{v 3} & J_{v 4} & J_{v 5} \\ J_{\omega 1} & J_{\omega 2} & J_{\omega 3} & J_{\omega 4} & J_{\omega 5}\end{array}\right]$     (11)

When:

$J_{v 1}=R_{0}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \times\left(o_{5}^{0}-o_{0}^{0}\right)$,

$J_{v 2}=R_{1}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \times\left(o_{5}^{0}-o_{1}^{0}\right)$

$J_{v 3}=R_{2}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \times\left(o_{5}^{0}-o_{2}^{0}\right)$

$J_{v 4}=R_{3}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \times\left(o_{5}^{0}-o_{3}^{0}\right)$

$J_{v 5}=R_{4}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] \times\left(o_{5}^{0}-o_{4}^{0}\right)$

$J_{\omega 1}=R_{0}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], J_{\omega 2}=R_{1}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right], J_{\omega 3}=R_{2}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]$

$J_{\omega 4}=R_{3}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] J_{\omega 5}=R_{4}^{0}\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right]$

Compensation value of $\left(R_{i}^{0}\right)$ refers to a rotational matrix of each joint that index finger consist of and it is meat the rotational of frame (0) to the frame (i), we can get it from a homogenous transformation matrix in the previous section, also compensation values of $\left(O_{i}^{0}\right)$ that refers to the displacement vector from the center of the frame (0) to the center of the frame (i), So we can find the Jacobian matrix as equation below:

$\left[\begin{array}{c}\dot{x} \\ \dot{y} \\ \dot{z} \\ \dot{\omega}_{x} \\ \dot{\omega}_{y} \\ \dot{\omega}_{z}\end{array}\right]=\left[\begin{array}{ccccc}J_{11} & J_{12} & J_{13} & J_{14} & J_{15} \\ J_{21} & J_{22} & J_{23} & J_{24} & J_{25} \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1\end{array}\right]\left[\begin{array}{c}\dot{\theta}_{1} \\ \dot{\theta}_{2} \\ \dot{\theta}_{3} \\ \dot{\theta}_{4} \\ \dot{\theta}_{5} \\ \dot{\theta}_{6}\end{array}\right]$      (12)

when:

$J_{11}=-\left(w a_{6} c \theta_{5}+y a_{6} s \theta_{5}+h\right)$,

$J_{12}=-\left(w a_{6} c \theta_{5}+y a_{6} s \theta_{5}+h-a_{2} s \theta_{1}\right)$

$J_{13}=w a_{6} c \theta_{5}+y a_{6} s \theta_{5}+h-\left(a_{7} c \theta_{2} s \theta_{1}\right.$

$\left.-a_{7} c \theta_{1} s \theta_{2}+a_{2} s \theta_{1}\right)$

$J_{14}=-\left(w a_{6} c \theta_{5}+y a_{6} s \theta_{5}+h-B\right)$

$J_{15}=l a_{5} c \theta_{4}+f a_{5} s \theta_{4}+B-\left(w a_{6} c \theta_{5}+y a_{6} s \theta_{5}\right.$

$+h)$

$J_{21}=r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g$

$J_{22}=r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g-a_{2} c \theta_{1}$

$J_{23}=r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g-\left(a_{7} c \theta_{2} c \theta_{1}+\right.$

$\left.a_{7} s \theta_{2} s \theta_{1}+a_{2} c \theta_{1}\right)$

$J_{24}=r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g-A$

$\begin{aligned} J_{25}=& r a_{6} c \theta_{5}+t a_{6} s \theta_{5}+g \\ &-\left(m a_{5} c \theta_{4}+n a_{5} s \theta_{4}+A\right) \end{aligned}$