**7. Direct kinematics for T-S robot**

For the direct kinematics analysis, point *o*, located in the origin of the coordinate of *A*1, reference system is regarded as the primary point and *o*′, point in the origin of the coordinate joined to the center of *C* point, is considered to be the final point.

The number of unknown parameters in each path is two. As a result, the total number of unknown parameter for each path is 6, but since the number of degree of freedom for each kinematics chains 3-PRP equals 9, the number of each path of the unknown parameters is 3, that is, the three parameters are related to other parameters that are geometrically obtained. In this case, the robotic degree of freedom is 3. If *S*11, *S*21, *S*<sup>31</sup> parameters are considered to be the inputs, the unknown parameters will be *S*13, *S*23, *S*33 and *θ*11, *θ*21, *θ*31.

What is stated is for the simplification and avoidance of mistakes in the number of direction index to the power of *k*. Therefore, to obtain the unknown parameters, final point transfor‐ mation matrix *o*′ is obtained in proportion to the primary point *o* and then relation (5, 11) is used.

$$\mathbb{E}\left[P\right]\_i = H^{k}\_{o,o}\mathbb{E}\left[P\right]\_{i\times 1} \tag{5}$$

$$H\_{o,o'}^{k} = H\_{0,1}^{K} \times H\_{1,2}^{K} \times H\_{2,3}^{K} \times H\_{3,4}^{K} \times \dots \times H\_{l,l+1}^{K} \tag{6}$$

$$H\_{o,o'}^1 = H\_{o,o'}^1 = H\_{o,o'}^3 \tag{7}$$

$$\mathbb{E}\left[P\_{O'}\right]\_{(O)(for\,\,pah1)} = \left[P\_{O'}\right]\_{(O)(for\,\,pah2)} = \left[P\_{O'}\right]\_{(O)(for\,\,pah3)}\tag{8}$$

$$\begin{aligned} \left[P\_i\right] &= \left[\left.H\_{i,i+1}^k\right] \cdot \left[P\_{i+1}\right]\right] \\ \left[P\_i\right] &= \left[x\_i; y\_i; z\_i; 1\right] \end{aligned} \tag{9}$$

In the above relation, [*P*]*<sup>i</sup>* is the position in relation to coordinate axes joined to member *i*, and [*P*]*<sup>i</sup>* + 1 is the position in relation to coordinate axes joined to member *i* + 1. The robot has three kinematics chains, and three different transformation matrixes are achieved. In this way, the three matrixes should be equalized according to relation (5). The unknown parameters are obtained through the solution of these equations.

Point *o*′ coordinate (*P*<sup>3</sup> origin of coordinate) located in the center of the star in *P*2 coordinate is as follows. Point *o*′ coordinates in *P*0 coordinate is as follows:

Kinematic Analysis of the Triangle-Star Robot with Telescopic Arm and Three Kinematics Chains as T-S Robot (3-PRP) http://dx.doi.org/10.5772/64556 77

$$\mathbb{E}\left[P\_{o'}\right]\_{(2)} = \left[P\_i\right] = \left[\mathbf{x}\_o; \mathbf{y}\_o; \mathbf{z}\_o; \ 1\right]\_{(2)} = \left[\mathbf{0}; \mathbf{0}; S\_2^k; 1\right]\_{(2)}\tag{10}$$

$$\begin{aligned} \mathbf{[[P\_{o'}]\_{(0)} = \begin{bmatrix} \mathbf{x\_{o'}} \\ \mathbf{y\_{o'}} \\ \mathbf{z\_{o'}} \\ 1 \end{bmatrix}\_{(0)} \end{aligned} \qquad \begin{aligned} \mathbf{x\_{o'}} \\ \mathbf{y\_{o'}} \\ \mathbf{z\_{o'}} \\ \mathbf{1} \end{bmatrix}\_{(0)} = \left[ \mathbf{H}\_{0,1}^{K} \right] \times \left[ \mathbf{H}\_{1,2}^{K} \right] \times \left[ \mathbf{H}\_{2,3}^{K} \right] \times \begin{bmatrix} \mathbf{0} \\ \mathbf{0} \\ \mathbf{S\_{z}^{K}} \\ \mathbf{1} \end{bmatrix}\_{(2)} \end{aligned} \tag{11}$$

$$\begin{aligned} \;^1H\_{O,\mathcal{O}'} = \begin{bmatrix} -Sin\theta & -Cas\theta & 0 & -Sin\theta \times \mathbf{x} - Sin\theta \times \mathbf{S}\_3^\mathbf{l} + \mathbf{S}\_1^\mathbf{l} \\\;Cos\theta & -Sin\theta & 0 & Cos\theta \times \mathbf{x} + Cos\theta \times \mathbf{S}\_3^\mathbf{l} \\\;0 & 0 & 1 & B \\\;0 & 0 & 0 & 1 \end{bmatrix} \end{aligned} \tag{12a}$$

$$\boldsymbol{H}\_{O,O'}^{2} = \begin{bmatrix} \wp & -\Re & 0 & \wp \times \mathbf{x} + \wp \times S\_{\mathfrak{z}}^{2} - 0.5 \times S\_{\mathfrak{z}}^{2} + \boldsymbol{e} \\ \Re & \wp & 0 & \Re \times \mathbf{x} + \Re \times S\_{\mathfrak{z}}^{2} + 0.866 \times S\_{\mathfrak{z}}^{2} \\ \mathbf{0} & \mathbf{0} & \mathbf{1} & \mathbf{b} \\ \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{1} \end{bmatrix}$$

3 3 3 1 3 3 2 3 1 , .0.866 0.5 0.5 0.866 , 0 0.5 0 0.866 0 01 0 00 1 *O O Sin Cos Sin Cos xS S xS S <sup>H</sup> b* q q q q ¢ Â=- - Ã= é ù à - à ´ +à ´ + ´ ¢¢ ¢ ¢ ê ú à ´ + ´ + ´ ¢¢ ¢ ¢ <sup>=</sup> ë û (12b)

0 Ã=- ¢ .5 0.866 0.866 0.5 *Sin Cos Sin Cos* q q q q + =- - ¢

**7. Direct kinematics for T-S robot**

76 Recent Advances in Robotic Systems

used.

the center of *C* point, is considered to be the final point.

the inputs, the unknown parameters will be *S*13, *S*23, *S*33 and *θ*11, *θ*21, *θ*31.

For the direct kinematics analysis, point *o*, located in the origin of the coordinate of *A*1, reference system is regarded as the primary point and *o*′, point in the origin of the coordinate joined to

The number of unknown parameters in each path is two. As a result, the total number of unknown parameter for each path is 6, but since the number of degree of freedom for each kinematics chains 3-PRP equals 9, the number of each path of the unknown parameters is 3, that is, the three parameters are related to other parameters that are geometrically obtained. In this case, the robotic degree of freedom is 3. If *S*11, *S*21, *S*<sup>31</sup> parameters are considered to be

What is stated is for the simplification and avoidance of mistakes in the number of direction index to the power of *k*. Therefore, to obtain the unknown parameters, final point transfor‐ mation matrix *o*′ is obtained in proportion to the primary point *o* and then relation (5, 11) is

> [ ] [ ] [ ] ,1 1

+ + = × é ù ë û

In the above relation, [*P*]*<sup>i</sup>* is the position in relation to coordinate axes joined to member *i*,

[*P*]*<sup>i</sup>* + 1 is the position in relation to coordinate axes joined to member *i* + 1. The robot has three kinematics chains, and three different transformation matrixes are achieved. In this way, the three matrixes should be equalized according to relation (5). The unknown parameters are

Point *o*′ coordinate (*P*<sup>3</sup> origin of coordinate) located in the center of the star in *P*2 coordinate is

*k i ii i i i ii*

*PH P P xyz*

[] ;;; 1

obtained through the solution of these equations.

as follows. Point *o*′ coordinates in *P*0 coordinate is as follows:

, 1 [] [] *<sup>k</sup> P HP i oo i* = ¢ <sup>+</sup> (5)

<sup>113</sup> *HHH oo oo oo* ,,, ¢¢¢ = = (7)

<sup>=</sup> (9)

and

, 0,1 1,2 2,3 3,4 , 1 ....... *k KKK K K H HHHH H o o*¢ =´´´ *i i*<sup>+</sup> (6)

[ ] [ ] [ ] *OO O* ( )( . 1) *O for path* ( )( . 2) *O for path* ( )( . 3) *O for path PPP* ¢¢ ¢ = = (8)

$$\begin{aligned} \text{path(I)}: \begin{bmatrix} \mathbf{x}\_{o^{r}} \\ \mathbf{y}\_{o^{r}} \\ z\_{o^{r}} \\ 1 \end{bmatrix}\_{\text{(0)}} &= \begin{bmatrix} -\mathbf{x} \times \text{Sin}\theta - S\_{3}^{1} \times \text{Sin}\theta + S\_{1}^{1} \\ \mathbf{x} \times \text{Cos}\theta + S\_{3}^{1} \times \text{Cos}\theta \\ \mathbf{b} \\ 1 \end{bmatrix} \\\\ \text{path(2)}: \begin{bmatrix} \mathbf{x}\_{o^{r}} \\ \mathbf{y}\_{o^{r}} \\ z\_{o^{r}} \\ 1 \end{bmatrix}\_{\text{(0)}} &= \begin{bmatrix} \boldsymbol{\wp} \times \mathbf{x} + \boldsymbol{\wp} \times S\_{3}^{1} - 0.5 \times S\_{1}^{2} + \mathbf{e} \\ \boldsymbol{\mathfrak{R}} \times \mathbf{x} + \boldsymbol{\mathfrak{R}} \times S\_{3}^{2} + 0.866 \times S\_{1}^{2} \\ \mathbf{e} \\ \mathbf{b} \\ 1 \end{bmatrix} \end{aligned} \tag{13ab}$$

$$path. \text{(3)}: \begin{bmatrix} \mathbf{x}\_{\mathcal{O}} \\ \mathbf{y}\_{\mathcal{O}} \\ \mathbf{z}\_{\mathcal{O}'} \\ \mathbf{l} \end{bmatrix}\_{\text{(0)}} = \begin{bmatrix} \wp' \times \mathbf{x} + \wp' \times \mathbf{S}\_{\text{3}}^{\text{3}} + \mathbf{0}.\mathbf{5} \times \mathbf{S}\_{\text{1}}^{\text{3}} \\ \mathfrak{R}' \times \mathbf{x} + \mathfrak{R}' \times \mathbf{S}\_{\text{3}}^{\text{3}} + \mathbf{0}.866 \times \mathbf{S}\_{\text{1}}^{\text{3}} \\ \mathbf{b} \\ \mathbf{l} \end{bmatrix}\_{\text{(13c)}} \tag{13c}$$

### **8. The motion geometry of T-S Robot**

$$\begin{aligned} \phi\_3 + \phi\_5 &= 180, \ \phi\_5 = \frac{\pi}{2} + \theta\_{\mathcal{G}}, \ \phi\_3 = \frac{\pi}{2} + \theta\_2 \Rightarrow \theta\_2 = -\theta\_3, \phi\_1 + \phi\_2 = 180, \ \phi\_1 = \frac{\pi}{2} + \theta\_{1'}, \ \phi\_2 = \frac{\pi}{2} - \theta\_2 \Rightarrow \theta\_1 = \theta\_2, \\\ \theta\_1^1, \ \theta\_1^2, \ \theta\_1^3 = \theta\_{\mathcal{G}} \\$int \theta &= \frac{2t}{1 + t^2}, \ \operatorname{Cov} \theta = \frac{1 - t^2}{1 + t^2}, \ t = \tan\frac{\theta}{2} \end{aligned}$$

From the motion geometry of the T-S robot and then relations (5,13), the motion equation are achieved.

$$-0.7\text{Sx} \times \text{Sin}\theta - \text{S}\_3^\text{l} \times \text{Sin}\theta + \text{S}\_1^\text{l} + 0.866\text{x} \times \text{Cas}\theta - 0.5\text{S}\_3^\text{z} \times \text{Sin}\theta + 0.866\text{S}\_3^\text{z} \text{Cas}\theta + 0.5\text{S}\_1^\text{z} - e = 0 \tag{14a}$$

$$0.75\mathbf{x} \times \mathbf{C} \mathbf{s} \theta + \mathbf{S}\_3^\mathrm{l} \times \mathbf{C} \mathbf{s} \theta + 0.866\mathbf{x} \times \mathbf{S} \|\mathbf{u}\theta + 0.866\mathbf{S}\_3^\mathrm{l} \times \mathbf{S} \|\mathbf{u}\theta + 0.5\mathbf{S}\_3^\mathrm{l} \mathbf{C} \mathbf{s} \theta + 0.866\mathbf{S}\_\mathrm{l}^\mathrm{2} = \mathbf{0} \tag{14b}$$

$$-0.7\text{Sx} \times \text{Sin}\theta - S\_\text{\text{\textdegree}}^\text{l} \times \text{Sin}\theta + S\_\text{\text{\textdegree}}^\text{l} - 0.866\text{x} \times \text{Cost}\theta - 0.5S\_\text{\textdegree}^\text{l} \times \text{Sin}\theta - 0.866S\_\text{\textdegree}^\text{l} \text{Cost}\theta - 0.5S\_\text{\textdegree}^\text{l} = 0\tag{14c}$$

$$0.75\mathbf{x} \times \mathbf{C} \mathbf{s} \theta + \mathbf{S}\_\circ^\dagger \times \mathbf{C} \mathbf{s} \theta - 0.866\mathbf{x} \times \mathbf{S} \hat{\mathbf{n}} \theta - 0.866\mathbf{S}\_\circ^\dagger \times \mathbf{S} \hat{\mathbf{n}} \theta + 0.5\mathbf{S}\_\circ^3 \mathbf{C} \mathbf{s} \theta - 0.866\mathbf{S}\_\circ^3 = \mathbf{0} \tag{14d}$$

$$-1.7\mathbf{x} \times \mathbf{C} \mathbf{s} \theta + 0.5\mathbf{S}\_{\circ}^2 \times \mathbf{S} \dot{m} \theta - 0.866\mathbf{S}\_{\circ}^2 \times \mathbf{C} \mathbf{s} \theta - 0.5\mathbf{S}\_{\circ}^2 - 0.5\mathbf{S}\_{\circ}^3 \text{S} \dot{m} \theta - 0.866\mathbf{S}\_{\circ}^3 \mathbf{C} \mathbf{s} \theta - 0.5\mathbf{S}\_{\circ}^3 = \mathbf{0} \tag{14e}$$

$$-1.7\mathbf{x} \times \text{Sin}\theta - 0.866\mathbf{S}\_\circ^2 \times \text{Sin}\theta - 0.5\mathbf{S}\_\circ^2 \times \text{Cas}\theta + 0.866\mathbf{S}\_\circ^2 - 0.866\mathbf{S}\_\circ^3 \text{Sin}\theta + 05\mathbf{S}\_\circ^3 \text{Cas}\theta - 0.866\mathbf{S}\_\circ^3 = 0 \tag{14f}$$

### **9. Case analysis**

If we consider the fixed angle of the triangle *A*1*A*2*A*3 in **Figure 8**, *e* = 1000 mm, *b* = 100 mm and each of the *S*11, *S*21, *S*31 inputs are applied to the direction of the triangle, then **Figure 9** (**a**, **b**) are obtained.

Kinematic Analysis of the Triangle-Star Robot with Telescopic Arm and Three Kinematics Chains as T-S Robot (3-PRP) http://dx.doi.org/10.5772/64556 79

**Figure 8.** Geometric relation between triangle and star sides.

3 3 3 1 3 3 3 1

(13c)

0.5

(0)

<sup>2</sup> + *θ*2⇒*θ*<sup>2</sup> = −*θ*3,

<sup>1</sup> <sup>+</sup> *<sup>t</sup>* <sup>2</sup> , *t* = tan

*θ* 2

From the motion geometry of the T-S robot and then relations (5,13), the motion equation are

1 1 2 22

1 22 2

1 1 3 33

1 33 3

2 2 23 3 3

2 2 23 3 3

3 1 3 31 - ´ - ´ + + ´ - ´ + + -= 0.75*x Sin S Sin S x Cos S Sin S Cos S e*

<sup>3</sup> 33 1 0.75*x Cos S Cos x Sin S Sin S Cos S* ´ +´ + ´ + ´ + + =

3 1 3 31 - ´ -´ +- ´ - ´ - - = 0.75*x Sin S Sin S x Cos S Sin S Cos S*

<sup>3</sup> 33 1 0.75*x Cos S Cos x Sin S Sin S Cos S* ´ +´ - ´ - ´ + - =

<sup>3</sup> <sup>3</sup> <sup>1</sup> <sup>3</sup> <sup>3</sup> <sup>13</sup> -´ + ´ - ´ - - - - = 1.7 0.5 0.866 0.5 0.5 0.866 0.5 0 *x Cos S Sin S Cos S S Sin S Cos S*

3 3 13 3 1 -´ - ´ - ´ + - + - = 1.7 0.866 0.5 0.866 0.866 05 0.866 0 *x Sin S Sin S Cos S S Sin S Cos S*

If we consider the fixed angle of the triangle *A*1*A*2*A*3 in **Figure 8**, *e* = 1000 mm, *b* = 100 mm and each of the *S*11, *S*21, *S*31 inputs are applied to the direction of the triangle, then **Figure 9** (**a**, **b**)

 q

 q  q

 q  q

*O O O*

ê ú

**8. The motion geometry of T-S Robot**

<sup>2</sup> <sup>+</sup> *<sup>θ</sup>*3, *<sup>ϕ</sup>*<sup>3</sup> <sup>=</sup> *<sup>π</sup>*

<sup>1</sup> <sup>+</sup> *<sup>t</sup>* <sup>2</sup> , *Cosθ* <sup>=</sup> <sup>1</sup> <sup>−</sup> *<sup>t</sup>* <sup>2</sup>

*Sinθ* <sup>=</sup> <sup>2</sup>*<sup>t</sup>*

qq

qq

qq

qq

qq

qq

q

**9. Case analysis**

are obtained.

*<sup>ϕ</sup>*<sup>3</sup> <sup>+</sup> *<sup>ϕ</sup>*<sup>5</sup> =180, *<sup>ϕ</sup>*<sup>5</sup> <sup>=</sup> *<sup>π</sup>*

78 Recent Advances in Robotic Systems

*θ*1 1 , *θ*<sup>1</sup> 2 , *θ*<sup>1</sup> <sup>3</sup> =*θ*,

achieved.

¢ ¢ ¢

0.866 .(3) :

*<sup>y</sup> xS S path <sup>z</sup> <sup>b</sup>*

*x xS S*

é ù é ù à ´ +à ´ + ´ ¢ ¢ ê ú ê ú ´ + ´ + ´ ¢ ¢ ê ú <sup>=</sup> ê ú

ê ú ë û ë û

*<sup>ϕ</sup>*<sup>1</sup> <sup>+</sup> *<sup>ϕ</sup>*<sup>2</sup> =180, *<sup>ϕ</sup>*<sup>1</sup> <sup>=</sup> *<sup>π</sup>*

 q

> q

> > q

 q

0.866 0.5 0.866 0.5 0 (14a)

0.866 0.866 0.5 0.866 0 (14b)

0.866 0.5 0.866 0.5 0 (14c)

0.866 0.866 0.5 0.866 0 (14d)

 q

q

<sup>2</sup> <sup>+</sup> *<sup>θ</sup>*1, *<sup>ϕ</sup>*<sup>2</sup> <sup>=</sup> *<sup>π</sup>*

 q

 q

 q  q

> q

 q

(14e)

(14f)

<sup>2</sup> −*θ*2⇒*θ*<sup>1</sup> =*θ*2,

1 1

**Figure 9.** (a) Variation *S*13, *S*23, *S*33 (output) versus actuator *S*11, *S*21, *S*31 (input) and (b) *X*–*Y* plots of the trajectory of the motion point *c*.

### **10. Inverse kinematics for T-S Robot**

The representation of the location and orientation of the robotic final manipulator can lead to the estimation of all the possible joint collections, which are used to transfer the robot and to obtain the assumed orientation. This process is called kinematics reverse estimation [2]. Therefore, in the reverse kinematics analysis, we have access to the coordinate components of the central point of the moving star (point *C* located in the reference system *O*′) which are relative to the primary reference system and which are the goal for obtaining the unknown parameters *S*11, *S*21, *S*31, (*S*<sup>1</sup> 1 , *S*<sup>1</sup> 2 , *S*<sup>1</sup> 3 ). The independent linear equations are obtained through equalizing the transformative matrixes related to the central point of the moving star.

Solving these equations in which the inputs are *θ*<sup>1</sup> 1 , *θ*<sup>1</sup> 2 , *θ*<sup>1</sup> <sup>3</sup> <sup>=</sup>*θ* and *S*13, *S*23, *S*33 (*S*<sup>3</sup> 1 , *S*<sup>3</sup> 2 , *S*<sup>3</sup> 3 ), then *S*11, *S*21, *S*31 (*S*<sup>1</sup> 1 , *S*<sup>1</sup> 2 , *S*<sup>1</sup> 3 ) are obtained.

$$-X \times \text{Sin}\theta - \text{Sin}\theta \times \text{S}\_3^\text{l} + \text{S}\_1^\text{l} - q\_x = \mathbf{0} \tag{15a}$$

$$X \times \text{Cos}\,\theta + \text{Cos}\,\theta \times S\_3^\text{l} - q\_y = 0 \tag{15b}$$

$$0.5X \times \text{Sin}\theta - 0.866X \times \text{Cas}\theta + 0.5S\_3^2 \times \text{Sin}\theta - 0.866S\_3^2 \times \text{Cas}\theta - 0.5S\_2^1 + e - q\_x = 0\tag{15c}$$

$$-0.866X \times \text{Sin}\theta - 0.5X \times \text{Cas}\theta - 0.866S\_\text{\text{\textquotedbl{}}}^2 \times \text{Sin}\theta - 0.5S\_\text{\textquotedbl{}}^2 \times \text{Cas}\theta + 0.866S\_\text{\textquotedbl{}} - q\_\text{\textquotedbl{}} = 0\tag{15d}$$

$$-0.5X \times \text{Sin}\theta + 0.866X \times \text{Cas}\theta - 0.5S\_\circ^3 \times \text{Sin}\theta + 0.866S\_\circ^3 \times \text{Cas}\theta + 0.5S\_\circ^3 - q\_x = 0\tag{15e}$$

$$-0.866X \times \text{Sin}\theta - 0.5X \times \text{Cas}\theta - 0.866S\_3^3 \times \text{Sin}\theta - 0.5S\_3^3 \times \text{Cas}\theta + 0.5S\_1^3 - q\_y = 0\tag{15f}$$

To conduct the reverse kinematics analysis, we should assume two directions: (1) liner, (2) circular, in which the center of the star (point located in *O*′ reference system) goes through the two mentioned directions as shown in **Figure 10** (**a**) and **Figure 11** (**a**, **e**). The outcome is respectively represented in **Figure 10** (**b**) and **Figure 11** (**b**, **c**, **d**, **e**).

Kinematic Analysis of the Triangle-Star Robot with Telescopic Arm and Three Kinematics Chains as T-S Robot (3-PRP) http://dx.doi.org/10.5772/64556 81

**10. Inverse kinematics for T-S Robot**

1 , *S*<sup>1</sup> 2 , *S*<sup>1</sup> 3

Solving these equations in which the inputs are *θ*<sup>1</sup>

) are obtained.

parameters *S*11, *S*21, *S*31, (*S*<sup>1</sup>

80 Recent Advances in Robotic Systems

1 , *S*<sup>1</sup> 2 , *S*<sup>1</sup> 3

q

q

*S*11, *S*21, *S*31 (*S*<sup>1</sup>

The representation of the location and orientation of the robotic final manipulator can lead to the estimation of all the possible joint collections, which are used to transfer the robot and to obtain the assumed orientation. This process is called kinematics reverse estimation [2]. Therefore, in the reverse kinematics analysis, we have access to the coordinate components of the central point of the moving star (point *C* located in the reference system *O*′) which are relative to the primary reference system and which are the goal for obtaining the unknown

equalizing the transformative matrixes related to the central point of the moving star.

3 1 0 -´ - ´ + - = *X Sin Sin S S q*

<sup>3</sup> 0 *X Cos Cos S q* ´ + ´-=

3 32 0.5 0.866 0.5 0.866 0.5 0 *X Sin X Cos S Sin S Cos S e q* ´ - ´ + ´ - ´ - +- =

33 2 0.866 0.5 0.866 0.5 0.866 0 - ´ - ´ - ´ - ´ + -= *X Sin X Cos S Sin S Cos S q*

3 31 0.5 0.866 0.5 0.866 0.5 0 - ´ + ´ - ´ + ´ + -= *X Sin X Cos S Sin S Cos S q*

33 1 0.866 0.5 0.866 0.5 0.5 0 - ´ - ´ - ´ - ´ + -= *X Sin X Cos S Sin S Cos S q*

qq

 q

> q

To conduct the reverse kinematics analysis, we should assume two directions: (1) liner, (2) circular, in which the center of the star (point located in *O*′ reference system) goes through the two mentioned directions as shown in **Figure 10** (**a**) and **Figure 11** (**a**, **e**). The outcome is

 q

 q

q

q

qq

respectively represented in **Figure 10** (**b**) and **Figure 11** (**b**, **c**, **d**, **e**).

qq

qq

1 , *θ*<sup>1</sup> 2 , *θ*<sup>1</sup>

1 1

1

2 21

 q

> q

> > q

 q

22 1

3 33

33 3

). The independent linear equations are obtained through

<sup>3</sup> <sup>=</sup>*θ* and *S*13, *S*23, *S*33 (*S*<sup>3</sup>

*<sup>x</sup>* (15a)

*<sup>y</sup>* (15b)

1 , *S*<sup>3</sup> 2 , *S*<sup>3</sup> 3 ), then

*<sup>x</sup>* (15c)

*<sup>y</sup>* (15d)

*<sup>x</sup>* (15e)

*<sup>y</sup>* (15f)

**Figure 10.** (a) *X*-*Y* plots of the path of the motion point c and (b) variation *S*11, *S*21, *S*<sup>31</sup> (output) versus *S*13, *S*23, *S*33 S13, S23, S33 (mm).

**Figure 11.** (a, e) *X*-*Y* plots of the path of the motion point c; (b, d) variation *S*11, *S*21, *S*<sup>31</sup> (outputs) versus *S*13, *S*23, *S*<sup>33</sup> for *θ* = 0; (c) variation *S*11, *S*21, *S*31 (outputs) versus *S*13, *S*23, *S*33 (inputs) for *θ* = *pi*/6.
