2.1. Proposed method

Three parallel robots are arranged in symmetrical order with vision sensors and hardware interfacing, as shown in Fig. 1. This paper is limited to the kinematic and dynamic analyses of the system to catch the cube. In this method, a new architecture that comprises of three delta robots and each of the end effectors behave like one finger, so it comprises three fingers that do the catching in the real-time scenario. Three pre-calibrated vision sensors (Basler scout cameras) are used to estimate the pose of the three cube faces. The forward kinematics for all three fixed delta robots is achieved to know the shared workspace (W_catch = Intersection (W1, W2, W3)).

Figure 1.

Experimental setup.

The delta robots are used as fingers because this architecture can go for a high payload and is dedicated to high-speed applications like manipulation. In this architecture, the moving frame has no rotation to the fixed frame, i.e., 3DOF has the only translation in the X, Y, Z axes, as shown in Fig. 2.

Figure 2.

Mechanical structure of individual manipulator [Reference Sachin, Rajesh and Sudipto26].

2.2. Mechanical structure

The delta robot is the three-DOF motion platform invented by a research team led by Prof. Reymond Clavel in the early 1980s as a manipulator for light and smaller objects. Since then, it has been used for various applications in industries and elsewhere. Their applications include driving simulators, medical devices, entertainment industries, vibration isolators, etc. Delta manipulator, as shown in Fig. 3, employs revolute joints and spherical joints to give platform output to translational motion [Reference Zsombor-Murray2]. A schematic diagram of the manipulator constructed with all revolute joints. The lower arm is labelled as links 1, 2, 3. The upper arms of the delta robot architecture comprise four parallelograms; links are shown in Fig. 2. Links 4, 7, 10, and 13 are marked as I^st limb, links 5, 8, 11, and 14 are marked as the II^nd limb, and links 6, 9, 12, and 15 are marked as III^rd limb.

Figure 3.

Three delta combined system architecture [Reference Sachin and Sudipto27].

A, B, C are the points where three parallel joints attach the lower arms, upper arms, the two surfaces, and each of the parallelograms has four revolute joints. Overall, there are 17 links and 21 revolute joints. The general degree of freedom equation is

(1) \begin{align}n = \,\,s\,\left( {r\, - \,p\, - \,1} \right) + \Sigma \,{n_i}\end{align}

where S represents dimensions of working space (For spatial, s = 6), R represents several rigid bodies or links in the system = 17, and P represents nunmber of kinematic pairs of joints in the system = 21

n _i = relative degree of freedom of each joint.

\begin{align*}DOF,\,n = \, = \,6^{*}(17\, - \,21 - \, - \,1) + 21\left( 1 \right) = - 9\end{align*}

The joints are redundant and have a 3DOF translation. Due to the four-bar parallelogram and the three revolute joints at points A, B, and C, the limb constrains the unfixed surface from rotating about the z and x-axis. The two limbs constraint the unfixed body from rotating about any axis. It is helpful in the x-y-z positioning device.

The delta manipulator used in this paper is shown in Fig. 4. The base platform was 0, and the top platform was 1, as shown in Fig. 4. The inclination of the second link due to a spherical joint cannot be shown in the X-Z plane. The angle between any two axes X, u₁, v₁ is 120 ^o, and the direction of axes Z, w₁, and w₂ is the same.

Figure 4.

Schematic diagram of delta mechanism.

3.1. System description and mathematical background

The three degrees of freedom delta robot is capable of 3D transitional of its moving surface. As in the proposed delta architecture, there are three identical Revolute-Universal-Universal (R-U-U) chains as legs. In the Figs. 5 and 6, points B_i, i = 1,2,3 are the hips, points, P_i, i = 1,2,3 are the ankles, and points, A_i, i = 1,2,3 are the knees. s_B represents the length of the equilateral triangle (base), and s_P represents the length of the equilateral triangle (moving). In our proposed delta architecture, the moving platform is inverted to the base. There is no rotation between the moving platform and the base platform, i.e., [R_3x3 is Identity matrix]. Thus, only the translational vector between the moving and the fixed platform is to be considered.

Figure 5.

Delta robot fixed base schematic.

Figure 6.

Delta robot moving platform.

‘B’ represents the cartesian reference frame (fixed). The origin is situated in the centre of the bottom triangle. $\text{'}$ P’ is not a fixed Cartesian reference frame. The origin is located in the centre of the platform triangle. The rotation matrix $ \begin{array}{c}B\\ P\end{array}\textit{R}$ = I ₃ always is constant as orientations of P and B are the same. The joint variables are θ = {θ_1, θ₂, θ₃}^T, and the Cartesian parameters are $ \begin{array}{c}B\\ P\end{array}\textit{R}$ = {x y z}^T..The three upper legs of lengths $\text{'}$ L’ and three lower legs of lengths $\text{'}$ l’ have high symmetry. The geometric details of the delta robot having a fixed bottom and platform are depicted in Figs. 5 and 6, respectively.

In the bottom frame $\text{'}$ B $\text{'}$ , the fixed-bottom joint points B_i are constant. In the top frame $\text{'}$ P $\text{'}$ , the platform-fixed joint points P_i is constant. In the kinematic diagram (in Fig. 7), the three vector-loop closure equations for the parallel architecture are given below:

\begin{align*} \begin{array}{c}B\\ \\[-7pt] 1\end{array}B&=\left\{\begin{array}{c}0\\ \\[-7pt] -{w}_{B}\\ \\[-7pt] 0\end{array}\right\} \textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\begin{array}{c}B\\ \\[-7pt] 2\end{array}B=\left\{\begin{array}{c}\dfrac{\sqrt{3}}{2}{w}_{B}\\ \\[-7pt] \dfrac{1}{2}{w}_{B}\\ \\[-7pt] 0\end{array}\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\begin{array}{c}B\\ \\[-7pt] 3\end{array}B=\left\{\begin{array}{c}\dfrac{-\sqrt{3}}{2}{w}_{B}\\ \\[-7pt] \dfrac{1}{2}{w}_{B}\\ \\[-7pt] 0\end{array}\right\}\\ \\[-7pt]\begin{array}{c}p\\ \\[-7pt] 1\end{array}P&=\left\{\begin{array}{c}0\\ \\[-7pt] -{u}_{P}\\ \\[-7pt] 0\end{array}\right\} \textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\begin{array}{c}p\\ \\[-7pt] 2\end{array}P=\left\{\begin{array}{c}\dfrac{{S}_{p}}{2}\\ \\[-7pt] {w}_{P}\\ \\[-7pt] 0\end{array}\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{} \textrm{}\begin{array}{c}p\\ 3\end{array}P = \left\{\begin{array}{c}\dfrac{-{S}_{p}}{2}\\ {w}_{p}\\ 0\end{array}\right\}\end{align*}

Figure 7.

Delta robot kinematic diagram.

The vertices of the bottom triangle (fixed) are represented as

\begin{align*} \begin{array}{c}B\\ \\[-7pt] 1\end{array}b=\left\{\begin{array}{c}\frac{{S}_{B}}{2}\\ \\[-7pt] -{w}_{B}\\ \\[-7pt] 0\end{array}\right\} \textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\begin{array}{c}B\\ \\[-7pt] 1\end{array}b=\left\{\begin{array}{c}0\\ \\[-7pt] {u}_{B}\\ \\[-7pt] 0\end{array}\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\begin{array}{c}B\\ \\[-7pt] 3\end{array}b=\left\{\begin{array}{c}\frac{-{S}_{B}}{2}\\ \\[-7pt] {-w}_{B}\\ \\[-7pt] 0\end{array}\right\}\end{align*}

where: $ {w}_{B}=\frac{\sqrt{3}}{6}{S}_{B}$ ; $ {w}_{p}=\frac{\sqrt{3}}{6}{S}_{p}$ ; $ {u}_{B}=\frac{\sqrt{3}}{3}{S}_{B};\ {u}_{p}=\frac{\sqrt{3}}{3}{S}_{p}$

In the kinematic diagram, the three vector-loop equations as depicted in Fig. 7;

(2) \begin{align} &\left\{\begin{array}{c}B\\ i\end{array}B\right\}+\left\{\begin{array}{c}B\\ i\end{array}L\right\}+\left\{\begin{array}{c}B\\ i\end{array}l\right\}=\left\{\begin{array}{c}B\\ i\end{array}P\right\}+\left[\begin{array}{c}B\\ P\end{array}R\right]\left\{\begin{array}{c}P\\ i\end{array}P\right\}=\left\{\begin{array}{c}B\\ P\end{array}P\right\}+\left\{\begin{array}{c}P\\ i\end{array}P\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\nonumber\\[3pt] i &=\textrm{1,2},3\end{align}

$ \left[\begin{array}{c}B\\ P\end{array}R\right]$ = $ \left[{I}_{3}\right]$ . Delta robot does not perform rotations with respect to the delta fixed frame. The length of the lower leg should be correct, constant length.

(3) \begin{align} {l}_{i}=|\left\{{}_{i}{}^{B}l\right\}|=|\left\{{}_{p}{}^{B}P\right\}+\left\{{}_{i}{}^{p}P\right\}-\left\{{}_{i}{}^{B}B\right\}-\left\{{}_{i}{}^{B}L\right\}|\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}i=\textrm{1,2},3\end{align}

It is appropriate to square constraint equations to stay away from the square-root in the Euclidean norms:

(4) \begin{align} {l}_{i}^{2}={|\left\{{}_{i}{}^{B}l\right\}|}^{2}={l}_{ix}^{2}+{l}_{iy}^{2}+{l}_{iz}^{2}\textrm{}\textrm{}\textrm{}i=\textrm{1,2},3\end{align}

The Cartesian variables are $ {}_{p}{}^{B}P={\left\{xyz\right\}}^{T}$ . The constant vector values for points $ {P}_{i}$ and $ {B}_{i}$ were given previously. The vectors $ \left\{{}_{i}{}^{B}L\right\}$ are dependent on the joint variables ʘ $ {\left\{{\theta }_{1}{\theta }_{2}{\theta }_{3}\right\}}^{T}$ :

\begin{align*} {}_{1}{}^{\textrm{B}}\textrm{L}=\left\{\begin{array}{c}0\\ \\[-7pt] -\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}\end{array}\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}{}_{2}{}^{\textrm{B}}\textrm{L}=\left\{\begin{array}{c}\dfrac{\sqrt{3}}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}\\ \\[-7pt] \dfrac{1}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}\end{array}\right\}\textrm{}{}_{3}{}^{\textrm{B}}\textrm{L}=\left\{\begin{array}{c}-\dfrac{\sqrt{3}}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}\\ \\[-7pt] \dfrac{1}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}\end{array}\right\}\end{align*}

After applying the values into the vector equations, yields equation below:

\begin{align*} &\left\{{}_{1}{}^{B}l\right\}=\left\{\begin{array}{c}\textrm{x}\\ \\[-7pt] \textrm{y}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}+a\\ \\[-7pt] \textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}\end{array}\right\}\textrm{}\textrm{};\textrm{}\left\{{}_{2}{}^{B}l\right\}=\left\{\begin{array}{c}x-\dfrac{\sqrt{3}}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+b\\ \\[-7pt] y-\dfrac{1}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+c\\ \\[-7pt] \textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}\end{array}\right\}\textrm{};\textrm{}\left\{{}_{3}{}^{B}l\right\}=\left\{\begin{array}{c}x+\dfrac{\sqrt{3}}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}-b\\ \\[-7pt] y-\dfrac{1}{2}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}+c\\ \\[-7pt] \textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}\end{array}\right\} \\ \\[-7pt] &a={\textrm{w}}_{\textrm{B}}-{\textrm{u}}_{\textrm{p}};\ b=\dfrac{{\textrm{S}}_{\textrm{p}}}{2}-\dfrac{\sqrt{3}}{2} {\textrm{w}}_{\textrm{B}};\ c={\textrm{w}}_{\textrm{p}}-\dfrac{1}{2}{\textrm{w}}_{\textrm{B}} \end{align*}

The three constraint equations result in the kinematics equations:

(5) \begin{align} 2L\left(y+a\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}+2zL\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}+{x}^{2}+{y}^{2}+{z}^{2}+{a}^{2}+{L}^{2}+2ya-{l}^{2}=0\end{align}

(6) \begin{align} -L\left(\sqrt{3}\left(x+b\right)+y+c\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+2zL\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}+{x}^{2}+{y}^{2}+{z}^{2}+{b}^{2}+{c}^{2}+{L}^{2}+2xb+2yc-{l}^{2}=0\textrm{}\dots \textrm{}\textrm{}\end{align}

(7) \begin{align} L\left(\sqrt{3}\left(x-b\right)-y-c\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}+2zL\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}+{x}^{2}+{y}^{2}+{z}^{2}+{b}^{2}+{c}^{2}+{L}^{2}-2xb+2yc-{l}^{2}=0\end{align}

The knee points are calculated using $ \begin{array}{c}B\\ i\end{array}\textit{A}$ = $ \begin{array}{c}B\\ i\end{array}\textit{B}$ + $ \begin{array}{c}B\\ i\end{array}\textit{L}$ ; i = 1, 2, and 3:

\begin{align*} \begin{array}{c}B\\ \\[-7pt] 1\end{array}\textit{A}=\left\{\begin{array}{c}0\\ \\[-7pt] -{\textrm{w}}_{\textrm{B}}-\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}\end{array}\right\};\begin{array}{c}B\\ \\[-7pt] 2\end{array}\textit{A}=\left\{\begin{array}{c}\dfrac{\sqrt{3}}{2}{(\textrm{}\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2})\\ \\[-7pt] \dfrac{1}{2}({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2})\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}\end{array}\right\};\begin{array}{c}B\\ \\[-7pt] 3\end{array}\textit{A}=\left\{\begin{array}{c}-\dfrac{\sqrt{3}}{2}{(\textrm{}\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3})\\ \\[-7pt] \dfrac{1}{2}({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3})\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}\end{array}\right\}\end{align*}

3.2. Geometric model

Thedelta robot is characterized by the Denavit–Hartenberg (DH) parameters, as shown in Table I. Voice coil actuator (VCA), i.e., the actuator having joint coordinates are given with variables θ1, θ2, θ3, θ4, θ5, θ6, θ7, θ8, θ9. Its geometric architecture is defined in Fig. 8 Table II.

Table I.

Parameters of the delta architecture.

Figure 8.

Geometric architecture for subsystem I.

Table II.

The DH parameters [Reference John28] of the k^th leg of the delta platform.

Inverse kinematics of parallel robots is a process where actuated or active joint trajectories are to be computed for the respective poses of the platform motion. The inverse kinematics of parallel robots calculated both active and passive joint trajectories. The calculated results of inverse kinematics are needed for inverse dynamics to determine the driving torques/forces at the active joints.

3.3. Forward position kinematics solution [active joints]

The forward position kinematics is defined as the three actuated joint angles θ = {θ_1, θ₂, θ₃} ^T are given then compute the resulting Cartesian position of the unfixed surface control point (P), $ \begin{array}{c}B\\ P\end{array}\textit{R}$ = {x y z} ^T. The forward position kinematics solution for parallel robots is tough.

Thecorrect solution set is selected for a straightforward analytical solution, as there is translation-only motion in the 3DOF delta robot. Given $ \Theta ={\left\{{\theta }_{1}{\theta }_{2}{\theta }_{3}\right\}}^{T}$ , then compute the three absolute vector knee points using $ \begin{array}{c}B\\ i\end{array}\textit{A}=\begin{array}{c}B\\ i\end{array}\textit{B}+\begin{array}{c}B\\ i\end{array}\textit{L},\textit{}\textit{i}=\textrm{1,2},3$ . The unfixed surface of the delta robot’s orientation is constant and horizontal with [ $ \begin{array}{c}B\\ i\end{array}\textit{A}]=[{\textrm{I}}_{3}]$ ; therefore, all the virtual sphere centres (three in number) are $ \begin{array}{c}B\\ iv\end{array}\textit{A}=\begin{array}{c}B\\ i\end{array}\textit{A}-\begin{array}{c}p\\ i\end{array}\textit{P},\textit{}\textit{i}=\textrm{1,2},3$ .

\begin{align*} &{}_{1\textrm{v}}{}^{\textrm{B}}\textrm{A}=\left\{\begin{array}{c}0\\ \\[-7pt] {-\textrm{w}}_{\textrm{B}}-\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}+{\textrm{u}}_{\textrm{p}}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}\end{array}\right\}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\\[3pt]&{}_{2\textrm{v}}{}^{\textrm{B}}\textrm{A}=\left\{\begin{array}{c}\dfrac{\sqrt{3}}{2}\left({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}\right)-\dfrac{{\textrm{s}}_{\textrm{p}}}{2}\\ \\[-7pt] \dfrac{1}{2}\left({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}\right)-{\textrm{w}}_{\textrm{p}}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}\end{array}\right\}\textrm{}\\[3pt]&{}_{3\textrm{v}}{}^{\textrm{B}}\textrm{A}=\left\{\begin{array}{c}-\dfrac{\sqrt{3}}{2}\left({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}\right)+\dfrac{{\textrm{s}}_{\textrm{p}}}{2}\\ \\[-7pt] \dfrac{1}{2}\left({\textrm{w}}_{\textrm{B}}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}\right)-{\textrm{w}}_{\textrm{p}}\\ \\[-7pt] -\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}\end{array}\right\}\end{align*}

Delta robot forward position kinematics is determined by its point of intersection of the three spheres. This sphere is represented by a centre point called c and the scalar radius defined by r.

In this section, an analytical solution is computed as the intersection point of the given three spheres. In this concern, it also needs to solve transcendental equations which is of coupled in nature. So, to solve the forward kinematics solution, an intersection-three-sphere algorithm exists. If this is the case when the entire three-sphere centres { $ \begin{array}{c}B\\ iv\end{array}\textit{A}$ } have the same height (Z), then the solution will be in the zone of the algorithmic singularity. To fix the problem rotating the coordinates, i.e., all { $ \begin{array}{c}B\\ iv\end{array}\textit{A}$ }, heights values are not the same.

As in this problem, two solutions were encountered when the three spheres intersect. When a sphere meeting is tangential, it results in one solution. The second case is an imaginary solution, and the data are inconsistent. The algorithm (spheres-intersection) compute solution sets, and the computer automatically chooses the solution below the base triangle. This approach to the forward position kinematics for the delta robot estimated results same as of solving the kinematics equations, depicted as $ \begin{array}{c}B\\ P\end{array}\textit{R}$ = {x y z} ^T given θ = {θ_1, θ₂, θ₃} ^T.

3.4. Inverse position kinematics solution [active joints]

The inverse position kinematics problem is defined as the Cartesian position of the unfixed surface control point (P), $ \begin{array}{c}B\\ P\end{array}\textit{R}$ = {x y z}T are given then compute the three actuated revolute joint angles θ = {θ1, θ2, θ3} T. The inverse position kinematics solution delta robot is not trivial, therefore calculated analytically. The inverse position kinematics problem is not dependent on the three RUU legs of a delta robot; therefore, it can be solved individually for each leg. As shown in the figure above, geometrically, inverse position kinematics solution of each leg, the point of intersection between the known sphere (defined by radius l, and it is centre on the moving triangle vertex $ \begin{array}{c}P\\ i\end{array}\textit{P}$ ) and known circle (defined by radius L, and it is centre on the fixed triangle R joint point $ \begin{array}{c}B\\ i\end{array}\textit{B}$ ).

The inverse position kinematics solved both geometrically and trigonometrically. The inverse position kinematics are now solved analytically. The three independent scalar inverse position kinematics equations are shown below:

\begin{align*} {\textrm{E}}_{\textrm{i}}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{\textrm{i}}+{\textrm{F}}_{\textrm{i}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{\textrm{i}}+{\textrm{G}}_{\textrm{i}}=0,i=\textrm{1,2},3\end{align*}

where

(8) \begin{align} {\textrm{E}}_{1}=2L\left(y+a\right) {\textrm{F}}_{1}=2zL \end{align}

(9) \begin{align} {\textrm{}\textrm{G}}_{1}={x}^{2}+{y}^{2}+{z}^{2}+{a}^{2}+{L}^{2}+2ya-{l}^{2}\end{align}

(10) \begin{align} {\textrm{E}}_{2}=-L\left(\sqrt{3}\left(x+b\right)+y+c\right) {\textrm{F}}_{2}=2zL \end{align}

(11) \begin{align} {\textrm{}\textrm{G}}_{2}={x}^{2}+{y}^{2}+{z}^{2}+{b}^{2}+{c}^{2}+{L}^{2}+2xb+2yc-{l}^{2}\end{align}

(12) \begin{align} {\textrm{E}}_{3}=L\left(\sqrt{3}\left(x-b\right)-y-c\right),\textrm{}{\textrm{F}}_{3}=2zL\end{align}

(13) \begin{align} {\textrm{}\textrm{G}}_{3}={x}^{2}+{y}^{2}+{z}^{2}+{b}^{2}+{c}^{2}+{L}^{2}-2xb+2yc-{l}^{2}\end{align}

The equation $ {\textrm{E}}_{\textrm{i}}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{\textrm{i}}+{\textrm{F}}_{\textrm{i}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{\textrm{i}}+{\textrm{G}}_{\textrm{i}}=0$ already used in robot, kinematics. A tangent half-angle solves it. If $ {\textrm{t}}_{\textrm{i}}=tan\frac{{{\theta }}_{\textrm{i}}}{2}$ , then $ \textrm{c}\textrm{o}\textrm{s}{{\theta }}_{\textrm{i}}=\frac{1-{t}_{i}^{2}}{1+{t}_{i}^{2}}$ and $ \textrm{s}\textrm{i}\textrm{n}{{\theta }}_{\textrm{i}}=\frac{2{t}_{i}}{1+{t}_{i}^{2}}$ . If tangent half-angle substituted into the EFG equation, then results below:

(14) \begin{align} &{\textrm{E}}_{\textrm{i}}\left(\frac{1-{t}_{i}^{2}}{1+{t}_{i}^{2}}\right)+ {\textrm{F}}_{\textrm{i}}\left(\frac{2{t}_{i}}{1+{t}_{i}^{2}}\right)+{\textrm{G}}_{\textrm{i}}=0\ {\textrm{E}}_{\textrm{i}}\left(1-{t}_{i}^{2}\right) + {\textrm{F}}_{\textrm{i}}\left(2{t}_{i}\right)+{\textrm{G}}_{\textrm{i}}(1+{t}_{i}^{2})=0 (1.37) \nonumber\\[3pt] &\quad {(\textrm{G}}_{\textrm{i}}-{\textrm{E}}_{\textrm{i}}){t}_{i}^{2}+(2{\textrm{F}}_{\textrm{i}}){t}_{i}+({\textrm{G}}_{\textrm{i}}+{\textrm{E}}_{\textrm{i}})=0\dots .\end{align}

Quadratic formula: $ {t}_{{i}_{\textrm{1,2}}}=\frac{-{\textrm{F}}_{\textrm{i}}\pm \sqrt{{E}_{i}^{2}+{F}_{i}^{2}-{G}_{i}^{2}}}{{\textrm{G}}_{\textrm{i}}-{\textrm{E}}_{\textrm{i}}}$ . The $ {{\theta }}_{\textrm{i}}$ solved by inverting the original tangent, as shown below:

\begin{align*} {{\theta }}_{\textrm{i}}=2{tan}^{-1}\left({\textrm{t}}_{\textrm{i}}\right)\end{align*}

Two solutions (positive and negative) are obtained from the quadratic formula. The first solution represents the knee left, and the second is the knee right of a delta robot. There are two possible solutions for each leg. Therefore, the eight possible solutions are there.

3.5. Delta robot velocity kinematics equations [active joints]

The velocity kinematics equations of the revolute-input delta robot are computed through the derivative of the position constraint equations.

(15) \begin{align} 2\textrm{L}\dot{\textrm{y}}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}-2\textrm{L}\left(\textrm{y}+\textrm{a}\right){\dot{\theta }}_{1}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}+2\textrm{L}\textrm{}\dot{\textrm{z}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}+2\textrm{L}\textrm{z}{\dot{\theta }}_{1}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}+2\textrm{x}\dot{\textrm{x}\textrm{}}+2\left(\textrm{y}+\textrm{a}\right)\textrm{}\dot{\textrm{y}}+2\textrm{z}\dot{\textrm{z}}=0\end{align}

(16) \begin{align} &-\textrm{L}\left(\sqrt{3}\dot{\textrm{x}\textrm{}}+\dot{\textrm{y}}\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+\textrm{L}\left(\sqrt{3(}\textrm{x}+\textrm{b}\right)+\textrm{y}+\textrm{c}){\dot{\theta }}_{2}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+2\textrm{L}\dot{\textrm{z}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}+2\textrm{L}\textrm{z}{\dot{\theta }}_{2}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+2(\textrm{x}+\textrm{b})\dot{\textrm{x}\textrm{}}\nonumber\\ &\quad +2(\textrm{y}+\textrm{c})\dot{\textrm{y}\textrm{}}+2\textrm{z}\dot{\textrm{z}\textrm{}}=0\end{align}

(17) \begin{align} &\textrm{L}\left(\sqrt{3}\dot{\textrm{x}\textrm{}}-\dot{\textrm{y}\textrm{}}\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}-\textrm{L}\left(\sqrt{3}\left(\textrm{x}-\textrm{b}\right)-\textrm{y}-\textrm{c}\right){\dot{\theta }}_{3}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}+2\textrm{L}\dot{\textrm{z}}{\dot{\theta }}_{3}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}+2\textrm{L}\textrm{z}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}-2\left(\textrm{x}-\textrm{b}\right)\dot{\textrm{x}\textrm{}}\textrm{}\nonumber\\ &\quad +2\left(\textrm{y}+\textrm{c}\right)\dot{\textrm{y}\textrm{}}\textrm{}+2\textrm{z}\dot{\textrm{z}\textrm{}}\textrm{}=0\end{align}

Equation 15, 16, and 17 can be re-written as

(18) \begin{align} \textrm{x}\dot{\textrm{x}\textrm{}}\textrm{}+\left(\textrm{y}+\textrm{a}\right)\dot{\textrm{y}\textrm{}}+\textrm{L}\dot{\textrm{y}\textrm{}}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}+\textrm{z}\dot{\textrm{z}\textrm{}}\textrm{}+\textrm{L}\dot{\textrm{z}\textrm{}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}=\textrm{L}\left(\textrm{y}+\textrm{a}\right){\dot{\theta }}_{1}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}-\textrm{L}\textrm{z}{\dot{\theta }}_{1}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}\end{align}

(19) \begin{align} &2\left(\textrm{x}+\textrm{b}\right)\dot{\textrm{x}\textrm{}}+2\left(\textrm{y}+\textrm{c}\right)\dot{\textrm{y}\textrm{}}-\textrm{L}\left(\sqrt{3}\dot{\textrm{x}\textrm{}}+\dot{\textrm{y}\textrm{}}\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}+2\textrm{z}\dot{\textrm{z}\textrm{}}+2\textrm{L}\dot{\textrm{z}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}\nonumber\\[3pt] &\quad =-\textrm{L}\left(\sqrt{3}\left(\textrm{x}+\textrm{b}\right)+\textrm{y}+\textrm{c}\right){\dot{\theta }}_{2}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}-2\textrm{L}\textrm{z}{\dot{\theta }}_{2}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}\end{align}

(20) \begin{align} & 2\left(\textrm{x}-\textrm{b}\right)\dot{\textrm{x}\textrm{}}+2\left(\textrm{y}+\textrm{c}\right)\dot{\textrm{y}}+\textrm{L}\left(\sqrt{3}\dot{\textrm{x}\textrm{}}-\dot{\textrm{y}}\right)\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}+2\textrm{z}\dot{\textrm{z}}+2\textrm{L}\dot{\textrm{z}}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}\nonumber\\[3pt] &\quad =-\textrm{L}\left(\sqrt{3}\left(\textrm{x}-\textrm{b}\right)-\textrm{y}-\textrm{c}\right){\dot{\theta }}_{3}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}-2\textrm{L}\textrm{z}{\dot{\theta }}_{3}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}\end{align}

In matrix-vector form: $ \left[\textrm{A}\right]\{\dot{\textrm{X}\}=\left[\textrm{B}\right]\{\dot{{\theta }\}}}$

(21) \begin{align} \left[\begin{array}{c@{\quad}c@{\quad}c}\textrm{x}& \textrm{y}+\textrm{a}+\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}& \textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}\\ \\[-7pt]2\left(\textrm{x}+\textrm{b}\right)-\sqrt{3}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2} &2\left(\textrm{y}+\textrm{c}\right)-\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}& 2(\textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2})\\ \\[-7pt] 2\left(\textrm{x}-\textrm{b}\right)+\sqrt{3}\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3} &2\left(\textrm{y}+\textrm{c}\right)-\textrm{L}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}& 2(\textrm{z}+\textrm{L}\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3})\end{array}\right]\left\{\begin{array}{c}\dot{\textrm{x}}\\ \\[-7pt] \dot{\textrm{y}}\\ \\[-7pt] \dot{\textrm{z}}\end{array}\right\}=\left[\begin{array}{c@{\quad}c@{\quad}c}{\textrm{b}}_{11}& 0& 0\\ \\[-7pt] 0& {\textrm{b}}_{22}& 0\\ \\[-7pt] 0& 0& {\textrm{b}}_{33}\end{array}\right]\left\{\begin{array}{c}\dot{{{\theta }}_{1}}\\ \\[-7pt] \dot{{{\theta }}_{2}}\\ \\[-7pt] \dot{{{\theta }}_{3}}\end{array}\right\}\end{align}

where $ {\textrm{b}}_{11}=\textrm{L}[(\textrm{y}+\textrm{a})\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{1}-\textrm{z}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{1}];\,\, {\textrm{b}}_{22}=-\textrm{L}[(\sqrt{3}(\textrm{x}+\textrm{b})+\textrm{y}+\textrm{c})\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{2}+2\textrm{z}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{2}]$ $ {\textrm{b}}_{33}=\textrm{L}[(\sqrt{3}(\textrm{x}-\textrm{b})-\textrm{y}-\textrm{c})\textrm{s}\textrm{i}\textrm{n}{{\theta }}_{3}-2\textrm{z}\textrm{c}\textrm{o}\textrm{s}{{\theta }}_{3}]$

3.6. Inverse position kinematics solution [active and passive joints]

In kinematic and dynamic analyses, the moving platform is assumed to be cut at three spherical joints and detached from the three identical legs. As shown in Fig. 6(a), the three-DOF delta platform has a fixed base connected to three identical legs through universal joints at three different points, namely, at Ai’s. These legs are connected to the platform (moving) through three spherical joints at Bi’s. Each leg consists of two links connected with a universal joint denoted as Ui’s. Therefore, each leg has three-DOF. Referring to Fig. 6(b), the loop-closure equation for the kth loop is given as

(22) \begin{align} \overrightarrow{{B}_{i}}+\overrightarrow{{L}_{i}}+\textrm{}\overrightarrow{{l}_{i}}=\overrightarrow{P}+\textrm{}{p}_{i} \end{align}

(23) \begin{align} \overrightarrow{{L}_{i}}+\textrm{}\overrightarrow{{l}_{i}}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]+\left[\begin{array}{c}0\\ -{\textrm{u}}_{\textrm{p}}\\ 0\end{array}\right]-\textrm{}\left[\begin{array}{c}0\\ -{\textrm{w}}_{\textrm{B}}\\ 0\end{array}\right]\end{align}

$ \overrightarrow{{\textrm{L}}_{\textrm{i}}}+\overrightarrow{{\textrm{l}}_{\textrm{i}}}$ can be found using the D-H parameter approach. $ \overrightarrow{{\textrm{L}}_{\textrm{i}}}$ is the vector from the base vertex and corresponding platform vertex, i.e., from Ai to Bi. The magnitude of this vector also represents leg length for the respective platform motion. Vector p is the measure of the displacement of the platform to the base. Here $\text{'}$ p’ represents the position vector of point $\text{'}$ P’ on the platform from the centre of the base, i.e., points O, $ \overrightarrow{{\textrm{L}}_{\textrm{i}}}\textrm{}and\textrm{}\overrightarrow{{\textrm{l}}_{\textrm{i}}}$ are the vectors locating the vertices of the base and platform in their respective frames, i.e., XYZ and Xp, Yp, Zp as depicted in Fig.6(a). Since the platform has three-DOF, its motion can be represented by three independent translations and about axes Xp, Yp, and Zp. If the values of xp, yp, zp, the vector $ {\textrm{a}}_{\textrm{k}}\textrm{}$ can be calculated from Eq. 2. The coordinate values of the base and platform vertices, i.e., A_I, A_III, and B_I…, B_IX are taken from the CAD model to calculate the active and passive joint trajectories. The homogeneous transformation matrix between frame 3 and frame 1 can be given as the homogeneous transformation matrix (HTM).

(24) \begin{align} {T}_{k}\equiv \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{C\theta }_{ki}& {-S\theta }_{ki}{C\alpha }_{ki} & {-S\theta }_{ki}{S\alpha }_{ki}& {{a}_{i}C\theta }_{ki}\\ \\[-7pt] {S\theta }_{ki}& {C\theta }_{ki}{C\alpha }_{ki} & {-C\theta }_{ki}{S\alpha }_{ki}& {{a}_{ki}S\theta }_{ki}\\ \\[-7pt] 0& {\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}S\alpha }_{ki}& {C\alpha }_{ki}&{b}_{ki}\\ \\[-7pt] 0& 0 & 0&1\end{array}\right]\end{align}

Consider $,{a}_{2}=0$ ; $ {T}_{1}=\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}c{\theta }_{1} &-s{\theta }_{1} &0& {a}_{1}c{\theta }_{1}\\s{\theta }_{1} &c{\theta }_{1} &0& {a}_{1}s{\theta }_{1}\\0& 0&1&0\\ 0 &0& 0&1 \end{array}\right]$ $ \textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}{T}_{2}=\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}c{\theta }_{2} &0 & -s{\theta }_{2}& {a}_{2}c{\theta }_{2}\\s{\theta }_{2}& 0 & c{\theta }_{2}& {a}_{2}s{\theta }_{2}\\ 0&0&0&0\\0&0&0&1\\\end{array}\right]$

\begin{align*} {T}_{1}{T}_{2}{T}_{3}=\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}c{\theta }_{12}c{\theta }_{3}& -c{\theta }_{12}s{\theta }_{3} & \textrm{}\textrm{}-s{\theta }_{12}& {\textrm{}\textrm{}\textrm{}a}_{1}c{\theta }_{1}+{a}_{2}c{\theta }_{12}\\ \\[-7pt]s{\theta }_{12}c{\theta }_{3} &-s{\theta }_{12}s{\theta }_{3} & \textrm{}\textrm{}c{\theta }_{12}& {\textrm{}\textrm{}\textrm{}a}_{1}s{\theta }_{1}+\textrm{}{a}_{2}s{\theta }_{12}\\ \\[-7pt]-s{\theta }_{3}& -c{\theta }_{3}& 0&{-a}_{3}s{\theta }_{3}\\ \\[-7pt]0& 0 &0&1\\\end{array}\right]\end{align*}

Using equation 24 and putting in the triangular frame, represented as

(25) \begin{align} \overrightarrow{{L}_{i}}+\textrm{}\overrightarrow{{l}_{i}}=\left[\begin{array}{c}{-a}_{3}s{\theta }_{3}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}a}_{1}c{\theta }_{1}+{a}_{2}c{\theta }_{12}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}a}_{1}s{\theta }_{1}+\textrm{}{a}_{2}s{\theta }_{12}\end{array}\right],\end{align}

(26) \begin{align} \overrightarrow{{L}_{i}}+\textrm{}\overrightarrow{{l}_{i}}=\textrm{}\textrm{}\left[\begin{array}{c}x\\ \\[-7pt] y1\textrm{}\\ \\[-7pt] z\end{array}\right] \end{align}

where $y1=y-{\textrm{u}}_{\textrm{p}}+{\textrm{w}}_{\textrm{B}}$

\begin{align*} {-a}_{3}s{\theta }_{3}=x\\[3pt] {\theta }_{3}=\textrm{}-{\textit{sin}}^{-1}(x/{a}_{3}) \end{align*}

Using equations 25 and 26, square and adding resulted in the computation of the $ {\theta }_{2}$

\begin{align*} {\theta }_{2}=\textrm{}-{\textit{sin}}^{-1}\left(k\right)\end{align*}

where k = (x² + y² + z² – a₁ ² – a₃ ²/2a₁a₃ $ c{\theta }_{3}$ )

(27) \begin{align} y1=\textrm{} {\textrm{}\textrm{}\textrm{}a}_{1}c{\theta }_{1}+{a}_{2}c{\theta }_{12}\end{align}

(28) \begin{align} z\textrm{}\, {\textrm{}\textrm{}\textrm{}a}_{1}s{\theta }_{1}+\textrm{}{a}_{2}s{\theta }_{12} .\end{align}

(29) \begin{align} y1\textrm{}\textrm{*}\textrm{}\left({\textrm{}\textrm{}\textrm{}a}_{3}s{\theta }_{2}c{\theta }_{3}\right)\textrm{}=\textrm{} {\textrm{}\textrm{}\textrm{}a}_{1}c{\theta }_{1}+{a}_{2}c{\theta }_{12} \,{}^{*}\,{\textrm{}\textrm{}\textrm{}a}_{3}s{\theta }_{2}c{\theta }_{3}) \end{align}

(30) \begin{align} z\textrm{*}\left({a}_{1}+{a}_{3}c{\theta }_{2}c{\theta }_{3}\right) {a}_{1}s{\theta }_{1}+\textrm{}{a}_{2}s{\theta }_{12} \,{}^{*}\, ({a}_{1}+{a}_{3}c{\theta }_{2}c{\theta }_{3})\end{align}

Multiplying $ {a}_{3}s{\theta }_{2}c{\theta }_{3}$ in equations 27 and $ {a}_{1}+{a}_{3}c{\theta }_{2}c{\theta }_{3}$ in equation 28 and termed the equations 29 and 30, respectively. Using equations 29 and 30 and subtracting equation 27 from equation 28 resulted in the computation of the $ {\theta }_{1}$ .

\begin{align*} {\theta }_{1}=\textrm{}{\textit{tan}}^{-1}\left(q\right)\end{align*}

(31) \begin{align} \textrm{where} q = (z ( {a}_{1}+{a}_{3}c{\theta }_{2}c{\theta }_{3})-y1 ( {a}_{3}s{\theta }_{2}c{\theta }_{3} )/z ( {a}_{1}+{a}_{3}c{\theta }_{2}c{\theta }_{3}) + y1 ( {a}_{3}s{\theta }_{2}c{\theta }_{3}))\end{align}

Kinematic loop closure equation inherited from the equations 25 and 26, represented as

(32) \begin{align} \left[\begin{array}{c}{-a}_{3}s{\theta }_{3}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}a}_{1}c{\theta }_{1}+{a}_{2}c{\theta }_{12}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}a}_{1}s{\theta }_{1}+\textrm{}{a}_{2}s{\theta }_{12}\end{array}\right]=\textrm{}\textrm{}\left[\textrm{}\begin{array}{c}x\\ \\[-7pt] y1\textrm{}\\ \\[-7pt] z\end{array}\right] \end{align}

Differentiate equation 32 with respect to the time

\begin{align*} &-{a}_{3}c{\theta }_{3}\dot{{\theta }_{3}}=\textrm{}\dot{\textrm{x}}\textrm{}\\ &{a}_{3}\left(-c{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{12}}-\textrm{}s{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{3}}\textrm{}\right)-\textrm{}{a}_{1}s{\theta }_{1}\dot{{\theta }_{1}=} \textrm{}\dot{\textrm{y}1}\\ & {a}_{3}\left(c{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{12}}-\textrm{}s{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{3}}\textrm{}\right)+\textrm{}{a}_{1}c{\theta }_{1}\dot{{\theta }_{1}=} \textrm{}\dot{\textrm{z}}\end{align*}

$ [J \dot{{\theta }}=\dot{\overrightarrow{X}}$ ] represented as

(33) \begin{align} \left[\begin{array}{c@{\quad}c@{\quad}c}0&-{a}_{3}c{\theta }_{3}\dot{{\theta }_{3}}&0\\ \\[-7pt] {-a}_{1}s{\theta }_{1}-{a}_{3}c{\theta }_{3}s{\theta }_{12}&-{a}_{3}c{\theta }_{3}s{\theta }_{12}&-{a}_{3}s{\theta }_{3}c{\theta }_{12}\\ \\[-7pt] \textrm{}{a}_{1}c{\theta }_{1}+{a}_{3}c{\theta }_{3}c{\theta }_{12}&{a}_{3}c{\theta }_{3}c{\theta }_{12}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}-{a}_{3}s{\theta }_{3}s{\theta }_{12}\end{array}\right]\left[\begin{array}{c}\dot{{\theta }_{1}}\\ \\[-7pt] \dot{{\theta }_{2}}\\ \\[-7pt] \dot{{\theta }_{3}}\end{array}\right] &= \left[\begin{array}{c}\dot{x}\\ \\[-7pt] \dot{y}\\ \\[-7pt] \dot{z}\end{array}\right]\textrm{}\textrm{}\dots. \nonumber\\[3pt] \left[J\right]\left[\begin{array}{c}\dot{{\theta }_{1}}\\ \\[-7pt] \dot{{\theta }_{2}}\\ \\[-7pt] \dot{{\theta }_{3}}\end{array}\right]&=\left[\begin{array}{c}\dot{x}\\ \\[-7pt] \dot{y}\\ \\[-7pt] \dot{z}\end{array}\right] \end{align}

Again, foracceleration level, differentiate equation 1.57 to the $ \dot{{\theta }}$ and represented as

(34) \begin{align} J\textrm{}\ddot{\theta }+\dot{J}\dot{\theta }=\ddot{\overrightarrow{X\textrm{}}}\end{align}

where $ \dot{{\theta }}={\left[{\dot{{\theta }}}_{\textrm{k}1}\textrm{}{\dot{{\theta }}}_{\textrm{k}2\textrm{}}{\dot{{\theta }}}_{\textrm{k}3}\right]}^{\textrm{T}}$ and J_k is the 3×3 Jacobian matrix related to the serial chain of the k^th leg. The Jacobian matrix J_k can be obtained by differentiating the first three elements of the last column of the homogeneous transformation matrix T_k.

(35) \begin{align} J_{k} \equiv \left[\begin{array}{c@{\quad}c@{\quad}c}0&-{a}_{3}c{\theta }_{3}\dot{{\theta }_{3}}&0\\ \\[-7pt] {-a}_{1}s{\theta }_{1}-{a}_{3}c{\theta }_{3}s{\theta }_{12}&-{a}_{3}c{\theta }_{3}s{\theta }_{12}&-{a}_{3}s{\theta }_{3}c{\theta }_{12}\\ \\[-7pt] {a}_{1}c{\theta }_{1}+{a}_{3}c{\theta }_{3}c{\theta }_{12}&{a}_{3}c{\theta }_{3}c{\theta }_{12}&-{a}_{3}s{\theta }_{3}s{\theta }_{12}\end{array}\right]\end{align}

Similarly differentiating Eq. 34, the joint acceleration can be derived as

(36) \begin{align} \ddot{\theta }={J}_{k}^{-1}(\ddot{{l}_{k}}-{\dot{J}}_{k}{\dot{\theta }}_{k})\end{align}

where $ {\ddot{\theta }}_{k}\equiv {\left[{\ddot{\theta }}_{k1}{\ddot{\theta }}_{k2}{\ddot{p}}_{k3}\right]}^{T}$ . The derivative of the Jacobian matrix,

$ {\dot{J}}_{k}\equiv \left[{\dot{J}}_{1k}{\dot{J}}_{2k}{\dot{J}}_{3k}\right]$ in Eq. 2 is as follows:

(37) \begin{align} {j}_{1k}\equiv \left[\begin{array}{c}0\textrm{}\textrm{}\\ \\[-7pt] {-a}_{1}c{\theta }_{1}\dot{{\theta }_{1}}-{a}_{3}(c{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{12}}-\textrm{}s{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\\ \\[-7pt] {-a}_{1}s{\theta }_{1}\dot{{\theta }_{1}}-{a}_{3}(c{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{12}}-\textrm{}s{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\textrm{}\end{array}\right]\end{align}

(38) \begin{align} {j}_{2k}\equiv \left[\begin{array}{c}0\textrm{}\textrm{}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}-a}_{3}(c{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{12}}-s{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\\ \\[-7pt] {-a}_{3}{(c\theta }_{3}s{\theta }_{12}\dot{{\theta }_{12}}+c{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\textrm{}\end{array}\right]\end{align}

(39) \begin{align} {j}_{3k}\equiv \left[\begin{array}{c}{\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}a}_{3}s{\theta }_{3}\dot{{\theta }_{3}}\textrm{}\textrm{}\\ \\[-7pt] {\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}-a}_{3}(-s{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{12}}+c{\theta }_{3}c{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\textrm{}\textrm{}\textrm{}\\ \\[-7pt] {-a}_{3}{(s\theta }_{3}c{\theta }_{12}\dot{{\theta }_{12}}+c{\theta }_{3}s{\theta }_{12}\dot{{\theta }_{3}})\textrm{}\textrm{}\end{array}\right]\end{align}

4.1. Dynamics of the legs

The dynamic analysis of the delta robot is performed using the DeNOC formulation [give reference]. Subsystem one, i.e., leg 1, has two links but three-DOF; therefore, to derive the dynamic equations, a virtual link was introduced to make the number of links equal to the degree of freedom of the leg. The unconstraint Newton–Euler equations for the i ^th link of the k ^th leg can be expressed as

(40) \begin{align} {I}_{ik}{\omega }_{ik}+{\omega }_{ik}\textrm{}\times \textrm{}{I}_{ik}{\omega }_{ik}={n}_{ik}\nonumber\\[3pt] {m}_{ik}{\ddot{c}}_{ik}={f}_{ik} \end{align}

where $ {I}_{ik}$ and $ {m}_{ik}$ are the inertia tensor and mass of the link, respectively, $ {\omega }_{ik}$ and $ {\ddot{c}}_{ik}$ are the angular velocity and linear velocity of centre-of-mass (COM), respectively, while $ {\omega }_{ik}$ and $ {\ddot{c}}_{ik}$ are their corresponding time derivative, $ {n}_{ik}$ and $ {f}_{ik}$ are the moment and force about/at the COM of the link, respectively. The above expression can be written in matrix-vector as given below

(41) \begin{align} {M}_{ik}{\dot{t}}_{ik}+{W}_{ik}{M}_{ik}{t}_{ik}={\omega }_{ik}\end{align}

where $ {M}_{ik}=\left[\begin{array}{cc}{I}_{ik} {O}_{3}\\ {O}_{3} {m}_{ik}{I}_{3}\end{array}\right]$ is the mass matrix, $ {t}_{ik}=\left[\begin{array}{c}{\omega }_{ik}\\ {\dot{c}}_{ik}\end{array}\right]$ is the twist vector, $ {W}_{ik}=\left[\begin{array}{cc}{\omega }_{ik}{I}_{3} {O}_{3}\\ {O}_{3} {O}_{3}\end{array}\right]$ is the angular velocity matrix, $ {\omega }_{ik}=\left[\begin{array}{c}{n}_{ik}\\ {f}_{ik}\end{array}\right]$ is the wrench vector, and $ {1}_{3}$ and $ {\textit{O}}_{3}$ are 3x3 identity and null matrix, respectively. The unconstraint equations for the k ^th leg can be written as

(42) \begin{align} {M}_{ik}{\dot{t}}_{ik}+{W}_{ik}{M}_{ik}{t}_{ik}={\omega }_{ik}\end{align}

where $ {M}_{k}=diag\left[\begin{array}{cc}{M}_{1k} \begin{array}{cc}{M}_{2k} {M}_{3k}\end{array}\end{array}\right]$

\begin{align*} \textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}{W}_{k}=diag\left[\begin{array}{cc}{W}_{1k} \begin{array}{cc}{W}_{2k} {W}_{3k}\end{array}\end{array}\right]\\[3pt]\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}\textrm{}{t}_{k}={\left[\begin{array}{cc}{t}_{1k} \begin{array}{cc}{t}_{2k} {t}_{3k}\end{array}\end{array}\right]}^{T}\\[3pt]{\textrm{}w}_{k}={\left[\begin{array}{cc}{w}_{1k} \begin{array}{cc}{w}_{2k} {w}_{3k}\end{array}\end{array}\right]}^{T}\end{align*}

4.1.1. Derivation of DeNOC matrices through kinematic constraints

The angular velocity and linear velocity of COM of the links in k ^th can be written in recursive as given below

(43) \begin{align} {\omega }_{1k}&={\dot{\theta }}_{1k}{e}_{1k} \nonumber\\[3pt] {\dot{c}}_{1k}&={\omega }_{1k}\times {d}_{1k}={\dot{\theta }}_{1k}{e}_{1k}\times {d}_{1k}\end{align}

(44) \begin{align} {\omega }_{2k}&={\omega }_{1k}+{\dot{\theta }}_{2k}{e}_{2k} \nonumber\\[3pt] {\dot{c}}_{2k}&={\textrm{}{\dot{c}}_{1k}+\textrm{}\omega }_{1k}\times {({r}_{1k}+d}_{2k})+{\dot{\theta }}_{2k}{e}_{2k}\times {d}_{2k}\end{align}

(45) \begin{align} {\omega }_{3k}&={\omega }_{3k}+{\dot{\theta }}_{3k}{e}_{3k} \nonumber \\[3pt] {\dot{c}}_{3k}&={\textrm{}{\dot{c}}_{2k}+\textrm{}\omega }_{2k}\times {({r}_{2k}+d}_{3k})+{\dot{\theta }}_{3k}{e}_{3k}\times {d}_{3k} \\[-5pt]\nonumber\end{align}

The corresponding twist vector for each link of the k ^th leg can be expressed as

(46) \begin{align} {t}_{1k}={p}_{1k}{\dot{\theta }}_{1k};\ \textrm{}{t}_{2k}={B}_{\textrm{2,1}k}{t}_{1k}+{p}_{2k}{\dot{\theta }}_{2k};\ {t}_{3k}={B}_{\textrm{3,2}k}{t}_{2k}+{p}_{2k}{\dot{\theta }}_{3k}\end{align}

where $ {P}_{ik}=\left[\begin{array}{c}{e}_{ik}\\ {e}_{ik}\times {d}_{ik}\end{array}\right];\ i=\textrm{1,2},3$

\begin{align*} {B}_{i,i-1k}=\left[\begin{array}{c@{\quad}c}{1}_{3}& {O}_{3}\\ {c}_{i,i-1k}\times {1}_{3}& {1}_{3}\end{array}\right] \nonumber\\[3pt] {\textrm{}c}_{i,i-1k}=-\left({r}_{i-1k}+{d}_{ik}\right);i=\textrm{2,3}. \end{align*}

The above twist vector can be written as $ {t}_{k}={N}_{k}{\dot{\theta }}_{k}$ with $ {N}_{k}={N}_{lk}{N}_{dk}$

where

\begin{align*} {N}_{lk}\equiv \left[\begin{array}{c@{\quad}c@{\quad}c}{1}_{6} &{O}_{6}& {O}_{6}\\ \\[-7pt] {B}_{\textrm{2,1}k}& {1}_{6}& {O}_{6}\\ \\[-7pt] {B}_{\textrm{3,1}k} &{B}_{\textrm{3,2}k}& {1}_{6}\end{array}\right];\ {N}_{dk}\equiv \left[\begin{array}{c@{\quad}c@{\quad}c}{P}_{1k}& \phi & \phi \\ \\[-7pt] \phi & {P}_{2k}& \phi \\ \\[-7pt] \phi & \phi & {P}_{3k}\end{array}\right];\ {\dot{\theta }}_{k}\equiv \left[\begin{array}{c}{\dot{\theta }}_{1k}\\ \\[-7pt] {\dot{\theta }}_{2k}\\ \\[-7pt] {\dot{\theta }}_{3k}\end{array}\right]\end{align*}

$ {1}_{6}$ and $ {O}_{6}$ are the $ 6\times 6$ identity and null matrix, respectively, while $\phi $ is the $ 6\times 1$ zero vector.The $ 6\times 6$ matrix $ {B}_{\textrm{3,1}k}$ is the matrix multiplication of $ {B}_{\textrm{3,2}k}{B}_{\textrm{2,1}k}$ . The above $ {N}_{k}$ matrix is known as the Natural Orthogonal Complement (NOC) matrix, and its decoupled form $ {N}_{lk}{N}_{dk}$ is known as Decoupled NOC, which would be used to project the unconstraint equations of motion on the kinematic constraint space as given below:

(47) \begin{align} {N}_{dk}^{T}{N}_{lk}^{T}\left\{{M}_{k}{\dot{t}}_{k}+{W}_{k}{M}_{k}{t}_{k}\right\}={N}_{dk}^{T}{N}_{lk}^{T}{w}_{k}\end{align}

(48) \begin{align} {I}_{k}\left({\theta }_{k}\right){\ddot{\theta }}_{k}+{C}_{k}\left({\theta }_{k},{\dot{\theta }}_{k}\right){\dot{\theta }}_{k}=-{\gamma }_{k}+{\tau }_{k}\\ \nonumber \end{align}

where $ {I}_{k}\left({\theta }_{k}\right)\equiv {N}_{k}^{T}{M}_{k}{N}_{k}$ is the generalized inertia matrix (GIM); $ {C}_{k}\left({\theta }_{k},{\dot{\theta }}_{k}\right)\equiv {N}_{k}^{T}\left[{M}_{k}{\dot{N}}_{k}+{W}_{k}{M}_{k}{N}_{k}\right]$ is the matrix of Coriolis forces; and $ -{\gamma }_{k}+{\tau }_{k}\equiv {N}_{k}^{T}{w}_{k}$ in which $ {\gamma }_{k}$ the generalized is the force due to gravity and $ {\tau }_{k}$ represents the actuator torques. The above equations of motion considering the Lagrange multipliers ${\bf{\lambda }}$ can be expressed as

(49) \begin{align} {I}_{k}\left({\theta }_{k}\right){\ddot{\theta }}_{k}+{C}_{k}\left({\theta }_{k},{\dot{\theta }}_{k}\right){\ddot{\theta }}_{k}+{\gamma }_{k}={J}_{k}^{T}{}_{k}+{\tau }_{k}\end{align}

where $ {J}_{k}$ is the Jacobian matrix and $k = 1,2,3.$

4.2. Dynamics of the platform

The platform is shown in Fig. 11. The local frame attached to the centroid $ {0}_{p}$ of platform is $ {X}_{p}-{Y}_{p}-{Z}_{p}$ . The platform can only translate in 3D space, and therefore, the rotation matrix between the base frame and the platform frame is the identity matrix. The platform is connected to the legs at points $ {p}_{1},{p}_{2}$ , and $ {p}_{3}$ . The position vector of the moving platform with respect to the base frame is given below

(50) \begin{align} p={\left[x\textrm{}y\textrm{}z\right]}^{T}\end{align}

The dynamics of the platform using Newton’s equation of motion can be expressed as

(51) \begin{align} {m}_{p}\ddot{p}+{\gamma }_{p}={\lambda}_{1}+{\lambda}_{2}+{\lambda}_{3}\end{align}

where $ {m}_{p}$ is the mass of the platform, $ \ddot{p}$ is the acceleration of the platform, $ {\gamma }_{p}$ is the gravity force acting on the platform, and $ {\lambda}_{1},{\lambda}_{2},{\lambda}_{3}$ representing the Lagrange’s multipliers corresponding to the three-legs.

4.3. Dynamics of the delta robot

The dynamics of the delta robot can be expressed by combining the dynamics of the legs and the platform. The dynamics of the delta robot using Lagrange’s multipliers and Jacobian matrices corresponding to each leg can be expressed in matrix-vector form as

(52) \begin{align} I\left(\theta \right)\ddot{\theta }+C\left(\theta ,\dot{\theta }\right)\ddot{\theta }+\gamma ={J}^{T}\lambda +\tau \end{align}

where $ I\left(\theta \right)\equiv \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{I}_{1}& {0}_{3} & {0}_{3}& {0}_{3} \\ {0}_{3}& {I}_{2} &{0}_{3} &{0}_{3}\\{0}_{3} &{0}_{3}& {I}_{3}& {0}_{3}\\{0}_{3}& {0}_{3} & {0}_{3}& {m}_{p}{I}_{3}\end{array}\right]$

\begin{align*} C(\theta ,\dot{\theta })\equiv \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{C}_{1}& {0}_{3}& {0}_{3}& {0}_{3}\\ \\[-7pt] {0}_{3}& {C}_{2} &{0}_{3}& {0}_{3}\\ \\[-7pt]{0}_{3} &{0}_{3}& {C}_{3} &{0}_{3}\\ \\[-7pt] {0}_{3}& {0}_{3}& {0}_{3}& {0}_{3}\end{array}\right]; \,\, \textrm{}\gamma \equiv \left[\begin{array}{c}{\gamma }_{1}\\ \\[-7pt] {\gamma }_{2}\\ \\[-7pt] \begin{array}{c}{\gamma }_{3}\\ \\[-7pt] {\gamma }_{p}\end{array}\end{array}\right]\end{align*}

\begin{align*} J\equiv \left[\begin{array}{c@{\quad}c@{\quad}c}{J}_{1}& {0}_{3}& {0}_{3}\\ \\[-7pt] {0}_{3}& {J}_{1} &{0}_{3}\\ \\[-7pt] {0}_{3}& {0}_{3}& {J}_{1} \\ \\[-7pt] {I}_{3} & {I}_{3} & {I}_{3} \end{array}\right];\,\,\, \textrm{}\lambda \equiv \left[\begin{array}{c}{\lambda }_{1}\\ \\[-7pt] {\lambda }_{2}\\ \\[-7pt] {\lambda }_{3}\end{array}\right]\,\, {\rm and}\,\, \tau \equiv \left[\begin{array}{c}{\tau }_{1}\\ \\[-7pt] {\tau }_{2}\\ \\[-7pt] {\tau }_{3}\end{array}\right]\end{align*}

The dynamic analyses of the multibody system are required to know the trajectory of different coordinates for a given torque or for a required trajectory what should be the torque at a different degree of freedom [Reference Gupta, Chaudhary and Saha29]. The dynamic modelling of the manipulator (parallel) is essential for their control because they are preferred in an application where precise positioning is required. Combining the equations of motion yields equation 53 and is represented as

(53) \begin{align} &\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{I}_{{1}_{3x3}}& {0}_{3x3}& 0& 0\\ \\[-11pt] 0& {I}_{2}& 0& 0\\ \\[-11pt]0& 0& {I}_{3} &0\\ \\[-11pt]0& 0 & 0& {M}_{pf}\end{array}\right]\left[\begin{array}{c}\ddot{{\theta }_{1}}\\ \\[-11pt] \begin{array}{c}\ddot{{\theta }_{2}}\\ \\[-11pt] \ddot{{\theta }_{3}}\\ \\[-11pt] \begin{array}{c}\ddot{{\theta }_{4}}\\ \\[-11pt] \ddot{{\theta }_{5}}\\ \\[-11pt] \begin{array}{c}\ddot{{\theta }_{6}}\\ \\[-11pt] \ddot{{\theta }_{7}}\\ \\[-11pt] \begin{array}{c}\ddot{{\theta }_{8}}\\ \\[-11pt] \ddot{{\theta }_{9}}\\ \\[-11pt] \begin{array}{c}\ddot{{x}_{p}}\\ \\[-11pt] \begin{array}{c}\ddot{{y}_{p}}\\ \\[-11pt] \ddot{{z}_{p}}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\right]+\left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{C}_{1}& 0&0 &0\\ \\[-11pt] 0& {C}_{2}& 0& 0\\ \\[-11pt] 0& 0& {C}_{3} &0\\ \\[-11pt] 0 &0& 0& 0\end{array}\right]+\left[\begin{array}{c}{\gamma }_{1}\\ \\[-11pt] {\gamma }_{2}\\ \\[-11pt] \begin{array}{c}{\gamma }_{3}\\ \\[-11pt] {\gamma }_{pf}\end{array}\end{array}\right] \nonumber\\[3pt] & \quad =\left[\begin{array}{c@{\quad}c@{\quad}c}{J}_{{c1}_{3x3}}^{T} &0& 0\\ \\[-11pt]0& {J}_{{c2}_{3x3}}^{T}& 0 \\ \\[-11pt]0&0&{J}_{{c3}_{3x3}}^{T}\\ \\[-11pt]{1}_{3x3}& {1}_{3x3}& {1}_{3x3}\end{array}\right]{\left[\begin{array}{c}{\lambda }_{1}x\dots z\\ \\[-11pt] {\lambda }_{2}x\dots z\\ \\[-11pt] {\lambda }_{3}x\dots z\end{array}\right]}_{9x1}+\left[\begin{array}{c}\begin{array}{c}{\tau }_{1}\\ \\[-11pt] 0\\ \\[-11pt] 0\end{array}\\ \\[-11pt] \begin{array}{c}{\tau }_{1}\\ \\[-11pt] 0\\ \\[-11pt] 0\end{array}\\ \\[-11pt] \begin{array}{c}{\tau }_{1}\\ \\[-11pt] 0\\ \\[-11pt] \begin{array}{c}0\\ \\[-11pt] 0\\ \\[-11pt] \begin{array}{c}0\\ \\[-11pt] 0\end{array}\end{array}\end{array}\end{array}\right]\end{align}

The RHS of equation 53 can also be re-written for inverse dynamics as

(54) \begin{align} \left[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c}1& {J}_{11}& {J}_{21}&{J}_{31} &0&0&0&0&0&0&0&0\\ \\[-11pt] 0& {J}_{12}& {J}_{22}&{J}_{32}&0&0&0&0&0&0&0&0\\ \\[-11pt] 0& {J}_{13}& {J}_{23}& {J}_{33}&0&0&0&0&0&0&0&0\\ \\[-11pt] 0&0&0&0&1& {J}_{11}& {J}_{21}&{J}_{31}&0&0&0&0\\ \\[-11pt] 0&0&0&0& 0& {J}_{12}& {J}_{22}&{J}_{32}&0&0&0&0\\ \\[-11pt] 0&0&0&0& 0& {J}_{13}& {J}_{23}& {J}_{33}&0&0&0&0\\ \\[-11pt]0&0&0&0&0&0&0&0&1&{J}_{11}& {J}_{21}&{J}_{31}\\ \\[-11pt]0&0&0&0&0&0&0&0&0& {J}_{12}& {J}_{22}&{J}_{32}\\ \\[-11pt]0&0&0&0&0&0&0&0&0& {J}_{13}& {J}_{23}& {J}_{33}\\ \\[-11pt]0&1&0&0&0&1&0&0&0&1&0&0\\ \\[-11pt]0&0&1&0&0&0&1&0&0&0&1&0\\ \\[-11pt]0&0&0&1&0&0&0&1&0&0&0&1 \\ \\[-11pt] \end{array}\right]\left[\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}\begin{array}{c}{\tau }_{1}\\ \\[-11pt] {\lambda }_{1x}\end{array}\\ \\[-11pt] {\lambda }_{1y}\\ \\[-11pt] {\lambda }_{1z}\end{array}\\ \\[-11pt] {\tau }_{2}\\ \\[-11pt] {\lambda }_{2x}\end{array}\\ \\[-11pt] {\lambda }_{2y}\\ \\[-11pt] {\lambda }_{2z}\end{array}\\ \\[-11pt] {\tau }_{3}\\ \\[-11pt] {\lambda }_{3x}\end{array}\\ \\[-11pt] {\lambda }_{3y}\\ \\[-11pt] {\lambda }_{3z}\end{array}\right]\end{align}

4.4. Numerical illustration

This section presents the implementation of the inverse dynamics methodology for the three – DOF delta robots proposed in section IV (B).

\begin{align*}{{\bf{J}}_{\tau k}} \equiv \left[ {\begin{array}{*{20}{c}}{{J_k}(1,3)}\\ \\[-7pt]{{J_k}(2,3)}\\ \\[-7pt]{{J_k}(3,3)}\end{array}} \right];\,{{\bf{\lambda }}_k} \equiv \left[ {\begin{array}{*{20}{c}}{{\lambda _{xk}}}\\ \\[-7pt]{{\lambda _{yk}}}\\ \\[-7pt]{{\lambda _{zk}}}\end{array}} \right]\end{align*}

Figure 10(a) shows the delta robot whose mass centre was offset to the geometric centre of the RecurDyn^TM model for demonstration. Referring to Fig. 10(b), each cut joint is represented by three components of the Lagrange multipliers vector ${\bf{\lambda }}$ , as the joints are spherical.

Figure 11

(a) Forward kinematics results [X axis] of the delta robot. (b) Forward kinematics results [Y axis] of the delta robot. (c) Forward kinematics results [Z axis] of the delta robot. (d) Inverse kinematics results (Joint angle profile (Leg1)). (e) Inverse kinematics results (joint angle profile (Leg2)). (f) Inverse kinematics results (joint angle profile (Leg3)).

In a way, there are nine components of the Lagrange multipliers to be computed due to three cut joints. Using the methodology explained in section IV (D), forces at the legs were calculated as given next.

(55) \begin{align}\phi _u^* \equiv \left[ {\begin{array}{*{20}{c}}{{\phi _{I1}}}\\ \\[-7pt]{{\phi _{I2}}}\\ \\[-7pt] \vdots \\ \\[-7pt]{{\phi _{VI1}}}\\ \\[-7pt]{{\phi _{VI2}}}\end{array}} \right];\,{{\bf{J}}_u} \equiv \left[ {\begin{array}{c@{\quad}c@{\quad}c@{\quad}c}{{{\bf{J}}_{Iu}}}&{\bf{O}} &\cdots& {\bf{O}}\\ \\[-7pt]{\bf{O}}&{{{\bf{J}}_{IIu}}} &\cdots &{\bf{O}}\\ \\[-7pt] \vdots & \vdots & \ddots& \vdots \\ \\[-7pt]{\bf{O}}&{\bf{O}} &\cdots &{{{\bf{J}}_{VIu}}}\end{array}} \right]\end{align}

where $\phi _u^*$ is the 12-dimensional vector related to unactuated joints of the legs, ${{\bf{J}}_u}$ is 12×18 matrix containing the elements of the Jacobian matrices associated with the unactuated joints of the legs, and ${\bf{O}}$ is the 2×3 matrix of zeros. The diagonal elements ${{\bf{J}}_u}$ have been represented using the k^th subsystem as

\begin{align*}{{\bf{J}}_{ku}} \equiv \left[ {\begin{array}{*{20}{c}}{{J_k}(1,1)}{{J_k}(2,1)}{{J_k}(3,1)}\\ \\[-7pt] {{J_k}(1,2)}{{J_k}(2,2)}{{J_k}(3,2)}\end{array}} \right]\end{align*}

6.1. Kinematics simulation results

The 3D trajectories with respect to the fixed base frame of all three delta robots are shown in Fig. 13(a).

Figure 13

(a) The 3D trajectory of delta robot {1, 2, 3} with respect to delta frame. (b) Joint angles of Delta1 [Leg1, Leg2, Leg3]. (c) Joint angles of Delta2 [Leg1, Leg2, Leg3]. (d) Joint angles of Delta3 [Leg1, Leg2, Leg3]. (e) The workspace of Delta1 [Leg1, Leg2, Leg3].

Given the position vectors to the fixed base frame and performing the inverse kinematics resulted in the joint angles of each of the Deltas as shown in Fig. 13(b)–(d). From Fig. 13(b)–(d), each leg of the delta robot starts from 0 degrees and reaches the final point at 30 degrees.

Given the input, the joint angles and forward kinematics resulted in the position vectors to the fixed base frame of each delta, as shown in Fig. 13(e).

6.2. Kinematics results validated using RecurDyn ^TM

The position and joint angles plots are given in Fig. 11(a)–(f), respectively. Note that to validate the results, the kinematic analyses were carried out using the commercial software RecurDyn^TM. The comparison in the above figures proves the correctness of the proposed methodology.

6.3. Dynamics analysis results

The dynamic analysis performed for the straight-line trajectory, i.e., initial x_i = 1.0384e-15, y_i = 0.00002479, z_i = 0.0878350; ${{\rm{P}}_1} = \left[ {\begin{array}{*{20}{c}}{{x_{\rm{i}}}}\\{{y_{\rm{i}}}}\\{{z_{\rm{i}}}}\end{array}} \right];{\rm{\;}}$ and the final x_f = 1.400e-15, y_f = 0.0000248, z_f = 0.1828063; ${{\rm{P}}_2} = \left[ {\begin{array}{*{20}{c}}{{x_f}}\\{{y_{\rm{f}}}}\\{{z_{\rm{f}}}}\end{array}} \right].{\rm{\;\;}}$

Case I. [From P1 to P2]

Figure 14(a)–(c) shows the position, velocity, and acceleration profile. The torque acting on the active joints (rotatory) is shown in Fig. 14(d). The torque acting on the passive joints (universal) are shown in Fig. 14(e)–(h).

Figure 14

(a) Forward kinematics results of the delta robot [P1 to P2]. (b) Velocity profile for the P1 to P2 trajectory. (c) Acceleration of the active joints (R1, R2, R3). (d) Joint angle. (e) The torque acting on the active joints (R1, R2, R3). (f) Lagrange multipliers acting on the passive joint on leg1 (P₁, P₂, P₃). (g) Lagrange multipliers acting on the passive joint on leg2 (P₄, P₅, P₆). (h) Lagrange multipliers acting on the passive joint on leg3 (P₇, P₈, P₉).

Case II. [From P2 to P1]

The joint angle, position, velocity, and acceleration profile are shown in Fig. 15(a)–(d). The torque acting on the active joints (rotatory) is shown in Fig. 15(e).

Figure 15

(a) Forward kinematics results of the delta robot [P2 to P1]. (b) Velocity profile for the P2 to P1 trajectory. (c) Acceleration of the active joints (R1, R2, R3). (d) Joint angle. (e) The torque acting on the active joints (R1, R2, R3).

Case III. Circular Trajectory

The joint angle, position, velocity, and acceleration profile are shown in Fig. 16(a)–(d). The torque acting on the active joints (rotatory) is shown in Fig. 16(e).

Figure 16

(a) Forward kinematics results of the delta robot [circular trajectory]. (b) Velocity profile for the P2 to P1 trajectory. (c) Acceleration of the active joints (R1, R2, R3). (d) Joint angle. (e) The torque acting on the active joints (R1, R2, R3).

Case IV. Helical Trajectory

The joint angle, position, velocity, and acceleration profile are shown in Fig. 17(a)–(d). The torque acting on the active joints (rotatory) is shown in Fig. 17(e).

Figure 17

(a) Forward kinematics results of the delta robot [helical trajectory]. (b) Velocity profile for the P2 to P1 trajectory. (c) Acceleration of the active joints (R1, R2, R3). (d) Joint angle. (e) The torque acting on the active joints (R1, R2, R3).

6.4. Real-time controlling of the delta manipulator

In this case, a planar circular path was input. The delta robot movement while following the path 02 times recorded by the encoders. The planar circular path is shown in Fig. 18(a).

Figure 18

(a) A planar circular path is provided to the delta robot. (b) Desired and measured end-effector (circle trajectory) position.

Figure 19.

Delta robot FPK diagram.

Figure 20

(a) A helical trajectory path is provided to the delta robot. (b) Desired and measured end-effector (helical trajectory) position. (c) The desired angle for Leg1, Leg2, and Leg3 in helical profile. (d) Measured angle for Leg1, Leg2, and Leg3 in helical profile.

In this case, a helical trajectory was given input. The delta robot movement following the path was recorded. As the helical path is provided to the delta robot, as shown in Fig. 19(a), to the inverse kinematics function and its corresponding theta angles estimated. The desired and measured end-effector (helical trajectory) position is shown in Fig. 19(b). The error in executing a circular profile is analysed. Desired angles in Leg1, Leg2 and Leg3 are shown in Fig. 19(c), and measured angles in Leg1, Leg2 and Leg3 are shown in Fig. 19(d).

I have hence tested the fundamental motions. Since the general motion of a rigid body in space is described by a screw (Chasal’s Theorem), I have also tested helical motion. I have hence validated that the error between the vision-based measurement and kinematics based on link length and encoder angles is of the order of a degree.

The result section (VI), sub-section A (kinematics simulation results), sub-section B (kinematics results validated using RecurDyn^TM and sub-section C (dynamics analysis results) presented the simulation-based implementation. The sub-section D (real-time controlling of the delta manipulator) presented the experimental results using the real robot(s) (delta robot).