A Generalised Quaternion and Clifford Algebra Based Mathematical Methodology to Effect Multi-stage Reassembling Transformations in Parallel Robots

Abstract

A robotic manipulator is a networked assemblage of kinematic pairs, mobile platform(s) and/or tool(s) that generate motion and forces at its end-effector. A manipulator’s theoretical dynamic, kinematic and workspace capabilities are governed by the design parameters (mass, geometry, dimensions, etc.) of its kinematic pairs and its architecture (number of limbs, degrees of freedom, actuation ability, etc.). By adjusting or changing the manipulator’s parameters (both architectural and kinematic pair-related), either from a design standpoint or physically, the robot undergoes what the authors broadly refer to as a ‘reassembling transformation’. By effecting reassembling transformations on a robot, we may generate desired changes in the workspace, kinematic and dynamic capabilities for a specific task manoeuvre. This work presents a novel methodology that marries quaternion rotors, the Lagrangian formulation, geometric algebra and associated mathematical apparatus, to tackle the problem of analysing and modelling robotic manipulators or mechanisms that undergo such ‘reassembling transformations’. Upon presenting a quaternion-based methodology for position, kinematics and Lagrangian dynamics, reassembling transformation types are introduced and detailed. We apply reassembling transformations to a Delta manipulator, in two separate case studies, each with unique target changes to the dynamic capabilities of the post transformed robot in its chosen task manoeuvres. The framework presented in this paper is generalised and its methods can be used to analyse and transform any parallel manipulator and also extended to the study of non-linear systems of equations for modelling multi-body dynamics.

Appendix

1.1 Appendix A: Derivation of the Quaternion Rotor based Lagrangian expressions for a Pre-Transformed Delta Manipulator

In Sect. 2.1, Eq. (18), the terms $E_l$,$\ F_l$,$\ G_l$, for $l=1,2,3$, are defined as follows:

$$\begin{aligned} E_1= & {} 2|X_{1\ 2}||X_{0\ 1}|- 2|X_{1\ 2}||X_{5\ E}|- 2|X_{1\ 2}|x, \end{aligned}$$

(109a)

$$\begin{aligned} F_1= & {} 2|X_{1\ 2}|z, \end{aligned}$$

(109b)

$$\begin{aligned} G_1= & {} {|X_{0\ 1}|}^2-2|X_{0\ 1}||X_{5\ E}|-2|X_{0\ 1}|x + {|X_{5\ E}|}^2\nonumber \\&+ 2|X_{5\ E}|x + x^2+ y^2+ z^2+ {|X_{1\ 2}|}^2- {|X_{3\ 4}|}^2, \end{aligned}$$

(109c)

$$\begin{aligned} E_2= & {} 2|X_{6\ 7}||X_{0\ 6}|- 2|X_{6\ 7}||X_{10\ E}|+ |X_{6\ 7}|x - \sqrt{3}|X_{6\ 7}|y, \end{aligned}$$

(109d)

$$\begin{aligned} F_2= & {} 2|X_{6\ 7}|z, \end{aligned}$$

(109e)

$$\begin{aligned} G_2= & {} {|X_{0\ 6}|}^2- 2|X_{0\ 6}||X_{10\ E}|+ |X_{0\ 6}|x - \sqrt{3}|X_{0\ 6}|y + {|X_{10\ E}|}^2\nonumber \\&- |X_{10\ E}|x +\sqrt{3}|X_{10\ E}|y + x^2+ y^2+ z^2+ {|X_{6\ 7}|}^2- {|X_{8\ 9}|}^2, \nonumber \\\end{aligned}$$

(109f)

$$\begin{aligned} E_3= & {} 2|X_{11\ 12}||X_{0\ 10}|- 2|X_{11\ 12}||X_{15\ E}|+ |X_{11\ 12}|x + \sqrt{3}|X_{11\ 12}|y, \nonumber \\\end{aligned}$$

(109g)

$$\begin{aligned} F_3= & {} 2|X_{11\ 12}|z, \end{aligned}$$

(109h)

$$\begin{aligned} G_3= & {} {|X_{0\ 10}|}^2- 2|X_{0\ 10}||X_{15\ E}|+ |X_{0\ 10}|x + \sqrt{3}|X_{0\ 10}|y \nonumber \\&+{|X_{15\ E}|}^2- |X_{15\ E}|x - \sqrt{3}|X_{15\ E}|y + x^2+ y^2+ z^2\nonumber \\&+ {|X_{11\ 12}|}^2- {|X_{13\ 14}|}^2=0. \end{aligned}$$

(109i)

1.2 Appendix B: Derivation of the Quaternion Rotor based Lagrangian expressions for a Pre-Transformed Delta Manipulator

We present expressions for the quaternion rotor based Lagrangian dynamics of a Delta manipulator, where the expressions for the inverse position solutions to find ${\theta }_{l,1}$ are given in Eq. (22), while the expression for the inverse kinematics solutions, ${\dot{\theta }}_{l,1}$ and ${\ddot{\theta }}_{l,1}$ are given in Eqs. (27a) and (27b), respectively. In this example, we set the six generalised coordinates to be $c_1,c_2,\dots ,c_6=x,y,z,{\theta }_{1,1},{\theta }_{2,1},{\theta }_{3,1}$, where x, y, z are redundant coordinates [50]. Since the distance between kinematic pairs 3 and 4 (${(X_{3\ 4})}^2$) is always equal to the length of the connecting rod of the upper arm $a_2=|H_3|=|X_{3\ 4}|$ (based on Fig. 5), we define the constraint function ${{\varGamma }}_l$ as follows [50]

$$\begin{aligned} {{\varGamma }}_l= & {} {\left( X^l_{3\ 4}\right) }^2-{\left( |X^l_{3\ 4}|\right) }^2\nonumber \\= & {} {\left( x+|X^l_{5\ E}|{\cos \left( {\psi }_l\right) }-|X^l_{0\ 1}|{\cos \left( {\psi }_l\right) } -|X^l_{1\ 2}|{\cos \left( {\psi }_l\right) }{\cos \left( {\theta }_{l,1}\right) }\right) }^2\nonumber \\&+ {\left( y+|X^l_{5\ E}|{\sin \left( {\psi }_l\right) }-|X^l_{1\ 2}|{\sin \left( {\psi }_l\right) } {\cos \left( {\theta }_{l,1}\right) }\right) }^2\nonumber \\&+ {\left( z-{|X^l_{1\ 2}|\sin \left( {\theta }_{l,1}\right) }\right) }^2-{\left( |X^l_{3\ 4}|\right) }^{2}=0 \end{aligned}$$

(110)

where $l=1,2,3$ and the symbol ${\psi }_l$ denotes the angles between the vector from the origin to the first joint of each l-th limb and the x-axis of the base reference frame (such that ${\psi }_1,{\psi }_2,{\psi }_3=0,\ \frac{2\pi }{3},-\frac{2\pi }{3}$). To simplify the analysis, we assume that the mass of each connecting rod in the forearm ($m_2$) is divided evenly and concentrated at the two endpoints ($X_{0\ 3}\$and$\ X_{0\ 4})$. As a result, we derive the Lagrangian function L by first defining the total kinetic energy of the manipulator as

$$\begin{aligned} K=K_E+\sum ^3_{l=1}{(K_{a^l_1}+K_{a^l_2})} \end{aligned}$$

(111)

where $K_E$ is the kinetic energy of the end-effector platform, $K_{a^l_1}$ is the kinetic energy of the bicep link of the l-th limb and $K_{a^l_2}$ is the kinetic energy of the forearm of the l-th limb. Noting that $|X_{0 \ 1}|=|X_{0 \ 6}|=|X_{0 \ 11}|$, $|X_{1 \ 2}|=|X_{6 \ 7}|=|X_{11 \ 12}|$, $|X_{3 \ 4}|=|X_{8 \ 9}|=|X_{13 \ 14}|$ and $|X_{5 \ E}|=|X_{10 \ E}|=|X_{15 \ E}|$ this gives (which is in agreement with [50])

$$\begin{aligned} K_E= & {} \frac{1}{2}m_E({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2), \end{aligned}$$

(112)

$$\begin{aligned} K_{a^l_1}= & {} \frac{1}{2}\left( I_{motor}+\frac{1}{3}m_{a_1}{\left( |X_{1\ 2}|\right) }^2\right) {\left( {\dot{\theta }}_{l,1}\right) }^2, \end{aligned}$$

(113)

$$\begin{aligned} K_{a^l_2}= & {} \frac{1}{2}m_{a_2}\left( {\dot{x}}^{\mathrm {2}}\mathrm {+}{\dot{y}}^{\mathrm {2}}\mathrm {+}{\dot{z}}^{\mathrm {2}}\right) +\frac{1}{2}m_{a_2}{\left( |X_{1\ 2}|\right) }^2{\left( {\dot{\theta }}_{l,1}\right) }^2, \end{aligned}$$

(114)

where $m_E$ is the mass of the end-effector platform, $m_{a_1}$ is the mass of the bicep and $m_{a_2}$ is the mass of one of the two connecting rods in the forearm. The mass of the link and its density distribution is assumed to be identical in each limb. $I_{motor}$ is the axial moment of inertia of the rotor mounted on the l-th limb. Assuming that gravitational acceleration is in the $-z$ direction, the total potential energy of the manipulator relative to the fixed $x-y$ plane is

$$\begin{aligned} U=U_E+\sum ^3_{l=1}{(U_{a^l_1}+U_{a^l_2})} \end{aligned}$$

(115)

where $U_E$ is the potential energy of the end-effector platform, $U_{a^l_1}$ is the potential energy of the bicep link of the l-th limb and $U_{a^l_2}$ is the potential energy of the forearm of the l-th limb, giving [50]

$$\begin{aligned}&\displaystyle U_E=\frac{1}{2}m_Eg_cz,&\end{aligned}$$

(116)

$$\begin{aligned}&\displaystyle U_{a^l_1} = \frac{1}{2}m_{a_1}g_c|X_{1\ 2}|\sin ({\theta }_{l,1}),&\end{aligned}$$

(117)

$$\begin{aligned}&\displaystyle U_{a^l_2} = m_{a_2}g_c\left( z+|X_{1\ 2}|\sin ({\theta }_{l,1})\right) ,&\end{aligned}$$

(118)

where $g_c$ is the symbol for gravitational acceleration. The Lagrangian function is defined, using Eqs. (111)–(118), as

$$\begin{aligned} L= & {} K-U\nonumber \\= & {} \frac{1}{2}\left( m_E+3m_{a_2}\right) \left( {\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2\right) \nonumber \\&+ \frac{1}{2}\left( I_{motor}+\frac{1}{3}m_{a_1}{\left( |X_{1\ 2}|\right) }^2+m_{a_2}{\left( |X_{1\ 2}|\right) }^2\right) \nonumber \\&\quad \left( {\left( {\dot{\theta }}_{1,1}\right) }^2+{\left( {\dot{\theta }}_{2,1}\right) }^2+{\left( {\dot{\theta }}_{3,1}\right) }^2\right) -\left( m_E+3m_{a_2}\right) g_cz \nonumber \\&-\left( \frac{1}{2}m_{a_1}+m_{a_2}\right) g_c|X_{1\ 2}|({\sin \left( {\theta }_{1,1}\right) }+{\sin \left( {\theta }_{2,1}\right) }+\sin ({\theta }_{3,1})). \end{aligned}$$

(119)

Taking the partial derivatives of the Lagrangian function (Eq. 119), with respect to the six generalised coordinates $(x,y,z,{\theta }_{1,1},{\theta }_{2,1},{\theta }_{3,1})$, we get

$$\begin{aligned}&\displaystyle \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{x}}}\right) = \left( m_E+3m_{a_2}\right) {\ddot{x}}_{},&\end{aligned}$$

(120a)

$$\begin{aligned}&\displaystyle \frac{\partial L}{\partial x}=0,&\end{aligned}$$

(120b)

$$\begin{aligned}&\displaystyle \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{y}}}\right) =\left( m_E+3m_{a_2}\right) {\ddot{y}}_{},&\end{aligned}$$

(121a)

$$\begin{aligned}&\displaystyle \frac{\partial L}{\partial y}=0,&\end{aligned}$$

(121b)

$$\begin{aligned}&\displaystyle \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{z}}}\right) =\left( m_E+3m_{a_2}\right) {\ddot{z}}_{},&\end{aligned}$$

(122a)

$$\begin{aligned}&\displaystyle \frac{\partial L}{\partial z}=-\left( m_E+3m_{a_2}\right) g_c,&\end{aligned}$$

(122b)

$$\begin{aligned} \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{\theta }}_{1,1}}\right)= & {} \left( I_{motor}+\frac{1}{3}m_{a_1}{\left( |X_{1\ 2}|\right) }^2+m_{a_2}{\left( |X_{1\ 2}|\right) }^2\right) {\ddot{\theta }}_{1,1}, \end{aligned}$$

(123)

$$\begin{aligned} \frac{\partial L}{\partial {\theta }_{1,1}}= & {} -\frac{1}{2}\left( m_{a_1}+m_{a_2}\right) g_c|X_{1\ 2}|\cos ({\theta }_{1,1}),\end{aligned}$$

(124)

$$\begin{aligned} \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{\theta }}_{2,1}}\right)= & {} \left( I_{motor}+\frac{1}{3}m_{a_1}{\left( |X_{1\ 2}|\right) }^2+m_{a_2}{\left( |X_{1\ 2}|\right) }^2\right) {\ddot{\theta }}_{2,1},\end{aligned}$$

(125)

$$\begin{aligned} \frac{\partial L}{\partial {\theta }_{2,1}}= & {} -\frac{1}{2}\left( m_{a_1}+m_{a_2}\right) g_c|X_{1\ 2}|\cos ({\theta }_{2,1}),\end{aligned}$$

(126)

$$\begin{aligned} \frac{d}{dt}\left( \frac{\partial L}{\partial {\dot{\theta }}_{3,1}}\right)= & {} \left( I_{motor}+\frac{1}{3}m_{a_1}{\left( |X_{1\ 2}|\right) }^2+m_{a_2}{\left( |X_{1\ 2}|\right) }^2\right) {\ddot{\theta }}_{3,1},\end{aligned}$$

(127)

$$\begin{aligned} \frac{\partial L}{\partial {\theta }_{3,1}}= & {} -\frac{1}{2}\left( m_{a_1}+m_{a_2}\right) g_c|X_{1\ 2}|\cos ({\theta }_{3,1}). \end{aligned}$$

(128)

Taking the partial derivatives of the constraint function (${{\varGamma }}_l$), with respect to the six generalised coordinates $(x,y,z,{\theta }_{1,1},{\theta }_{2,1},{\theta }_{3,1})$, we get

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial x}=2\left( x+|X_{5\ E}|{\cos \left( {\psi }_l\right) }- |X_{0\ 1}|{\cos \left( {\psi }_l\right) }-|X_{1\ 2}|\cos ({\psi }_l\right) \cos ({\theta }_{l,1}))\nonumber \\&\displaystyle \text {for} \ l=1,2,3,\end{aligned}$$

(129)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial y}=2\left( y+|X_{5\ E}|{\cos \left( {\psi }_l\right) }- |X_{0\ 1}|{\cos \left( {\psi }_l\right) }-|X_{1\ 2}|\sin ({\psi }_l\right) \cos ({\theta }_{l,1}))\nonumber \\&\displaystyle \text {for} \ l=1,2,3,\end{aligned}$$

(130)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial z}=2(z-|X_{1\ 2}|\sin ({\theta }_{l,1}))\ \text{ for }\ l=1,2,3,\end{aligned}$$

(131)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_1}{\partial {\theta }_{1,1}}=2|X_{1\ 2}|((x{\cos ({\psi }_1)}+y{\sin ({\psi }_1})+|X_{5\ E}|\nonumber \\&\displaystyle -|X_{0\ 1}|)\sin ({\theta }_{1,1})-z\cos ({\theta }_{1,1}))\end{aligned}$$

(132)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial {\theta }_{1,1}}=0 \ \text {for}\ l=2,3,\end{aligned}$$

(133)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_2}{\partial {\theta }_{2,1}}=2|X_{1\ 2}|((x{\cos ({\psi }_2)}+y{\sin ({\psi }_2)}+|X_{5\ E}|\nonumber \\&\displaystyle -|X_{0\ 1}|)\sin ({\theta }_{2,1})-z\cos ({\theta }_{2,1}))\end{aligned}$$

(134)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial {\theta }_{2,1}}=0\ \text {for}\ l=1,3,\end{aligned}$$

(135)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_3}{\partial {\theta }_{3,1}}=2|X_{1\ 2}|((x{\cos ({\psi }_3)}+y{\sin ({\psi }_3)}+|X_{5\ E}|\nonumber \\&\displaystyle -|X_{0\ 1}|)\sin ({\theta }_{3,1})-z\cos ({\theta }_{3,1}))\end{aligned}$$

(136)

$$\begin{aligned}&\displaystyle \frac{\partial {{\varGamma }}_l}{\partial {\theta }_{3,1}}=0 \ \text {for}\ l=1,2. \end{aligned}$$

(137)

By substituting the partial derivatives $\frac{d}{dt}(\frac{\partial L}{\partial {\dot{c}}_j})$ and $\frac{\partial L}{\partial c_j}$ (Eqs. (120)–(128)) and $\frac{\partial {{\varGamma }}_i}{\partial c_j}$ (Eqs. (129)–(137)), into Eq. (53) and by setting $l=1,2,3$, we get the system of equations shown below, which we solve simultaneously to find ${\lambda }_l$

$$\begin{aligned}&2\sum ^3_{l=1}{{\lambda }_l(x+|X_{5\ E}|{\cos \left( {\psi }_l\right) }-|X_{0\ 1}|\cos ({\psi }_l)}-|X_{1\ 2}|{\cos \left( {\psi }_l\right) }{\cos \left( {\theta }_{l,1}\right) )}\nonumber \\&\quad =\left( m_E+3m_{a_2}\right) \ddot{x}-f_x,\end{aligned}$$

(138a)

$$\begin{aligned}&2\sum ^3_{l=1}{{\lambda }_l(y+|X_{5\ E}|{\sin \left( {\psi }_l\right) }-|X_{0\ 1}|\sin ({\psi }_l)}- |X_{1\ 2}|{\sin \left( {\psi }_l\right) }{\cos \left( {\theta }_{l,1}\right) )}\nonumber \\&\quad =\left( m_E+3m_{a_2}\right) \ddot{y}-f_y,\end{aligned}$$

(138b)

$$\begin{aligned}&2\sum ^3_{l=1}{{\lambda }_l}(z-|X_{1\ 2}|{\sin \left( {\theta }_{l,1}\right) )}=\left( m_E+3m_{a_2}\right) \ddot{z}+ \left( m_E+3m_{a_2}\right) g_c-f_z,\nonumber \\ \end{aligned}$$

(138c)

where ($f_x,f_y,f_z$) are the x, y and z components of the external force exerted on the end effector and the gravitational acceleration $g_c$ is set to $-9.81\ ms^{-2}$. As mentioned, $(f_x,f_y,f_z) \rightarrow -(f_x,f_y,f_z)$ if the convention is inverted to evaluate the end-effector wrench applied to the environment. The expressions for ${\tau }_l$, a quaternion based torque representation of the torques of the active joint of the l-th limb, is shown below, which agrees with [50]

$$\begin{aligned} {\tau }_l= & {} \left( I_{motor}+\frac{1}{3}m_{{a}_{1}}{\left| X_{1\ 2}\right| }^2+ m_{{a}_{2}}{\left| X_{1\ 2}\right| }^2\right) {\ddot{\theta }}_{l,1}\nonumber \\&+\left( \frac{1}{2}m_{{a}_{1}}+m_{{a}_{2}}\right) g_c|X_{1\ 2}|\cos ({\theta }_{l,1})\nonumber \\&-2|X_{1\ 2}|{\lambda }_l\left( \left( x \cos \left( {\psi }_l\right) +y \sin \left( {\psi }_l\right) \right. \right. \nonumber \\&\left. \left. + |X_{5\ E}|-|X_{0\ 1}|\right) {\sin \left( {\theta }_{l,1}\right) }-z{\cos \left( {\theta }_{l,1}\right) }\right) \nonumber \\&\qquad \text {for}\ l=1,2,3. \end{aligned}$$

(139)

1.3 Appendix C: Example of an Investigation of the OptimalTransformation Type using Affectation Indices

In this example, we investigate the optimal transformation type and its corresponding kinematic pair, for a 2R planar manipulator, using arbitrary values of ${{}^{\varvec{F}}\varvec{V}}_{(\varvec{p},\varvec{F})}(t)$ and ${\varvec{}}^{\varvec{R}}{\varvec{V}}^{\varvec{}}_{(\varvec{p},\varvec{F})}[i]\varvec{(}t)$, chosen for the sake of brevity, presented in Tables 13, 14, 15 and 16. We demonstrate the computation of the affectation indices ${\varvec{A}}_{\varvec{d}\varvec{1}}({\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{i}\}},E_T)$ and ${\varvec{A}}_{\varvec{d}\varvec{2}}({\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{i}\}},E_T)$ under a Type 1a transformation. We set $Y_1=3$, $Y_2=2$ and ${{}^{\varvec{F}}\varvec{V}}_{(\varvec{p},\varvec{F})}(t)$ and ${\varvec{}}^{\varvec{R}}{\varvec{V}}^{\varvec{}}_{(\varvec{p},\varvec{F})}[i]\varvec{(}t)$ to be the values shown in Tables 13, 14, 15 and 16 for the pre-transformed and post-transformed manipulators.

Table 13 The values of $\varvec{|}{{\varvec{}}^{\varvec{R}}\varvec{V}}_{(\varvec{p},\varvec{F})}[i](t )\varvec{|}$, at three selected end-effector poses, in the trajectory $(t_1,t_2,t_3)$, for the pre-transformed manipulator, where $i=1,2$

Table 14 The values of $\varvec{|}{{\varvec{}}^{\varvec{R}}\varvec{V}}_{(\varvec{p},\varvec{F})}[i](t )|$, at three selected end-effector poses, in the trajectory $(t_1,t_2,t_3)$, under a Type 1a transformation ${\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{1}\}}(\Delta H_{1})$, where $i=1,2$, $\varepsilon =1$ and $\Delta H_{1}=0.1$

Table 15 The values of $|{{}^{\varvec{R}}\varvec{V}}_{(\varvec{p},\varvec{F})}[i](t)|$, at three selected end-effector poses, in the trajectory $\varvec{(}t_1,t_2,t_3\varvec{)}$, under a Type 1a transformation ${\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{1}\}}\varvec{(}\Delta H^{}_1\varvec{)}$, where $i=1,2$, $\varepsilon =2$ and $\Delta H_1=0.2$

Table 16 The values of $|{{}^{\varvec{F}}\varvec{V}}_{(\varvec{p},\varvec{F})}(t)|$ at three selected end-effector poses in the trajectory $\varvec{(}t_1,t_2,t_3\varvec{)}$, for the pre-transformed manipulator and for the post-transformed manipulator, under a Type 1a transformation ${\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{1}\}}\varvec{(}\Delta H^{}_1\varvec{)}$, for $\varepsilon =1$ and 2 and $\Delta H^{}_1=0.1$ and 0.2

By referring to Tables 14, 15 and 16 and employing Eqs. (92)–(93), we compute ${\varvec{A}}_{\varvec{d}\varvec{1}}({\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{1}\}},E_T)$ and ${\varvec{A}}_{\varvec{d}\varvec{2}}({\varvec{P}}^{\varvec{1}\varvec{a}}_{\{\varvec{1}\}},E_T)$ to be

$$\begin{aligned} {\varvec{A}}_{\varvec{d}\varvec{1}}\left( {\varvec{P}}^{\varvec{1}\varvec{a}}_{\left\{ \varvec{1}\right\} },E_T\right)= & {} \frac{1000}{3*2}\left( \frac{\mathrm {1.5}-1.4}{1.4}+\frac{\mathrm {2.5}-2.4}{2.4}+\frac{\mathrm {3.5}-3.4}{3.4} \right. \nonumber \\&+\frac{\mathrm {3.1}-3.0}{3.0}+\frac{\mathrm {4.2}-4.1}{4.1}+\frac{\mathrm {5.4}-5.3}{5.3}+\frac{\mathrm {2.5}-1.4}{1.4}\nonumber \\&+\frac{\mathrm {3.5}-2.4}{2.4}+\frac{\mathrm {4.5}-3.4}{3.4}+\frac{\mathrm {5.1}-3.0}{3.0}\nonumber \\&\left. +\frac{\mathrm {6.2}-4.1}{4.1}+\frac{\mathrm {7.4}-5.3}{5.3}\right) \nonumber \\= & {} 510.53, \end{aligned}$$

(140)

$$\begin{aligned} {\varvec{A}}_{\varvec{d}\varvec{2}}\left( {\varvec{P}}^{\varvec{1}\varvec{a}}_{\left\{ \varvec{1}\right\} },E_T\right)= & {} \frac{1000}{3*2}\left( \frac{\mathrm {0.15}-0.14}{0.14}+\frac{\mathrm {0.25}-0.24}{0.24}\right. \nonumber \\&+\frac{\mathrm {0.35}-0.34}{0.34}+\frac{0.2-0.14}{0.14}+\frac{\mathrm {0.4}-0.24}{0.24}\nonumber \\&\left. +\frac{\mathrm {0.8}-0.34}{0.34}\right) \nonumber \\= & {} 393.5. \end{aligned}$$

(141)