# Spatial Parallel Robotic Machines with Revolute Actuators

• Dan Zhang
Chapter

## Abstract

In this chapter, first, a six degrees of freedom fully parallel robotic machine with revolute actuators is presented and analyzed. Then, a serial of parallel manipulators with 3-dof, 4-dof, and 5-dof whose degree of freedom is dependent on an additional passive leg, this passive leg is connecting the center between the base and the moving platform. Together with the inverse kinematics and velocity equations for both rigid-link and flexible-link mechanisms, a general kinetostatic model is established for the analysis of the structural rigidity and accuracy of this family of mechanisms, case studies for 3-dof, 4-dof, and 5-dof mechanisms are given in detail to illustrate the results.

## Keywords

Jacobian Matrix Parallel Mechanism Kinematic Chain Jacobian Matrice Rigid Link
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

## 6.1 Preamble

In this chapter, first, a six degrees of freedom fully parallel robotic machine with revolute actuators is presented and analyzed. Then, a serial of parallel manipulators with 3-dof, 4-dof, and 5-dof whose degree of freedom is dependent on an additional passive leg, this passive leg is connecting the center between the base and the moving platform. Together with the inverse kinematics and velocity equations for both rigid-link and flexible-link mechanisms, a general kinetostatic model is established for the analysis of the structural rigidity and accuracy of this family of mechanisms, case studies for 3-dof, 4-dof, and 5-dof mechanisms are given in detail to illustrate the results.

## 6.2 Six Degrees of Freedom Parallel Robotic Machine with Revolute Actuators

### 6.2.1 Geometric Modeling

Figures 6.1 and 6.2 represent a 6-dof parallel mechanism with revolute actuators. This mechanism consists of six actuated legs with identical topology, connecting the fixed base to a moving platform. The kinematic chains consist – from base to platform – of an actuated revolute joint, a moving link, a Hooke joint, a second moving link, and a spherical joint attached to the platform. A fixed reference frame O − xyz is connected to the base of the mechanism and a moving coordinate frame P − x′y′z′ is connected to the platform.

### 6.2.2 Global Velocity Equation

#### 6.2.2.1 Rigid Model

The global velocity equation for rigid model can be expressed as
$$\mathbf{A}\mathbf{t} = \mathbf{B}\dot{\theta },$$
(6.1)
where vectors $$\dot{\theta }$$ and t are defined as
$$\dot{\theta } ={ \left [\begin{array}{ccc} \dot{{\theta }}_{1} & \cdots &\dot{{\theta }}_{6} \end{array} \right ]}^{\mathrm{T}},$$
(6.2)
$$\mathbf{t} ={ \left [\begin{array}{cc} {\omega }^{\mathrm{T}},&{\mathbf{\dot{p}}}^{\mathrm{T}} \end{array} \right ]}^{\mathrm{T}}$$
(6.3)
and
$$\mathbf{A} ={ \left [\begin{array}{cccccc} {\mathbf{m}}_{1} & {\mathbf{m}}_{2} & {\mathbf{m}}_{3} & {\mathbf{m}}_{4} & {\mathbf{m}}_{5} & {\mathbf{m}}_{6} \end{array} \right ]}^{\mathrm{T}}$$
(6.4)
$$\begin{array}{rcl} & & \mathbf{B} = \mathrm{diag}[{({\mathbf{p}}_{1} -{\mathbf{b}'}_{1})}^{\mathrm{T}}{\mathbf{e}}_{ 1},{({\mathbf{p}}_{2} -{\mathbf{b}'}_{2})}^{\mathrm{T}}{\mathbf{e}}_{ 2},{({\mathbf{p}}_{3} -{\mathbf{b}'}_{3})}^{\mathrm{T}}{\mathbf{e}}_{ 3}, \\ & & {({\mathbf{p}}_{4} -{\mathbf{b}'}_{4})}^{\mathrm{T}}{\mathbf{e}}_{ 4},{({\mathbf{p}}_{5} -{\mathbf{b}'}_{5})}^{\mathrm{T}}{\mathbf{e}}_{ 5},{({\mathbf{p}}_{6} -{\mathbf{b}'}_{6})}^{T}{\mathbf{e}}_{ 6}] \end{array}$$
(6.5)
and m i is a six-dimensional vector expressed as
$${ \mathbf{m}}_{i} = \left [\begin{array}{c} (\mathbf{Q}{\mathbf{r'}}_{i}) \times ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i}) \\ ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i}) \end{array} \right ],\quad i = 1,\ldots ,6$$
(6.6)
and again, the Jacobian matrix J can be written as
$$\mathbf{J} ={ \mathbf{B}}^{-1}\mathbf{A}.$$
(6.7)

#### 6.2.2.2 Compliant Model

For the case of compliant model, one can obtain the global velocity equation as
$$\mathbf{A}\mathbf{t} = \mathbf{B}\dot{\theta },$$
(6.8)
where vector $$\dot{\theta }$$ is defined as
$$\dot{\theta } ={ \left [\begin{array}{cccccccccccc} \dot{{\theta }}_{11} & \dot{{\theta }}_{12} & \dot{{\theta }}_{21} & \dot{{\theta }}_{22} & \dot{{\theta }}_{31} & \dot{{\theta }}_{32} & \dot{{\theta }}_{41} & \dot{{\theta }}_{42} & \dot{{\theta }}_{51} & \dot{{\theta }}_{52} & \dot{{\theta }}_{61} & \dot{{\theta }}_{62} \end{array} \right ]}^{\mathrm{T}}$$
(6.9)
matrix A and its terms are as given in (6.4) and (6.6) and
$${ \mathbf{B}}_{6\times 12} = \left [\begin{array}{llllllllllll} {b}_{11} & {b}_{12} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & {b}_{21} & {b}_{22} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & {b}_{31} & {b}_{32} & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & {b}_{41} & {b}_{42} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {b}_{51} & {b}_{52} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & {b}_{61} & {b}_{62} \end{array} \right ],$$
(6.10)
where
$${b}_{ij} = {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}{\mathbf{d}}_{ ij},\quad i = 1,\ldots ,6,\;j = 1,2$$
(6.11)
The derivation of the relationship between Cartesian velocities and joint rates is thereby completed.

### 6.2.3 Stiffness Model

Again, the stiffness of the structure has been obtained as
$$\mathbf{K} ={ \mathbf{J}}^{\mathrm{T}}{\mathbf{K}}_{\mathrm{ J}}\mathbf{J}.$$
(6.12)
One obtains
$$\mathbf{t} = \mathbf{J}'\dot{\theta },$$
(6.13)
where
$$\mathbf{J'} ={ \mathbf{A}}^{-1}\mathbf{B}$$
(6.14)
according to the principle of virtual work, one has
$${\tau }^{\mathrm{T}}\dot{\theta } ={ \mathbf{w}}^{\mathrm{T}}\mathbf{t},$$
(6.15)
where τ is a vector of the actuator torques applied at each actuated joint or joint with spring. If we assume that no gravitational forces act on any of the intermediate links and w is a vector composed of forces and moments (hereafter called wrench) applied by the end-effector. Substituting (6.13) into (6.15) one can obtain
$$\tau ={ \mathbf{J'}}^{\mathrm{T}}\mathbf{w}.$$
(6.16)
The joint forces and displacements of each joint can be related by Hooke’s law, i.e.,
$$\tau ={ \mathbf{K}}_{\mathrm{J}}\Delta \theta.$$
(6.17)
Δθ only includes the actuated joints and joint with springs, i.e.,
$${ \mathbf{K}}_{\mathrm{J}}\Delta \theta ={ \mathbf{J'}}^{\mathrm{T}}\mathbf{w}$$
(6.18)
hence
$$\Delta \theta ={ \mathbf{K}}_{\mathrm{J}}^{-1}{\mathbf{J'}}^{\mathrm{T}}\mathbf{w}.$$
(6.19)
premultiplying by J on both sides, one obtains
$$\mathbf{J'}\Delta \theta = \mathbf{J'}{\mathbf{K}}_{\mathrm{J}}^{-1}{\mathbf{J'}}^{\mathrm{T}}\mathbf{w}.$$
(6.20)
Substituting (6.13) into (6.20), one obtains
$$\mathbf{t} = \mathbf{J'}{\mathbf{K}}_{\mathrm{J}}^{-1}{\mathbf{J'}}^{\mathrm{T}}\mathbf{w},$$
(6.21)
therefore, one obtains the compliance matrix of the mechanism κ as follow
$$\kappa = \mathbf{J'}{\mathbf{K}}_{\mathrm{J}}^{-1}{\mathbf{J'}}^{\mathrm{T}}$$
(6.22)
and the system stiffness matrix is
$$\mathbf{K} ={ [\mathbf{J'}{\mathbf{K}}_{\mathrm{J}}^{-1}{\mathbf{J'}}^{\mathrm{T}}]}^{-1},$$
(6.23)
where
$${ \mathbf{K}}_{\mathrm{J}} = \mathrm{diag}[{k}_{11},{k}_{12},{k}_{21},{k}_{22},{k}_{31},{k}_{32},{k}_{41},{k}_{42},{k}_{51},{k}_{52},{k}_{61},{k}_{62}]$$
(6.24)
where k i1 is stiffness of the ith actuator and k i2 is the lumped stiffness of each leg.
In order to illustrate the effect of the flexible links on the parallel mechanism, an example of 6-dof mechanism is presented. The parameters are given as
$$\begin{array}{rcl} & {\theta }_{\mathrm{p}} = 22.3{4}^{\circ },{\theta }_{\mathrm{b}} = 42.88{3}^{\circ }, & \\ & {R}_{\mathrm{p}} = 6\,\mathrm{cm},{R}_{\mathrm{b}} = 15\,\mathrm{cm}, & \\ & {l}_{i1} = 46\,\mathrm{cm},{l}_{i2} = 36\,\mathrm{cm},\quad i = 1,\ldots ,6& \\ & {k}_{i1} = 1,000\,\mathrm{Nm},\quad i = 1,\ldots ,6, & \\ \end{array}$$
where k i1 is the stiffness of each leg, l i1, l i2 are the link lengths for the 1st and 2nd link of each leg, and the Cartesian coordinates are given by
$$\begin{array}{rcl} & x \in [-3,3]\,\mathrm{cm},y \in [-3,3]\,\mathrm{cm},z = 68\,\mathrm{cm},& \\ & \phi = 0,\theta = 0,\psi = 0. & \\ \end{array}$$

Figure 6.3 shows the variation of the stiffness for the above example. The comparison between the parallel mechanism with rigid link and the parallel mechanism with flexible links is given in Table 6.1. The results are similar to those obtained in previous cases.

Table 6.1

Comparison of the mechanism stiffness between the mechanism with rigid links and the mechanism with flexible links

 K actuator K link K x K y K z $${K}_{{\theta }_{x}}$$ $${K}_{{\theta }_{y}}$$ $${K}_{{\theta }_{z}}$$ 1,000 1,000 3,700.65 3,700.65 10,082.1 18.1478 18.1478 20.633 1,000 101 K a 6,967.15 6,967.15 18,981.4 34.1665 34.1665 38.8454 1,000 102 K a 7,641.67 7,641.67 20,819.1 37.4743 37.4743 42.6062 1,000 103 K a 7,716.37 7,716.37 21,022.6 37.8406 37.8406 43.0227 1,000 104 K a 7,723.92 7,723.92 21,043.2 37.8777 37.8777 43.0648 1,000 105 K a 7,724.68 7,724.68 21,045.2 37.8814 37.8814 43.069 1,000 106 K a 7,724.76 7,724.76 21,045.4 37.8818 37.8818 43.0694 1,000 107 K a 7,724.76 7,724.76 21,045.4 37.8818 37.8818 43.0695 1,000 Rigid 7,724.76 7,724.76 21,045.4 37.8818 37.8818 43.0695
From Fig. 6.4, one can find that K x and K y , $${K}_{{\theta }_{x}}$$ and $${K}_{{\theta }_{y}}$$ are symmetric with respect to each other. In Fig. 6.4(a), the stiffness in X becomes higher when the platform moves further from the Y -axis. This was to be expected because when the platform moves aside along the X-axis, the projection of the legs on this axis becomes larger, and the mechanism is stiffer in Y. And the same reasoning applies to Fig. 6.4(b) for the stiffness in Y.

In Fig. 6.4d, e, the torsional stiffnesses in θ x and θ y are shown, the stiffness is larger when the platform moves further from the Y -axis. However, in the center of the workspace, the K z is at its minimum, and the stiffness in the Z becomes higher when the platform moves further from the center of the workspace. On the other hand, from Fig. 6.4(f), the stiffness in θ z is higher near the center of the workspace, which is the best position for supporting torsional loads around Z-axis. All these are in accordance with what would be intuitively expected.

## 6.3 General Kinematic Model of n Degrees of Freedom Parallel Mechanisms with a Passive Constraining Leg and Revolute Actuators

### 6.3.1 Geometric Modeling and Lumped Compliance Model

An example of parallel mechanism belonging to the family of mechanisms studied in this chapter is shown in Figs. 6.5 and 6.6. It is a 5-dof parallel mechanism with revolute actuators.

In order to obtain a simple kinetostatic model, link compliances are lumped at the joints should be considered. In this framework, link bending stiffnesses are replaced by equivalent torsional springs located at virtual joints.

### 6.3.2 Inverse Kinematics

#### 6.3.2.1 Solution for the Case of Mechanisms with Rigid Links

In order to solve the inverse kinematic problem, one must first consider the passive constraining leg as a serial n-dof mechanism whose n Cartesian coordinates are known, which is a well-known problem. Once the solution to the inverse kinematics of this n-dof serial mechanism is found, the complete pose (position and orientation) of the platform can be determined using the direct kinematic equations for this serial mechanism. Figure 6.7 illustrates the configuration of the ith actuated joint of the mechanism with revolute actuators. Point B i is defined as the center of the Hooke joint connecting the two moving links of the ith actuated leg. Moreover, the Cartesian coordinates of point B i expressed in the fixed coordinate frame are represented as (b′ ix , b′ iy , b′ iz ) and the position vector of point B′ i in the fixed frame is given by vector b i . Since the axis of the fixed revolute joint of the ith actuated leg is assumed to be parallel to the xy plane of the fixed coordinate frame, one can write
$${b'}_{ix} = {b}_{ix} - {l}_{i1}\sin ({\theta }_{bi} + {\beta }_{i})\cos {\theta }_{i1},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},\quad$$
(6.25)
$${b'}_{iy} = {b}_{iy} + {l}_{i1}\cos ({\theta }_{bi} + {\beta }_{i})\cos {\theta }_{i1},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},\quad$$
(6.26)
$${b'}_{iz} = {b}_{iz} + {l}_{i1}\sin {\theta }_{i1},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},\quad$$
(6.27)
where θ bi is the angle between the positive direction of the x-axis of the base coordinate frame and the line connecting points O and B i and θ i1 is the joint variable – rotation angle around the fixed revolute joint – associated with the ith actuated leg, β i is the angle between the positive direction of the line connecting points O and B i and the axis of the ith actuated joint. Moreover, l i1 is the length of the first link of the ith actuated leg. From the configuration of Fig. 6.7, the relationships between the parameters can be written as
$${({b'}_{ix} - {x}_{i})}^{2} + {({b'}_{ iy} - {y}_{i})}^{2} + {({b'}_{ iz} - {z}_{i})}^{2} = {l}_{ i2}^{2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.28)
where x i , y i , z i are the coordinates of point P i and l i2 is the length of the second link of the ith actuated leg.
Substituting (6.25) – (6.27) into (6.28), one has
$${E}_{i}\cos {\theta }_{i1} + {F}_{i}\sin {\theta }_{i1} = {G}_{i},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.29)
where
$${E}_{i} = ({y}_{i} - {b}_{iy})\cos ({\theta }_{bi} + {\beta }_{i}) - ({x}_{i} - {b}_{ix})\sin ({\theta }_{bi} + {\beta }_{i}),$$
(6.30)
$${F}_{i} = {z}_{i} - {b}_{iz},$$
(6.31)
$${G}_{i} = \frac{{({x}_{i} - {b}_{ix})}^{2} + {({y}_{i} - {b}_{iy})}^{2} + {({z}_{i} - {b}_{iz})}^{2} + {l}_{i1}^{2} - {l}_{i2}^{2}} {2{l}_{i1}}$$
(6.32)
and angle θ i1 can be obtained by
$$\sin {\theta }_{i1} = \frac{{F}_{i}{G}_{i} + {K}_{i}{E}_{i}\sqrt{{H}_{i}}} {{E}_{i}^{2} + {F}_{i}^{2}} ,\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.33)
$$\cos {\theta }_{i1} = \frac{{E}_{i}{G}_{i} - {K}_{i}{F}_{i}\sqrt{{H}_{i}}} {{E}_{i}^{2} + {F}_{i}^{2}} ,\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.34)
where K i  =  ± 1 is the branch index of the mechanism associated with the configuration of the ith leg and
$${H}_{i} = {E}_{i}^{2} + {F}_{ i}^{2} - {G}_{ i}^{2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.35)
Finally, the solution of the inverse kinematic problem is completed by performing
$${\theta }_{i1} = \mathrm{atan}2[\sin {\theta }_{i1},\cos {\theta }_{i1}],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.36)
Meanwhile, referring to Fig. 6.7, the vector of leg length can be written as
$${ \mathbf{b}'}_{i} ={ \mathbf{b}}_{i} + {l}_{i1}{\mathbf{Q}}_{ti1}{\mathbf{d}}_{i},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.37)
with
$${ \mathbf{Q}}_{\mathit{ti}1} = \left [\begin{array}{ccc} \cos ({\theta }_{\mathrm{b}i} + {\beta }_{i})& -\sin ({\theta }_{\mathrm{b}i} + {\beta }_{i})&0 \\ \sin ({\theta }_{\mathrm{b}i} + {\beta }_{i})& \cos ({\theta }_{\mathrm{b}i} + {\beta }_{i}) &0 \\ 0 & 0 &1 \end{array} \right ],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.38)
and
$${ \mathbf{d}}_{i1} = \left [\begin{array}{c} 0\\ \cos {\theta }_{ i1} \\ \sin {\theta }_{i1} \end{array} \right ],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.39)
assuming that the distance between points P i and B′ i is noted l i2, then one has
$${l}_{i2}^{2} = {({\mathbf{p}}_{ i} -{\mathbf{b}}_{i})}^{\mathrm{T}}({\mathbf{p}}_{ i} -{\mathbf{b}}_{i}),\quad i = 1,\ldots ,n,\quad n = \mbox{ 3, 4, or 5}.$$
(6.40)

#### 6.3.2.2 Solutions for the Mechanisms with Flexible Links

In order to uniquely describe the architecture of a kinematic chain, i.e., the relative location and orientation of its neighboring pair axes, the Denavit–Hartenberg notation is used to define the nominal geometry of each of the serial kinematic chains of the parallel mechanism. A coordinate frame F i is defined with the origin O i and axes X i , Y i , Z i , this frame is attached to the (i − 1)th link. Figure 6.8 represents one of the identical kinematic chains for the n-dof parallel mechanism discussed above. Joint 2 is a virtual joint used to model the compliance of the driven link. From Fig. 6.8, one has θ i2 = 0, when there is no deflection. Angles θ i3 and θ i4 can be obtained by writing the coordinates of point P i in Frame 3 as
$${x}_{i3} = {l}_{i2}\cos {\theta }_{i4}\cos {\theta }_{i3},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.41)
$${y}_{i3} = {l}_{i2}\cos {\theta }_{i4}\sin {\theta }_{i3},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.42)
$${z}_{i3} = {l}_{i2}\sin {\theta }_{i4},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.43)
and
$${[\mathbf{p}]}_{3} ={ \mathbf{Q}}_{i2}^{T}{\mathbf{Q}}_{ i1}^{T}{\mathbf{Q}}_{ i0}^{T}[{\mathbf{p}}_{ i} -{\mathbf{b'}}_{i}],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.44)
then, combining (6.41) – (6.43) and (6.44), one can find θ i3 and θ i4 easily.
From Fig. 6.8, one can express the position of point B′ i as
$${ \mathbf{b}'}_{i} ={ \mathbf{b}}_{i} +{ \mathbf{Q}}_{i0}{\mathbf{a}}_{i1} +{ \mathbf{Q}}_{i0}{\mathbf{Q}}_{i1}{\mathbf{a}}_{i2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.45)
where Q i0, a i1, a i2 and Q i1 can be expressed as
$${ \mathbf{a}}_{i1} = \left [\begin{array}{c} 0\\ 0 \\ 0 \end{array} \right ],\;{\mathbf{a}}_{i2} = \left [\begin{array}{c} {l}_{i1}\cos {\theta }_{i2} \\ {l}_{i1}\sin {\theta }_{i2} \\ 0 \end{array} \right ],\;{\mathbf{Q}}_{i1} = \left [\begin{array}{ccc} \cos {\theta }_{i1} & 0 & \sin {\theta }_{i1} \\ \sin {\theta }_{i1} & 0 & -\cos {\theta }_{i1} \\ 0 & 1 & 0 \end{array} \right ],$$
(6.46)
$${ \mathbf{Q}}_{i0} = \left [\begin{array}{ccc} -\sin ({\theta }_{\mathrm{b}i} + {\beta }_{i}) & 0 & \cos ({\theta }_{\mathrm{b}i} + {\beta }_{i}) \\ \cos ({\theta }_{\mathrm{b}i} + {\beta }_{i}) & 0 & \sin ({\theta }_{\mathrm{b}i} + {\beta }_{i}) \\ 0 & 1 & 0 \end{array} \right ].$$
(6.47)

### 6.3.3 Jacobian Matrices

#### 6.3.3.1 Rigid Mechanisms

The parallel mechanisms studied here comprise two main components, namely, the passive constraining leg – which can be thought of as a serial mechanism – and the actuated legs acting in parallel.

Considering the passive constraining leg, one can write
$${ \mathbf{J}}_{n+1}{\dot{\theta }}_{n+1} = \mathbf{t},\;n = \mbox{ 3, 4, or 5},$$
(6.48)
where J n + 1 consists of e i and r i , $$\mathbf{t} ={ \left [\begin{array}{cc} {\omega }^{\mathrm{T}} & {\mathbf{\dot{p}}}^{\mathrm{T}} \end{array} \right ]}^{\mathrm{T}}$$is the twist of the platform, ω is the angular velocity of the platform, and $${\dot{\theta }}_{n+1} ={ \left [\begin{array}{ccccc} \dot{{\theta }}_{n+1,1} & \cdots &\dot{{\theta }}_{n+1,n} \end{array} \right ]}^{\mathrm{T}},(n = \mbox{ 3, 4, or 5})$$ is the joint velocity vector associated with the passive constraining leg. Matrix J n + 1 is the Jacobian matrix of the passive constraining leg which is taken as a serial n-dof mechanism.

#### 6.3.3.2 Compliant Model

If the compliances of the links and joints are included, (6 − n) virtual joints will then be added to the passive constraining leg in order to account for the compliance of the links [62]. Hence, the Jacobian matrix of the passive constraining leg becomes
$${ \mathbf{J}'}_{n+1}{\dot{\theta }'}_{n+1} = \mathbf{t},\quad n = \mbox{ 3, 4, or 5},$$
(6.49)
where
$${ \dot{\theta }'}_{n+1} ={ \left [\begin{array}{ccc} \dot{{\theta }}_{n+1,1} & \cdots &\dot{{\theta }}_{n+1,6} \end{array} \right ]}^{\mathrm{T}},\quad n = \mbox{ 3, 4, or 5}.$$
(6.50)

#### 6.3.3.3 Global Velocity Equations

1. 1.

Rigid Model:

Now, considering the parallel component of the mechanism, the parallel Jacobian matrix can be obtained by differentiating (6.37), (6.39), and (6.40) with respect to time. One has
$$\dot{{\mathbf{b}'}_{i}} = {l}_{i1}{\mathbf{Q}}_{ti1}\dot{{\mathbf{d}}_{i}},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.51)
$$\dot{{\mathbf{d}}}_{i1} = \left [\begin{array}{c} 0\\ -\sin {\theta }_{ i1} \\ \cos {\theta }_{i1}\end{array} \right ]\dot{{\theta }_{i1}},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.52)
$${({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}\dot{{\mathbf{b}}'}_{ i} - {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}\dot{{\mathbf{p}}}_{ i} = 0,\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.53)
One obtains
$$\dot{{\mathbf{p}}}_{i} =\dot{ \mathbf{p}} +\dot{ \mathbf{Q}}{\mathbf{r'}}_{i},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.54)
assuming
$${ \mathbf{e}}_{i} = {l}_{i1}{\mathbf{Q}}_{ti1}\left [\begin{array}{c} 0\\ -\sin {\theta }_{ i1} \\ \cos {\theta }_{i1}\end{array} \right ],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.55)
then
$$\dot{{\mathbf{b}}'}_{i} ={ \mathbf{e}}_{i}\dot{{\theta }}_{i1},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}$$
(6.56)
therefore (6.53) can be rewritten as (for i = 1, , n, n = 3, 4, or 5)
$${({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}{\mathbf{e}}_{ i}\dot{{\theta }_{i1}} = {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}\dot{\mathbf{p}} + {[(\mathbf{Q}{\mathbf{r'}}_{ i}) \times ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})]}^{\mathrm{T}}\omega.$$
(6.57)
Hence, one has the velocity equation as
$$\mathbf{A}\mathbf{t} = \mathbf{B}\dot{\theta },$$
(6.58)
where vector $$\dot{\theta }$$ and t are defined as
$$\dot{\theta } ={ \left [\begin{array}{ccccc} \dot{{\theta }}_{1} & \cdots &\dot{{\theta }}_{n}\end{array} \right ]}^{\mathrm{T}},\quad n = \mbox{ 3, 4, or 5},$$
(6.59)
$$\mathbf{t} ={ \left [\begin{array}{cccccc} {\omega }_{1} & {\omega }_{2} & {\omega }_{3} & \dot{x}&\dot{y}&\dot{z}\end{array} \right ]}^{\mathrm{T}},$$
(6.60)
vector ω is the angular velocity of the platform, and
$$\mathbf{A} = \left [\begin{array}{c} {\mathbf{a}}_{1}^{\mathrm{T}} \\ {\mathbf{a}}_{2}^{\mathrm{T}}\\ \vdots \\ {\mathbf{a}}_{n}^{\mathrm{T}}\end{array} \right ],\quad \mathbf{B} = \mathrm{diag}[{({\mathbf{p}}_{1}-{\mathbf{b}'}_{1})}^{\mathrm{T}}{\mathbf{e}}_{ 1},\ldots ,{({\mathbf{p}}_{n}-{\mathbf{b}'}_{n})}^{\mathrm{T}}{\mathbf{e}}_{ n}],$$
(6.61)
where a i is a six-dimensional vector, which can be expressed as
$${ \mathbf{a}}_{i} = \left [\begin{array}{c} (\mathbf{Q}{\mathbf{r'}}_{i}) \times ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i}) \\ ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})\end{array} \right ],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.62)

2. 2.

Compliant Model:

Differentiating (6.45) and (6.46) with respect to time, one has
$$\dot{{\mathbf{b}}'}_{i} ={ \mathbf{Q}}_{i0}\dot{{\mathbf{Q}}}_{i1}{\mathbf{a}}_{i2} +{ \mathbf{Q}}_{i0}{\mathbf{Q}}_{i1}\dot{{\mathbf{a}}}_{i2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.63)
$$\dot{{\mathbf{a}}}_{i2} = \left [\begin{array}{c} - {l}_{i1}\sin {\theta }_{i2} \\ {l}_{i1}\cos {\theta }_{i2} \\ 0\end{array} \right ]\dot{{\theta }}_{i2},\quad i = 1,\ldots ,n,$$
(6.64)
$$\dot{{\mathbf{Q}}}_{i1} = \left [\begin{array}{ccc} -\sin {\theta }_{i1} & 0&\cos {\theta }_{i1} \\ \cos {\theta }_{i1} & 0&\sin {\theta }_{i1} \\ 0 &0& 0\end{array} \right ]\dot{{\theta }_{i1}},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.65)
For i = 1, , n, n = 3, 4, or 5, assuming
$${ \mathbf{d}}_{i1} ={ \mathbf{Q}}_{i0}\left [\begin{array}{ccc} -\sin {\theta }_{i1} & 0&\cos {\theta }_{i1} \\ \cos {\theta }_{i1} & 0&\sin {\theta }_{i1} \\ 0 &0& 0\end{array} \right ]{\mathbf{a}}_{i2},\;{\mathbf{d}}_{i2} ={ \mathbf{Q}}_{i0}{\mathbf{Q}}_{i1}\left [\begin{array}{c} - {l}_{i1}\sin {\theta }_{i2} \\ {l}_{i1}\cos {\theta }_{i2} \\ 0\end{array} \right ]$$
(6.66)
then one has
$$\dot{{\mathbf{b}}'}_{i} ={ \mathbf{d}}_{i1}\dot{{\theta }}_{i1} +{ \mathbf{d}}_{i2}\dot{{\theta }}_{i2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.67)
Differentiating (6.40) with respect to time, one obtains (6.53), and following a derivation similar to the one presented above for the mechanism with rigid links, for $$i = 1,\ldots ,n,\quad n = \mbox{ 3, 4, or 5}$$, one obtains
$${({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}({\mathbf{d}}_{ i1}\dot{{\theta }}_{i1} +{ \mathbf{d}}_{i2}\dot{{\theta }}_{i2}) = {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}\dot{\mathbf{p}} + {[(\mathbf{Q}{\mathbf{r'}}_{ i}) \times ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})]}^{\mathrm{T}}\omega.$$
(6.68)
Hence one has the velocity equation as
$$\mathbf{A}\mathbf{t} ={ \mathbf{B}}_{1}\dot{{\theta }}_{1} +{ \mathbf{B}}_{2}\dot{{\theta }}_{2},$$
(6.69)
where vectors $${\dot{\theta }}_{1}$$ and $${\dot{\theta }}_{2}$$ are defined as
$${ \dot{\theta }}_{1} ={ \left [\begin{array}{ccccc} \dot{{\theta }}_{11} & \cdots &\dot{{\theta }}_{n1}\end{array} \right ]}^{\mathrm{T}},\quad n = \mbox{ 3, 4, or 5},$$
(6.70)
$${ \dot{\theta }}_{2} ={ \left [\begin{array}{ccccc} \dot{{\theta }}_{12} & \cdots &\dot{{\theta }}_{n2}\end{array} \right ]}^{\mathrm{T}},\quad n = \mbox{ 3, 4, or 5},$$
(6.71)
matrices A, B 1 and B 2 are given as
$$\mathbf{A} ={ \left [\begin{array}{cccccc} {\mathbf{a}}_{1} & {\mathbf{a}}_{2} & {\mathbf{a}}_{3} & {\mathbf{a}}_{4} & {\mathbf{a}}_{5} & {\mathbf{a}}_{6}\end{array} \right ]}^{\mathrm{T}}$$
(6.72)
$${ \mathbf{B}}_{1} = \mathrm{diag}[{b}_{11},\ldots ,{b}_{n,2n-1}],\quad n = \mbox{ 3, 4, or 5},$$
(6.73)
$${ \mathbf{B}}_{2} = \mathrm{diag}[{b}_{12},\ldots ,{b}_{n,2n}],\quad n = \mbox{ 3, 4, or 5},$$
(6.74)
where a i is a six-dimensional vector and b i, 2i − 1, b i, 2i are the diagonal items of B 1 and B 2, respectively. They can be expressed as
$${ \mathbf{a}}_{i} = \left [\begin{array}{c} (\mathbf{Q}{\mathbf{r'}}_{i}) \times ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i}) \\ ({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})\end{array} \right ],\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.75)
$${b}_{i,2i-1} = {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}{\mathbf{d}}_{ i,1},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5},$$
(6.76)
$${b}_{i,2i} = {({\mathbf{p}}_{i} -{\mathbf{b}'}_{i})}^{\mathrm{T}}{\mathbf{d}}_{ i,2},\quad i = 1,\ldots ,n,\;n = \mbox{ 3, 4, or 5}.$$
(6.77)

### 6.3.4 Kinetostatic Model for the Mechanism with Rigid Links

According to the principle of virtual work, one can finally obtain the Cartesian compliance matrix with the same approach as in Chap. 5.
$${ \mathbf{C}}_{\mathrm{c}} ={ \mathbf{J}}_{n+1}{({\mathbf{AJ}}_{n+1})}^{-1}\mathbf{B}\mathbf{C}{\mathbf{B}}^{\mathrm{T}}{({\mathbf{AJ}}_{ n+1})}^{-\mathrm{T}}{\mathbf{J}}_{ n+1}^{\mathrm{T}}$$
(6.78)
with
$$\Delta \mathbf{c} ={ \mathbf{C}}_{\mathrm{c}}\mathbf{w},$$
(6.79)
where C c is a symmetric positive semidefinite (6 ×6) matrix, as expected.

### 6.3.5 Kinetostatic Model for the Mechanism with Flexible Links

Again, based on the principle of virtual work, one can write
$${ \mathbf{w}}^{\mathrm{T}}\mathbf{t} = {\tau }_{ n+1}^{\mathrm{T}}{\dot{\theta }'}_{ n+1} + {\tau }_{1}^{\mathrm{T}}{\dot{\theta }}_{ 1} + {\tau }_{2}^{\mathrm{T}}{\dot{\theta }}_{ 2},$$
(6.80)
where τ1 and τ2 correspond to a partition of vector τ, in components associated with $${\dot{\theta }}_{1}$$ and $${\dot{\theta }}_{2}$$, respectively, i.e., the first and second joint of each leg. τ is the vector of actuator forces, $$\dot{\theta }$$ is the vector of actuator velocities (actuated joints and joints with virtual springs), and τ n + 1 is the vector of joint torques in the passive constraining leg. This vector is defined as follows, where K n + 1 is the stiffness matrix of the passive constraining leg,
$${\tau }_{n+1} ={ \mathbf{K}}_{n+1}\Delta {\theta '}_{n+1},$$
(6.81)
$${\tau }_{1} ={ \mathbf{K}}_{j1}\Delta {\theta }_{1},$$
(6.82)
$${\tau }_{2} ={ \mathbf{K}}_{j2}\Delta {\theta }_{2},$$
(6.83)
$${ \mathbf{K}}_{j1} = \mathrm{diag}[{k}_{11},\ldots ,{k}_{n1}],$$
(6.84)
$${ \mathbf{K}}_{j2} = \mathrm{diag}[{k}_{12},\ldots ,{k}_{n2}].$$
(6.85)
Matrix K n + 1 is a diagonal 6 ×6 matrix in which the ith diagonal entry is zero if it is associated with a real joint or it is equal to k i if it is associated with a virtual joint, where k i is the stiffness of the virtual spring located at the ith joint. k 11, , k n1 are the compound stiffnesses of actuators and first links stiffnesses while k 12, , k n2 are the first links stiffnesses. One can rewrite (6.69) as
$$\dot{{\theta }}_{1} ={ \mathbf{B}}_{1}^{-1}\mathbf{A}\mathbf{t} -{\mathbf{B}}_{ 1}^{-1}{\mathbf{B}}_{ 2}\dot{{\theta }}_{2}.$$
(6.86)
Substituting (6.86) and (6.49) into (6.80), one can obtain
$${ \mathbf{w}}^{\mathrm{T}}{\mathbf{J}'}_{ n+1}{\dot{\theta }'}_{n+1} = {\tau }_{n+1}^{\mathrm{T}}{\dot{\theta }'}_{ n+1} + {\tau }_{2}^{\mathrm{T}}{\dot{\theta }}_{ 2} + {\tau }_{1}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-1}\mathbf{A}{\mathbf{J}'}_{ n+1}{\dot{\theta }'}_{n+1} -{\tau }_{1}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-1}{\mathbf{B}}_{ 2}\dot{{\theta }}_{2}.$$
(6.87)
Since there are 11 degrees of freedom in the compliant mechanism, this equation must be satisfied for any value of $${\dot{\theta }'}_{n+1}$$ and $${\dot{\theta }}_{2}$$. Therefore, one can equate the coefficients of the terms in $${\dot{\theta }'}_{n+1}$$ and the terms in $${\dot{\theta }}_{2}$$, hence one can obtain
$${({\mathbf{J}'}_{n+1})}^{\mathrm{T}}\mathbf{w} = {\tau }_{ n+1} + {({\mathbf{J}'}_{n+1})}^{\mathrm{T}}{\mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\tau }_{ 1},$$
(6.88)
$${\tau }_{2} ={ \mathbf{B}}_{2}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\tau }_{ 1}.$$
(6.89)
Substituting (6.81), (6.82), and (6.83) into (6.88) and (6.89), one obtains
$${({\mathbf{J}'}_{n+1})}^{\mathrm{T}}\mathbf{w} ={ \mathbf{K}}_{ n+1}\Delta {\theta '}_{n+1} + {({\mathbf{J}'}_{n+1})}^{\mathrm{T}}{\mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}\Delta {\theta }_{1},$$
(6.90)
$$\Delta {\theta }_{2} ={ \mathbf{K}}_{j2}^{-1}{\mathbf{B}}_{ 2}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}\Delta {\theta }_{1}.$$
(6.91)
Substituting (6.91) into (6.69), one obtains
$$\mathbf{A}\mathbf{t} = \mathbf{W}\dot{{\theta }}_{1},$$
(6.92)
where
$$\mathbf{W} ={ \mathbf{B}}_{1} +{ \mathbf{B}}_{2}{\mathbf{K}}_{j2}^{-1}{\mathbf{B}}_{ 2}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}.$$
(6.93)
Substituting (6.92) into (6.90), one obtains
$${({\mathbf{J}'}_{n+1})}^{\mathrm{T}}\mathbf{w} ={ \mathbf{K}}_{ n+1}{({\mathbf{J}'}_{n+1})}^{-1}\Delta \mathbf{c} + {({\mathbf{J}'}_{ n+1})}^{\mathrm{T}}{\mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A}\Delta \mathbf{c},$$
(6.94)
i.e.,
$$\mathbf{w} = ({({\mathbf{J}'}_{n+1})}^{-\mathrm{T}}{\mathbf{K}}_{ n+1}{({\mathbf{J}'}_{n+1})}^{-1} +{ \mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A})\Delta \mathbf{c},$$
(6.95)
which is in the form
$$\mathbf{w} = \mathbf{K}\Delta \mathbf{c},$$
(6.96)
where K is the stiffness matrix, which is equal to
$$\mathbf{K} = [{({\mathbf{J}'}_{n+1})}^{-\mathrm{T}}{\mathbf{K}}_{ n+1}{({\mathbf{J}'}_{n+1})}^{-1} +{ \mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A}].$$
(6.97)

Matrix K is a symmetric (6 ×6) positive semidefinite matrix, as expected. Matrix K will be of full rank in nonsingular configurations. Indeed, the sum of the two terms in (6.97) will span the complete space of constraint wrenches.

### 6.3.6 Examples

#### 6.3.6.1 3-dof Parallel Mechanism

This mechanism is illustrated is Fig. 6.5, the compliance matrix for the mechanism with rigid links can be written as
$${ \mathbf{C}}_{c} ={ \mathbf{J}}_{6}{({\mathbf{AJ}}_{6})}^{-1}\mathbf{B}\mathbf{C}{\mathbf{B}}^{\mathrm{T}}{({\mathbf{AJ}}_{ 6})}^{-\mathrm{T}}{\mathbf{J}}_{ 6}^{\mathrm{T}},$$
(6.98)
where
$$\mathbf{C} = \mathrm{diag}[{c}_{1},{c}_{2},{c}_{3},{c}_{4},{c}_{5}]$$
(6.99)
with c 1, c 2, c 3, c 4 and c 5 the compliances of the actuators and J 6 is the Jacobian matrix of the passive constraining leg in this 5-dof case. Matrices A and B are the Jacobian matrices of the structure without the passive constraining leg.
Similarly, the stiffness matrix for the mechanism with flexible links can be written as
$$\mathbf{K} = {({\mathbf{J}'}_{6})}^{-\mathrm{T}}{\mathbf{K}}_{ 6}{({\mathbf{J}'}_{6})}^{-1} +{ \mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A},$$
(6.100)
where
$${ \mathbf{K}}_{6} = \mathrm{diag}[0,{k}_{62},0,0,0,0],$$
(6.101)
$${ \mathbf{K}}_{j1} = \mathrm{diag}[{k}_{11},{k}_{21},{k}_{31},{k}_{41},{k}_{51}],$$
(6.102)
$${ \mathbf{K}}_{j2} = \mathrm{diag}[{k}_{12},{k}_{22},{k}_{32},{k}_{42},{k}_{52}],$$
(6.103)
where k 62 is the stiffness of the virtual joint of the passive constraining leg, and J 6 is the Jacobian matrix of the passive constraining leg in this 5-dof case. Matrices A and B are the Jacobian matrices of the structure without the passive constraining leg.
The comparison between the parallel mechanism with rigid links (without virtual joints) and the parallel mechanism with flexible links (with virtual joints) is given in Table 6.2.
Table 6.2

Comparison of the 5-dof mechanism compliance between the mechanism with flexible links and the mechanism with rigid links

 K actuator K link $${\kappa }_{{\theta }_{x}}$$ $${\kappa }_{{\theta }_{y}}$$ $${\kappa }_{{\theta }_{z}}$$ κ x κ y κ z 1,000 1,000 0.255808 0.478997 0.766154 0.00479741 0.0116413 0.00169207 1,000 101 K a 0.137552 0.257339 0.412528 0.00257412 0.00625667 0.000910745 1,000 102 K a 0.125726 0.235173 0.377165 0.0023518 0.0057182 0.000832613 1,000 103 K a 0.124543 0.232956 0.373629 0.00232956 0.00566435 0.000824799 1,000 104 K a 0.124425 0.232734 0.373275 0.00232734 0.00565897 0.000824018 1,000 105 K a 0.124413 0.232712 0.37324 0.00232712 0.00565843 0.00082394 1,000 106 K a 0.124412 0.23271 0.373236 0.00232709 0.00565838 0.000823932 1,000 107 K a 0.124412 0.23271 0.373236 0.00232709 0.00565837 0.000823931 1,000 Rigid 0.124412 0.23271 0.373236 0.00232709 0.00565837 0.000823931

#### 6.3.6.2 5-dof Parallel Mechanism

This mechanism is illustrated is Fig. 6.9, the compliance matrix for the mechanism with rigid links will be
$${ \mathbf{C}}_{\mathrm{c}} ={ \mathbf{J}}_{5}{({\mathbf{AJ}}_{5})}^{-1}\mathbf{B}\mathbf{C}{\mathbf{B}}^{\mathrm{T}}{({\mathbf{AJ}}_{ 5})}^{-\mathrm{T}}{\mathbf{J}}_{ 5}^{\mathrm{T}}$$
(6.104)
where
$$\mathbf{C} = \mathrm{diag}[{c}_{1},{c}_{2},{c}_{3},{c}_{4}]$$
(6.105)
with c 1, c 2, c 3, and c 4 are the compliances of the actuators and J 5 is the Jacobian matrix of the constraining leg in this 4-dof case. Matrices A and B are the Jacobian matrices of the structure without the passive constraining leg. Similarly, the stiffness matrix for the mechanism with flexible links can be written as
$$\mathbf{K} = {({\mathbf{J}'}_{5})}^{-\mathrm{T}}{\mathbf{K}}_{ 5}{({\mathbf{J}'}_{5})}^{-1} +{ \mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A},$$
(6.106)
where
$${ \mathbf{K}}_{5} = \mathrm{diag}[0,{k}_{52},0,{k}_{54},0,0],$$
(6.107)
$${ \mathbf{K}}_{j1} = \mathrm{diag}[{k}_{11},{k}_{21},{k}_{31},{k}_{41}],$$
(6.108)
$${ \mathbf{K}}_{j2} = \mathrm{diag}[{k}_{12},{k}_{22},{k}_{32},{k}_{42}],$$
(6.109)
where k 52 and k 54 are the stiffnesses of the virtual joints of the passive constraining leg, J 5 is the Jacobian matrix of the passive constraining leg in this 4-dof case, while A and B 1, B 2 are the Jacobian matrices of the structure without the passive constraining leg.
The comparison between the parallel mechanism with rigid links (without virtual joints) and the parallel mechanism with flexible links (with virtual joints) is given in Table 6.3. Again, the effect of link flexibility is clearly demonstrated.
Table 6.3

Comparison of the 4-dof mechanism compliance between the mechanism with flexible links and the mechanism with rigid links

 K actuator K link $${\kappa }_{{\theta }_{x}}$$ $${\kappa }_{{\theta }_{y}}$$ $${\kappa }_{{\theta }_{z}}$$ κ x κ y κ z 1,000 1,000 3.46122 0.138363 1. 5 ×10 − 3 0.0146575 1. 32691 ×10 − 3 0.000271569 1,000 101 K a 1.49876 0.0703633 1. 5 ×10 − 4 0.00717397 1. 32691 ×10 − 4 0.000106446 1,000 102 K a 1.30251 0.0635633 1. 5 ×10 − 5 0.00642561 1. 32691 ×10 − 5 0.0000899335 1,000 103 K a 1.28289 0.0628833 1. 5 ×10 − 6 0.00635078 1. 32691 ×10 − 6 0.0000882822 1,000 104 K a 1.28093 0.0628153 1. 5 ×10 − 7 0.0063433 1. 32691 ×10 − 7 0.0000881171 1,000 105 K a 1.28073 0.0628085 1. 5 ×10 − 8 0.00634255 1. 32691 ×10 − 8 0.0000881006 1,000 106 K a 1.28071 0.0628079 1. 5 ×10 − 9 0.00634247 1. 32691 ×10 − 9 0.000088099 1,000 107 K a 1.28071 0.0628078 1. 5 ×10 − 10 0.00634246 1. 32691 ×10 − 10 0.0000880988 1,000 Rigid 1.28071 0.0628078 0.0 0.00634246 0.0 0.0000880988

#### 6.3.6.3 3-dof Parallel Mechanism

This mechanism is illustrated is Fig. 6.10, the compliance matrix for the rigid mechanism can be written as
$${ \mathbf{C}}_{c} ={ \mathbf{J}}_{4}{({\mathbf{AJ}}_{4})}^{-1}\mathbf{B}\mathbf{C}{\mathbf{B}}^{\mathrm{T}}{({\mathbf{AJ}}_{ 4})}^{-\mathrm{T}}{\mathbf{J}}_{ 4}^{\mathrm{T}},$$
(6.110)
where C c = diag[c 1, c 2, c 3], with c 1, c 2 and c 3 are the compliances of the actuators and J 4 is the Jacobian matrix of the passive constraining leg in this 3-dof case. A and B are the Jacobian matrices of the structure without the passive constraining leg.
Similarly, the stiffness matrix for the mechanism with flexible links will be written as
$$\mathbf{K} = [{({\mathbf{J}'}_{4})}^{-\mathrm{T}}{\mathbf{K}}_{ 4}{({\mathbf{J}'}_{4})}^{-1} +{ \mathbf{A}}^{\mathrm{T}}{\mathbf{B}}_{ 1}^{-\mathrm{T}}{\mathbf{K}}_{ j1}{\mathbf{W}}^{-1}\mathbf{A}]$$
(6.111)
where W is defined in (6.93) and
$${ \mathbf{K}}_{4} = \mathrm{diag}[{k}_{41},{k}_{42},{k}_{43},0,0,0],$$
(6.112)
$${ \mathbf{K}}_{j1} = \mathrm{diag}[{k}_{11},{k}_{21},{k}_{31}],$$
(6.113)
$${ \mathbf{K}}_{j2} = \mathrm{diag}[{k}_{12},{k}_{22},{k}_{32}]$$
(6.114)
and J 4 is the Jacobian matrix of the passive constraining leg with virtual joints.
The comparison between the parallel mechanism with rigid links (without virtual joints) and the parallel mechanism with flexible links (with virtual joints) is given in Table 6.4. The Cartesian compliance in each of the directions is given for a reference configuration of the mechanism, for progressively increasing values of the link stiffnesses.
Table 6.4

Comparison of the 3-dof mechanism compliance between the mechanism with flexible links and the mechanism with rigid links

 K actuator K link $${\kappa }_{{\theta }_{x}}$$ $${\kappa }_{{\theta }_{y}}$$ $${\kappa }_{{\theta }_{z}}$$ κ x κ y κ z 1,000 1,000 0.09937 0.09937 10 − 3 1. 06152 ×10 − 3 1. 06152 ×10 − 3 0.000186579 1,000 101 K a 0.02904 0.02904 10 − 4 1. 06152 ×10 − 4 1. 06152 ×10 − 4 0.0000975989 1,000 102 K a 0.02201 0.02201 10 − 5 1. 06152 ×10 − 5 1. 06152 ×10 − 5 0.0000887009 1,000 103 K a 0.02131 0.02131 10 − 6 1. 06152 ×10 − 6 1. 06152 ×10 − 6 0.0000878111 1,000 104 K a 0.02123 0.02123 10 − 7 1. 06152 ×10 − 7 1. 06152 ×10 − 7 0.0000877221 1,000 105 K a 0.02123 0.02123 10 − 8 1. 06152 ×10 − 8 1. 06152 ×10 − 8 0.0000877132 1,000 106 K a 0.02123 0.02123 10 − 9 1. 06152 ×10 − 9 1. 06152 ×10 − 9 0.0000877123 1,000 107 K a 0.02123 0.02123 10 − 10 1. 06152 ×10 − 10 1. 06152 ×10 − 10 0.0000877122 1,000 Rigid 0.02123 0.02123 0.0 0.0 0.0 0.0000877122

From Table 6.4, one can find that with the improvement of link stiffness, the mechanism’s compliance with flexible links is very close to that of mechanism with rigid links. This means that one can assume the flexible mechanism to be rigid only if the link stiffness reaches a high value.

## 6.4 Conclusions

In this chapter, mechanisms with revolute actuators (whose degrees of freedom are 3, 4, 5 and 6) have been considered. Solutions for the inverse kinematic problem have been given. The Jacobian matrices obtained have been used to establish the kinetostatic model of the mechanisms. The lumped models of the link and joint compliances have been used for the study of the Cartesian compliance. It has been shown that the kinetostatic analysis can be used to assess the stiffness properties of this family of mechanisms. Finally, examples have been investigated and numerical results have been obtained and the results clearly demonstrate the relevance of the kinetostatic analysis in the context of design of such mechanisms.

## References

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Gosselin CM, Zhang D (1999) Stiffness analysis of parallel mechanisms using a lumped model. Technical report, Département de Génie Mécanique, Université LavalGoogle Scholar