An enhanced inverse dynamic and joint force analysis of multibody systems using constraint matrices

Abstract

In this paper, a new way of computing the constraint transfer matrix for the inverse dynamic and joint force analysis of multibody systems is developed. The method is based on the Newton–Euler method and the screw theory notations. This method is first developed in (Taghvaeipour et al. in Multibody Syst. Dyn. 29(2):139–168, 2013), however, in this study, it is efficiently modified by incorporating a unified constraint transfer matrix for all types of joints. This change makes both the derivation of the equations and the computations less time consuming. Moreover, in the foregoing procedure, the constraint wrenches of a system are obtained in one reference frame, namely, the global reference frame. As a case study, the proposed method is carried out on the agile wrist which is a three-legged spherical parallel robot with three degrees of freedom. At the end, the results obtained by the modified method are verified with the ones calculated by the original procedure and a software package.

Appendix: Derivation of screw matrix for different types of kinematic pairs

In this section, the screw matrices will be derived and evaluated for different types of kinematic pairs. The following screw matrices can also be found in [3, 30]; however, the following derivations can give bright insight about the unit screws of kinematic pairs and the development of the formulation proposed.

Revolute joint

Consider two rigid links, 1 and 2, are connected by means of a revolute joint 2, as depicted in Fig. 14. Links 1 and 2 can rotate with respect to each other about the axis of rotation with unit vector $\mathbf{u}_{2}$. The point $P$ is considered at the center of the joint on the axis of rotation. $P_{1}$ and $P_{2}$ has the same position as point $P$ but are considered attached to the links 1 and 2, respectively. Writing Eq. (8) for the revolute joint will yield

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{\mathbf{t}}_{2}-\bar{ \mathbf{t}}_{1} )= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 $$

(28)

in which $\boldsymbol{\omega }_{i}$ is the angular velocity of link $i$ and, $\bar{\mathbf{v}}_{i}$ is the velocity of point $P_{i}$; $i=1,2$. Since points $P_{1}$ and $P_{2}$ have no relative motion, $\bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}=\mathbf{0}$. Moreover, $\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2, namely $\boldsymbol{\omega }_{\mathit{rel}}={\omega }_{\mathit{rel}}\mathbf{u}_{2}$, in which ${\omega }_{\mathit{rel}}$ is the magnitude of vector $\boldsymbol{\omega } _{\mathit{rel}}$. Thus, Eq. (28) becomes

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \mathbf{0} \end{bmatrix} ={\omega }_{\mathit{rel}} \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ \mathbf{0} \end{bmatrix} =0. $$

Fig. 14

Links 1 and 2 are connected by a revolute joint

Regarding Eq. (9) and noting that ${\omega }_{\mathit{rel}} \neq 0$, it can be concluded that the unit screw of a revolute joint with the axis of rotation along unit vector $\mathbf{u}$ is,

$$ \mathbf{s}_{\mathrm{R}}= \begin{bmatrix} \mathbf{u} \\ \mathbf{0} \end{bmatrix} . $$

(29)

Prismatic joint

The rigid links 1 and 2, are now connected by means of a prismatic joint 2, as depicted in Fig. 15. Link 1 can slide along the axis $U_{2}$ with the unit vector $\mathbf{u}_{2}$. The point $P$ is considered on the axis of sliding. $P_{1}$ and $P_{2}$ has the same position but attached to the links 1 and 2, respectively. Writing Eq. (28) for the prismatic joint results,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 $$

(28)

in which $\boldsymbol{\omega }_{i}$ is the angular velocity of link $i$ and, $\bar{\mathbf{v}}_{i}$ is the velocity of point $P_{i}$; $i=1,2$. As links 1 and 2 do not have relative angular velocity, $\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}=\mathbf{0}$. Moreover, $\bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}$ is the relative motion of the points $P_{1}$ and $P_{2}$, namely $\bar{\mathbf{v}} _{\mathit{rel}}=\bar{v}_{\mathit{rel}}\mathbf{u}_{2}$, in which $\bar{v}_{\mathit{rel}}$ is the magnitude of vector $\bar{\mathbf{v}}_{\mathit{rel}}$. Thus,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \bar{\mathbf{v}}_{\mathit{rel}} \end{bmatrix} =\bar{v}_{\mathit{rel}} \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \mathbf{u}_{2} \end{bmatrix} =0 . $$

(30)

Fig. 15

Links 1 and 2 are connected by a prismatic joint

Regarding Eq. (9) and noting that $\bar{v}_{\mathit{rel}} \neq 0$, it can be concluded that the unit screw of a prismatic joint with the axis of sliding along the unit vector $\mathbf{u}$ is

$$ \mathbf{s}_{\mathrm{P}}= \begin{bmatrix} \mathbf{0} \\ \mathbf{u} \end{bmatrix} . $$

(31)

Helical joint

As it is depicted in Fig. 16, link 2 is attached to link 1 by a helical joint in which link 2 can rotate around the axis $U_{2}$ while it moves linearly along the same axis. The point $P$ is considered on the axis $U_{2}$ and, $P_{1}$ and $P_{2}$ are attached to the links 1 and 2, respectively. Writing Eq. (28) for the helical joint will yield

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 . $$

(28)

The terms $\boldsymbol{\omega }_{i}$ and $\bar{\mathbf{v}}_{i}$ are defined as before. $\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2, i.e. $\boldsymbol{\omega }_{\mathit{rel}}={\omega }_{\mathit{rel}}\mathbf{u}_{2}$. Moreover, $\bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}$ is the relative motion of the points $P_{1}$ and $P_{2}$, namely $\bar{\mathbf{v}}_{\mathit{rel}}= \bar{v}_{\mathit{rel}}\mathbf{u}_{2}$. Hence,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \bar{\mathbf{v}}_{\mathit{rel}} \end{bmatrix} =0 . $$

(32)

In a helical joint, the translational velocity of link 2 is proportional to its angular velocity, i.e.,

$$ \bar{\mathbf{v}}_{\mathit{rel}}=p\boldsymbol{\omega }_{\mathit{rel}} $$

where $p$ is called the joint’s pitch. Thus,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \bar{\mathbf{v}}_{\mathit{rel}} \end{bmatrix} =\boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ p\boldsymbol{\omega }_{\mathit{rel}} \end{bmatrix} ={\omega }_{\mathit{rel}}\boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ p\mathbf{u}_{2} \end{bmatrix} =0 . $$

(33)

Fig. 16

Links 1 and 2 are connected by a helical joint

Regarding Eq. (9) and noting that ${\omega }_{\mathit{rel}} \neq 0$, it can be concluded that the unit screw for a helical joint with the axis of rotation along the unit vector $\mathbf{u}$, and pitch $p$, is:

$$ \mathbf{s}_{\mathrm{H}}= \begin{bmatrix} \mathbf{u} \\ p\mathbf{u} \end{bmatrix} . $$

(34)

Cylindrical joint

Regarding Fig. 17, link 2 is connected to link 1 by a cylindrical joint in which the link 2 can move along the axis $U_{2}$ while it rotates freely about the same axis. $P_{1}$ and $P_{2}$ has the same position but attached to the links 1 and 2, respectively. Writing Eq. (28) for the cylindrical joint will result in

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 ; $$

(28)

$\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2, i.e. $\boldsymbol{\omega } _{\mathit{rel}}={\omega }_{\mathit{rel}}\mathbf{u}_{2}$. Moreover, $\bar{\mathbf{v}} _{2}-\bar{\mathbf{v}}_{1}$ is the relative motion of the points $P_{1}$ and $P_{2}$, namely $\bar{\mathbf{v}}_{\mathit{rel}}=\bar{v}_{\mathit{rel}} \mathbf{u}_{2}$. Thus,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \bar{\mathbf{v}}_{\mathit{rel}} \end{bmatrix} ={\omega }_{\mathit{rel}} \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ \mathbf{0} \end{bmatrix} +\bar{v}_{\mathit{rel}}\boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \mathbf{u}_{2} \end{bmatrix} =0 ; $$

(35)

Equation (35) is of the form

$$ ax+by=0. $$

(36)

In Eq. (36), if $x$ and $y$ are independent, non-zero variables, then it can be concluded that $a=b=0$. Similarly in Eq. (35), ${\omega }_{\mathit{rel}}$ and $\bar{v}_{\mathit{rel}}$ are non-zero and independent variables as the cylindrical joint is a 2-DOF kinematic pair. Hence, it can be concluded that coefficients in Eq. (35) are equal to zero. So,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ \mathbf{0} \end{bmatrix} =0, \qquad \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \mathbf{u}_{2} \end{bmatrix} =0. $$

Casting the above equations in matrix form yields

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2}&\mathbf{0} \\ \mathbf{0}&\mathbf{u}_{2} \end{bmatrix} =\mathbf{0}. $$

According to Eq. (11), it can be concluded that the screw matrix of a cylindrical joint with the axis of rotation and translation along the unit vector $\mathbf{u}$ is

$$ \mathbf{S}_{\mathrm{CY}}= \begin{bmatrix} \mathbf{u}&\mathbf{0} \\ \mathbf{0}&\mathbf{u} \end{bmatrix} . $$

(37)

Fig. 17

Links 1 and 2 are connected by a cylindrical joint

Spherical joint

Figure 18 shows a spherical joint connecting links 1 and 2. The point $P$ is considered at the intersection of links’ axes. Writing Eq. (28) for the spherical joint will culminate in,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 . $$

(28)

Since points $P_{1}$ and $P_{2}$ have no relative motion, $\bar{ \mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}=\mathbf{0}$. Moreover, $\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2. As the spherical joint can rotate in every direction, $\boldsymbol{\omega }_{\mathit{rel}}$ has components in three perpendicular directions, i.e.

$$ \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}= \boldsymbol{\omega }_{\mathit{rel}}={}^{\mathbf{u}_{2}}{{\omega }} _{\mathit{rel}}\mathbf{u}_{2}+{}^{\mathbf{e}_{2}}{{\omega }}_{\mathit{rel}} \mathbf{e}_{2}+{}^{\mathbf{q}_{2}}{{\omega }}_{\mathit{rel}} \mathbf{q}_{2} $$

where $\mathbf{u}_{2}$ and $\mathbf{e}_{2}$ are two arbitrary perpendicular unit vectors of the spherical joint as shown in Fig. 18. Also, $\mathbf{q}_{2}=\mathbf{u}_{2} \times \mathbf{e}_{2}$. Moreover, ${}^{\mathbf{u}_{2}}{{\omega }}_{\mathit{rel}}$, ${}^{\mathbf{e}_{2}}{{\omega }}_{\mathit{rel}}$ and ${}^{\mathbf{q}_{2}}{{\omega }}_{\mathit{rel}}$ are the components of the relative angular velocity along the unit vectors $\mathbf{u}_{2}$, $\mathbf{e}_{2}$ and $\mathbf{q}_{2}$, respectively. Thus,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \mathbf{0} \end{bmatrix} ={}^{\mathbf{u}_{2}}{{\omega }}_{\mathit{rel}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ \mathbf{0} \end{bmatrix} +{}^{\mathbf{e}_{2}}{{\omega }}_{\mathit{rel}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{e}_{2} \\ \mathbf{0} \end{bmatrix} +{}^{\mathbf{q}_{2}}{{\omega }}_{\mathit{rel}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{q}_{2} \\ \mathbf{0} \end{bmatrix} =0 ; $$

(38)

${}^{\mathbf{u}_{2}}{{\omega }}_{\mathit{rel}}$, ${}^{\mathbf{e}_{2}}{{\omega }}_{\mathit{rel}}$ and ${}^{\mathbf{q}_{2}}{{\omega }}_{\mathit{rel}}$ are completely independent as the spherical joint is a 3-DOF kinematic pair. Also, ${}^{\mathbf{u}_{2}}{{\omega }}_{\mathit{rel}}$, ${}^{\mathbf{e}_{2}}{{\omega }}_{\mathit{rel}}$, ${}^{\mathbf{q}_{2}}{{\omega }}_{\mathit{rel}}\neq 0$. Similar to what argued for the cylindrical joint, it can be concluded that coefficients in Eq. (38) are equal to zero. Thus,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2}&\mathbf{e}_{2}&\mathbf{q}_{2} \\ \mathbf{0}&\mathbf{0}&\mathbf{0} \end{bmatrix} =\mathbf{0} . $$

(39)

By comparing Eqs. (11) and (39), it can be concluded that the screw matrix of a spherical joint with two arbitrary perpendicular axes along the unit vectors $\mathbf{u}$ and $\mathbf{e}$ is,

$$ \mathbf{S}_{\mathrm{S}}= \begin{bmatrix} \mathbf{u}&\mathbf{e}&\mathbf{q} \\ \mathbf{0}&\mathbf{0}&\mathbf{0} \end{bmatrix} $$

(40)

where $\mathbf{q}=\mathbf{u}\times \mathbf{e}$.

Fig. 18

Links 1 and 2 are connected by a spherical joint

Planar joint

A planar joint is shown in Fig. 19. The point $P$ is considered on the intersection of two arbitrary axes, i.e. $U_{2}$ and $E_{2}$, on the plane of contact. Equation (28) for a planar joint will be of the following form:

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 ; $$

(28)

$\bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}$ is the relative motion of the points $P_{1}$ and $P_{2}$, which is in the two-dimensional space. So, $\bar{\mathbf{v}}_{\mathit{rel}}$ has components in two perpendicular axes which both are in the plane of contact. The unit vectors of these arbitrary axes are shown in Fig. 19 by $\mathbf{u} _{2}$ and $\mathbf{e}_{2}$. Moreover, $\boldsymbol{\omega }_{2}- \boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2 which is about the axis perpendicular to the plane of contact, namely, it is directed along $\mathbf{q}_{2}=\mathbf{u}_{2} \times \mathbf{e}_{2}$. Thus,

$$\begin{aligned} \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} &=\bar{ \mathbf{v}}_{\mathit{rel}}= {}^{{\mathbf{u}_{2}}}{\bar{{v}}_{\mathit{rel}}} \mathbf{u}_{2}+ {}^{{\mathbf{e}_{2}}}{\bar{{v}}_{\mathit{rel}}} \mathbf{e}_{2}, \\ \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} &= \boldsymbol{\omega }_{\mathit{rel}}={\omega }_{\mathit{rel}} \mathbf{q}_{2}, \end{aligned}$$

where ${}^{{\mathbf{u}_{2}}}{\bar{{v}}_{\mathit{rel}}}$ and ${}^{{\mathbf{e}_{2}}}{\bar{{v}}_{\mathit{rel}}}$ are the components of the relative translational velocity along unit vectors $\mathbf{u} _{2}$ and $\mathbf{e}_{2}$, respectively. So,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \bar{\mathbf{v}}_{\mathit{rel}} \end{bmatrix} ={}^{{\mathbf{u}_{2}}}{ \bar{{v}}_{\mathit{rel}}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \mathbf{u}_{2} \end{bmatrix} +{}^{{\mathbf{e}_{2}}}{ \bar{{v}}_{\mathit{rel}}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0} \\ \mathbf{e}_{2} \end{bmatrix} +{\omega }_{\mathit{rel}} \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{q}_{2} \\ \mathbf{0} \end{bmatrix} =0 ; $$

(41)

${}^{{\mathbf{u}_{2}}}{\bar{{v}}_{\mathit{rel}}}$, ${}^{{\mathbf{e}_{2}}}{\bar{{v}}_{\mathit{rel}}}$ and ${\omega }_{\mathit{rel}}$ are independent as the planar joint is a 3-DOF kinematic pair. Also, ${}^{{\mathbf{u}_{2}}}{\bar{{v}}_{\mathit{rel}}}$, ${}^{{\mathbf{e}_{2}}}{\bar{{v}}_{\mathit{rel}}}$, ${\omega }_{\mathit{rel}} \neq 0$. Thus, it can be concluded that coefficients in Eq. (41) are equal to zero, namely,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{0}&\mathbf{0}&\mathbf{q}_{2} \\ \mathbf{u}_{2}&\mathbf{e}_{2}&\mathbf{0} \end{bmatrix} =\mathbf{0} . $$

(42)

By comparing Eqs. (11) and (42), it can be concluded that the screw matrix of a planar joint with two arbitrary perpendicular unit vectors $\mathbf{u}$ and $\mathbf{e}$ in its plane is

$$ \mathbf{S}_{\mathrm{PL}}= \begin{bmatrix} \mathbf{0}&\mathbf{0}&\mathbf{q} \\ \mathbf{u}&\mathbf{e}&\mathbf{0} \end{bmatrix} $$

(43)

where $\mathbf{q}=\mathbf{u}\times \mathbf{e}$.

Fig. 19

Links 1 and 2 are connected by a planar joint

Universal joint

A universal joint is shown in Fig. 20. The links 1 and 2 can rotate about the axes $U_{2}$ and $E_{2}$. The point $P$ is considered at the intersection point of the two axes. Writing Eq. (28) for the universal joint will yield,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \Delta \bar{ \mathbf{t}}_{2}=\boldsymbol{\lambda }^{\mathrm{T}}_{2} (\bar{ \mathbf{t}}_{2}-\bar{\mathbf{t}}_{1} )= \boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1} \\ \bar{\mathbf{v}}_{2}-\bar{\mathbf{v}}_{1} \end{bmatrix} =0 . $$

(28)

Since points $P_{1}$ and $P_{2}$ have no relative motion, $\bar{ \mathbf{v}}_{2}-\bar{\mathbf{v}}_{1}=\mathbf{0}$. Moreover, $\boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}$ is the relative angular velocity of the links 1 and 2 which can occur about both axes $U_{2}$ and $E_{2}$. So, $\boldsymbol{\omega }_{\mathit{rel}}$ has components along two perpendicular unit vectors $\mathbf{u}_{2}$ and $\mathbf{e}_{2}$. Thus,

$$ \boldsymbol{\omega }_{2}-\boldsymbol{\omega }_{1}= \boldsymbol{\omega }_{\mathit{rel}}= {}^{{\mathbf{u}_{2}}}{{\omega }_{\mathit{rel}}}\mathbf{u}_{2}+ {}^{{\mathbf{e}_{2}}}{{\omega }_{\mathit{rel}}}\mathbf{e}_{2} $$

where ${}^{{\mathbf{u}_{2}}}{{\omega }_{\mathit{rel}}}$ and ${}^{{\mathbf{e}_{2}}}{{\omega }_{\mathit{rel}}}$ are the components of the relative angular velocity of links 1 and 2 along the unit vectors $\mathbf{u}_{2}$ and $\mathbf{e}_{2}$, respectively. Accordingly,

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2}\Delta \bar{\mathbf{t}}_{2}= \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \boldsymbol{\omega }_{\mathit{rel}} \\ \mathbf{0} \end{bmatrix} ={}^{{\mathbf{u}_{2}}}{{\omega }_{\mathit{rel}}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2} \\ \mathbf{0} \end{bmatrix} +{}^{{\mathbf{e}_{2}}}{{\omega }_{\mathit{rel}}}\boldsymbol{\lambda } ^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{e}_{2} \\ \mathbf{0} \end{bmatrix} =0 ; $$

(44)

${}^{{\mathbf{u}_{2}}}{{\omega }_{\mathit{rel}}}$ and ${}^{{\mathbf{e}_{2}}}{{\omega }_{\mathit{rel}}}$ are independent variables as the universal joint is a 2-DOF kinematic pair. Also, ${}^{{\mathbf{u}_{2}}}{{\omega }_{\mathit{rel}}}$, ${}^{{\mathbf{e}_{2}}}{{\omega }_{\mathit{rel}}}\neq 0$. Similar to the preceding parts, it can be concluded that

$$ \boldsymbol{\lambda }^{\mathrm{T}}_{2} \begin{bmatrix} \mathbf{u}_{2}&\mathbf{e}_{2} \\ \mathbf{0}&\mathbf{0} \end{bmatrix} =\mathbf{0} . $$

(45)

By comparing Eqs. (11) and (45), it can be concluded that the screw matrix of a universal joint with the axes of rotation along the unit vectors $\mathbf{u}$ and $\mathbf{e}$ is

$$ \mathbf{S}_{\mathrm{U}}= \begin{bmatrix} \mathbf{u}&\mathbf{e} \\ \mathbf{0}&\mathbf{0} \end{bmatrix} . $$

(46)

Fig. 20

Links 1 and 2 are connected by a universal joint