Degeneracy conditions of the dynamic model of parallel robots

Abstract

Despite their well-known advantages in terms of higher intrinsic rigidity, larger payload-to-weight ratio, and higher velocity and acceleration capacities, parallel robots have drawbacks. Among them, the most important one is surely the presence of singularities in the workspace, which divide the workspace into different aspects (each aspect corresponding to one or more assembly modes) and near which the performance is considerably reduced.

In order to increase the reachable workspace of parallel robots, a promising solution consists in the definition of optimal trajectories passing through the singularities to change either the leg working modes or the robot assembly modes. Previous works on the field have shown that it is possible to define optimal trajectories that allow the passing through the robot type 2 singularities. Such trajectories must respect a physical criterion that can be obtained through the analysis of the degeneracy conditions of the parallel robot inverse dynamic model.

However, the mentioned works were not complete: they lacked a degeneracy condition of the parallel robot inverse dynamic model, which is not due to type 2 singularity anymore, but to a serial singularity. Crossing a serial singularity is appealing as in that case we can change the robot leg working mode and then potentially access to other workspace zones. This absence is due to the fact that the authors used a reduced dynamic model, which was not taking into account all link dynamic parameters.

The present paper aims to fill this gap by providing a complete study of the degeneracy conditions of the parallel robot dynamic model and by demonstrating that it is possible to cross the type 2, but also serial singularity, by defining trajectories that respect some given criteria obtained from the analysis of the degeneracy of the robot dynamic model. It also aims to demonstrate that the serial singularities have impacts on the robot effort transmission, which is a point that is usually bypassed in the literature. All theoretical developments are validated through simulations and experiments.

Appendices

Appendix A: Kinematics of the five-bar mechanism

For the five-bar mechanism, the loop-closure equations (10) can be written as ($i=1,2$)

$$ \mathbf{0} = x \mathbf{x}_{0} + y \mathbf{y}_{0} - d_{2i} \mathbf{x}_{1i} - d_{3i} \mathbf{x}_{2i}, $$

(66)

which can be expanded in the base frame as

$$ \begin{aligned} 0 &= x - x_{A_{i}} - d_{2i}\cos q_{1i} - d_{3i}\cos(q_{1i}+q_{2i}), \\ 0 &= y - y_{A_{i}} - d_{2i}\sin q_{1i} - d_{3i}\sin(q_{1i}+q_{2i}), \end{aligned} $$

(67)

where $x$ and $y$ are the end-effector coordinates, and

$$ 0 = \pi- q_{11}-q_{21}-q_{31} + q_{12}+q_{22}, $$

(68)

where $x_{A_{i}}$ and $y_{A_{i}}$ are the position coordinates along $\mathbf{x}_{0}$ and $\mathbf{y}_{0}$ axes for the point $A_{i}$.

From (67), the reduced loop-closure equations (11), which directly relate the displacements of the actuated joints to the moving platform coordinates, can be obtained after deleting from (67) the terms in $\cos(q_{1i}+q_{2i})$ or $\sin (q_{1i}+q_{2i})$ (for $i=1, 2$):

$$ d_{3i}^{2} = (x - x_{B_{i}} )^{2} + (y - y_{B_{i}} )^{2}, $$

(69)

where $x_{B_{i}}=x_{A_{i}} + d_{2i}\cos q_{1i}$ and $y_{B_{i}}=y_{A_{i}} + d_{2i}\sin q_{1i}$ are the position coordinates of point $B_{i}$.

Then,

$$ x= f_{i} y + k_{i},\qquad y= \frac{-p_{i}\pm\sqrt{p_{i}^{2} - 4 g_{i} r_{i}}}{2 g_{i} }, $$

(70)

where

$$ \begin{aligned} f_{i}&=-\frac{y_{B_{2}}-y_{1}}{x_{B_{2}}-x_{B_{1}}},\qquad g_{i}=f_{i}^{2}+1, \\ k_{i}&=\frac{x_{B_{1}}^{2}+y_{B_{1}}^{2}-y_{B_{1}}^{2}-y_{B_{2}}^{2}}{2(x_{B_{2}}-x_{B_{1}})}, \\ p_{i}&=2f_{i}(k_{i}-x_{B_{1}})-2y_{B_{1}}, \\ r_{i}&=x_{B_{1}}^{2}+y_{B_{1}}^{2}-d_{3i}^{2}+k_{1}^{2}-2k_{1}x_{B_{1}}. \end{aligned} $$

(71)

In (70), the sign “±” denotes the two robot assembly modes.

Then, it comes easily from (67) and (68) that:

$$\begin{aligned} q_{2i} =& \tan^{-1} \biggl(\frac{y-y_{B_{i}}}{x-x_{B_{i}}} \biggr), \end{aligned}$$

(72)

$$\begin{aligned} q_{31} =& \pi- q_{11}-q_{21} + q_{12}+q_{22}. \end{aligned}$$

(73)

Now, differentiating (69) with respect to time and simplifying, we can find the matrices $\mathbf{A}_{p}$ and $\mathbf{B}_{p}$ of (15):

$$ \mathbf{A}_{p}= \left[ \textstyle\begin{array}{c@{\quad}c} c_{121} & s_{121} \\ c_{122} & s_{122} \end{array}\displaystyle \right] = \begin{bmatrix} \mathbf{x}_{21}^{T} \\ \mathbf{x}_{22}^{T} \end{bmatrix} , $$

(74)

where $c_{12i}=\cos(q_{1i}+q_{2i})$ and $s_{12i}=\sin(q_{1i}+q_{2i})$ ($i=1,2$),

$$ \mathbf{B}_{p}= -d_{2i} \left[ \textstyle\begin{array}{c@{\quad}c} \sin q_{21} & 0 \\ 0 & \sin q_{22} \end{array}\displaystyle \right] , $$

(75)

leading thus to

$$ \mathbf{A}_{p} \begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} + \mathbf{B}_{p} \begin{bmatrix} \dot{q}_{11} \\ \dot{q}_{12} \end{bmatrix} = \mathbf{0}. $$

(76)

Now, differentiating (66) and (68) with respect to time, we can find that

$$\begin{aligned} \mathbf{0} =& \dot{x} \mathbf{x}_{0} + \dot{y} \mathbf{y}_{0} - d_{2i} \mathbf {y}_{1i} \dot{q}_{1i} - d_{3i} \mathbf{y}_{2i} (\dot{q}_{1i}+\dot{q}_{2i}), \end{aligned}$$

(77)

$$\begin{aligned} 0 =& - \dot{q}_{11}- \dot{q}_{21}- \dot{q}_{31} + \dot{q}_{12}+ \dot{q}_{22}. \end{aligned}$$

(78)

Projecting these equations in the frame of the link $2i$ and developing, we get that

$$ \left[ \textstyle\begin{array}{c@{\quad}c} c_{12i} & s_{12i} \\ -s_{12i} & c_{12i} \end{array}\displaystyle \right] \left[ \textstyle\begin{array}{c} \dot{x} \\ \dot{y} \end{array}\displaystyle \right] = \left[ \textstyle\begin{array}{c@{\quad}c} d_{2i} \sin q_{2i} & 0 \\ d_{2i} \cos q_{2i}+d_{3i} & d_{3i} \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{1i} \\ \dot{q}_{2i} \end{bmatrix} $$

(79)

for $i=1, 2$.

Combining (78) and (79) and noticing that the first line of (79) can be disregarded as the velocity $\dot{q}_{2i}$ of the passive joints at $B_{i}$ is not included in this equation, we get

$$\begin{aligned} \left[ \textstyle\begin{array}{c@{\quad}c} -s_{121} & c_{121} \\ -s_{122} & c_{122} \\ 0 & 0 \end{array}\displaystyle \right] \begin{bmatrix} \dot{x} \\ \dot{y} \end{bmatrix} =& \left[ \textstyle\begin{array}{c@{\quad}c} d_{21} \cos q_{21}+d_{31} & 0 \\ 0& d_{22} \cos q_{22}+d_{32} \\ -1 & 1 \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{11} \\ \dot{q}_{12} \end{bmatrix} \\ &{}+ \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} d_{31} & 0 & 0 \\ 0 & d_{32} & 0 \\ -1 & 1 & -1 \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{21} \\ \dot{q}_{31} \\ \dot{q}_{22} \end{bmatrix} , \end{aligned}$$

(80)

which can be rewritten as

$$ \mathbf{J}_{tk}\mathbf{v} - \mathbf{J}_{k_{a}}\dot{ \mathbf{q}}_{a} - \mathbf{J}_{k_{d}}\dot{\mathbf{q}}_{d} = \mathbf{0} $$

(81)

with

$$\begin{aligned} \mathbf{J}_{tk} =& \left[ \textstyle\begin{array}{c@{\quad}c} -s_{121} & c_{121} \\ -s_{122} & c_{122} \\ 0 & 0 \end{array}\displaystyle \right] , \end{aligned}$$

(82)

$$\begin{aligned} \mathbf{J}_{k_{a}} =& \left[ \textstyle\begin{array}{c@{\quad}c} d_{21} \cos q_{21}+d_{31} & 0\\ 0 & d_{22} \cos q_{22}+d_{32} \\ -1 & 1 \end{array}\displaystyle \right] , \end{aligned}$$

(83)

$$\begin{aligned} \mathbf{J}_{k_{d}} =& \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} d_{31} & 0 & 0 \\ 0 & 0 & d_{32} \\ -1 & -1 & 1 \end{array}\displaystyle \right] , \end{aligned}$$

(84)

and $\mathbf{v}^{T} = [\dot{x} \,\, \dot{y}]$, $\dot{\mathbf{q}}_{a}^{T} = [\dot{q}_{11} \,\, \dot{q}_{12}]$, and $\dot{\mathbf{q}}_{d}^{T} = [\dot{q}_{21} \,\, \dot{q}_{31} \,\, \dot{q}_{22}]$.

From (74) and (84) it is possible to observe that

• the matrix $\mathbf{J}_{k_{d}}$ is constant and never singular; as a result, the robot does not encounter LPJTS singularities,

• the matrix $\mathbf{A}_{p}$ is singular when $\mathbf{x}_{21}$ and $\mathbf{x}_{22}$ are collinear, which is the condition of type 2 singularity mentioned in Sect. 3.1.

All velocities and accelerations quantities can be then computed from (76) and (81) by using relations (12) and (25) given in Sect. 2.3.

Appendix B: Dynamics of the five-bar mechanism

The inverse dynamic model of the open-loop virtual structure of the five-bar mechanism can be obtained by noticing that:

• leg 1 is a planar 3R robot in which the last body is massless,

• leg 2 is a planar 2R robot.

Its inverse dynamic model may be found in [38]:

$$\begin{aligned} \tau_{t_{11}} =& \bigl(zz_{11} + ia_{11} + d_{2i}^{2} m_{21} \bigr)\ddot {q}_{11}+zz_{21}(\ddot{q}_{11}+ \ddot{q}_{21}) \\ &{}+d_{2i}mx_{21} \bigl((2\ddot{q}_{11}+ \ddot{q}_{21})\cos q_{21}-\dot{q}_{21}(2\dot {q}_{11}+\dot{q}_{21})\sin q_{21} \bigr) \\ &{}+d_{2i}my_{21} \bigl((2\ddot {q}_{11}+ \ddot{q}_{21})\sin q_{21}+\dot{q}_{21}(2 \dot{q}_{11}+\dot {q}_{21})\cos q_{21} \bigr) \\ &{}+fs_{11}\,\mathrm{sign}(\dot{q}_{11}) + fv_{11}\dot{q}_{11}, \\ \tau_{t_{21}} =& zz_{21}(\ddot{q}_{11}+ \ddot{q}_{21}) +d_{2i}mx_{21} \bigl( \ddot{q}_{11}\cos q_{21}+\dot{q}_{11}^{2} \sin q_{21} \bigr) \\ &{}+d_{2i}my_{21} \bigl(\ddot{q}_{11}\sin q_{21}-\dot {q}_{11}^{2}\cos q_{21}\bigr) \\ &{}+ fs_{21}\,\mathrm{sign}(\dot{q}_{21}) + fv_{21} \dot{q}_{21}, \\ \tau_{t_{31}} =& fs_{31}\,\mathrm{sign}(\dot{q}_{31}) + fv_{31} \dot{q}_{31}, \\ \tau_{t_{12}} =& \bigl(zz_{12} + ia_{12} + d_{2i}^{2} m_{22} \bigr)\ddot {q}_{12}+zz_{22}( \ddot{q}_{12}+\ddot{q}_{22}) \\ &{}+d_{2i}mx_{22} \bigl((2\ddot{q}_{12}+ \ddot{q}_{22})\cos q_{22}-\dot{q}_{22}(2\dot {q}_{12}+\dot{q}_{22})\sin q_{22} \bigr) \\ &{}+d_{2i}my_{22} \bigl((2\ddot {q}_{12}+ \ddot{q}_{22})\sin q_{22}+\dot{q}_{22}(2 \dot{q}_{12}+\dot {q}_{22})\cos q_{22} \bigr) \\ &{}+fs_{12}\,\mathrm{sign}(\dot{q}_{12}) + fv_{12} \dot{q}_{12}, \\ \tau_{t_{22}} =& zz_{22}(\ddot{q}_{12}+ \ddot{q}_{22}) +d_{2i}mx_{22} \bigl( \ddot{q}_{12}\cos q_{22}+\dot{q}_{12}^{2} \sin q_{22} \bigr) \\ &{}+d_{2i}my_{22} \bigl(\ddot{q}_{12}\sin q_{22}-\dot {q}_{12}^{2}\cos q_{22} \bigr) \\ &{}+ fs_{22}\,\mathrm{sign}(\dot{q}_{22}) + fv_{22} \dot{q}_{22}, \end{aligned}$$

(85)

where

• the parameters $zz_{ji}$, $ia_{ji}$, $m_{ji}$, $mx_{ji}$, $my_{ji}$, $fs_{ji}$, $fv_{ji}$ are defined in Sect. 2.2 ($j=1,2,3$),

• the angles $q_{ji}$ and length $d_{2i}$ are defined in Table 1 and Fig. 4,

• $\tau_{t_{1i}}$ is the torque of the virtual actuator located at point $A_{i}$, $\tau_{t_{2i}}$ is the torque of the virtual actuator located at point $B_{i}$, and $\tau_{t_{3i}}$ is the torque of the virtual actuator located at point $C_{i}$. The vector $\boldsymbol{\tau }_{t_{a}}$ of (3) stacks all components $\boldsymbol{\tau }_{t_{a}}= [\tau_{t_{11}}\ \ \tau_{t_{12}} ]^{T}$ whereas the vector $\boldsymbol{\tau}_{t_{d}}$ of (4) stacks all vectors $\boldsymbol{\tau}_{t_{d}}= [\tau_{t_{21}}\ \ \tau_{t_{31}}\ \ \tau _{t_{22}} ]^{T}$.

The inverse dynamic model of the free body corresponding to the end-effector (body 4) in the virtual system is

$$ \begin{aligned} \tau_{p_{1}} &= m_{4}\ddot{x}, \\ \tau_{p_{2}} &= m_{4}\ddot{y}, \end{aligned} $$

(86)

with $\tau_{p_{j}}$ being the $j$th components of the vector $\boldsymbol {\tau}_{pr}$ of (9); $m_{4}$ is the end-effector mass.

Combining these expressions with those of Appendix A into the equations of Sect. 2.3, we can straightforwardly compute the inverse dynamic model of the five-bar mechanism.

Then, for analyzing the degeneracy conditions of expression (37), let us compute the term $\mathbf{w}_{p}$. For that, let us rewrite the vector $\boldsymbol{\tau}_{t_{d}}$ under the form

$$ \boldsymbol{\tau}_{t_{d}}=\mathbf{M}_{d}(\mathbf{q_{t}}) \ddot{\mathbf {q}}_{t}+\mathbf{c}_{d}(\mathbf{q}_{t}, \dot{\mathbf{q}}_{t}), $$

(87)

where

$$ \mathbf{M}_{d}= \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} m_{d}^{11} & 0 & zz_{21} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & m_{d}^{32} & 0 & 0 & zz_{22} \end{array}\displaystyle \right] $$

(88)

with $m_{d}^{11}=zz_{21}+d_{2i}(mx_{21}\cos q_{21} + my_{21}\sin q_{21})$, $m_{d}^{32}=zz_{22}+d_{2i}(mx_{22}\cos q_{22} + my_{22}\sin q_{22})$, and

$$\begin{aligned} \mathbf{c}_{d} =& \left[ \textstyle\begin{array}{c@{\quad}c} d_{21}mx_{21}\sin q_{21} - d_{21}my_{21}\cos q_{21} & 0 \\ 0 & 0 \\ 0 & d_{22}mx_{22}\sin q_{22} - d_{22}my_{22}\cos q_{22} \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{11}^{2}\\ \dot{q}_{12}^{2} \end{bmatrix} \\ &{} + \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} fv_{21} & 0 & 0 \\ 0 & fv_{31} & 0 \\ 0 & 0 & fv_{22} \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{21}\\ \dot{q}_{31} \\ \dot{q}_{22} \end{bmatrix} + \begin{bmatrix} fs_{21} \,\mathrm{sign}(\dot{q}_{21})\\ fs_{31} \,\mathrm{sign}(\dot{q}_{31}) \\ fs_{22} \,\mathrm{sign}(\dot{q}_{22}) \end{bmatrix} \\ =& \mathbf{C}_{d}^{r} \begin{bmatrix} \dot{q}_{11}^{2}\\ \dot{q}_{12}^{2} \end{bmatrix} + \mathbf{F}_{v_{d}} \dot{\mathbf{q}}_{d} + \mathbf{F}_{s_{d}}. \end{aligned}$$

(89)

Introducing (14), (20), (22), and (25) into (88) and simplifying and skipping all mathematical derivations, we get

$$ \boldsymbol{\tau}_{t_{d}}=\mathbf{M}_{d}^{x}( \mathbf{x},\mathbf{q}_{t}) \dot {\mathbf{v}} + \mathbf{c}_{d}^{x}( \mathbf{x},\mathbf{q}_{t},\mathbf{v}), $$

(90)

where

$$ \mathbf{M}_{d}^{x}=\mathbf{M}_{d} \begin{bmatrix} \mathbf{J}_{p}^{-1} \\ \mathbf{J}_{q_{d}} \end{bmatrix} $$

(91)

and

$$ \mathbf{c}_{d}^{x}=\mathbf{M}_{d} \begin{bmatrix} \mathbf{J}_{p}^{d} \\ \mathbf{J}_{q_{d}}^{d} \end{bmatrix} \mathbf{v} + \mathbf{C}_{d}^{r} \begin{bmatrix} (\mathbf{j}_{1}^{\mathrm{inv}}\mathbf{v})^{2}\\ (\mathbf {j}_{2}^{\mathrm{inv}}\mathbf{v})^{2} \end{bmatrix} + \mathbf{F}_{v_{d}} \mathbf{J}_{q_{d}} \mathbf{v} + \mathbf{F}_{s_{d}} $$

(92)

with $\mathbf{j}_{i}^{\mathrm{inv}}$ the $i$th line of the matrix $\mathbf {J}_{p}^{-1}=-\mathbf{B}_{p}^{-1} \mathbf{A}_{p}= \bigl[ {\scriptsize\begin{matrix}{} \mathbf{j}_{1}^{\mathrm{inv}} \cr \mathbf{j}_{2}^{\mathrm{inv}} \end{matrix}}\bigr]$ defined in Appendix A, and $\mathbf{J}_{q_{d}}$, $\mathbf{J}_{p}^{d}$, and $\mathbf{J}_{q_{d}}^{d}$ three matrices defined at (20), (22), and (25).

Appendix C: Kinematics of the Tripteron

For the Tripteron, the loop-closure equations (10) can be written as ($i=1,2,3$)

$$ \mathbf{0} = x \mathbf{x}_{0} + y \mathbf{x}_{0} + z \mathbf{z}_{0} - \mathbf {x}_{A_{i}} - \mathbf{x}_{D_{i}P}- q_{1i}\mathbf{x}_{1i} - d_{3i} \mathbf {x}_{2i} - d_{4i} \mathbf{x}_{3i}, $$

(93)

which can be expanded in the leg $i$ frame (Fig. 6(b)) as

$$ \begin{aligned} 0 &= {}^{i} x_{D_{i}} - {}^{i} x_{A_{i}} - d_{2i}\cos q_{2i} - d_{3i}\cos (q_{2i}+q_{3i}), \\ 0 &= {}^{i} y_{D_{i}} - {}^{i} y_{A_{i}} - d_{2i}\sin q_{2i} - d_{3i}\sin (q_{2i}+q_{3i}), \\ 0 &= {}^{i} z_{D_{i}} - r_{1i}, \end{aligned} $$

(94)

and

$$ 0 = q_{2i}+q_{3i}+q_{4i}, $$

(95)

where ${}^{i} x_{D_{i}}$, ${}^{i} y_{D_{i}}$, and ${}^{i} z_{D_{i}}$ are the point $D_{i}$ coordinates expressed in the frame of the leg $i$,

$$\begin{aligned} &{}^{1} x_{D_{1}} = x, \qquad {}^{1} y_{D_{1}} = y, \qquad {}^{1} z_{D_{1}} = z, \end{aligned}$$

(96)

$$\begin{aligned} &{}^{2} x_{D_{2}} = y, \qquad {}^{2} y_{D_{2}} = z, \qquad {}^{2} z_{D_{2}} = x, \end{aligned}$$

(97)

$$\begin{aligned} &{}^{3} x_{D_{3}} = z, \qquad {}^{3} y_{D_{3}} = x, \qquad {}^{3} z_{D_{3}} = y, \end{aligned}$$

(98)

${}^{i} x_{A_{i}}$, ${}^{i} y_{A_{i}}$, and ${}^{i} z_{A_{i}}$ are the point $A_{i}$ coordinates (also regrouped in the vector $\mathbf{x}_{A_{i}}$) expressed in the frame of the leg $i$, $\mathbf{x}_{D_{i}P}=\overrightarrow{D_{i}P}$ ($P$ is the platform center), and $r_{1i}$ is defined in the Table 5.

From the last line of (94) we directly get:

$$ \begin{aligned} x &= q_{12} - a, \\ y &= q_{13} + a, \\ z &= q_{11}. \end{aligned} $$

(99)

From (94), by deleting the terms in $\cos (q_{2i}+q_{3i})$ or $\sin(q_{2i}+q_{3i})$, it is possible to obtain (for $i=1,\ldots, 3$):

$$ d_{4i}^{2} = (x_{B_{i}D_{i}} - d_{3i}\cos q_{2i} )^{2} + (y_{B_{i}D_{i}} - d_{3i}\sin q_{2i} )^{2}, $$

(100)

where $x_{A_{i}D_{i}} = {}^{i} x_{D_{i}}-{}^{i} x_{A_{i}}$ and $y_{A_{i}D_{i}} = {}^{i} y_{D_{i}}-{}^{i} y_{A_{i}}$.

Then, expanding (100), we have

$$ 0 = a_{i}\cos q_{2i} +b_{i} \sin q_{2i} + c_{i}, $$

(101)

where

$$ \begin{aligned} a_{i}&= -2 d_{3i} x_{A_{i}D_{i}}, \\ b_{i}&=- 2 d_{3i} y_{A_{i}D_{i}}, \\ c_{i}&=x_{A_{i}D_{i}}^{2} + y_{A_{i}D_{i}}^{2} + d_{3i}^{2} - d_{4i}^{2}. \end{aligned} $$

(102)

Finally, by using the tangent half-angle formula, we can obtain

$$ q_{2i} = 2 \tan^{-1} \biggl(\frac{-b_{i} \pm\sqrt {b_{i}^{2}-c_{i}^{2}+a_{i}^{2}}}{c_{i}-a_{i}} \biggr). $$

(103)

In (103), the sign “±” denotes the two robot leg working modes.

Then, it comes easily from (94) and (95) that

$$ q_{3i}=\tan^{-1} \biggl(\frac{{}^{i} y_{D_{i}}-{}^{i} y_{C_{i}}}{{}^{i} x_{D_{i}}-{}^{i} x_{C_{i}}} \biggr) $$

(104)

with ${}^{i} x_{C_{i}}={}^{i} x_{A_{i}} + d_{2i}\cos q_{2i}$, ${}^{i} y_{C_{i}}={}^{i} y_{A_{i}} + d_{2i}\sin q_{2i}$, and

$$ q_{4i} = -q_{2i}-q_{3i}. $$

(105)

Now, differentiating (99) with respect to time and simplifying, we can find the matrices $\mathbf{A}_{p}$ and $\mathbf{B}_{p}$ of (15):

$$ \mathbf{A}_{p}=\mathbf{I}_{3},\qquad \mathbf{B}_{p}= \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\displaystyle \right] , $$

(106)

where $\mathbf{I}_{3}$ is the identity matrix of dimension 3, leading thus to

$$ \mathbf{A}_{p} \begin{bmatrix} \dot{x} \\ \dot{y}\\ \dot{z} \end{bmatrix} + \mathbf{B}_{p} \begin{bmatrix} \dot{q}_{11} \\ \dot{q}_{12}\\ \dot{q}_{13} \end{bmatrix} = \begin{bmatrix} \dot{x} \\ \dot{y}\\ \dot{z} \end{bmatrix} - \begin{bmatrix} \dot{q}_{11} \\ \dot{q}_{12}\\ \dot{q}_{13} \end{bmatrix} = \mathbf{0}. $$

(107)

Now, differentiating (94) and (95) with respect to time, we can find that:

$$\begin{aligned} 0 =& {}^{i} \dot{x}_{D_{i}} + d_{2i}\sin q_{2i} \dot{q}_{2i} + d_{3i}\sin (q_{2i}+q_{3i}) (\dot{q}_{2i}+\dot{q}_{3i}), \\ 0 =& {}^{i} \dot{y}_{D_{i}} - d_{2i}\cos q_{2i} \dot{q}_{2i} - d_{3i}\cos (q_{2i}+q_{3i}) (\dot{q}_{2i}+\dot{q}_{3i}), \end{aligned}$$

(108)

$$\begin{aligned} 0 =& {}^{i} \dot{z}_{D_{i}} - \dot{q}_{1i}, \\ 0 =& \dot{q}_{2i}+\dot{q}_{3i}+\dot{q}_{4i} \end{aligned}$$

(109)

for $i=1, 2,3$, where ${}^{i} \dot{x}_{D_{i}}$, ${}^{i} \dot{y}_{D_{i}}$, and ${}^{i} \dot{z}_{D_{i}}$ are the point $D_{i}$ velocities along the axes of the frame of the leg $i$,

$$\begin{aligned} &{}^{1} \dot{x}_{D_{1}} = \dot{x} , \qquad {}^{1} \dot{y}_{D_{1}} = \dot{y} , \qquad {}^{1} \dot{z}_{D_{1}} = \dot{z}, \end{aligned}$$

(110)

$$\begin{aligned} &{}^{2} \dot{x}_{D_{2}} = \dot{y} , \qquad {}^{2} \dot{y}_{D_{2}} = \dot{z} , \qquad {}^{2} \dot{z}_{D_{2}} = \dot{x}, \end{aligned}$$

(111)

$$\begin{aligned} &{}^{3} \dot{x}_{D_{3}} = \dot{z} , \qquad {}^{3} \dot{y}_{D_{3}} = \dot{x} , \qquad {}^{3} \dot{z}_{D_{3}} = \dot{y}. \end{aligned}$$

(112)

Combining (109), (109), and (112) and noticing that the last line of (109) can be disregarded as the velocities of the passive joints are not included in this equation, we get

$$\begin{aligned} \mathbf{J}_{tk_{i}} \begin{bmatrix} \dot{x} \\ \dot{y} \\ \dot{z} \end{bmatrix} =& \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \dot{q}_{1i} + \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} d_{2i}\sin q_{2i} + d_{3i}\sin(q_{2i}+q_{3i}) & d_{3i}\sin(q_{2i}+q_{3i}) & 0 \\ - d_{2i}\cos q_{2i} - d_{3i}\cos(q_{2i}+q_{3i}) & - d_{3i}\cos(q_{2i}+q_{3i}) & 0 \\ 1 & 1 & 1 \end{array}\displaystyle \right] \\ &{}\times \begin{bmatrix} \dot{q}_{2i} \\ \dot{q}_{3i} \\ \dot{q}_{4i} \end{bmatrix} , \end{aligned}$$

(113)

which can be rewritten as

$$ \mathbf{J}_{tk_{i}}\mathbf{v} - \mathbf{J}_{k_{ai}}\dot{q}_{1i} - \mathbf {J}_{k_{di}}\dot{\mathbf{q}}_{di} = \mathbf{0} $$

(114)

with

$$\begin{aligned} \mathbf{J}_{tk_{1}} =& \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right] ,\quad \quad \mathbf{J}_{tk_{2}}= \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\displaystyle \right] ,\quad \quad \mathbf{J}_{tk_{3}}= \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right] , \end{aligned}$$

(115)

$$\begin{aligned} \mathbf{J}_{k_{ai}} =& \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & 0 & 0 \end{array}\displaystyle \right] ^{T}, \end{aligned}$$

(116)

$$\begin{aligned} \mathbf{J}_{k_{di}} =& \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} d_{2i}\sin q_{2i} + d_{3i}\sin(q_{2i}+q_{3i}) & d_{3i}\sin(q_{2i}+q_{3i}) & 0 \\ - d_{2i}\cos q_{2i} - d_{3i}\cos(q_{2i}+q_{3i}) & - d_{3i}\cos(q_{2i}+q_{3i}) & 0 \\ 1 & 1 & 1 \end{array}\displaystyle \right] , \end{aligned}$$

(117)

and $\mathbf{v}^{T} = [\dot{x} \,\, \dot{y}\,\, \dot{z}]$ and $\dot{\mathbf {q}}_{di}^{T} = [\dot{q}_{2i} \,\, \dot{q}_{3i} \,\, \dot{q}_{4i}]$.

Now, considering the legs 1 to 3, we obtain

$$ \mathbf{J}_{tk}\mathbf{v} - \mathbf{J}_{k_{a}}\dot{ \mathbf{q}}_{a} - \mathbf{J}_{k_{d}}\dot{\mathbf{q}}_{d} = \mathbf{0} $$

(118)

with

$$\begin{aligned} \mathbf{J}_{tk} =& \begin{bmatrix} \mathbf{J}_{tk_{1}} \\ \mathbf{J}_{tk_{2}} \\ \mathbf{J}_{tk_{3}} \end{bmatrix} , \end{aligned}$$

(119)

$$\begin{aligned} \mathbf{J}_{k_{a}} =& \mathbf{0}_{9\times3} \end{aligned}$$

(120)

with $\mathbf{0}_{9\times3}$ a $(9\times3)$ zero matrix, and

$$ \mathbf{J}_{k_{d}}= \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} \mathbf{J}_{k_{d1}} & \mathbf{0}_{3\times3} & \mathbf{0}_{3\times3} \\ \mathbf{0}_{3\times3} & \mathbf{J}_{k_{d2}} & \mathbf{0}_{3\times3} \\ \mathbf{0}_{3\times3} & \mathbf{0}_{3\times 3} & \mathbf{J}_{k_{d3}} \end{array}\displaystyle \right] $$

(121)

with $\mathbf{0}_{3\times3}$ a $(3\times3)$ zero matrix and $\dot {\mathbf{q}}_{d}^{T} = [\dot{\mathbf{q}}_{d1}^{T} \,\, \dot{\mathbf {q}}_{d2}^{T} \,\, \dot{\mathbf{q}}_{d3}^{T}]$.

From (106) and (121) it is possible to observe that

• the matrix $\mathbf{J}_{k_{d}}$ is singular if one block matrix $\mathbf{J}_{k_{di}}$ is singular; $\mathbf{J}_{k_{di}}$ is singular if and only if $q_{3i}=0$ or $\pi$ (i.e., $\mathbf{x}_{2i}$ is collinear to $\mathbf{x}_{3i}$),

• the matrix $\mathbf{A}_{p}$ is constant and never singular; as a result, the robot does not encounter type 2 singularities.

All velocities and accelerations quantities can be then computed from (107) and (118) by using relations (12) and (25) given in Sect. 2.3.

Appendix D: Dynamics of the Tripteron

As mentioned in Appendix C, the Tripteron encounters only LPJTS singularities. Thus, let us now compute the criterion (48).

The inverse dynamic model of the open-loop virtual structure of the Tripteron can be obtained by noticing that each leg is composed

• of a first active prismatic joint,

• followed by a planar 3R robot in which the last body is massless.

The inverse dynamic model of the leg $i$ is:

$$\begin{aligned} \tau_{t_{1i}} =& (m_{1i}+m_{2i}+m_{3i} + ia_{1i})\ddot{q}_{1i}+fs_{1i} \,\mathrm{sign}(\dot{q}_{1i}) + fv_{1i} \dot{q}_{1i} + \tau_{g_{1i}}, \\ \tau_{t_{2i}} =& \bigl(zz_{2i} + d_{3i}^{2} m_{3i} \bigr)\ddot {q}_{2i}+zz_{3i}(\ddot{q}_{2i}+\ddot{q}_{3i}) \\ &{}+d_{3i}mx_{3i} \bigl((2\ddot{q}_{2i}+ \ddot{q}_{3i})\cos q_{3i}-\dot{q}_{3i}(2\dot {q}_{2i}+\dot{q}_{3i})\sin q_{3i} \bigr) \\ &{}+d_{3i}my_{3i} \bigl((2\ddot {q}_{2i}+ \ddot{q}_{3i})\sin q_{3i}+\dot{q}_{3i}(2 \dot{q}_{2i}+\dot {q}_{3i})\cos q_{3i} \bigr) \\ &{}+fs_{2i} \,\mathrm{sign}(\dot{q}_{2i}) + fv_{2i} \dot{q}_{2i} + \tau_{g_{2i}}, \\ \tau_{t_{3i}} =& zz_{3i}(\ddot{q}_{2i}+ \ddot{q}_{3i}) +d_{3i}mx_{3i} \bigl( \ddot{q}_{2i}\cos q_{3i}+\dot{q}_{2i}^{2} \sin q_{3i} \bigr) \\ &{}+d_{3i}my_{3i} \bigl(\ddot{q}_{2i}\sin q_{3i}-\dot {q}_{2i}^{2}\cos q_{3i} \bigr) \\ &{}+ fs_{3i} \,\mathrm{sign}(\dot{q}_{3i}) + fv_{3i} \dot{q}_{3i} + \tau_{g_{3i}}, \\ \tau_{t_{4i}} =& fs_{4i} \,\mathrm{sign}(\dot{q}_{4i}) + fv_{4i} \dot{q}_{4i}, \end{aligned}$$

(122)

where

$$\begin{aligned} &\tau_{g_{11}} = g\,(m_{11}+m_{21}+m_{31}), \quad \quad \tau_{g_{12}} = \tau _{g_{13}} = 0, \end{aligned}$$

(123)

$$\begin{aligned} &\tau_{g_{21}} = 0, \quad \quad \tau_{g_{2i}} = g \,(mx_{2i}+m_{3i}d_{3i})\cos q_{2i} - g\,my_{2i} \sin q_{2i} + \tau_{g_{3i}} \quad \mbox{for }i=2,3, \end{aligned}$$

(124)

$$\begin{aligned} &\tau_{g_{31}} = 0, \quad \quad \tau_{g_{3i}} = g\,mx_{3i}\cos (q_{2i}+q_{3i}) - g\,my_{3i}\sin(q_{2i}+q_{3i}) \quad \mbox{for }i=2,3, \end{aligned}$$

(125)

and

• the parameters $zz_{ji}$, $ia_{ji}$, $m_{ji}$, $mx_{ji}$, $my_{ji}$, $fs_{ji}$, $fv_{ji}$ are defined in Sect. 2.2 ($j=1\ldots4$),

• the parameters $q_{ji}$ and length $d_{3i}$ are defined in Tables 5, 6 and Figs. 6(b) and 17 ($j=1\ldots4$),

• $\tau_{t_{1i}}$ is the torque of the virtual actuator located in the prismatic pair, $\tau_{t_{2i}}$ is the torque of the virtual actuator located at point $B_{i}$, $\tau_{t_{3i}}$ is the torque of the virtual actuator located at point $C_{i}$, and $\tau_{t_{4i}}$ is the torque of the virtual actuator located at point $D_{i}$. The vector $\boldsymbol{\tau}_{t_{a}}$ of (3) stacks all vectors $\boldsymbol{\tau}_{t_{a}}= [\tau_{t_{11}}\ \ \tau_{t_{12}}\ \ \tau _{t_{13}} ]^{T}$, whereas the vector $\boldsymbol{\tau}_{t_{d}}$ of (4) stacks all vectors $\boldsymbol{\tau}_{t_{d}}= [\tau_{t_{d1}}\ \ \tau_{t_{d2}}\ \ \tau_{t_{d3}} ]^{T}$ with $\tau _{t_{di}}= [\tau_{t_{2i}}\ \ \tau_{t_{3i}}\ \ \tau_{t_{4i}} ]^{T}$.

The inverse dynamic model of the free body corresponding to the end-effector (body 5) in the virtual system is

$$ \begin{aligned} \tau_{p_{1}} &= m_{5}\ddot{x}, \\ \tau_{p_{2}} &= m_{5}\ddot{y}, \\ \tau_{p_{3}} &= m_{5}(\ddot{z} +g), \end{aligned} $$

(126)

with $\tau_{p_{j}}$ being the $j$th components of the vector $\boldsymbol {\tau}_{pr}$ of (9); $m_{5}$ is the end-effector mass.

Combining these expressions with those of Appendix C into the equations of Sect. 2.3, we can straightforwardly compute the inverse dynamic model of the Tripteron.

Then, for analyzing the degeneracy conditions of expression (44), let us rewrite the vector $\boldsymbol{\tau}_{t_{d}}$ under the form

$$ \boldsymbol{\tau}_{t_{d}}=\mathbf{M}_{d}(\mathbf{q_{t}}) \ddot{\mathbf {q}}_{t}+\mathbf{c}_{d}(\mathbf{q}_{t}, \dot{\mathbf{q}}_{t}), $$

(127)

where

$$ \mathbf{M}_{d} = \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \mathbf{0}_{3\times3} & \mathbf{M}_{d1} & \mathbf{0}_{3\times3} & \mathbf{0}_{3\times3} \\ \mathbf{0}_{3\times3} & \mathbf{0}_{3\times3} & \mathbf{M}_{d2} & \mathbf{0}_{3\times3} \\ \mathbf{0}_{3\times3} & \mathbf{0}_{3\times3} & \mathbf{0}_{3\times 3} & \mathbf{M}_{d3} \end{array}\displaystyle \right] $$

(128)

and

$$ \mathbf{c}_{d} = \begin{bmatrix} \mathbf{c}_{d1} \\ \mathbf{c}_{d2} \\ \mathbf{c}_{d3} \end{bmatrix} , $$

(129)

in which

$$ \mathbf{M}_{di} = \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} m_{di}^{11} & m_{di}^{12} & 0 \\ m_{di}^{12} & zz_{3i} & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right] $$

(130)

with $m_{di}^{11}=zz_{2i} + d_{3i}^{2} m_{3i} +zz_{3i}+2d_{3i}(mx_{3i}\cos q_{3i} +my_{3i}\sin q_{3i})$, $m_{di}^{12}=zz_{3i}+d_{3i}(mx_{3i}\times\cos q_{3i}+my_{3i}\sin q_{3i})$, and

$$\begin{aligned} \mathbf{c}_{di} =& \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} 0 & c_{di}^{12} & 2c_{di}^{12} \\ d_{3i}mx_{3i}-d_{3i}my_{3i}\cos q_{3i}\sin q_{3i} & 0 & 0 \\ 0 & 0 & 0 \end{array}\displaystyle \right] \left[ \textstyle\begin{array}{c} \dot{q}_{2i}^{2} \\ \dot{q}_{3i}^{2} \\ \dot{q}_{3i}\dot{q}_{2i} \end{array}\displaystyle \right] \\ &{}+ \left[ \textstyle\begin{array}{c@{\quad}c@{\quad}c} fv_{2i} & 0 & 0 \\ 0 & fv_{3i} & 0 \\ 0 & 0 & fv_{4i} \end{array}\displaystyle \right] \begin{bmatrix} \dot{q}_{2i} \\ \dot{q}_{3i} \\ \dot{q}_{4i} \end{bmatrix} + \begin{bmatrix} fs_{2i} \,\mathrm{sign}(\dot{q}_{2i}) \\ fs_{3i} \,\mathrm{sign}(\dot{q}_{3i}) \\ fs_{4i} \,\mathrm{sign}(\dot{q}_{4i}) \end{bmatrix} , \\ =& \mathbf{C}_{di}^{r} \begin{bmatrix} \dot{q}_{2i}^{2} \\ \dot{q}_{3i}^{2} \\ \dot{q}_{3i}\dot{q}_{2i} \end{bmatrix} + \mathbf{F}_{v_{di}} \dot{\mathbf{q}}_{di} + \mathbf{F}_{s_{di}}, \end{aligned}$$

(131)

with $c_{di}^{12}=-d_{3i}(mx_{3i}\sin q_{3i}+my_{3i}\cos q_{3i})$.

Introducing (14), (20), (22), and (25) into (127), simplifying and skipping all mathematical derivations, we get

$$ \boldsymbol{\tau}_{t_{d}}=\mathbf{M}_{d}^{x}( \mathbf{x},\mathbf{q}_{t}) \dot {\mathbf{v}} + \mathbf{c}_{d}^{x}( \mathbf{x},\mathbf{q}_{t},\mathbf{v}), $$

(132)

where

$$ \mathbf{M}_{d}^{x}=\mathbf{M}_{d} \begin{bmatrix} \mathbf{J}_{p}^{-1} \\ \mathbf{J}_{q_{d}} \end{bmatrix} $$

(133)

and

$$ \mathbf{c}_{d}^{x}=\mathbf{M}_{d} \begin{bmatrix} \mathbf{J}_{p}^{d} \\ \mathbf{J}_{q_{d}}^{d} \end{bmatrix} \mathbf{v} + \begin{bmatrix} \mathbf{c}_{d1}^{x} \\ \mathbf{c}_{d2}^{x} \\ \mathbf {c}_{d3}^{x} \end{bmatrix} $$

(134)

with

$$ \mathbf{c}_{di}^{x}=\mathbf{C}_{di}^{r} \begin{bmatrix} (\mathbf{j}_{q_{di}}^{1}\mathbf{v})^{2} \\ (\mathbf{j}_{q_{di}}^{2}\mathbf{v})^{2} \\ (\mathbf{j}_{q_{di}}^{1}\mathbf{v})(\mathbf{j}_{q_{di}}^{2}\mathbf{v}) \end{bmatrix} + \mathbf{F}_{v_{di}} \mathbf{J}_{q_{di}} \mathbf{v} + \mathbf{F}_{s_{di}}, $$

(135)

in which:

• $\mathbf{J}_{q_{d}}$, $\mathbf{J}_{p}^{d}$, and $\mathbf{J}_{q_{d}}^{d}$ are three matrices defined at (20), (22), and (25).

• $\mathbf{j}_{q_{di}}^{j}$ is the line of the matrix $\mathbf {J}_{q_{d}}$ corresponding to the variable $q_{d_{ij}}$.

Thus, for one given robot configuration, $\boldsymbol{\tau}_{t_{d}}$ is a function of $\dot{\mathbf{v}}$ and $\mathbf{v}$ only.