Reduced Dynamics of the Non-holonomic Whipple Bicycle

Abstract

Though the bicycle is a familiar object of everyday life, modeling its full nonlinear three-dimensional dynamics in a closed symbolic form is a difficult issue for classical mechanics. In this article, we address this issue without resorting to the usual simplifications on the bicycle kinematics nor its dynamics. To derive this model, we use a general reduction-based approach in the principal fiber bundle of configurations of the three-dimensional bicycle. This includes a geometrically exact model of the contacts between the wheels and the ground, the explicit calculation of the kernel of constraints, along with the dynamics of the system free of any external forces, and its projection onto the kernel of admissible velocities. The approach takes benefits of the intrinsic formulation of geometric mechanics. Along the path toward the final equations, we show that the exact model of the bicycle dynamics requires to cope with a set of non-symmetric constraints with respect to the structural group of its configuration fiber bundle. The final reduced dynamics are simulated on several examples representative of the bicycle. As expected the constraints imposed by the ground contacts, as well as the energy conservation, are satisfied, while the dynamics can be numerically integrated in real time.

Appendices

Appendix 1: Calculation of the Kernel of Constraints

In this appendix, we calculate $H = \ker (A,B)$ by inverting symbolically the system that we rewrite in the form $D=(A,B)(\eta ^\mathrm{T},\dot{r}^\mathrm{T})^\mathrm{T}=0_6$ where D is the $6\times 1$ vector whose components $D_i$, $i=1,2\ldots 6$, form the left-hand side of (38–43). Firstly, $V_1$, $V_2$ and $V_3$ can be extracted from $D_4$, $D_5$ and $D_6$, respectively, as follows:

$$\begin{aligned}&V_1 = F_{16} {{\varOmega }_{3}} + F_{19} \dot{r_3} , \end{aligned}$$

(77)

$$\begin{aligned}&V_2 = F_{26} {{\varOmega }_{3}} + F_{29} \dot{r_3} , \end{aligned}$$

(78)

$$\begin{aligned}&V_3 = F_{34} {{\varOmega }_{1}} + F_{35} {{\varOmega }_{2}} , \end{aligned}$$

(79)

where we introduced the notations:

$$\begin{aligned}&F_{16} = -A_{46} ,\quad F_{19} = -B_{43} , \end{aligned}$$

(80)

$$\begin{aligned}&F_{26} = -A_{56} ,\quad F_{29} = -B_{53} , \end{aligned}$$

(81)

$$\begin{aligned}&F_{34} = -A_{64} ,\quad F_{35} = -A_{65} . \end{aligned}$$

(82)

After inserting (79) in $D_3$, one can express $\varOmega _1$ as:

$$\begin{aligned} \varOmega _1 = F_{45} \varOmega _2 + F_{47} \dot{r_1} + F_{48} \dot{r_2} , \end{aligned}$$

(83)

with:

$$\begin{aligned} F_{45} = -\frac{A_{35} + F_{35}}{F_{34} + A_{34}} , F_{47} = -\frac{B_{31}}{F_{34} + A_{34}} , F_{48} = -\frac{B_{32}}{F_{34} + A_{34}} . \end{aligned}$$

(84)

In a similar way, inserting (77) in $D_1$ gives:

$$\begin{aligned} \dot{r_2} = F_{85} \varOmega _2 + F_{86} \varOmega _3 + F_{87} \dot{r_1} + F_{89} \dot{r_3} , \end{aligned}$$

(85)

with:

$$\begin{aligned} F_{85} = -\frac{A_{15}}{B_{12}} ,\quad F_{86} = -\frac{A_{16} + F_{16}}{B_{12}} ,\quad F_{87} = -\frac{B_{11}}{B_{12}} , \quad F_{89} = -\frac{F_{19}}{B_{12}} . \end{aligned}$$

(86)

Inserting (85), (77) and (83) in $D_2$ allows rewriting $\varOmega _3$ as:

$$\begin{aligned} \varOmega _3 = H_{65} \varOmega _2 + H_{67} \dot{r_1} + H_{69} \dot{r_3} , \end{aligned}$$

(87)

with:

$$\begin{aligned}&H_{65} = -\frac{A_{24} F_{45} + ( A_{24} F_{48} + B_{22} ) F_{85}}{( B_{22} + A_{24} F_{48} ) F_{86} + F_{26} + A_{26}} ,\nonumber \\&H_{67} = -\frac{B_{21} + A_{24} F_{47} + ( A_{24} F_{48} + B_{22} ) F_{87}}{( B_{22} + A_{24} F_{48} ) F_{86} + F_{26} + A_{26}} , \end{aligned}$$

(88)

$$\begin{aligned}&H_{69} = -\frac{F_{29} + ( A_{24} F_{48} + B_{22} ) F_{89}}{( B_{22} + A_{24} F_{48} ) F_{86} + F_{26} + A_{26}} . \end{aligned}$$

(89)

Now, inserting (87) in (77), (78) and (85) allows us to rewrite them in the following form:

$$\begin{aligned}&V_1 = H_{15} \varOmega _2 + H_{17} \dot{r_1} + H_{19} \dot{r_3} , \end{aligned}$$

(90)

$$\begin{aligned}&V_2 = H_{25} \varOmega _2 + H_{27} \dot{r_1} + H_{29} \dot{r_3} , \end{aligned}$$

(91)

$$\begin{aligned}&\dot{r_2} = H_{85} \varOmega _2 + H_{87} \dot{r_1} + H_{89} \dot{r_3} , \end{aligned}$$

(92)

with:

$$\begin{aligned}&H_{15} = F_{16} H_{65} ,\quad H_{15} = F_{16} H_{65} ,\quad H_{19} = F_{16} H_{69} + F_{19} , \end{aligned}$$

(93)

$$\begin{aligned}&H_{25} = F_{26} H_{65} ,\quad H_{27} = F_{26} H_{67} ,\quad H_{29} = F_{26} H_{69} + F_{29} , \end{aligned}$$

(94)

$$\begin{aligned}&H_{85} = F_{86} H_{65} + F_{85} ,\quad H_{87} = F_{86} H_{67} + F_{87} ,\quad H_{89} = F_{86} H_{69} + F_{89} . \qquad \end{aligned}$$

(95)

Then, using (92) in (83) gives:

$$\begin{aligned} \varOmega _1 = H_{45} \varOmega _2 + H_{47} \dot{r_1} + H_{49} \dot{r_3} , \end{aligned}$$

(96)

with:

$$\begin{aligned} H_{45} = F_{48} H_{85} + F_{45} ,\quad H_{47} = F_{48} H_{87} + F_{47} , \quad H_{49} = F_{48} H_{89} . \end{aligned}$$

(97)

Finally, inserting (96) into (79), we find:

$$\begin{aligned} V_3 = H_{35} \varOmega _2 + H_{37} \dot{r_1} + H_{39} \dot{r_3} , \end{aligned}$$

(98)

with:

$$\begin{aligned} H_{35} = F_{34} H_{45} + F_{35} ,\quad H_{37} = F_{34} H_{47} , \quad H_{39} = F_{34} H_{49} . \end{aligned}$$

(99)

Finally, any vector in the kernel of the constraints can be written in form (62) which require the ordered calculations of (80)–(82), (84), (86), (88), (89), (93), (94), (95), (97), (99).

Appendix 2: Free Bicycle Dynamics

We first start by calculating the acceleration of the four bodies that compose the bicycle in the configuration space $SE(3)\times S$. This is done by removing all the accelerations $\dot{\eta }_j$, for $j\ne 0$ in the recursion on accelerations (64). After straightforward algebra we find:

$$\begin{aligned}&\dot{\eta _0} = \dot{\eta _0} , \end{aligned}$$

(100)

$$\begin{aligned}&\dot{\eta _1} = {{\mathrm{Ad}}}_{{}^1g_0} \dot{\eta _0} + \ddot{r_1} {{\mathrm{A}}}_1+ \zeta _1 , \end{aligned}$$

(101)

$$\begin{aligned}&\dot{\eta _2} = {{\mathrm{Ad}}}_{{}^2g_0} \dot{\eta _0} + {{\mathrm{Ad}}}_{{}^2g_1} {{\mathrm{A}}}_1\ddot{r_1} + {{\mathrm{A}}}_2 \ddot{r_2}+ {{\mathrm{Ad}}}_{{}^2g_1} \zeta _1 + \zeta _2 , \end{aligned}$$

(102)

$$\begin{aligned}&\dot{\eta _3} = {{\mathrm{Ad}}}_{{}^3g_0} \dot{\eta _0} + {{\mathrm{A}}}_3\ddot{r_3} + \zeta _3 , \end{aligned}$$

(103)

which need to use the detail expressions of the adjoint map and its time derivative:

$$\begin{aligned} {{\mathrm{Ad}}}_{{}^jg_{i}} = \begin{pmatrix} {}^{j}R_{i} &{} {}^{j}R_{i} \,\,\hat{p}_i(O_j)^\mathrm{T}\\ 0 &{} {}^{j}R_{i}\\ \end{pmatrix} , \quad \zeta _j = \begin{pmatrix} ({}^jV_{i} + {}p_{j}(O_i) \times {}^j\varOmega _{i}) \times \dot{r}_{j}a_j\\ \varOmega _{i} \times \dot{r}_j a_j\\ \end{pmatrix} ,\nonumber \\ \end{aligned}$$

(104)

and where ${{\mathrm{A}}}_j = (0_3^\mathrm{T},a_j^\mathrm{T})^\mathrm{T}$ is the $(6 \times 1)$ unit vector supporting the joint axis j. Now writing the top row of (63) with k the indexes of all the bodies just after $\mathcal {B}_{j}$ when descending the structure from $\mathcal {B}_0$ to its tips, we have for each of the four bodies of the bicycle:

$$\begin{aligned}&f_3 = \mathcal {M}_3 \dot{\eta _3} + f_{\mathrm{in}, 3} + f_{\mathrm{ext}, 3} , \end{aligned}$$

(105)

$$\begin{aligned}&f_2 = \mathcal {M}_2 \dot{\eta _2} + f_{\mathrm{in}, 2} + f_{\mathrm{ext}, 2} , \end{aligned}$$

(106)

$$\begin{aligned}&f_1 = \mathcal {M}_1 \dot{\eta _1} + f_{\mathrm{in}, 1} + f_{\mathrm{ext}, 1} + {{\mathrm{Ad}}}_{{}^2g_1}^\mathrm{T} f_2 , \end{aligned}$$

(107)

$$\begin{aligned}&f_0 = \mathcal {M}_0 \dot{\eta _0} + f_{\mathrm{in}, 0} + f_{\mathrm{ext}, 0} + {{\mathrm{Ad}}}_{{}^3g_0}^\mathrm{T} f_3 + {{\mathrm{Ad}}}_{{}^1g_0}^\mathrm{T} f_1 , \end{aligned}$$

(108)

which need the detailed expressions:

$$\begin{aligned} \mathcal {M}_{j}= & {} \begin{pmatrix} m_j1_{3\times 3} &{}\quad \hat{ms}_{j}^\mathrm{T}\\ \hat{ms}_{j} &{}\quad I_{j}\\ \end{pmatrix} , \quad f_{\mathrm{in},j} = \begin{pmatrix} (ms_j\times \varOmega _j)\times \varOmega _j + \varOmega _j \times (m_jV_j) \\ \varOmega _{j} \times (I_{j} \varOmega _j) + ms_j\times (\varOmega _j\times V_j) \\ \end{pmatrix},\nonumber \\ \end{aligned}$$

(109)

where $m_j 1_{3 \times 3}$ and $I_{j}$ are the matrices of linear and angular inertia, while $ms_j = m_j p_{j}(G_j)$ is the vector of first inertia moments that couple linear and angular accelerations, all being related to ${\mathcal {B}_j}$. Then, inserting (107), (106), (105), (101), (102) and (103) into (108) that we identify with the first row of (65), we obtain:

$$\begin{aligned} \mathcal {M}= & {} \sum _{i=0}^{3} {{\mathrm{Ad}}}_{{}^ig_0}^\mathrm{T} \mathcal {M}_i {{\mathrm{Ad}}}_{{}^ig_0} , \end{aligned}$$

(110)

$$\begin{aligned} M^\mathrm{T}= & {} \left( \sum _{i=1}^{2} {{\mathrm{Ad}}}_{{}^ig_0}^\mathrm{T} \mathcal {M}_i {{\mathrm{Ad}}}_{{}^ig_1} {{\mathrm{A}}}_1 , {{\mathrm{Ad}}}_{{}^2g_0}^\mathrm{T} \mathcal {M}_2 {{\mathrm{A}}}_2 , {{\mathrm{Ad}}}_{{}^3g_0}^\mathrm{T} \mathcal {M}_3 {{\mathrm{A}}}_3\right) , \end{aligned}$$

(111)

$$\begin{aligned} f_\mathrm{in}= & {} \sum _{i=0}^{3} {{\mathrm{Ad}}}_{{}^ig_0}^\mathrm{T} f_{\mathrm{in}, i} + \sum _{j=1}^{2}(\sum _{i=j}^{2} {{\mathrm{Ad}}}_{{}^ig_0}^\mathrm{T} \mathcal {M}_i {{\mathrm{Ad}}}_{{}^ig_j}) \zeta _j + {{\mathrm{Ad}}}_{{}^3g_0}^\mathrm{T} \mathcal {M}_3 \zeta _3 , \end{aligned}$$

(112)

$$\begin{aligned} f_\mathrm{ext}= & {} \sum _{i=0}^{3} {{\mathrm{Ad}}}_{{}^ig_0}^\mathrm{T} f_{\mathrm{ext}, i} . \end{aligned}$$

(113)

Then introducing (105)–(108) into $\tau _j = {{\mathrm{A}}}_j^\mathrm{T} f_j$ gives :

$$\begin{aligned}&\tau _1 = {{\mathrm{A}}}_1^\mathrm{T} (\mathcal {M}_1 \dot{\eta _1} + f_{\mathrm{in}, 1} + f_{\mathrm{ext}, 1} + {{\mathrm{Ad}}}_{{}^2g_1}^\mathrm{T} (\mathcal {M}_2 \dot{\eta _2} + f_{\mathrm{in}, 2} + f_{\mathrm{ext}, 2})) , \end{aligned}$$

(114)

$$\begin{aligned}&\tau _2 = {{\mathrm{A}}}_2^\mathrm{T} (\mathcal {M}_2 \dot{\eta _2} + f_{\mathrm{in}, 2} + f_{\mathrm{ext}, 2}) , \end{aligned}$$

(115)

$$\begin{aligned}&\tau _3 = {{\mathrm{A}}}_3^\mathrm{T} (\mathcal {M}_3 \dot{\eta _3} + f_{\mathrm{in}, 3} + f_{\mathrm{ext}, 3}) , \end{aligned}$$

(116)

that once identified with the second row of (65) gives the expressions below:

$$\begin{aligned} m= & {} \begin{pmatrix} \overset{2}{\underset{i=1}{\sum }} {{\mathrm{A}}}_1^T {{\mathrm{Ad}}}_{{}^ig_1}^T \mathcal {M}_i {{\mathrm{Ad}}}_{{}^ig_1} {{\mathrm{A}}}_1 &{} {{\mathrm{A}}}_1^T {{\mathrm{Ad}}}_{{}^2g_1}^T \mathcal {M}_2 {{\mathrm{A}}}_2 &{} 0 \\ {{\mathrm{A}}}_2^T \mathcal {M}_2 {{\mathrm{Ad}}}_{{}^2g_1} {{\mathrm{A}}}_1 &{} {{\mathrm{A}}}_2^T \mathcal {M}_2 {{\mathrm{A}}}_2 &{} 0 \\ 0 &{} 0 &{} {{\mathrm{A}}}_3^T \mathcal {M}_3 {{\mathrm{A}}}_3 \end{pmatrix} ,\nonumber \\ Q_{in}= & {} \begin{pmatrix} \overset{2}{\underset{i=1}{\sum }} {{\mathrm{A}}}_1^T {{\mathrm{Ad}}}_{{}^ig_1}^T f_{in, i} + \overset{2}{\underset{j=1}{\sum }} (\overset{2}{\underset{i=j}{\sum }} {{\mathrm{A}}}_1^T {{\mathrm{Ad}}}_{{}^ig_1}^T \mathcal {M}_i {{\mathrm{Ad}}}_{{}^ig_j}) \zeta _j \\ {{\mathrm{A}}}_2^T f_{in, 2} + \overset{2}{\underset{i=1}{\sum }} {{\mathrm{A}}}_2^T \mathcal {M}_2 {{\mathrm{Ad}}}_{{}^2g_i} \zeta _i \\ {{\mathrm{A}}}_3^T f_{in, 3} + {{\mathrm{A}}}_3^T \mathcal {M}_3 \zeta _3 \end{pmatrix} , \quad Q_{ext} \nonumber \\= & {} \begin{pmatrix} \overset{2}{\underset{i=1}{\sum }} {{\mathrm{A}}}_1^T {{\mathrm{Ad}}}_{{}^ig_1}^T f_{ext, i} \\ {{\mathrm{A}}}_2^T f_{ext, 2} \\ {{\mathrm{A}}}_3^{T} f_{ext, 3} \end{pmatrix} . \end{aligned}$$

(117)

Finally, once supplemented with the recursive geometric model $^{e}g_{j}=\text {}^{e}g_i\text {}^{i}g_{j}(r_j)$ and kinematic one (63-bottom), the above expressions along with (110–113) define all the matrices of the free dynamics, or equivalently, of the left-hand side of (3) and (4).