2. System dynamics

In this work, a slider crank-based reconfiguration mechanism is used in the two front wheels of the robot. Location of these wheels can be seen in the actual picture of the robot as shown in Figure 1 along with the corresponding CAD model in Figure 2. The rear two wheels of the robots are passive caster wheels. The reconfigurable wheel used in this study has two modes of operation. One is wheel mode and another is leg mode. In wheel mode, the front two wheels and rear two caster wheels work together as a four-wheeled mobile robot. However, in legged mode, the front two wheels will turn into a five-spoked wheel and behave as if five legs are rotating around the center of the wheel. The actuation of these legs (wheg mode) is the main role of the reconfiguration mechanism and is shown in Figure 3. There are many variations of reconfiguration mechanisms, through which a normal wheel can be transformed into a spoked one, as discussed in the introduction section. In general, spoke wheel design variations can be broadly classified into out-of-the-wheel-plane reconfiguration and in-the-wheel-plane reconfiguration. In the in-plane-wheel reconfiguration design, spokes of the wheel before reconfiguration and after reconfiguration remain in the same plane, as shown in Figure 4. The advantage of this design is that it is compact, less complex, and requires less number of intermediate links for the transformation, when compared with the out-of-wheel-plane reconfiguration design. However, this design may not be able to sustain the lateral load coming onto the wheel during locomotion in an undulated terrain and may fail due to the absence of any supporting link in this direction. On the contrary, in out-of-wheel-plane reconfiguration design, the transformation of the wheel from a normal wheel to a spoked wheel takes place in a three-dimensional space and requires a larger number of links. Particularly in the present study, five sets of slider crank mechanisms are used in one wheel, which jointly make the one full reconfigurable wheel when assembled, as shown in Figure 3. In each of the wheels, the crank is connected to the corresponding coupler link, the coupler link is connected to the slider, and the slider is connected to the two racks (rack gears). These racks are engaged with the pinion gear, and the pinion gear is directly mounted onto the output shaft of the reconfiguration motor. When the motor rotates the pinion gear, it converts the rotary motion into linear motion of the rack and consequently the slider. For more understanding, an animation of the working mechanism can be seen with this link. Once the sliders on either side of the wheel start moving outward, the wheel expands into the spoked configuration in a three-dimensional space like an umbrella and confirms the legged configuration of the system. Since in this configuration there are multiple links like couplers of each mechanism, which can easily handle the lateral load coming on the wheel due to the terrain undulation, and hence can provide sturdiness to the wheel. Also, to make sure that the wheel can sustain those external loads, particularly lateral loads, the full dynamics of the wheel and its supporting mechanism will be discussed in the reconfiguration dynamics section. This section explores the dynamics of the robot as a four-wheeled vehicle traversing longitudinal and lateral slopes. This study will elucidate the relationship between the contact reaction force of the wheel and the ground under the vehicle’s rolling and pitching circumstances. The normal reaction force will primarily be used to calculate the motor torque necessary for the wheel reconfiguration process, which will be addressed in the reconfiguration dynamics section. Furthermore, in the robot dynamics section, the vehicle’s motion will be simulated along a 3D trajectory, allowing for a comparison of the torque requirements of the actuator under longitudinal and lateral gradient circumstances, so facilitating the finalization of motor selection.

Figure 3.

Out of the plane reconfiguration of the spoked wheel. (a) The front two reconfigurable legged-wheels are in legged mode, (b) wheel in compact form. (c) One set of a two-dimensional slider crank mechanism. (d) Close-up view of the rack and pinion arrangement for both front right and front left sides of the wheel with one reconfiguration motor.

Figure 4.

Planar reconfiguration of the spoked wheel [Reference Zheng and Lee24].

2.1. Longitudinal dynamics

2.1.1. Robot is in wheel mode

In this case, the front wheels of the robot are not reconfigured, and the robot is considered in the wheel mode. Here, it is assumed that the vehicle or robot is moving on an inclined plane, with a wedge angle of ‘ $\theta$ ’. In addition, this angle can be considered as the vehicle’s pitch motion about the Y-axis of the inertial frame. Here, it is assumed that there is equal normal reaction (i.e. ${\mathbf{N_{2}}}$ ) on a pair of two caster wheels located on the rear side; similarly, the normal reaction (i.e ${\mathbf{N_{1}}}$ ) is equal on a pair of front reconfigurable wheels located on the front side of the robot. So, considering the dynamic equilibrium of the system in the x-z plane as shown in Figure 5.

(1) \begin{equation} \sum F_{x}=m\dot {u} \,:\, 2F_{x1}-2F_{x2}-mgsin\theta = m\dot {u} \end{equation}

where $F_{x1}$ is the traction force in the positive $X$ direction on the front powered wheels. $F_{x2}$ is the friction force on the rear caster wheels and will be in the negative $X$ direction. $m$ is the total mass of the system, and $\dot {u}$ is the linear acceleration component of the robot in the X direction with respect to the body frame of the robot. Force, $F_{x1}=F_{1x}=F_{2x}=\mu _{x}{\mathbf{N_{1}}}$ , where $\mu _{x}$ is the coefficient of friction between the wheel and surface contact point along the $X$ axis of the vehicle frame. Since in our system, only the front two wheels are powered by motors, the rear caster wheels receive the traction or driving force indirectly from the front two wheels, so for the rear caster wheels, we have $F_{x2} = F_{3x}=F_{4x}\leq \mu _{x}{\mathbf{N_{2}}}$ , which ensures no slip condition or perfect rolling condition. Therefore, we can write

(2) \begin{equation} \sum F_{z}=0 \,:\, 2{\mathbf{N_{1}}}+2{\mathbf{N_{2}}}-mgcos\theta =0 \end{equation}

(3) \begin{equation} \sum M_y^{COM}=0 \,:\, 2{\mathbf{N_{1}}}l_{1}-2{\mathbf{N_{2}}}l_{2}+2F_{x1}h-2F_{x2}h=0 \end{equation}

Figure 5.

Longitudinal dynamics in wheel mode.

$M_{y}^{COM}$ denotes the moment at the center of mass (COM) about $Y$ -axis of the vehicle frame. $l_{1}$ and $l_{2}$ denote the distances from the COM to the center of the front wheel and the center of the rear caster, respectively, along the $X$ -axis of the vehicle frame. Likewise, $h$ denotes the height of the COM above the ground. Eqs. (1), (2) and (3) are reduced into two independent equations in two variables ${\mathbf{N_{1}}}$ and ${\mathbf{N_{2}}}$ , which can be represented in matrix form as:

(4) \begin{equation} \begin{bmatrix} 2 & 2\\[2pt]2l_{1} & -2l_{2} \end{bmatrix}\begin{bmatrix} {\mathbf{N_{1}}}\\[2pt]{\mathbf{N_{2}}} \end{bmatrix}=\begin{bmatrix} mg\cos {\theta }\\[2pt]-mgh\sin {\theta }-m\dot {u}h \end{bmatrix} \end{equation}

which on solving gives

(5) \begin{equation} {\mathbf {N}_{1}}={\mathbf {N}}_{FRW}^{lng}={\mathbf {N}}_{FLW}^{lng}=\frac {mgl_{2}\cos {\theta }}{2l}-\frac {m\dot {u}h+mgh\sin {\theta }}{2l} \end{equation}

and

(6) \begin{equation} {\mathbf {N}}_{2}={\mathbf {N}}_{RRC}^{lng}={\mathbf {N}}_{RLC}^{lng}=\frac {m\dot {u}h+mgh\sin {\theta }}{2l}+\frac {mgl_{1}\cos {\theta }}{2l} \end{equation}

${\mathbf {N}}_{FRW}^{lng}$ denotes the normal reaction on the front right wheel in longitudinal dynamics, ${\mathbf {N}}_{FLW}^{lng}$ signifies the normal reaction on the front left wheel in longitudinal dynamics, ${\mathbf {N}}_{RRC}^{lng}$ represents the normal reaction on the rear right caster in longitudinal dynamics, and ${\mathbf {N}}_{RLC}^{lng}$ indicates the normal reaction on the rear left caster in longitudinal dynamics. Furthermore, $ l_{1} + l_{2} = l$ , representing the entire length from the center of the front wheels to the center of the rear casters along the longitudinal direction of the vehicle. Moreover, here it is important to note that the normal reactions on the front two pair wheels are assumed to be equal, and so is with the rear two castors. Since in the present analysis, gradient of the terrain is considered only about the Y-axis of the inertial frame; however, if the gradient is present in both about X- and Y-axes of the inertial frame, in case of a highly unstructured terrain, then the normal reactions obtained in longitudinal dynamics and lateral dynamics (discussed in lateral dynamics subsection) must be compared and whichever appears higher in magnitude, should be considered as final.

Figure 6.

Longitudinal dynamics in legged mode.

2.1.2. Robot is in leg mode

The front two wheels of the robot are regarded in the legged mode upon reconfiguration, as seen in Figure 6. Due to the reconfiguration of the wheel, altering the position of the COM and therefore the normal reaction forces on the wheels, the normal reaction on the front and rear wheels will be recalculated, based on the new COM location for the leg mode. The COM shift arises from the unfolding of the front two wheels, resulting in the robot body rotating around the contact point of the rear caster wheels and the ground by an angle of $\beta =10.25^\circ$ , in an anti-clockwise direction around the vehicle frame’s $Y$ -axis. Consequently, to get the new coordinates of the COM, we may multiply the previous coordinates of the COM by the rotation matrix corresponding to the angle ( $-\beta$ ) around the $Y$ -axis of the vehicle frame. So, we now possess two frames, old and new, corresponding to the robot’s previous and current COM. The updated coordinates of the new COM, relative to the world frame, may be computed as follows:

(7) \begin{equation} \begin{bmatrix} X_{new}\\[2pt]Y_{new}\\[2pt]Z_{new} \end{bmatrix}=\begin{bmatrix} \cos {(-\beta )} &\quad 0 &\quad \sin {(-\beta )}\\[2pt] 0 &\quad 1 &\quad 0\\[2pt] -\sin {(-\beta )} &\quad 0 &\quad \cos {(-\beta )} \end{bmatrix}\begin{bmatrix} l_{2}\\[2pt]0\\[2pt]h \end{bmatrix} \end{equation}

through which we can obtain the expression for $l_{2}^{\prime}$ , $l_{1}^{\prime}$ , and $h^{\prime}$ as follows:

(8) \begin{equation} l^{\prime}_{2}=l_{2}cos\beta -hsin\beta \end{equation}

(9) \begin{equation} h^{\prime}=l_{2}sin\beta +hcos\beta \end{equation}

where $l_{1}^{\prime}=l-l^{\prime}_{2}$ , since $l_{1}+l_{2}=l=l_{1}^{\prime}+l^{\prime}_{2}$ . Now substituting the expression of $l_{1}^{\prime}$ , $l^{\prime}_{2}$ , and $h^{\prime}$ in place of $l_{1}$ , $l_{2}$ , and $h$ , respectively, from Eqs. (8) and (9) into Eqs. (5) and (6). Which gives us new expressions for the normal reactions in leg mode, as given below

(10) \begin{equation} {\mathbf {N}}_{1}={\mathbf {N}}_{FRW}^{lng}={\mathbf {N}}_{FLW}^{lng}=\frac {mgl_{2}^{\prime}\cos {\theta }}{2l}-\frac {m\dot {u}h^{\prime}+mgh^{\prime}\sin {\theta }}{2l} \end{equation}

(11) \begin{equation} {\mathbf {N}}_{2}={\mathbf {N}}_{RRC}^{lng}={\mathbf {N}}_{RLC}^{lng}=\frac {m\dot {u}h^{\prime}+mgh^{\prime}\sin {\theta }}{2l}+\frac {mgl_{1}^{\prime}\cos {\theta }}{2l} \end{equation}

Figure 7.

Robot is in wheel mode and resting on the banked surface.

2.2. Lateral dynamics

This section aims to elucidate the robot’s rolling dynamics. In this case, since the robot is inclined on a banked surface laterally which the normal reaction on the right side wheel and left side wheel will be different. Additionally, robot’s front wheels and rear wheels, which are casters in the present case, are not inline like an ordinary passenger vehicle, therefore reaction on the left side (both front and rear) wheels cannot be considered equal to the right side (both front and rear) wheels, and hence all four wheels will have different normal reactions. Moreover, in the lateral dynamics case, the robot can be considered in the Y-Z plane, and only three independent equations can be obtained through equilibrium equations corresponding to four unknowns, and hence closed-loop analytical solution for all four normal reactions is not possible. Therefore, it is decided to consider the present case of lateral dynamics by assuming the four wheel robot model as a three wheel tricycle model, since rear two passive caster wheels of the robot are very close to the centerline (6 cm away) or the longitudinal axis, compared to front two wheels, which are 21.6 cm away from the longitudinal axis. So in the lateral dynamics study, the rear two casters are considered as one caster located along the centerline or the longitudinal axis of the system. Additionally, we need three independent equations to find three normal reactions, for which the robot’s top view is also considered to get the third independent equation required for the solution. Now, the height of the COM from the ground in wheel mode is denoted as $h$ , whereas in legged mode it is represented as $h'$ , as seen in Figs. 7 and 8, respectively. Moreover, as previously shown in the longitudinal dynamics, the reconfiguration of the robot’s wheel will result in changes only to $l_{1}$ , $l_{2}$ , and $h$ , while the remaining expressions will stay unchanged. Therefore, in this section, we will derive the mathematical expressions for the normal reaction in legged mode directly, and by substituting the angle $\beta$ with $\beta =0$ , the result will correspond to the wheel mode. So, now considering the equilibrium conditions:

(12) \begin{equation} \sum {F_{y}}= 0 \,:\, \mu _{y}({\mathbf{N_{1}}}+{\mathbf{N_{2}}}+\mathbf{N_{3}})-mg\sin {\phi }= 0 \end{equation}

(13) \begin{equation} \sum F_{z}= 0 \,:\, {\mathbf{N_{1}}}+{\mathbf{N_{2}}}+{\mathbf{N_{3}}}-mg\cos {\phi }= 0 \end{equation}

(14) \begin{equation} \sum M_{x}^{COM}= 0 \,:\, -{\mathbf{N_{1}}}d_{1}+{\mathbf{N_{2}}}d_{1}+mgh^{\prime}\cos {\phi }= 0 \end{equation}

Figure 8.

Robot is in the leg mode and resting on the banked surface.

Eqs. (12), (13), and (14) are two independent equations in three variables ${{\mathbf {N}}_{1}}$ (normal reaction on the front right wheel), ${\mathbf {N}}_{2}$ (normal reaction on the front left wheel) and ${\mathbf {N}}_{3}$ (normal reaction on the two rear casters). So one more independent equation is required to solve for three variables, and this equation can be found from the sub-figure (b) of the Figure 7, by taking the moment about the Z axis of the vehicle frame at the COM, as given below:

(15) \begin{equation} \sum M_{z}^{COM}=0 \,:\, \mu _{y}({\mathbf{N_{1}}}+{\mathbf{N_{2}}})l_{1}-\mu _{y}\mathbf{N_{3}}l_{2}=0 \end{equation}

Now, arranging the three independent equations in the matrix form as:

(16) \begin{equation} \begin{bmatrix} 1 &\quad 1 &\quad 1\\[2pt]-d_{1} &\quad d_{1} &\quad 0\\[2pt] \mu _{y}l_{1} &\quad \mu _{y}l_{1} &\quad -\mu _{y}l_{2} \end{bmatrix}\begin{bmatrix} {\mathbf{N_{1}}}\\[2pt]{\mathbf{N_{2}}}\\[2pt] {\mathbf{N_{3}}} \end{bmatrix}=\begin{bmatrix} mg\cos {\phi }\\[2pt]-mgh'\sin {\phi }\\[2pt] 0 \end{bmatrix} \end{equation}

On solving Eq. (16), normal reactions on all three wheels can be represented as given below:

(17) \begin{equation} {{\mathbf {N}}_{1}}={\mathbf {N}}_{FRW}^{lat}= \frac {mgd_{1}l_{2}\cos {\phi }+h^{\prime}l_{1}\sin {\phi }+h^{\prime}l_{2}\sin {\phi }}{2d_{1}(l_{1}+l_{2})} \end{equation}

(18) \begin{equation} {\mathbf {N}}_{2}={\mathbf {N}}_{FLW}^{lat} = \frac {-mgh^{\prime}l_{1}\sin {\phi }+d_{1}l_{2}\cos {\phi }+h^{\prime}l_{2}\sin {\phi }}{2d_{1}(l_{1}+l_{2})} \end{equation}

(19) \begin{equation} {\mathbf {N}}_{RRC}^{lat} = {\mathbf {N}}_{RLC}^{lat}=\frac {{\mathbf {N}}_{3}}{2}=\frac {mgl_{1}\cos {\phi }}{2(l_{1}+l_{2})} \end{equation}

where ${\mathbf {N}}_{FRW}^{lat}$ , ${\mathbf {N}}_{FLW}^{lat}$ , ${\mathbf {N}}_{RRC}^{lat}$ , and ${\mathbf {N}}_{RLC}^{lat}$ are the normal reaction on the front right wheel, front left wheel, rear right caster, and rear left caster in the lateral dynamics, respectively. Additionally, $\mu _{y}$ is the coefficient of static friction between the wheel and ground along the Y-axis of the vehicle frame. Normal reaction calculated in lateral and longitudinal dynamics, as derived above, will be useful to assess the stability of the robot in motion and to calculate the amount of input force required in the reconfiguration mechanism, used in both front wheels of the robot, which is discussed in the next section.

Figure 9.

Reconfiguration mechanism in assembled and free state.

Figure 10.

Kinematic diagram of slider crank mechanism.

3. Reconfiguration dynamics

This section elucidates the dynamics of the mechanism used for wheel reconfiguration. This study employs the slider crank mechanism for reconfiguration, as seen in Figure (9d). Five such identical mechanisms are used in each wheel to provide smooth opening and closing of the same. Each mechanism is activated by the external force applied to the slider, facilitated by a motor and a rack and pinion mechanism housed inside the main body. The output shaft of the reconfiguration motor is connected to two pinion gears, which are connected to the same shaft since a single motor operates both side wheels. The slider obtains the input force from the reconfigurable motor through the rack and pinion mechanism connecting the slider and the motor. The coupler link converts the slider’s linear motion into the crank’s angular motion, as seen in Figure 10. The crank’s design integrates the characteristics of the wheel and the legs, relying only on the crank’s angular position. The angular position of the crank is dependent upon the slider. Consequently, this dynamic study aims to determine the input force necessary for the actuation of the reconfiguration mechanism, considering the normal reaction on the wheel and the frictional force between the wheel and the ground. During the reconfiguration process, the frictional force acting between the wheel and the ground will counteract the motion generated by the mechanism. Consequently, it is essential to determine the input force necessary at the slider for the effective implementation of the reconfiguration mechanism. Upon calculating this force, it will facilitate the selection of the appropriate motor for the reconfiguration procedure. In the dynamic analysis, each link is analyzed individually to determine the forces acting on the respective joints. From Figure (9a), we may formulate the equilibrium equations for the crank as follows:

(20) \begin{equation} F_{12y}-F_{32y}=m_{2}a_{2y}+\mu _{y}{\mathbf{F_{N}}} \end{equation}

(21) \begin{equation} F_{12z}+F_{32z}=m_{2}a_{2z}-{\mathbf{F_{N}}} \end{equation}

(22) \begin{equation} R_{21y}F_{12z}-R_{21z}F_{12y}+R_{22y}F_{32z}-R_{22z}F_{32y}=\mu R_{23z}{\mathbf{F_{N}}}-R_{23y}F_{N}+I_{2}\boldsymbol{\alpha _{2}} \end{equation}

where ${\mathbf{F_{12}}}$ is the force on link 2 due to link 1, ${\mathbf{F_{32}}}$ is the force on link 2 due to link 3, $I_{2}$ is the moment of inertial of link 2, ${\mathbf{a_{2}}}$ and $\boldsymbol{\alpha _{2}}$ are the linear and angular acceleration of the link 2, and ${\mathbf{F_{N}}}$ is the normal contact reaction force at the wheel. Now, corresponding to the Figure (9b), for coupler, we can write as:

(23) \begin{equation} F_{23y}-F_{43y}=m_{3}a_{3y} \end{equation}

(24) \begin{equation} F_{43z}-F_{23z}=m_{3}a_{3z} \end{equation}

(25) \begin{equation} R_{31y}F_{43z}-R_{31z}F_{43y}-R_{32y}F_{32z}+R_{32z}F_{32y}=I_{3}\boldsymbol{\alpha _{3}} \end{equation}

where ${\mathbf{F_{43}}}$ is the force on link 3 due to link 4, ${\mathbf{F_{23}}}$ is the force on link 3 due to link 2, $I_{3}$ is the moment of inertia of link 3, and ${\mathbf{a_{3}}}$ and $\boldsymbol{\alpha _{3}}$ are the linear and angular acceleration of link 3. Likewise, corresponding to Figure 9c, for slider, we can write as:

(26) \begin{equation} {\mathbf{F_{in}}}-\mu _{y}sign(\boldsymbol{\dot {s}})|F_{14z}|+F_{34y}=m_{4}a_{4y} \end{equation}

(27) \begin{equation} F_{34z}+F_{14z}=0 \end{equation}

where ${\mathbf{F_{in}}}$ is the input force to the slider, ${\mathbf{F_{14}}}$ is the force on link 4 due to link 1, ${\mathbf{F_{34}}}$ is the force on link 4 due to link 3, $\boldsymbol{\dot {s}}$ is the slider velocity, and ${\mathbf{a_{4}}}$ is the slider acceleration. In the case of a slider, due to its exclusively linear motion, its angular acceleration is therefore zero, i.e., $\boldsymbol{\alpha _{4}}$  = 0. Furthermore, owing to its constrained motion along the y-axis, there will be no acceleration in the z-direction; hence, $a_{4z}=0$ . All the aforementioned dynamic equations for the mechanism are articulated within the same reference frame as the vehicle frame, meaning that the individual axes of the vehicle frame align with the respective local coordinate frames of the links. Consequently, no additional transformation is necessary, allowing for the direct resolution of these equations. Now, since Eqs. (20) through (27) are required to be solved simultaneously. Therefore, these equations are arranged in the matrix form as shown below:

(28) \begin{equation} \begin{bmatrix} 1 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\[3pt] 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0\\[3pt] -R_{21z} & R_{21y} & -R_{22z} & R_{22y} & 0 & 0 & 0 & 0\\[3pt] 0 & 0 & -1 & 0 & -1 & 0 & 0 & 0\\[3pt] 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0\\[3pt] 0 & 0 & R_{32z} & -R_{32y} & -R_{31z} & R_{31y} & 0 & 0\\[3pt] 0 & 0 & 0 & 0 & -1 & 0 & -\mu _{y} sign(\dot {\mathbf {s}}) & 1\\[3pt] 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} F_{12y}\\[3pt]F_{12z}\\[3pt]F_{32y}\\[3pt]F_{32z}\\[3pt]F_{43y}\\[3pt]F_{43z}\\[3pt]F_{14z}\\[3pt] {\mathbf{F_{in}}} \end{bmatrix}=\begin{bmatrix} m_{2}a_{2y}+\mu _{y} {\mathbf{F_{N}}}\\[3pt]m_{2}a_{2z}-{\mathbf{F_{N}}}\\[3pt]R_{23z}\mu _{y} {\mathbf{F_{N}}}-R_{23y}{\mathbf{F_{N}}}+I_{2}\boldsymbol{\alpha _{2}}\\[3pt]m_{3}a_{3y}\\[3pt] m_{3}a_{3z}\\[3pt]I_{3}\boldsymbol{\alpha _{3}}\\[3pt]m_{4}a_{4y}\\[3pt]0 \end{bmatrix} \end{equation}

Upon resolving Eq. (28), we may ascertain all the joint forces inside the mechanism, including the ${\mathbf{F_{in}}}$ , which represents the input force necessary at the slider. Additional joint forces derived from this equation can be used to design the individual link by specifying the necessary cross-sectional area, suitable material, and safety factor. The expression for ${\mathbf{F_{in}}}$ is shown below:

(29) \begin{eqnarray} {\mathbf{F_{in}}} &=& \frac {\mathbf{F_{N}}(\sigma _{1}-\mu _{y}\sigma _{2})+I_{3}\boldsymbol{\alpha _{3}}\sigma _{3}+m_{2}\sigma _{4}+m_{3}\sigma _{5}+m_{4}\sigma _{6}+sign(\dot {\mathbf {s}})\mu _{y}\sigma _{7}}{2} \nonumber\\[8pt] && +\frac {\mathbf{F_{N}}\mu _{y}sign(\dot {\mathbf {s}})\sigma _{8}+\mu _{y}sign({\dot {\mathbf {s}}})\sigma _{9}+{\mathbf{F_{N}}}\mu _{y}^{2}sign(\dot {\mathbf {s}})\sigma _{10}}{2} \end{eqnarray}

where

(30) \begin{equation} \sigma _{1}=R_{32y}^2-R_{21y}R_{31y}-R_{21y}R_{32y}+R_{31y}R_{32y} \end{equation}

(31) \begin{equation} \sigma _{2}=R_{21z}R_{31y}+R{21z}R_{32y}-R_{31y}R_{32z}-R_{32y}R_{32z} \end{equation}

(32) \begin{equation} \sigma _{3}=R_{21y}-R_{22y}-R_{31y}-R_{32y} \end{equation}

(33) \begin{equation} \sigma _{4}=-R_{21z}R_{31y}a_{2y}+R_{21y}R_{31y}a_{2z}+R_{21y}R_{32y}a_{2z}-R_{21z}R_{32y}a_{2y} \end{equation}

(34) \begin{eqnarray} \sigma _{5} &=& -R_{21y}R_{31y}a_{2y}+R_{21z}R_{31y}a_{3y}+R_{21y}R_{32z}a_{3y}+R_{21z}R_{32y}a_{3y}+R_{22y}R_{31y}a_{3z} \nonumber\\[8pt] && +R_{22z}R_{31y}a_{3y}+R_{22y}R_{32z}a_{3y}+R_{22z}R_{32y}a_{3y} \end{eqnarray}

(35) \begin{eqnarray} \sigma _{6} &=& -R_{21y}R_{31z}a_{4y}-R_{21z}R_{31y}a_{4y}-R_{21y}R_{32z}a_{4y}-R_{21z}R_{32y}a_{4y} \nonumber\\[8pt] && +R_{22y}R_{31z}a_{4y}-R_{22z}R_{31y}a_{4y}+R_{22y}R_{32z}a_{4y}-R_{22z}R_{32y}a_{4y} \end{eqnarray}

(36) \begin{eqnarray} \sigma _{7} &=& -R_{21y}R_{31z}a_{2z}+R_{21z}R_{31z}a_{2y}-R_{21y}R_{32z}a_{2z}m_{2}+R_{21z}R_{32z}a_{2y}m_{2}+R_{21y}R_{31z}a_{3z}m_{3} \notag \\[8pt] && -R_{21z}R_{31z}a_{3y}m_{3}+R_{21y}R_{32z}a_{3z}m_{3}+R_{21z}R_{32y}a_{3y}m_{3}\\[8pt] \notag && -R_{22y}R_{31z}a_{3z}m_{3}-R_{22z}R_{31z}a_{3y}m_{3}-R_{22y}R_{32z}a_{3z}m_{3}+R_{22z}R_{32y}a_{3z}m_{3} \end{eqnarray}

(37) \begin{equation} \sigma _{8}=R_{21y}R_{31z}+R_{21y}R_{32z}-R_{31z}R_{32y}-R_{32y}R_{32z} \end{equation}

(38) \begin{equation} \sigma _{9}=I_{3}\boldsymbol{\alpha _{3}}(R_{21z}R_{22z})+I_{2}\boldsymbol{\alpha _{2}}(R_{31z}+R_{32z}) \end{equation}

(39) \begin{equation} \sigma _{10}=R_{21z}R_{31z}+R_{21z}R_{32z}+R_{31z}R_{32z} \end{equation}

Since the expression for the input force ( ${\mathbf{F_{in}}}$ ) contains the linear and angular acceleration terms, which have already been determined from the study of position, velocity, and acceleration analysis. These accelerations and angular velocities are also provided below:

(40) \begin{equation} \boldsymbol{\alpha _{2}}=\frac {a\cos ({\theta _{3}-\theta _{2}})\boldsymbol{\omega _{2}}^{2}-b\boldsymbol{\omega _{3}}^{2}-\ddot {\mathbf {s}}\cos {\theta _{3}}}{a\sin ({\theta _{3}-\theta _{2}})} \end{equation}

(41) \begin{equation} \boldsymbol{\alpha _{3}}=\frac {-a\boldsymbol{\omega _{2}}^{2}+b\cos ({\theta _{2}-\theta _{3}})\boldsymbol{\omega _{3}}^{2}-\ddot {\mathbf {s}}\cos {\theta _{2}}}{b\sin ({\theta _{2}-\theta _{3}})} \end{equation}

(42) \begin{equation} a_{2y}=-R_{21}\boldsymbol{\omega _{2}}^{2}\cos {\theta _{2}}-R_{21}\boldsymbol{\alpha _{2}}\sin {\theta _{3}} \end{equation}

(43) \begin{equation} a_{2z}=-R_{21}\boldsymbol{\omega _{2}}^{2}\sin {\theta _{2}}+R_{21}\boldsymbol{\alpha _{2}}\cos {\theta _{2}} \end{equation}

(44) \begin{equation} a_{3y}=-a\boldsymbol{\omega _{2}}^{2}\cos {\theta _{2}}-a\boldsymbol{\alpha _{2}}\sin {\theta _{2}}-R_{32}\boldsymbol{\omega _{3}}^{2}\cos {\theta _{3}}-R_{32}\boldsymbol{\alpha _{3}}\sin {\theta _{3}} \end{equation}

(45) \begin{equation} a_{3z}=-a\boldsymbol{\omega _{2}}^{2}\sin {\theta _{2}}+a\boldsymbol{\alpha _{2}}\cos {\theta _{2}}-R_{32}\boldsymbol{\omega _{3}}^2\sin {\theta _{3}}+R_{32}\boldsymbol{\alpha _{3}}\cos {\theta _{3}} \end{equation}

Similarly, acceleration terms depend on the angular velocities of the links. Therefore, expressions for the angular velocities are given below:

(46) \begin{equation} \boldsymbol{\omega _{2}}=\frac {b\cos {\theta _{3}}\dot {\mathbf {s}}}{a\cos {\theta _{2}(b\tan {\theta _{2}}\cos {\theta _{3}}}- b\sin {\theta _{3}})} \end{equation}

(47) \begin{equation} \boldsymbol{\omega _{3}}=\frac {\dot {\mathbf {s}}}{b\tan {\theta _{2}}\cos {\theta _{3}} - b\sin {\theta _{3}}} \end{equation}

In this section, the objective of calculating the input force ( $F_{in}$ ) is to find out the torque required at the reconfiguration motor end, which provides the required input force for executing reconfiguration. Therefore, as we have the expression for the force required at the slider end of the mechanism, we can now calculate the minimum torque required at the motor, as shown below:

(48) \begin{equation} \boldsymbol{\tau _{m}}=2(r_{p}{\mathbf{F_{in}}}) \end{equation}

where $\boldsymbol{\tau _{m}}$ is the torque required by the reconfiguration motor, and $r_{p}$ is the radius of the pinion gear (pitch circle radius of the gear) attached to the reconfiguration motor. Since we have two reconfigurable wheels on either side of the motor, and only one motor is used for this purpose, therefore, two times. Now, as the input force required at the slider in Eq. (29) is a function of slider acceleration and the normal reaction at the wheel, which indicates that, if normal reaction onto the wheel increases, so is the motor torque, or vice versa. Furthermore, increasing the slider acceleration can result in reduced reconfiguration time while simultaneously elevating motor torque. This we can verify from the Figure 11. From the figure, we can observe that as the slider acceleration is increased from 0.01 m/s² to 1 m/s², reconfiguration time is decreased from 197.78 s to 2 s. Similarly, motor torque corresponding to the 0.01 m/s² is 0.79 Nm, and corresponding to 1 m/s², it is 1.58 Nm. For this analysis, a normal reaction of 15 Newtons on each of the front two wheels is considered, and other physical parameters of the reconfiguration mechanism used for this verification are provided in Table VI.

Figure 11.

Graphical representation of the actuator selection criterion based on slider acceleration and transformation time.

4. Robot dynamics

In this section, we will be describing the robot’s motion using the Newton–Euler approach for three-dimensional space. There is plenty of work in the two-dimensional planar dynamic modeling of wheeled mobile robots, but very limited work on 3D dynamic modeling [Reference Jaskot, Posiadała and ’Spiewak25, Reference Zeidis and Zimmermann26]. Therefore, in this work, robot is modeled for a 3-dimensional space to get the normal contact reaction force observed at the wheel and ground contact frame due to the rolling and pitching motion of the vehicle. Consequently, it will help to calculate the torque required at the actuator level precisely. For this modeling, certain assumptions are taken, like:

1. 1. There is no slipping between the surface and the wheel, i.e., perfect rolling.

2. 2. The ground is rigid enough to support the weight of the robot, without any deformation in the surface or in the wheel.

3. 3. Aerodynamic forces are very small due to the small frontal area and low velocity, and hence are neglected.

Figure 12.

Position, orientation, and contact forces on the robot with respect to the Inertial frame.

Position and orientation of the robot body frame, with respect to the inertial frame, can be described by a vector, $\boldsymbol{\eta }=[x,y,z,\phi ,\theta ,\psi ]^{\text{T}}$ , as shown in Figure (12a). The linear and angular velocity vector of the robot is denoted by $\boldsymbol{\nu }=[u,v,w,p,q,r]^{\text{T}}$ . The vectors $\boldsymbol{\eta }$ and $\boldsymbol{\nu }$ are in the vector space ${\mathbb{R}}^{6}$ . Now the equation of motion for the robot can be written as:

(49) \begin{equation} {\mathbf {M}}\boldsymbol{\dot {\nu }}+{\mathbf {C}}(\boldsymbol{\nu })\boldsymbol{\nu }+{\mathbf {g}}(\boldsymbol{\eta })=\boldsymbol{\tau } \end{equation}

where ${\mathbf {M}}\in {\mathbb{R}}^{6X6}$ is the mass matrix, ${\mathbf {C}}(\boldsymbol{\nu })\in {\mathbb{R}}^{6X1}$ is the vector of centripetal and Coriolis forces, ${\mathbf {g}}(\boldsymbol{\eta }) \in {\mathbb{R}}^{6X1}$ is the gravity vector and $\boldsymbol{\tau }\in {\mathbb{R}}^{6X1}$ is the vector of input forces and moments on the robot, in body frame. From Eq. (49), the position and velocity of the robot obtained are in the body frame {B}, which needs to be mapped with respect to the inertial frame {I}. For this mapping, we need two transformation operators ${\mathbf {J}}_{1}({\boldsymbol{\eta _{2}}})$ for linear velocity transformation and $ {\mathbf {J}}_{2}(\boldsymbol{\eta _{2}})$ for angular velocity transformations, which are given below:

(50) \begin{equation} {\mathbf {J}}_{1}(\boldsymbol{\eta _{2}})=\begin{bmatrix} \cos {\psi }\cos {\theta } & -\sin {\psi }\cos {\phi }+\cos {\psi }\sin {\theta }\sin {\phi } & \sin {\psi }\sin {\phi }+\cos {\psi }\cos {\phi }\sin {\theta }\\[3pt] \sin {\psi }\cos {\theta } & \cos {\psi }\cos {\phi }+\sin {\phi }\sin {\theta }\sin {\psi } & -\cos {\psi }\sin {\phi }+\sin {\theta }\sin {\psi }\cos {\phi }\\[3pt] -\sin {\theta } & \cos {\theta }\sin {\phi } & \cos {\theta }\cos {\phi } \end{bmatrix} \end{equation}

(51) \begin{equation} {\mathbf {J}}_{2}(\boldsymbol{\eta _{2}})=\begin{bmatrix} 1 &\quad \sin {\phi }\tan {\theta } &\quad \cos {\phi }\tan {\theta }\\[3pt]0 &\quad \cos {\phi } &\quad -\sin {\phi }\\[3pt]0 &\quad \sin {\phi }/\cos {\theta } &\quad \cos {\phi }/\cos {\theta } \end{bmatrix} \end{equation}

In Eq. (49), the input vector is defined as $\tau =[F_{x},F_{y},F_{z},M_{x},M_{y},M_{z}]^{T}$ , where $F_{x}, F_{y},$ and $F_{z}$ represent the input forces along the $X, Y,$ and $Z$ axes, respectively, of the vehicle frame. Similarly, $M_{x}, M_{y}$ , and $M_{z}$ represent the moments around the $X$ , $Y$ , and $Z$ axes, respectively, of the vehicle frame. The equation of motion is defined in the robot’s body frame, which is situated at the vehicle’s COM. The force vector, including normal contact reaction, traction forces, and frictional forces, is located at the wheel and the ground contact frame and is in inertial frame {I}, but with a different orientation than the inertial frame. However, the moment vector is already computed with respect to the vehicle frame {B}, in the system dynamics section, and thus requires no further transformation. Therefore, before incorporating these contact forces into the $\boldsymbol{\tau }$ vector, it is necessary to transform them to the COM position or the body frame, via suitable transformations. In the present scenario, the contact force vector located at the wheel and ground contact frame will first be transformed to the Inertial frame using the required rotation matrices. After the force vector has been determined in the inertial frame, it can be transformed to the vehicle frame utilizing the ${\mathbf {J}}_{1}(\boldsymbol{\eta _{2}})$ transformation function, as delineated in Eq. (50). Additionally, the rotation matrices required to transform force vectors from their respective contact frame to the inertial frame can be derived from the DH parameter Tables I, II, III, and IV, with the associated kinematic frame diagram illustrated in Figure 13.

Table I.

DH Table for front right wheel.

Table II.

DH Table for front left wheel.

Table III.

DH Table for rear right caster wheel.

Table IV.

DH Table for rear left caster wheel.

Figure 13.

Kinematic frame diagram of the proposed robot.

Now since the contact frame of all four wheels, i.e. {FR2}, {FL2}, {RR2}, and {RL2} have the same orientation with respect to the body frame, which is frame {7} corresponding to the Figure 13. Therefore, the rotation matrix required to transform all four wheel contact frames with respect to the inertial frame ( ${\mathbf {R}}_{w}^{I}$ ) is given by Eq. (52).

(52) \begin{equation} {\mathbf {R}}_{w}^{I} = \begin{bmatrix} 1 &\quad 0 &\quad 0\\[3pt]0 &\quad -1 &\quad 0\\[3pt]0 &\quad 0 &\quad -1 \end{bmatrix} \end{equation}

Now, the components of the vector of input force with respect to the inertial frame can be computed as follows:

(53) \begin{equation} \begin{bmatrix} F_{x}\\[3pt]F_{y}\\[3pt]F_{z} \end{bmatrix}={\mathbf {R}}_{w}^{I}\begin{bmatrix} F_{x1}\\[3pt]F_{y1}\\[3pt]F_{z1} \end{bmatrix}+{\mathbf {R}}_{w}^{I}\begin{bmatrix} F_{x2}\\[3pt]F_{y2}\\[3pt]F_{z2} \end{bmatrix}+{\mathbf {R}}_{w}^{I}\begin{bmatrix} F_{x3}\\[3pt]F_{y3}\\[3pt]F_{z3} \end{bmatrix}+{\mathbf {R}}_{w}^{I}\begin{bmatrix} F_{x4}\\[3pt]F_{y4}\\[3pt]F_{z4} \end{bmatrix} \end{equation}

where

(54) \begin{equation} \begin{bmatrix} F_{x1}\\[3pt]F_{y1}\\[3pt]F_{z1} \end{bmatrix}=\begin{bmatrix} \frac {\tau _{m}}{r_{w}}-\mu _{x}{\mathbf {N}}_{FRW}\\[3pt]\mu _{y}{\mathbf {N}}_{FRW}\\[3pt] {\mathbf {N}}_{FRW} \end{bmatrix} \end{equation}

(55) \begin{equation} \begin{bmatrix} F_{x2}\\[3pt]F_{y2}\\[3pt]F_{z2} \end{bmatrix}=\begin{bmatrix} \frac {\tau _{m}}{r_{w}}-\mu _{x}{\mathbf {N}}_{FLW}\\[3pt]\mu _{y}{\mathbf {N}}_{FLW}\\[3pt] {\mathbf {N}}_{FLW} \end{bmatrix} \end{equation}

(56) \begin{equation} \begin{bmatrix} F_{x3}\\[3pt]F_{y3}\\[3pt]F_{z3} \end{bmatrix}=\begin{bmatrix} -\mu _{x}{\mathbf {N}}_{RRC}\\[3pt]\mu _{y}{\mathbf {N}}_{RRC}\\[3pt] {\mathbf {N}}_{RRC} \end{bmatrix} \end{equation}

(57) \begin{equation} \begin{bmatrix} F_{x4}\\[3pt]F_{y4}\\[3pt]F_{z4} \end{bmatrix}=\begin{bmatrix} -\mu _{x}{\mathbf {N}}_{RLC}\\[3pt]\mu _{y}{\mathbf {N}}_{RLC}\\[3pt] {\mathbf {N}}_{RLC} \end{bmatrix} \end{equation}

Utilizing Eqs. (54) through (57), in conjunction with Eq. (53), we can determine the necessary input force vector in the body frame as delineated in Eq. (58).

(58) \begin{equation} \begin{bmatrix} F_{x}\\[3pt]F_{y}\\[3pt]F_{z} \end{bmatrix}={\mathbf {J}}_{1}(\eta _{2})'\begin{bmatrix} \frac {2\tau _{m}}{r_{w}}-\mu _{x}({\mathbf {N}}_{FRW}+{\mathbf {N}}_{FLW}+{\mathbf {N}}_{RRC}+{\mathbf {N}}_{RLC})\\[3pt] -\mu _{y}({\mathbf {N}}_{FRW}+{\mathbf {N}}_{FLW}+{\mathbf {N}}_{RRC}+{\mathbf {N}}_{RLC})\\[3pt] {\mathbf {N}}_{FRW}+{\mathbf {N}}_{FLW}+{\mathbf {N}}_{RRC}+{\mathbf {N}}_{RLC} \end{bmatrix} \end{equation}

Similarly, we can articulate the input moment vector to the body frame as:

(59) \begin{equation} \begin{bmatrix} M_{x}\\[3pt]M_{y}\\[3pt]M_{z} \end{bmatrix}=\begin{bmatrix} d_{1}({\mathbf {N}}_{FLW}-{\mathbf {N}}_{FRW})-d_{2}({\mathbf {N}}_{RRC}-{\mathbf {N}}_{RLC})\\[3pt] -l_{1}({\mathbf {N}}_{FRW}+{\mathbf {N}}_{FLW})+l_{2}({\mathbf {N}}_{RRC}+{\mathbf {N}}_{RLC})-h(F_{x1}+F_{x2}-F_{x3}-F_{x4})\\[3pt] F_{x1}d_{1}-F_{x2}d_{1}-F_{x3}d_{2}+F_{x4}d_{2} \end{bmatrix} \end{equation}

In Eq. (59), if the system operates in wheel mode, we may use $l_{1}, l_{2}$ , and $h$ as specified in the equation. However, if the system is in legged mode, these lengths will be altered to $l_{1}^{\prime}, l^{\prime}_{2}$ , and $h'$ , as previously elucidated in Section 2. Additionally, $r_{w}$ denotes the radius of the front wheel, which varies across leg mode and wheel mode, i.e, 0.09 m in wheel mode and 0.16 m in legged mode. Additionally, $r_{c}$ denotes the radius of the caster wheel, whereas $\tau _{m}$ represents the motor torque supplied for propelling the front wheels.

5. Robot control

To examine the robot’s mobility, performance, and stability in various operating modes (legged or wheeled) over varied trajectories, a control plan must be developed to integrate with the system. This section will outline the control approach for managing the robot’s mobility over various trajectories. Moreover, this study primarily aims to examine the system’s dynamics under a certain control scheme, rather than to investigate the performance of many kinds of control systems on the robot [Reference Raikwar, Fehrmann and Herlitzius28–Reference Chen, Zhan, Jiang, Zhang, Rao, Wang and Li31]. Consequently, for this project, a computed input control system is used to control the motion of the robot, which fundamentally employs a Proportional Derivative (PD) control mechanism. To implement computed input control, it is assumed that the system exhibits exponential stability over time. Which means, as $ t \rightarrow \infty$ , it follows that $\boldsymbol{\tilde {\eta }} \rightarrow 0$ and $ \boldsymbol{\dot {\tilde {\eta }}} \rightarrow 0$ . where, $\boldsymbol{\tilde {\eta }}=\eta _{\text{desired}}-\eta _{\text{actual}}$ and $\boldsymbol{\dot {\tilde {\eta }}}=\dot {\eta }_{\text{desired}}-\dot {\eta }_{\text{actual}}$ . From the second-order error dynamic model, the basic error model can be articulated as:

(60) \begin{equation} \ddot {\boldsymbol{\eta }}+k_{d}\dot {\tilde {\boldsymbol{\eta }}}+k_{p}\tilde {\boldsymbol{\eta }}=0 \end{equation}

where $k_{p}\gt 0$ and $k_{d}\gt 0$ are the constants of proportional and derivative gains, respectively. However, these two are positive definite diagonal matrices, but for simplicity, they are represented as scalars here.

(61) \begin{equation} \ddot {\boldsymbol{\eta }}_{d}-\ddot {\boldsymbol{\eta }}+k_{d}\dot {\tilde {\boldsymbol{\eta }}}+k_{p}\tilde {\boldsymbol{\eta }}=0 \end{equation}

(62) \begin{equation} \ddot {\boldsymbol{\eta }}=\ddot {\boldsymbol{\eta }}_{d}+k_{d}\dot {\tilde {\boldsymbol{\eta }}}+k_{p}\tilde {\boldsymbol{\eta }} \end{equation}

but,

(63) \begin{equation} \ddot {\eta }={\mathbf {J}}(\boldsymbol{\eta })\dot {\boldsymbol{\nu }}+\dot {\mathbf {J}}(\boldsymbol{\eta })\boldsymbol{\nu } \end{equation}

and,

(64) \begin{equation} \dot {\boldsymbol{\nu }}={\mathbf {J}}^{-1}(\boldsymbol{\eta })[\ddot {\boldsymbol{\eta }}-\dot {\mathbf {J}}(\boldsymbol{\eta })\boldsymbol{\nu }] \end{equation}

using Eq. (49) we can write Eq. (64) as:

(65) \begin{equation} {\mathbf {M}}^{-1}[\boldsymbol{\tau } - {\mathbf {C}(\boldsymbol{\nu })\boldsymbol{\nu } + {\mathbf {g}}(\boldsymbol{\eta })}] = {\mathbf {J}}^{-1}(\boldsymbol{\eta })[\ddot {\boldsymbol{\eta }}-\dot {\mathbf {J}}(\boldsymbol{\eta })\boldsymbol{\nu }] \end{equation}

and now using Eq. (62) and (65), input vector $\tau$ can be written as:

(66) \begin{equation} \boldsymbol{\tau } = {\mathbf {M}}{\mathbf {J}}(\boldsymbol{\eta })^{-1}[\ddot {\boldsymbol{\eta }}_{d} + k_{d}\dot {\tilde {\boldsymbol{\eta }}} + k_{p}\tilde {\boldsymbol{\eta }}-\dot {\mathbf {J}}(\boldsymbol{\eta })\boldsymbol{\nu }]+ [{\mathbf {C}}(\boldsymbol{\nu })\boldsymbol{\nu } + {\mathbf {g}}(\boldsymbol{\eta })] \end{equation}

Eq. (66) represents the necessary control input supplied by the controller to the robot dynamics block, corresponding to the Figure 14, which illustrates a comprehensive control architecture. In this context, the forward dynamics is used to compute the system’s acceleration, which is then utilized by kinematics to determine the system’s position and orientation. The trajectory planner block supplies the intended position and orientation vector, in addition to the desired velocity and acceleration vector. The controller block computes the positional and velocity error, and using the corresponding $k_{p}$ and $k_{d}$ gains, it determines the necessary torque vector, which is then fed to the robot dynamics block, therefore completing the loop.

Figure 14.

Control scheme of the robot.

6. System design and integration

This section addresses the system’s physical design. The robot design is developed using SolidWorks 2019 software, as seen in Figure 2. The robot comprises a total of four wheels: two passive caster wheels positioned at the back and two reconfigurable powered wheels situated at the front. The wheel reconfiguration employs a slider crank mechanism coupled with a rack and pinion arrangement on both front wheels. Three motors are used on the actuator side, two for operating the front wheels and one for triggering the reconfiguration mechanism. The primary chassis of the robot consists of 20 mm $\times$ 20 mm aluminum profiles and 4 mm thick transparent acrylic sheets, as seen in Figure 1. The robot is equipped with two reconfigurable wheels and two standard caster wheels. The reconfigurable wheel consists of five sets of slider-crank mechanisms, functioning as the spokes of the wheel. These reconfigurable wheels are produced using 3D printing with PLA material. The reconfiguration mechanism of both wheels is connected to the reconfiguration motor via a rack and pinion assembly, which is fabricated using 3D printing technology. The robot comprises four racks and two pinions, with one pinion and two racks allocated to each wheel. Additional minor components, such as bearing holders and rack clamps, are also produced by 3D printing. The ESP-32 microcontroller serves as the primary controller for robot operation. To operate the robot, two 12V, 30 RPM DC geared motors equipped with encoders (Johnson motors from Orange company) are used. A 12V, 20 RPM DC geared motor (Johnson motor from Orange company) serves as the reconfiguration motor. Two CYTRON MDD3A motor driver boards are used to regulate the energy requirements of the motors and a 10,000 mAh Li-ion battery is used to power the robot. The robot utilizes micro-ROS for control, and the IMU MPU6050 is used for collecting accelerometer and gyroscopic data for determining the robot’s position. Following the assembly of all components into a final reconfigurable wheeled robot, it was evaluated inside the college campus in both wheel and legged modes across various indoor and outdoor environments, whose testing video is available at this link.

Figure 15.

Temporal variation in normal reaction (bottom) and positional error (top) in wheel mode of longitudinal dynamics.