Whole-body kinematic and dynamic modeling for quadruped robot under different gaits and mechanism topologies

Definition of coordinate system

The link’s rotation axis, velocity, and inertia are expressed differently in different coordinates. So, it is essential to define a unified coordinate system. Figure 3 shows the definition of the home position and the coordinate of body center, leg, and ground. The leg frame is defined at the intersection of the joint HAA (hip abduction/adduction) and HFE (Hip flexion/extension) axes in the same direction as the body frame.

Inverse kinematic

The inverse kinematics is given based on the leg frame’s representation $\overrightarrow{x}$ in the leg frame to calculate the joint rotation angle $\overrightarrow{q}$. Figure 3 shows the coordinates and steering definitions for each joint of the leg. Before solving the problem, we simplify the calculation by limiting the actual walking condition of the robot. Assume that the robot’s toes will not reach the top of the body, that is:

(1) $$\left|{\mathit{q}}_2\right|<{\displaystyle \frac{\pi }{2}\iff {}^{L}y<0}$$

First, q₁ is calculated by projecting onto the YOZ plane. In the YOZ plane, the projection of point C is D. We can obtain Eq. (2) by considering the geometrical relationship:

(2) $$\alpha =\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{\left|z\right|}{\sqrt{y^2+z^2}},\beta =\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{l_1}{\sqrt{y^2+z^2}}}}$$

Here α is the angle between $\overrightarrow{DO}$ and Z, and β is the angle between $\overrightarrow{AO}$ and $\overrightarrow{DO}$.

Figure 4 shows the geometric relationship of the ends under different positions. We can use this to calculate q₁ according to Eq. (3).

(3) $$q_1=\{\begin{array}{ll}\alpha -\beta & y<0,z>0\\ \pi -\alpha -\beta & y<0,z<0\end{array}$$

During the robot locomotion, the upper and lower links are always in the same plane. This plane is used as a new study plane to calculate q₂ and q₃. The new origin of the coordinate system was transferred to joint HFE, as shown in Fig. 5.

The conversion formula of the old and new coordinate systems is Eq. (4):

(4) $$\begin{array}{l}{x}^{\mathrm{\prime }}=x\\ y^{\mathrm{\prime }}=-AD=-\sqrt{y^2+z^2-l_1^2}\end{array}$$

Then the problem was transformed into solving the inverse kinematics of the planar two-link mechanism. According to the trigonometric relationship, we know that:

(5) $$\varphi =\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{\left|x\mathrm{\prime }\right|}{\sqrt{x{\mathrm{\prime }}^2+y{\mathrm{\prime }}^2}},\phi =\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{l_2^2+x{\mathrm{\prime }}^2+y{\mathrm{\prime }}^2-l_3^2}{2l_2\sqrt{x{\mathrm{\prime }}^2+y{\mathrm{\prime }}^2}}}}$$

Here ϕ is the angle between $\overrightarrow{AC}$ and the x-axis and φ is the angle between $\overrightarrow{AC}$ and $\overrightarrow{AB}$. As shown in Fig. 6, when the end effectors are in a specific position, there will be different solutions due to different ways to bend the legs.

(6) $$\begin{array}{l}if{q}_3>0\\ q_2=\{\begin{array}{c}{\displaystyle \frac{\pi }{2}-\phi -\varphi }{x}^{\mathrm{\prime }}>0,y\mathrm{\prime }<0\\ -{\displaystyle \frac{\pi }{2}-\phi +\varphi }{x}^{\mathrm{\prime }}<0,y\mathrm{\prime }<0\end{array}{q}_3=\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{l_2^2+l_3^2-x{\mathrm{\prime }}^2-y{\mathrm{\prime }}^2}{2l_2{l}_3}}\\ if{q}_3<0\\ q_2=\{\begin{array}{c}{\displaystyle \frac{\pi }{2}+\phi -\varphi }{x}^{\mathrm{\prime }}>0,y\mathrm{\prime }<0\\ -{\displaystyle \frac{\pi }{2}+\phi +\varphi }{x}^{\mathrm{\prime }}<0,y\mathrm{\prime }<0\end{array}{q}_3=-\mathrm{a}\mathrm{r}\mathrm{c}\cos {\displaystyle \frac{l_2^2+l_3^2-x{\mathrm{\prime }}^2-y{\mathrm{\prime }}^2}{2l_2{l}_3}}\end{array}$$

Forward kinematic

Forward kinematics is to calculate the expression of all links relative to the ground frame given the initial position and joint rotation angle. Because the robot is a floating base system, we cannot directly find the expression of the relevant parameters of the robot in the ground frame. So we first take the center of the robot body as the reference frame to solve the presentation of all links. And then, we transfer them to the ground coordinate according to the conversion formula of the body and the ground coordinate system. Screw theory is used to build a kinematics model. See Appendix A for the meaning of all symbols used in this paper.

In the screw theory, the velocity screw V and screw axis S of the rigid body is defined by a pair of vectors that is:

(7) $$S=\left(T;R\right)V=\left(v,w\right)$$

The screw axis S defines the position and positive rotation direction of the revolute joint. The first vector is the position on the ground coordinate frame, and the second vector is the direction of rotation. V represents the velocity and angular velocity of the rigid body.

In this paper, all the joint and link parameters are transferred to the body frame B for analysis. So, we assume that the body coordinate frame is relatively stationary when we solve the forward kinematic. The forward position model solves the end-effector position on the frame B from the inputs angle of every joint. Before computing the angle, the initial position of the robot should be determined to determine the initial velocity screw as follows:

(8) $$\begin{array}{c}{}^B{S}_{11}=\left(B_l/2,0,-B_w/2,-1,0,0\right)\hfill \\ {}^B{S}_{12}=\left(B_l/2,0,-B_w/2-l_1,0,0,-1\right)\hfill \\ {}^B{S}_{13}=\left(B_l/2,-l_2,-B_w/2-l_1,0,0,-1\right)\hfill \\ {}^B{S}_{21}=\left(-B_l/2,0,-B_w/2,-1,0,0\right);\hfill \\ {}^B{S}_{22}=\left(-B_l/2,0,-B_w/2-l_1,0,0,-1\right);\hfill \\ {}^B{S}_{23}=\left(-B_l/2,-l_2,-B_w/2-l_1,0,0,-1\right)\hfill \\ {}^B{S}_{31}=\left(-B_l/2,0,B_w/2,1,0,0\right);\hfill \\ {}^B{S}_{32}=\left(-B_l/2,0,B_w/2+l_1,0,0,1\right);\hfill \\ {}^B{S}_{33}=\left(-B_l/2,-l_2,B_w/2+l_1,0,0,1\right)\hfill \\ {}^B{S}_{41}=\left(B_l/2,0,B_w/2,1,0,0\right);\hfill \\ {}^B{S}_{42}=\left(B_l/2,0,B_w/2+l_1,0,0,1\right);\hfill \\ {}^B{S}_{43}=\left(B_l/2,-l_2,B_w/2+l_1,0,0,1\right)\hfill \end{array}$$

We can obtain the initial velocity screw of the joint computed according to the screw axis S_ij using Eq. (9). Where $\hat{\mathbf{s}}$ is the unit velocity screw. Such as the joint HAA of leg1 is $\hat{\mathbf{s}}=\left(0,0,0,1,0,0\right)$.

(9) $$J_{vso}^{ij}=T_v\left(S_{ij}\right)\cdot \hat{s}$$

Then we can get the end-effector position as Eq. (10) and the Jacobian matrix in Eq. (11).

(10) $$e{e}_i=P\left(J_{vso}^{i1}\cdot {\theta }_{i1}\right)P\left(J_{vso}^{i2}\cdot {\theta }_{i2}\right)P\left(J_{vso}^{i3}\cdot {\theta }_{i3}\right)e{e}_{io}$$

(11) $$J_i=\left[T_v\left(P_0\right)J_{vso}^{1i}{T}_v\left(P_{1i}\right)J_{vso}^{2i}{T}_v\left(P_{2i}\right)J_{vso}^{3i}\right]$$where i = 1, 2, 3, 4 is the leg number of the robot; P is the homogeneous transformation matrix when the joint rotation θ. S_ij is the initial velocity screw of the joint computed according to the screw axis in home position.

Stand

Show in Fig. 7D, When the robot stand, there is only one end-effector, the body. It can realize the action of squatting and standing up and the three-axis attitude transformation of roll, pitch, and yaw. In this case, each robot’s leg is connected to the ground as a passive spherical joint, and the three-dimensional translation is constrained. The body is the only end effector, and the robot can be regarded as a parallel mechanism. The force of each joint on its parent connecting link is defined as negative, and the force on the child connecting link is defined as positive so that the constraint matrix can be established.

Once we have the constraint matrix for each joint, write the constraint matrix for all joints and links as a larger matrix C.

(15) $$C=\begin{array}{c}ground\\ body\\ L_{11}\\ L_{12}\\ L_{13}\\ L_{21}\\ L_{22}\\ L_{23}\\ L_{31}\\ L_{32}\\ L_{33}\\ L_{41}\\ L_{42}\\ L_{43}\end{array}\left[\begin{array}{ccccccccccccccccccccccccccccc}fix& s_1& s_2& s_3& s_4& r_{11}& r_{12}& r_{13}& r_{21}& r_{22}& r_{23}& r_{31}& r_{32}& r_{33}& r_{41}& r_{42}& r_{43}& m_{11}& m_{12}& m_{13}& m_{21}& m_{22}& m_{23}& m_{31}& m_{32}& m_{33}& m_{41}& m_{42}& m_{43}\\ 1& 1& 1& 1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0\\ 0& 0& 0& 0& 0& 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0\\ 0& 0& -1& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0\\ 0& 0& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1\\ 0& 0& 0& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$

According to the formula of constraint matrix, C_ij represents the force of joint j on link i. For constraint: $C_{ij}={}^G{J}_{cmo}^{ij}$; For motion: $C_{ij}={}^G{M}_{cmo}^{ij}$.

Each of these rows is a linkage, there are 14 of them, and each of these columns is a joint. For the same joint, constraints and motions are considered separately. We knew that C is a sparse matrix, and we can use unique algorithms to calculate it to improve the computational efficiency of dynamics. We know that by conservation of energy in Eq. (16).

(16) $$C^T\cdot v=C_v=\left(\begin{array}{c}0\\ \dot{\theta }\end{array}\right)$$where, C _v represents the power of joint rotation, and v represents the speed of all the links. Take the derivative of both sides:

(17) $$C^T\cdot \dot{v}=\left(\begin{array}{c}0\\ \ddot{\theta }\end{array}\right)-{\dot{C}}^T\cdot v$$

Define: $C_a=\left(\begin{array}{c}0\\ \ddot{\theta }\end{array}\right)-{\dot{C}}^T\cdot v$ , then C^Ta = C_a

For any linkage, the force equilibrium condition satisfies Eq. (18):

(18) $$f_p=-Ia+f_c=-f_e-Ig+v{\times }^{\ast }Iv$$

By combining the equilibrium equations of all the links and Eq. (17), we can take the dynamic Eq. (19) of the whole-body dynamic:

(19) $$\left[\begin{array}{cc}-I& C\\ C^T\end{array}\right]\left[\begin{array}{c}a\\ \eta \end{array}\right]=\left[\begin{array}{c}{f}_p\\ c_a\end{array}\right]$$

Here a is the acceleration of all links and η is the forces of all joints, including constraint and driving force.

Walk

Quadruped robots can walk in various gaits, such as walk, trot, pace, gallop. The main differences between gaits are the order and the time of the stride and the duty cycle of the swing phase. Regardless of the gait, there always exists a supporting phase and a swinging phase at any given moment. The dynamic model was constructed with the contact point between the support and ground as a spherical joint and the swinging leg and body as the end-effectors. This section introduces how to build a constraint matrix by taking Walk-gait as an example. The methods are the same for other gaits. We need to analyze which leg is in the swing and which portion supports and modifies the constraints.

When the robot walks in Walk-gait, at any time, there are three legs in contact with the ground and one leg in the air. There are four different topologies. Figure 7B shows the third leg in the air. This section analyzes only this case, and other issues are similar. Among them, the 1st, 2nd, and 4th legs contribute to the support contact with the ground and support body movement. The contact points between the toe and the ground can be regarded as spherical joints, which constraining three-dimensional translation, but three-dimensional rotation is not restricted. Therefore, the constraint of a spherical joint can be obtained as follows:

$$s_i=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

Furthermore, the constraint matrices of all the joints can be combined into a larger matrix C to represent the robot’s force. The establishment and solution of the dynamic model are consistent with the above Stand topology, which will not be described in detail.

(20) $$C=\begin{array}{c}\\ ground\\ body\\ L_{11}\\ L_{12}\\ L_{13}\\ L_{21}\\ L_{22}\\ L_{23}\\ L_{31}\\ L_{32}\\ L_{33}\\ L_{41}\\ L_{42}\\ L_{43}\end{array}\left[\begin{array}{ccccccccccccccccccccccccccc}fix& s_1& s_3& r_{11}& r_{12}& r_{13}& r_{21}& r_{22}& r_{23}& r_{31}& r_{32}& r_{33}& r_{41}& r_{42}& r_{43}& m_{11}& m_{12}& m_{13}& m_{21}& m_{22}& m_{23}& m_{31}& m_{32}& m_{33}& m_{41}& m_{42}& m_{43}\\ 1& 1& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0\\ 0& 0& 0& 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& -1& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0\\ 0& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$

Bound

The Bound-gait differs from the above two. It is a floating base system, including five end effectors. The robot belongs to an unconstrained mechanism at this moment. Note that, the constraint matrix corresponding to the force of the robot is a singular matrix. That makes it impossible to calculate the velocities of all the links by calculating the constraint matrix.

(21) $$C=\begin{array}{c}\\ body\\ L_{11}\\ L_{12}\\ L_{13}\\ L_{21}\\ L_{22}\\ L_{23}\\ L_{31}\\ L_{32}\\ L_{33}\\ L_{41}\\ L_{42}\\ L_{43}\end{array}\left[\begin{array}{cccccccccccccccccccccccc}{r}_{11}& r_{12}& r_{13}& r_{21}& r_{22}& r_{23}& r_{31}& r_{32}& r_{33}& r_{41}& r_{42}& r_{43}& m_{11}& m_{12}& m_{13}& m_{21}& m_{22}& m_{23}& m_{31}& m_{32}& m_{33}& m_{41}& m_{42}& m_{43}\\ -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0& -1& 0& 0\\ 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& -1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& -1\\ 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]$$

Here we are going to use integrals to calculate. Then, when the robot starts to move, its position is known, and its velocity is zero. Then, we can establish constraint matrix Eq. (21) to solve the spatial accelerate.

And then, in the next loop, we can obtain the spatial velocity by numerical integration. Although the acceleration of the link changes from time to time, we use high-frequency real-time control, which can be calculated at the frequency of 1,000 Hz. The speed can be refreshed quickly according to the change of the link acceleration. The control effect has been achieved continuously and steadily.

Motion plan

According to bionics research from Polet & Bertram (2019), for tetrapod, there are different gaits at different speeds. The walking gait is used when motion at a slow pace, using the trot gait when moving faster. Furthermore, use the bound gait when chasing prey or running away. The main difference between different gaits is the order of the stride and the time cycle of the support and swing legs. This paper has mainly introduced the planning of the walk and trot gaits at a slow speed to verify the correctness and feasibility of the model. Figure 8 shows the stride order, stride length, and time to duty ratio in these two gaits. An elliptical trajectory transition is used between two steps, where step length, step height, and step time are adjustable. During the acceleration and deceleration, the displacement of the body is as half as the length of other times. So, we set the step length at the beginning and the end to be half a step, but the duration is the same.

To make the robot walk smoothly and stably, the trajectory needs to satisfy certain constraints, such as the position and velocity should be continuously differentiable. For this purpose, the planning uses the trapezoidal curve, making each step of walking experience a process of acceleration, uniform speed, and deceleration. Equation (22) is the functional expression of the trapezoidal curve, and all following trajectories are planned on this basis. In addition, different curves can be generated by setting different velocity and acceleration.

(22) $$s\left(t\right)=\{\begin{array}{ll}{\displaystyle \frac{1}{2}a{t}^2}& 0\le t\le t_a\\ vt-{\displaystyle \frac{v^2}{2a}}& t_a<t\le T-t_a\\ {\displaystyle \frac{2avT-2v^2-a^2{\left(t-T\right)}^2}{2a}}& T-t_a<t\le T\end{array}$$

A legged robot is a floating base system that needs to plan its four toes and body simultaneously. First, as shown in Eq. (23), is the trajectory planning of feet. Between every two footholds, the robot use elliptical trajectory transitions. Second, the body’s trajectory is determined by the number of steps and the position of the foothold shown in Eq. (24). Here L is the step length of forwarding direction; W is the step length of right and left directions.

(23) $$\{\begin{array}{ll}{x}_{leg}=x_{pre}+L+Lcos\left(\pi -s\left(t\right)\right)& 0\le t\le T\\ y_{leg}=Hsin\left(\pi -s\left(t\right)\right)& 0\le t\le T\\ z_{leg}=z_{pre}+W+Wcos\left(\pi -s\left(t\right)\right)& 0\le t\le T\\ \end{array}$$

(24) $$\{\begin{array}{c}{x}_{body}=\{\begin{array}{c}{x}_{prepos}+{\displaystyle \frac{L{t}^2}{4T^2}}\\ x_{prepos}+{\displaystyle \frac{L}{4}+{\displaystyle \frac{L}{2T\left(t-T\right)}}}\\ x_{prepos}-{\displaystyle \frac{L{\left(t-{\displaystyle \frac{2n-1}{T^2}}\right)}^2}{T^2}+L\left(n-{\displaystyle \frac{1}{2}}\right)}\\ \end{array}\\ y_{body}=0\hfill \\ z_{body}=\{\begin{array}{c}{z}_{prepos}+{\displaystyle \frac{W{t}^2}{4T^2}}\\ z_{prepos}+{\displaystyle \frac{W}{4}+{\displaystyle \frac{W}{2T\left(t-T\right)}}}\\ z_{prepos}-{\displaystyle \frac{L{\left(t-{\displaystyle \frac{2n-1}{T^2}}\right)}^2}{T^2}+W\left(n-{\displaystyle \frac{1}{2}}\right)}\\ \end{array}\\ \end{array}$$

Figure 9 shows the trajectory of leg and body in Cartesian space when robot walking with trot and walk gait. These trajectories are expressed in-ground coordinate systems. We can use Eq. (25) to transfer it from G frame to Leg frame. Then the joint rotation angle was calculated by the inverse kinematics. Sent the angle to the corresponding motor can make the robot walking with a planned gait.

(25) $${}^L\overrightarrow{x}={}^L{P}_B^B{P}_G{\left(t\right)}^G\overrightarrow{x}$$

Kinematic simulation

This paper use ADAMS to verify the kinematic and dynamic models. The physics engine of ADAMS has high computational precision. It can accurately reflect the ground reaction force of the robot and joint output torque. It can provide a theoretical basis for the mechanism design and hardware selection. The body and links are made of aluminum alloy in the simulation, close to the prototype used. The only difference is that the prototype has actuators, IMU, force sensors, batteries but look at them as a whole with the body in the simulation. We simplified the model and equivalently added all of these masses to body mass, ignoring the influence of its inertia. Figure 10 shows the schematic diagram of a quadruped robot walking in the trot and walk gait.

Dynamic simulation

When the robot is in motion with a walking gait, three legs are always in the support phase and another leg in the swing phase. At this time, the body is stationary relative to the ground. Figure 11 shows the torque curve of the knee joint of the swing leg during the switching between the swing phase and the support phase. It can be seen that the calculation results of the dynamic model are the same as the simulation results. It is shown that the dynamic model has high computational accuracy and veracity under this topology structure. In addition, the torque of the swing leg’s joint will suddenly change at the moment of switching. That is due to the disappearance and appearance of external forces, leading to a sudden change in joint acceleration.

Experiment in prototype

Our experiments were conducted on “XiLing”, a high dynamic response quadrupedal robot used an embedded PC with an Intel Core i7 4500U, running Ubuntu 16.04 (Linux-4.9.90 kernel) with Xenomai3 patch. The level communicates at 1 kHz over EtherCAT and control signals are generated in a 1 ms control loop that runs on a dedicated onboard industrial PC. The robot is driven by 12 identical motors, which are actuated by Elmo driver and are powered by a 48 V lithium battery. Table 1 shows motor performance parameters, with a continuous torque of 40 N·M and maximum torque of 108 N·M. In the prototype, the shank is driven by a gear belt with a reduction ratio of 1:2. The torque bearing capacity of the knee joint can reach 216 N·M, which completely satisfied the maximum torque requirements of the simulation. In other periods, the joint force is below 40 NM, which is lower than the continuous output torque of the motor.

The robot can quickly switch any posture in its workspace while standing. Figure 12 shows the basic movements such as roll, pitch, yaw. Respectively rotation angles of them are ±5°, ±10°, ±20°. The trapezoidal curve is used to complete the planning during the movement, and the time of the switching process can be adjusted by modifying the acceleration and the maximum speed. Figure 13 illustrates a snapshot of the yaw angle changing from −15° to 15° within 0.9 s. Figure 14 shows the tracking curve of knee position and velocity during this movement. With 1,000 points interpolated per second, both allow for fast and accurate tracking. It is worth noting that the measured velocity curve has burrs because of the gear backlash and the friction between the belt and pulley. Burrs can be eliminated by choosing more precise gears and optimizing the planning control algorithm. We will prioritize this problem in the next step.