1 Introduction

Recently, with the development of industrial robots, the mobile robots have been used in various industries. Compared to traditional mobile robots, the omnidirectional mobile robots have a broad application prospect in aerospace and other fields, as it can move in any direction and its turning radius can be zero. Mecanum omnidirectional mobile platform is a popular omnidirectional mobile robot. It can flexibly complete various tasks in crowded space.

The mobile platform built for riveting the rocket skin is shown in Fig. 1, which includes a mecanum omnidirectional mobile platform, laser sensors, a manipulator with six degrees of freedom. In order to make the mobile platform move precisely, we need to revise its movement equation. Muir and Neuman [1] have developed a kinematic model of mecanum robot using matrix theory. Wang and Chang [2] carried out error analysis interns of distribution with four mecanum wheels. Shimada et al. [3] introduced a position corrective feedback control method using a vision sensor on mecanum-wheel omnidirectional vehicles. Qian et al. [4] developed a more detailed analysis on the installation angle of roller. This paper does a further study from the perspective of the deformation of the roller and provides two methods to revise its movement equation.

2 System description and problem statement 2.1 Variables in motion equation

Figure 2 shows the prototype robotic system built at Shanghai University. According to kinematic analysis [5], its motion equation is shown as

where , .L is half of the distance from the front axle to the rear axle, l₀ the transverse distance from the wheel centers which are also the contact points on the ground to the platform center line, α the mounting angle of the roller, R₀ the radius of four wheels; ω₁, ω₂, ω₃ and ω₄ are angular velocities of the four wheels, respectively. All the parameters in Eq. (1) are defined in machine design, L=750 mm, l₀=615 mm, R₀=187.5 mm, α=45°.

Figure 3 is a view of a mecanum wheel and a roller. The roller is fitted on the edge of the mecanum wheel in angle a with its central axis, which can rotate freely around its central axis. The wheel relies on the friction that the roller acts on the ground to move. Because the material of its outer rim is usually rubbers, it will deform under pressure.

Figure 4 is a view of distribution of roller outer rim deformation in its transverse section. F is the pressure on the roller, T the driving torque. The radiuses of wheels reduce differently under different pressures. The greater the pressure, the greater the deformation is. The deformation zone shifts with the action of driving torque and the shifting causes the position of wheel center to change. From the force analysis of mecanum wheel [6], we know that with the wheel rotating, a driving torque is generated by the force in roller axial direction, so the position of the contact points will shift in axial direction. With the deformation of the roller, the R₀ and l₀ will change.

2.2 Motion model with changeable parameters

Based on the kinematic analysis of omnidirectional mobile system and deformation of the roller [7-16], Eq. (1) can be revised to individual parameters.

The model is established in Fig. 5, a coordinate system XOY is set at the center O of the model; V_X, V_Y, ω₀ are defined as the linear and angular speeds of platform movement, respectively. ω₁, ω₂, ω₃ and ω₄ are angular velocities of the wheels; l₁, l₂, l₃ and l₄ are the distances from the wheel centers O₁, O₂, O₃ and O₄ to the platform center O in X-direction; R₁, R₂, R₃ and R₄ are the radiuses of four wheels, respectively. The direction of the arrow indicates the positive direction.

In order to improve the system motion accuracy, we should solve the relative errors ΔR_i between R_i and R₀, and relative errors Δl_i between l_i and l0 (i=1, 2, 3, 4) to revise matrix D₁ in Eq. (2).

3 Method of solution 3.1 Relative error matrix

After multiplying time t, Eq. (2) becomes displacement equation

where Individual geometric parameters including l₁; l₂; l₃; l₄; R₁, R₂, R₃ and R₄ are variable in Eq. (3).

Then the relative errors can be determined by

Differentiating Eq. (4) leads to

Then the relative errors can be determined as

among them, M_i=l_i + L cosα, θ_i=ω_it (i=1, 2, 3, 4).

In order to carry out simulation, we need to verify whether the above method is feasible. From manufacturing and assembly tolerances to define ΔR_i the deformation of the roller and Δl_i in [-h, h] and [-j, j] (h=3; j=5).

The l_i and R_i can be written as

where l_i and R_i are in [l_o-j, l_o + j] and [R_o-h, R_o + h], respectively; r is a random number between 0 and 1.

Define ω₁=ω₂=ω₃=ω₄=20 rad; t=1 s. The rank of matrix G is 3. G is a full rank matrix. So ΔB satisfies Eq. (6).

3.2 Solution of relative errors 3.2.1 Displacement error measurement

In order to obtain matrix ΔB, we need to obtain displacement error matrix ΔX firstly. Assuming that the deformation of the rollers exists and the ground condition is good, four wheels of the platform can touch the ground [17].

Set matrix . Here the mobile system in a straight line in Y-direction, stop every 2 s (t=2 s) to measure its location coordinates X_k; Y_k and rotation angles θ_k (k=1, 2, 3, 4) [18-21].

The principle of measurement is shown in Fig. 6 and matrix ΔX can be obtained as

According to Eq. (1), X, Y and θ can be obtained as X=V_Xt=0, Y=V_Yt=2R₀ω₁=654 mm, θ=2ω₀=0.

Finally, through the experiment we obtain the displacement errors

Take Eq. (10) to Eq. (6) and use interval analysis and Monte Carlo analysis to obtain matrix ΔB:

3.2.1.1 Monte Carlo analysis

This method comes from Monte Carlo method. In Eqs. (7) and (8), l_i and R_i are the assigned random values. Take them into matrix G and Eq. (6) to obtain matrix ΔB.

As the value of ΔX has been obtained in Eq. (10), take the random values of l_i and R_i into Eq. (6), we can obtain the corresponding matrix ΔB which also needs to satisfy that -3≤ΔR_i≤3 and -5≤Dl_i≤5, solve matrix ΔB N times (N → ∞) until obtain 50 groups of matrix ΔB. It also means that through the displacement error ΔX in Eq. (10), we find 50 groups of matrix ΔB and its corresponding random matrix G.

We also obtain 50 groups of ΔR₁; ΔR₂; ΔR₃; ΔR₄; ΔR₁; ΔR₂; ΔR₃; ΔR₄ (see Fig. 7).

Average the 50 groups of ΔR₁; ΔR₂; ΔR₃; ΔR₄; ΔR₁; ΔR₂; ΔR₃; ΔR₄ to form a new matrix ΔB₁=[-0.541 1.237-0.605 1.055 0.458 0.683-0.048-0.683]^T. Now we use process capability index to estimate the value of matrix ΔB₁. Process capability, also known as working procedure capability, refers to the ability that process machining meets the machining quality. It is a measure of the minimum fluctuation of the internal consistency of process machining in the stablest state. When the process is stable, the quality characteristic value of product is in [μ-3σ, μ + 3σ] (μ is ensemble average of quality characteristic value of product, σ standard deviation of quality characteristic value of product), so this can be used to estimate the stability and accuracy of quality characteristic value of product.

There are two kinds of situations to solve process capability index. Firstly, μ=M is shown in Fig. 8. U_SL is the maximum machining error of machining quality; L_SL is the minimum machining error of machining quality; M=(U_SL + L_SL)/2; μ is the average value of process machining. C_p indicates process capability index, C_p=(U_SL -L_SL)/6σ.

Secondly, μ ≠ M is shown in Fig. 9.

C_pk indicates process capability index, C_pk=(U_SL -L_SL)/6σ -|M -μ|/3σ.

Only process capability index is greater than 1, and the value of process machining is valid.

As -3≤ΔR_i≤3 and -5≤Δl_i≤5, M=0 and μ ≠ M, we can obtain: C_pk(ΔR₁)=1.58＞1, C_pk(Δl₁)=0.57＜1, C_pk(ΔR₂)=1.06＞1, C_pk(Δl₂)=0.58＜1, C_pk (ΔR₃)=1.54＞1, C_pk(Δl₃)=0.63＜1, C_pk(ΔR₄)=1.17＞1, C_pk(Δl₄)=0.5＜1.

So the values of ΔR₁, ΔR₂, ΔR₃ and ΔR₄ in ΔB1 are closer to the real values than the values of ΔR₁, ΔR₂, ΔR₃ and ΔR₄ in ΔB.

Now take matrix ΔB₁(a)=[-0.541 1.237-0.605 1:055 0 0 0 0]^T; ΔB₁(b)=[0 0 0 0 0.458 0.683-0.048-0.683]^T into Eq. (5), we obtain ΔX₁(a)=[-0.4119 4-0.003]^T, ΔX₁(b)=[0 0 0.0032]^T.

According to the change rate of them relative to Eq. (10), the influence of ΔR₁, ΔR₂, ΔR₃ and ΔR₄ is much bigger than that of Δl₁, Δl₂, Δl₃ and Δl₄. Although the values of Δl₁, Δl₂, Δl₃ and Δl₄ are not accurate enough for the real values, matrix DB1 is valid.

3.2.1.2 Interval analysis

In Eq. (6), matrix G is uncertain because of the change of ΔR_i and Δl_i in [-3, 3] and [-5, 5], respectively. Now we set the increment of ΔR_i and Δl_i is 1 mm, there are 7 values of ΔR_i, and 11 values of Δl_i, so there are 7⁴ × 11⁴ groups of combination of matrix G. Take all the combination of matrix G to Eq. (6) to obtain matrix ΔB and exclude the combination of matrix ΔB that ΔR_i and Δl_i are not in [-3, 3] and [-5, 5], respectively. So we obtain 277 groups of matrix ΔB.

Average the values in Fig. 10 and obtain the average values of ΔR₁; ΔR₂; ΔR₃; ΔR₄; Δl₁; Δl₂; Δl₃; Δl₄. Use the average values to combine a new matrix ΔB₂=[-0.233 1.467-0.923 0.824 0.2-0.213 -0.068 -0.039]^T.

Use the same way to solve the process capability index of ΔB₂, we can obtain: C_pk(ΔR₁)=3.34＞1, C_pk (Δl₁)=0.68＜1, C_pk(ΔR₂)=1.22＞1, C_pk(Δl₂)=0.59＜1, C_pk(ΔR₃)=2.5＞1, C_pk(Δl₃)=0.63＜1, C pk (ΔR₄)=1.73＞1, C_pk(Δl₄)=0.6＜1.

So the values of ΔR₁, ΔR₂, ΔR₃ and ΔR₄ in ΔB₂ are closer to the real ones than the values of Δl₁, Δl₂, Δl₃ and Δl₄ in ΔB₂. As the influence of ΔR₁, ΔR₂, ΔR₃ and ΔR₄ is much bigger than Δl₁, Δl₂, Δl₃ and Δl₄, matrix ΔB₂ is valid.

3.2.2 Result validation

Use the matrix ΔB₁ to revise the parameters of matrix D₁ in Eq. (2), we obtain Set W₁= W₂= to calculate the displacement error matrices. S₁= S₂=

Use the same way to calculate with matrix ΔB₂, we can achieve , . The results of the two correction methods are almost identical in theory.

Set t=2 to measure the displacement errors, compare the values of measuring with S₁ and S₂.

Figure 11(a) is the entire experiment with 4 movements. Figure 11(b) is the image of actual movement. From the comparison, we find that S₁, S₂ are close to the measured displacement errors, so ΔB₁ and ΔB₂ are satisfied to revise the matrix D₁.

4 Conclusions

Through analyzing the roller deformation of mecanum wheel, we can find the changed parameters of the motion equation of mobile system. In this paper, the relative variation of the parameters in the motion equation of the mecanum motion platform is solved by Monte Carlo analysis and interval analysis. Using the relative variation of the parameters to revise the motion equation, then solve the displacement errors in different spaces in theory and compare them with the measured displacement errors in fact. From the comparison, both the two methods are satisfied to requirement.

Acknowledgements This work was supported by the Shanghai Municipal Science and Technology Commission (Grant Nos. 14111104502 and 15550721900).