Flexible wheel displacements for electromagnetic harmonic movable tooth drive system

Abstract

In this paper, the structure and operating principle of the electromagnetic harmonic movable tooth drive system is introduced. Equations of the magnetic density and the electromagnetic force are given. The force model on the flexible wheel under magnetic force is presented. The equivalent concentrated force applied to the flexible wheel is determined. Based on it, the displacement equations of the flexible wheel are deduced. Using these equations, the displacement distribution of the flexible wheel and its changes along with the main parameters are investigated. Relationship between coil current and the maximum radial displacement is determined. The limit current corresponding to the limit radial displacement is obtained and its changes along with the main parameters are analyzed. The results show that the displacements of flexible wheel increase with the radius of flexible wheel and the coil current, and decrease with the initial air gap and the thickness of flexible wheel. A limit current under which the flexible ring will be buckled occurs under electromagnetic force. The limit current corresponding to the limit displacement increases with the thickness of flexible wheel, the pole pair number and the air gap, and decreases with the radius of flexible wheel. The results can be used to design and analyze the drive system.

Keywords

Electromagnetic drive harmonic drive displacement force model limit current

1. Introduction

Electromechanical integrated system has advantage such as small sizes, light weight and easy to control which has been widely concerned by engineers and scholars [1–3]. Among existing electromechanical integrated system, electromagnetic harmonic drive system has compact structure and large speed ratio and is especially suitable for aerospace and robot technology [4,5]. Herdeg first proposed electromagnetic harmonic drive and developed the experimental prototype of the drive system [6]. Janes developed an electromagnetic harmonic gear drive system with inter coils, which is small in size [7,8]. Loechel studied a micro harmonic gear drive reducer for semiconductor chip packaging and micro robot drives [9]. Degen proposed a micro-harmonic gear drive and developed the model machine with the minimum mass of 3.5 g [10]. Rens developed a new permanent magnet harmonic drive prototype suitable for the requirements of large speed ratio [11]. Tjahjowidodo proposed a dynamics model of the harmonic drive and analyzed its nonlinear torsional behavior [12]. Man proposed a new structure of magnetic gear without direct contact between elements [13]. Yu designed a scheme to control the output torque of harmonic drivers by torque sensors [14]. Jose proposed a low-temperature non-contact harmonic drive made with magnetic superconductor [15]. Xu proposed an electromagnetic harmonic movable tooth drive system and studied its output torque [16–18]. Park proposed a dual-stage magnetic gear drive using two magnetic gears to allocate speed ratio [19]. Li studied a kind of coaxial magnetic gear for variable speed drive [20]. Chigira designed a harmonic gear with a stackable structure, and its maximum output torque can be improved by adjusting the structure of the gear [21]. Using linear transformation method, Zhang established an analytical model of magnetic field for the harmonic gear and analyzed the eccentric magnetic field [22]. Jing proposed a new type of eccentric harmonic magnetic gear drive [23].

Among above electromagnetic harmonic drives, the electromagnetic harmonic movable tooth drive system can give a large output torque. In the drive system, the displacements of the flexible ring are the key factor to drive movable teeth to move which decides on the operation performance of the drive. However, the displacements of the flexible ring for the drive system have not been investigated yet.

2. Operation principle

The electromagnetic harmonic movable tooth drive system is mainly composed of a wave generator part and a movable tooth drive part Solid model of the drive system is shown in Fig. 1. The wave generator consists of a flexible wheel and an electromagnetic steel core. There is an air gap between them. The movable tooth drive part consists of a center wheel, a movable tooth frame and several movable teeth. In the movable tooth drive part, the movable tooth is ball, and the radial guide groove with the same number of movable teeth is arranged on the movable tooth frame. The center wheel is fixed, and the movable tooth frame is directly connected to the output shaft for power output. The drive system is driven by elastic deformation of the flexible wheel. Under coil current, the magnetic forces are applied in sequence to two sectors which causes the deformation of the flexible wheel to push the movable tooth engaging with the central wheel. The coils are switched on sequentially, and the rotational electromagnetic field is formed which causes periodic elastic deformation of the flexible ring. So, the movable tooth is pushed to move in the radial guide groove of the movable tooth frame. The movable tooth is restrained by the inner tooth profile of the center wheel, and it pushes the moving tooth frame to rotate.

Fig. 1.

3D model of an electromagnetic harmonic movable tooth drive system. 1. flexible ring; 2. housing; 3. ion core; 4. coils; 5. air gap; 6. center wheel; 7. movable tooth; 8. movable tooth frame (output shaft).

3. Equivalent air gap and electromagnetic force

When the flexible wheel is not deformed, assuming that the average magnetic density in the initial air gap δ₀ is B ₀, then the average magnetic density in air gap increases to B _δ= _kn B ₀ after the flexible wheel is deformed. Here, k _n is the shape factor coefficient.

After deformation of the flexible wheel, the length of air gap decreases and the magnetic density increases. But, the magnetomotive force F _δ does not change, so $\begin{eqnarray}\displaystyle F_{{\delta}}={\displaystyle \frac{B_{0}}{{\mu}_{0}}}{\delta}_{0}={\displaystyle \frac{B_{{\delta}}}{{\mu}_{0}}}{\delta}={\displaystyle \frac{B_{0}}{{\mu}_{0}}}k_{n}{\delta} & & \displaystyle\end{eqnarray}$ (1) where μ₀ is the vacuum permeability; δ is the equivalent air gap length after the flexible wheel is deformed.

From Eq. (1), it is known $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle {\delta}={\delta}_{0}/k_{n}\end{eqnarray}$ (2) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle B_{{\delta}}={\displaystyle \frac{{\mu}_{0}F_{{\delta}}}{{\delta}}}={\displaystyle \frac{{\mu}_{0}k_{n}F_{{\delta}}}{{\delta}_{0}}}.\end{eqnarray}$ (3)

When the flexible wheel is working, the amplitude of the air gap magnetomotive force will be sinusoidal with the excitation current.

When the main action line of air gap sinusoidal magnetomotive force rotates at any angle ωt, the magnetomotive force amplitude at any point on the flexible wheel is $\begin{eqnarray}\displaystyle F(x)=F_{{\delta}}\cos ({\beta}-{\omega}t) & & \displaystyle\end{eqnarray}$ (4) where (𝛽 − ωt) is the angle between the main action line of the magnetomotive force and position vector at any point.

At the point, the magnetic density is $\begin{eqnarray}\displaystyle B(x)={\displaystyle \frac{{\mu}_{0}F(x)}{{\delta}(x)}}={\displaystyle \frac{{\mu}_{0}F_{{\delta}}\cos ({\beta}-{\omega}t)}{{\delta}_{0}-d_{m}\cos 2{\beta}}} & & \displaystyle\end{eqnarray}$ (5) where d _m is the maximum deformation of the flexible wheel.

According to the electromagnetic force formula, the electromagnetic force per unit area on the flexible wheel is $\begin{eqnarray}\displaystyle F={\displaystyle \frac{{\mu}_{0}F_{{\delta}}^{2}\cos ^{2}({\beta}-{\omega}t)}{2({\delta}_{0}-d_{m}\cos 2{\beta})^{2}}}. & & \displaystyle\end{eqnarray}$ (6)

The total electromagnetic force is $\begin{eqnarray}\displaystyle F_{z}=2\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}Frl\text{d}{\beta}=2\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}F_{{\delta}}^{2}\cos ^{2}({\beta}-{\omega}t)}{2({\delta}_{0}-d_{m}\cos 2{\beta})^{2}}}rl\text{d}{\beta} & & \displaystyle\end{eqnarray}$ (7) where l is the effective length of flexible ring; r is radius of the flexible wheel.

Substituting δ = δ₀∕k _n into Eqs ((6)) and ((7)), Eqs ((6)) and ((7)) can be simplified as $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle F\approx {\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{2}({\beta}-{\omega}t)}{2{\delta}_{0}^{2}}}\end{eqnarray}$ (8) $\begin{eqnarray}\displaystyle \text{} & \text{} & \displaystyle F_{z}\approx {\displaystyle \frac{{\pi}rl{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}}{4{\delta}^{2}}}\end{eqnarray}$ (9) where F _δ = 1.35 NI/p; N is the number of coil turns, I is the current in coil, p is the number of the coil pole pair.

4. Equivalent electromagnetic force

In the electromagnetic harmonic movable tooth drive system, the electromagnetic force on the flexible wheel is the radial distributed force. For simplifying calculation, the radial distributed force can be equivalent to radial concentrated force. The equivalent model is shown in Fig. 2.

Fig. 2.

Equivalent model of electromagnetic force. (a) concentrated force on flexible wheel; (b) distributed electromagnetic force on the flexible wheel.

Under concentrated force F, the displacements u, v and w of any point on the flexible wheel along the x, y and z coordinates can be calculated as follows $\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}u={\displaystyle \frac{Fr^{3}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{rc}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\cos n{\theta}\\ v={\displaystyle \frac{Fr^{3}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}cx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\sin n{\theta}\\ w={\displaystyle \frac{Fr^{3}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}cx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\cos n{\theta}\end{array}\right\} & & \displaystyle\end{eqnarray}$ (10) where θ is the angle between the displacement point and force F; K is the bending stiffness of flexible wheel, $K=\frac{E{\tau}^{3}}{12(1-{\gamma})}$ ; E is the modulus of elasticity of flexible wheel material; 𝛾 is Poisson’s ratio; τ is thickness of flexible wheel.

Figure 2 shows that θ = 𝛽 − ωt. Letting ωt = 0, the static displacements of flexible wheel under distributed force can be given by $\begin{eqnarray}\displaystyle \left.\begin{array}{@{}l@{}}u={\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \int _{c_{1}}^{c_{2}}\displaystyle \int _{0}^{{\displaystyle \frac{{\pi}}{2}}}F({\beta},c)\left[\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{rc}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\cos {\beta}\cos n{\theta}\right]\text{d}{\beta}dc\\ v={\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \int _{c_{1}}^{c_{2}}\displaystyle \int _{0}^{{\displaystyle \frac{{\pi}}{2}}}F({\beta},c)\\ \qquad \left\{\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}cx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\cos {\beta}\sin n{\theta}\right\}\text{d}{\beta}\text{d}c\\ w={\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \int _{c_{1}}^{c_{2}}\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}F({\beta},c)\\ \qquad \left\{\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}cx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\cos {\beta}\cos n{\theta}\right\}\text{d}{\beta}\text{d}c.\end{array}\right\} & & \displaystyle\end{eqnarray}$ (11)

The flexible wheel in the electromagnetic harmonic movable tooth transmission system is a cup-shaped shell with one end closed which is fixed. Here, the displacements u and w should be zero. According to it, following condition can be given from Eq. (11) $\begin{eqnarray}\displaystyle \displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}cx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]=0 & & \displaystyle\end{eqnarray}$ (12a) i.e. $\begin{eqnarray}\displaystyle x=x_{0}={\displaystyle \frac{-\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{(n^{2}-1)^{2}}}}{c\cdot \displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{n^{2}}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}}}. & & \displaystyle\end{eqnarray}$ (12b)

Therefore, when calculating the displacement of the flexible wheel, the origin of its coordinates should be at the position of x ₀ from the bottom of the flexible wheel. Thus, the equivalent model of electromagnetic force can be simplified as shown in Fig. 3.

Fig. 3.

Simple equivalent model. (a) concentrated force on flexible wheel; (b) distributed electromagnetic force on the flexible wheel.

Figure 3 shows that the total length of the flexible ring L = |x ₀| + l and c ₀ gives position of the equivalent concentrated force, c ₁ and c ₂ depends on the length of ion core. From Eq. (10), the maximum radial displacement of flexible wheel at x = c ₀ under concentrated force can be given by $\begin{eqnarray}\displaystyle w_{0}={\displaystyle \frac{Fr^{3}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}^{2}}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]. & & \displaystyle\end{eqnarray}$ (13)

From Eq. (11), the maximum radial displacement of flexible wheel at x = c ₀ under distributed force can be given by $\begin{eqnarray}\displaystyle \hspace{-15.0pt}w_{0}={\displaystyle \frac{2r^{4}}{{\pi}Kl}}\int _{c_{1}}^{c_{2}}\displaystyle \int _{0}^{{\displaystyle \frac{{\pi}}{2}}}F({\beta},c)\left\{\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}c}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\cos {\beta}\right\}\text{d}{\beta}\text{d}c. & & \displaystyle\end{eqnarray}$ (14)

From Eqs (13) and (14), the equivalent concentrated force F can be determined as below $\begin{eqnarray} \displaystyle F={\displaystyle \frac{2r\displaystyle \int _{c_{1}}^{c_{2}}\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}F({\beta},c)\left\{\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}c}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]\cos {\beta}\right\}\text{d}{\beta}\text{d}c}{\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}^{2}}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]}}. & & \displaystyle \end{eqnarray}$ (15)

5. Displacements of flexible ring

Substituting Eqs (9) and (15) into (1), the displacements u, v and w of any point on the flexible wheel along the x, y and z coordinates can be determined $\begin{eqnarray}\displaystyle u\text{} & =\text{} & \displaystyle {\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{r}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\cos n{\theta}\displaystyle \int _{c_{1}}^{c_{2}}c\displaystyle \int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{3}{\beta}}{2{\delta}_{0}^{2}}}\text{d}{\beta}\text{d}c\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle {\displaystyle \frac{2{\mu}_{0}r^{4}k_{n}^{2}F_{{\delta}}^{2}}{3{\pi}Kl{\delta}_{0}^{2}}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{r}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}(c_{2}^{2}-c_{1}^{2})\cos n{\theta}\end{eqnarray}$ (16) $\begin{eqnarray}\displaystyle v\text{} & =\text{} & \displaystyle {\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{n(n^{2}-1)^{2}}}\cdot \sin n{\theta}\displaystyle \int _{c_{1}}^{c_{2}}\displaystyle \int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{3}{\beta}}{2{\delta}_{0}^{2}}}\text{d}{\beta}\text{d}c\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad +∼{\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{nx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\cdot \sin n{\theta}\displaystyle \int _{c_{1}}^{c_{2}}c\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{3}{\beta}}{2{\delta}_{0}^{2}}}\text{d}{\beta}\text{d}c\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle \left[A\cdot {\displaystyle \frac{4}{3}}(c_{2}-c_{1})+B\cdot {\displaystyle \frac{2}{3}}(c_{2}^{2}-c_{1}^{2})\right]\sin n{\theta}\end{eqnarray}$ (17) $\begin{eqnarray}\displaystyle w\text{} & =\text{} & \displaystyle {\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{(n^{2}-1)^{2}}}\cos n{\theta}\displaystyle \int _{c_{1}}^{c_{2}}\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{3}{\beta}}{2{\delta}_{0}^{2}}}\text{d}{\beta}\text{d}c\nonumber\\ \displaystyle \text{} & \text{} & \displaystyle \quad +∼{\displaystyle \frac{2r^{4}}{{\pi}Kl}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{nx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\cos n{\theta}\displaystyle \int _{c_{1}}^{c_{2}}c\int _{0}^{{\displaystyle \frac{{\pi}}{2}}}{\displaystyle \frac{{\mu}_{0}k_{n}^{2}F_{{\delta}}^{2}\cos ^{3}{\beta}}{2{\delta}_{0}^{2}}}\text{d}{\beta}\text{d}c\nonumber\\ \displaystyle \text{} & =\text{} & \displaystyle \left[A\cdot {\displaystyle \frac{4}{3}}(c_{2}-c_{1})+B\cdot {\displaystyle \frac{2}{3}}(c_{2}^{2}-c_{1}^{2})\right]n\cos n{\theta}\end{eqnarray}$ (18) where $\begin{eqnarray}\displaystyle A={\displaystyle \frac{{\mu}_{0}r^{4}k_{n}^{2}F_{{\delta}}^{2}}{{\pi}Kl{\delta}_{0}^{2}}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{1}{n(n^{2}-1)^{2}}};\quad B={\displaystyle \frac{{\mu}_{0}r^{4}k_{n}^{2}F_{{\delta}}^{2}}{{\pi}Kl{\delta}_{0}^{2}}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }{\displaystyle \frac{nx}{(n^{2}-1)\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}. & & \displaystyle \nonumber\end{eqnarray}$

6. Results and discussion

Using Eqs (16)–(18), the displacement distributions of the flexible wheel for the drive system are calculated (see Figs 4 and 5). The main parameters of the drive system are given in Table 1. Here, k _n is 1.25.

Table 1
Parameters of the drive system

Parameter Value Parameter Value

l _T/mm 90 N 81

p 3 r/mm 39.5

l/mm 120 τ/mm 0.2

δ₀/mm 0.5 I/A 3

Parameter	Value	Parameter	Value
l _T/mm	90	N	81
p	3	r/mm	39.5
l/mm	120	τ/mm	0.2
δ₀/mm	0.5	I/A	3

Figure 4 gives the influence of term number n in Eqs (16)–(18) on calculation accuracy of the displacements. It shows that calculated results of the three displacements for the flexible wheel are different for different term number n in Eqs (16)–(18). When the term number n is changed from 8 to 10, changes of the calculated results of the three displacements become quite small which shows that the calculated error is quite small at n = 10. So, in Eqs (16)–(18), the first five terms (n = 2, 4, 6, 8, 10) are taken to obtain enough calculated accuracy of the flexible wheel displacements.

Fig. 4.

Influence of term number n on calculation accuracy of the displacements.

Figure 5 gives the displacement distribution of the flexible wheel. It shows:

Among three displacements of flexible wheel, the radial displacement w is the maximum, and the axial one is the minimum. Changes of the three displacements along with the position angles are similar to each other. The axial and radial displacements vary with the law of cosines, and the tangent displacement varies with the law of sine. The period of the changes is identical to each other and equal to π.

The amplitude of the axial displacement u is small and independent of the selected section. The axial displacement u maximizes at even times of π∕4, and it is equal to zero at odd times of π∕4.

The tangential displacement v of the flexible wheel varies with the position of the axial section. The farther from the bottom of the flexible wheel is, the greater the tangential displacement v. The tangential displacement v maximizes at odd times of π∕4, and it is equal to zero at even times of π∕4.

The radial displacement w of the flexible wheel varies with the position of the axial section. The farther from the bottom of the flexible wheel is, the greater the radial displacement w. The radial displacement w maximizes at even times of π∕4, and it is equal to zero at odd times of π∕4.

Effects of the main parameters on the displacements of flexible wheel are investigated. The calculations are done for section c ₀. The calculated results are given in Figs 6–8. Figures 6–8 show:

As the thickness τ of flexible wheel is increased, the three displacements u, v and w of flexible ring are decreased. When the thickness τ increases from 0.2 mm to 0.3 mm, the three displacements u, v and w are decreased by 3.37 times. When the thickness τ increases from 0.3 mm to 0.4 mm, the three displacements u, v and w are decreased by 2.37 times. The change of the flexible wheel displacements is inversely proportional to the cubic square of its thickness.

As the radius r of flexible wheel is increased, the three displacements u, v and w of flexible ring are increased. When the radius r increases from 30 mm to 40 mm, the displacement u is increased by 4.12 times, and the displacements v and w are increased by 3.06 times. When the radius r increases from 40 mm to 50 mm, the displacement u is increased by 3.98 times, and the displacements v and w are increased by 2.35 times. The change of the axial displacement u is proportional to the five power of its radius and the change of the tangent and radial displacements v and w are proportional to the four power of its radius.

As the initial air gap δ₀ between flexible wheel and stator is increased, the three displacements u, v and w of flexible ring are decreased. When the initial air gap δ₀ increases from 0.4 mm to 0.5 mm, the three displacements u, v and w are decreased by 1.56 times. When the air gap δ₀ increases from 0.5 mm to 0.6 mm, the three displacements u, v and w are decreased by 1.44 times. The change of the flexible wheel displacements is inversely proportional to the square of its initial air gap.

As the coil current I is increased, the three displacements u, v and w of flexible ring are increased. When the coil current I increases from 3 A to 4 A, the three displacements u, v and w are increased by 1.77 times. When the coil current I increases from 4 A to 5 A, the three displacements u, v and w are increased by 1.56 times. The change of the flexible wheel displacements is proportional to the cubic square of its coil current.

In a word, the thickness τ and the radius r of flexible wheel, the initial air gap δ₀, and the coil current I have important effects on the displacements of flexible wheel. The displacements increase with the radius r of flexible wheel and the coil current I, and decrease with the initial air gap δ₀ and the thickness τ of flexible wheel.

Fig. 5.

Displacement distribution of the flexible wheel.

Fig. 6.

Changes of axial displacement u with parameters.

Fig. 7.

Changes of tangent displacement v with parameters.

Fig. 8.

Changes of radial displacement w with parameters.

Among above four parameters, the thickness τ and the radius r of flexible wheel have more obvious effects on the displacements of flexible wheel than the initial air gap δ₀ and the coil current.

From Figs 6–8, it can be known that under electromagnetic force, the radial displacement of the flexible wheel is the maximum. It pays an important role in driving movable teeth to move. So, the radial displacement of the flexible wheel is analyzed in detail here.

Based on Eq. (18), it can be seen that at θ = 0, the radial displacement is the maximum. Taking section C ₀ as example, the maximum radial displacement is $\begin{eqnarray}\displaystyle w_{m}={\displaystyle \frac{2r^{4}B_{0}^{2}(c_{2}-c_{1})}{3{\mu}_{0}{\pi}Kl(1-K_{w})}}\displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}^{2}}{(n^{2}-1)^{2}\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right] & & \displaystyle\end{eqnarray}$ (19) where K _w = w _m∕δ₀.

Letting $\begin{eqnarray}\displaystyle Y={\displaystyle \frac{2r^{4}(c_{2}-c_{1})}{3{\pi}{\mu}_{0}Kl}}\cdot \displaystyle \mathop{\sum }_{n=2,4,6\ldots }^{\infty }\left[{\displaystyle \frac{1}{(n^{2}-1)^{2}}}+{\displaystyle \frac{n^{2}c_{0}^{2}}{(n^{2}-1)^{2}\left[{\displaystyle \frac{1}{3}}n^{2}l^{2}+2(1-{\gamma})r^{2}\right]}}\right]. & & \displaystyle \nonumber\end{eqnarray}$ Thus $\begin{eqnarray}\displaystyle w_{m}={\displaystyle \frac{B_{0}^{2}Y}{(1-K_{w})^{2}}}={\displaystyle \frac{1.8255N^{2}I^{2}Y}{p^{2}({\delta}_{0}-w_{m})^{2}}}. & & \displaystyle\end{eqnarray}$ (20) Substituting w _m = K _wδ₀(0 ≤ K _w ≤ 1) into Eq. ((20)), yields $\begin{eqnarray}\displaystyle I=\sqrt{{\displaystyle \frac{{\delta}_{0}^{3}p^{2}K_{w}(1-K_{w})^{2}}{1.8225N^{2}Y}}}. & & \displaystyle\end{eqnarray}$ (21)

Equation (21) gives relationship between coil current and displacement factor K _w. Substituting data in Table 1 into Eq. (21), the relationship between the maximum radial displacement and coil current and its changes along with main parameters are given (see Fig. 9). Figure 9 shows:

As the maximum radial displacement is increased, the coil current I required is first increased, and gets to a maximum value, and then it is decreased. The limit current I _max = 12.60 A corresponds to displacement factor K _w = 0.3333. It means that the maximum radial displacement w _m of the flexible wheel can get to 33.33% of the air gap δ₀. Above the limit displacement, the flexible wheel will be buckled.

As the air gap δ₀ is increased, the coil current required achieving the corresponding displacement increases. At δ₀ = 0.3 mm, I _max = 5.86 A. At δ₀ = 0.8 mm, I _max = 25.52 A. It can be seen that the limit current corresponding to the limit displacement is increased significantly with the air gap.

As the thickness τ of flexible wheel is increased, the coil current required achieving the corresponding displacement increases. At τ = 0.2 mm, I _max = 4.46 A. At τ = 0.4 mm, I _max = 23.16 A. It can be seen that the limit current corresponding to the limit displacement is increased significantly with the thickness τ of flexible wheel.

As the pole pair number p is increased, the coil current required achieving the corresponding displacement increases as well. At p = 2, I _max = 8.41 A. At δ₀ = 0.8 mm, I _max = 16.81 A. It can be seen that the limit current corresponding to the limit displacement is increased linearly with the air gap.

As the radius r of flexible wheel is increased, the coil current required achieving the corresponding displacement decreases. At r = 30 mm, I _max = 21.26 A. At r = 50 mm, I _max = 7.98 A. It can be seen that the limit current corresponding to the limit displacement is decreased significantly with the thickness τ of flexible wheel.

In a word, the limit current corresponding to the limit displacement increases with the thickness τ, the pole pair number p and the air gap δ₀, it decreases with the radius r of flexible wheel.

Fig. 9.

IK _w relationship under different parameters.

Using FEM software, ANSYS, a FEM model of the flexible wheel under magnetic force is given. Using the same boundary conditions as those for theoretical calculation, the displacements of the flexible wheel are determined (see Fig. 10). Table 2 shows comparison between maximum radial displacements of the flexible wheel obtained by theoretical calculation and FEM simulation. It shows:

The maximum radial displacements of the flexible wheel on three main sections obtained by the theoretical calculation are in good agreement with ones given by FEM. The maximum error between them is 6.68% which illustrates the theoretical calculation given by the work.

Fig. 10.

Displacements of the flexible wheel from FEM.

Table 2

Comparison of the maximum radial displacements

Section	Method	w _max (mm)	Errors
C ₁	Theoretical calculation	0.102	4.90%
	FEM	0.097
C ₀	Theoretical calculation	0.198	4.54%
	FEM	0.207
C ₂	Theoretical calculation	0.314	6.68%
	FEM	0.335

7. Conclusions

Among three displacements of flexible wheel, the radial displacement is the maximum, and the axial one is the minimum. The axial and radial displacements vary with angle position in law of cosines, and the tangent displacement varies with angle position in the law of sine. The period of the changes is π.

The radial displacement of the flexible wheel varies with the position of the axial section. The farther from the bottom of the flexible wheel is, the greater the radial displacement w. The radial displacement w maximizes at even times of π∕4, and it is equal to zero at odd times of π∕4.

The thickness and the radius of flexible wheel, the initial air gap between flexible wheel and stator, and the coil current have important effects on the displacements of flexible wheel. The displacements increase with the radius of flexible wheel and the coil current, and decrease with the initial air gap and the thickness of flexible wheel.

A limit current under which the flexible ring will be buckled occurs under electromagnetic force.

The limit current corresponding to the limit displacement increases with the thickness of flexible wheel, the pole pair number and the air gap, and decreases with the radius of flexible wheel.

The electromagnetic harmonic movable tooth drive system has a compact structure and can give a large output torque. So, it could be used in fields requiring small sizes and large load-carrying ability such as aerospace and robotics

Footnotes

Acknowledgements

This project is supported by National key R & D Program of China (No. 2018YFB1304800) and the Hebei Province Natural Science Foundation in China (No. E2017203021).

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