Study on movement forming methods for the machining of elliptic-shaped holes

Abstract

The movement forming methods for the machining of elliptic-shaped holes are presented. These methods can make the tool nose orbit, which is the synthesis of revolution of the hydrostatic shaft center around bearing center and rotation of the tool nose around shaft center, become an ellipse. Thus, the machining of elliptical holes can be achieved. Necessary conditions and sufficient conditions for forming elliptic tool nose orbit, under the condition that multiple shaft center orbits and the tool nose’s rotation orbit take synchronous forward synthesis and synchronous backward synthesis, are studied. The influence of initial phase angle of the tool nose on the formation of elliptic tool nose orbit has also been investigated. Characteristics and control equations of elliptic tool nose orbit under the conditions of multiple shaft center orbits and different initial phase angles of the tool nose are given. These studies will lay a theoretical foundation for the realization of the movement forming methods. Traditional machining methods for elliptic-shaped holes require auxiliary feed mechanism installed on the machine tool, and machining efficiency is limited by the frequency response characteristic of the servo bar. The forming methods presented in this article can overcome these shortcomings and provide a new approach for precision machining of elliptic-shaped holes.

Introduction

Currently, the load strength endured by the holes in mechanical equipment is increasingly high due to heavy load and high power output of mechanical equipment. The noncircular holes have some advantages, such as homogenizing loading, reducing stress concentration, and improving lubrication condition, and thus greatly increase the life span of parts. Whitacre and Trainer ¹ have investigated in detail the good carrying capacity of piston pin holes. The fatigue strength of noncircular pin holes is 30% higher than that for cylinder, which has significant technical and economic benefits. Elliptic-shaped hole is one kind of valuable noncircular holes. We propose a new forming method, and using this method, the ovality of elliptical hole can be freely adjusted, which is the greatest advantage compared with other methods.

At present, the commonly used machining methods for elliptic-shaped holes can be classified as follows: slope–intercept form, mechanical copying type, electromagnetic driven type, intelligent boring bar based on piezoelectric ceramic, and so on. The Italian DAMUS corporation adopts the elastic deformation-type feed mechanism and develops the noncircular pin hole machine tool. The German KS corporation applies mechanical copying structure in the machining of elliptic-shaped holes. ² The literature ³ realizes the radial micro-displacement feed through controlling shaft’s elastic deformation, and the method can be used for boring the noncircular holes. These methods have common disadvantages, that is, not high frequency response and poor flexibility.

The key to the precision machining of elliptic-shaped holes lies in the precision micro-displacement radical feed mechanism. ⁴ Zhang and colleagues^5–8 presented a revolving electromagnetic actuation mechanism composed of an electromagnetic stator and an electromagnetic rotor. The rotor’s main component is a flexure-hinged-based flexible body. There are four coils in the stator supplying actuation currents. The micro-displacement between the stator and rotor can be controlled by changing the currents applied to the coils. Wu and Xiang ⁹ used the characteristics of giant magnetostrictive materials (GMMs) to control the deformation of tool rod. And the developed equipment for boring the noncircular hole can reach the accuracy at micron level. Weck et al.^10,11 and Pasek et al.^12,13 developed different speedy tool servo systems based on piezoelectric ceramic. Xiang et al. ¹⁴ also studied the piezoelectric actuator-driven micro-displacement mechanism for achieving the feeding of boring tool. The experiment results indicate that 3 µm machining accuracy can be achieved. The researches mentioned above have some application cases, but most have complicated structure and difficulties in manufacture and assembling. And the researches lie in the stage of theoretical research and experiment in the laboratory.

Moller ¹⁵ investigated the application of a magnetic bearing spindle in the machining of noncircular hole. He succeeded in manufacturing a noncircular hole at a rotational speed of 22,500 r/min. Afterward, Kim et al. ¹⁶ also used a magnetic bearing to control the displacement of the rotor for boring the noncircular hole and carried out some experiments for a noncircular piston pin hole using the developed boring machine. The cutting error is within 2 µm at the rotational speed of 1500 r/min. However, active electromagnet bearing is unstable because of open loop. That complicates the structure of system and results in high requirements for controlling and costs. Therefore, in the case of heavy load and low (or medium) speed, electromagnet bearing is not a good option. ¹⁷

At present, there exist a lot of traditional machining methods for elliptic holes. But all these methods, except Moller’s ¹⁵ and Kim et al.’s, ¹⁶ require the auxiliary feed mechanism installed on the cutter bar. And thus, they have some shortcomings, such as complicated mechanical structure, limited installing space of cutter bar, and high difficulty in precision control. Therefore, it is urgent to seek a new machining method. The innovative forming methods presented here utilize the synthesis of the motion (the revolution) of the hydrostatic shaft center around bearing center and the motion (the rotation) of the tool nose around shaft center to obtain an elliptic tool nose orbit, and thus, the machining of elliptic-shaped holes can be achieved. And the hydrostatic bearing’s clearance can meet the ovality requirement ¹⁶ of most noncircular holes at present.

Liu et al.^18,19 find that characteristics of shaft center orbit, the initial phase angle of the tool nose, and movement synthesis methods have important effects on control realization of elliptic tool nose orbit. And such kinds of movement forming analyses have not been found in other articles. The literature^15,16 has not done research in such field, either.

The forming methods of elliptic tool nose orbit based on hydrostatic shaft are presented in this article. The implementation scheme and movement forming principle of elliptic tool nose orbit based on hydrostatic shaft are given. Necessary conditions and sufficient conditions for forming elliptic tool nose orbit, under the condition that multiple shaft center orbits and the tool nose’s rotation orbit take synchronous forward synthesis and synchronous backward synthesis, are studied. The influence of initial phase angle of the tool nose on the formation of elliptic tool nose orbit has also been investigated. Characteristics and controlling equations of elliptic tool nose orbit under the conditions of multiple shaft center orbits and different initial phase angles of the tool nose are provided. These studies are expected to establish the foundation for realization of these movement forming methods.

The implementation scheme and forming principle of elliptic tool nose orbit based on the rotation of shaft and the revolution of shaft center

The implementation scheme

In consideration of feasibility and generality, the physical model applied is represented in Figure 1. The rear bearing of the shaft is a high-precision rolling bearing with axial positioning, which guarantees that the shaft has neither radial displacement nor axial displacement in this point but allows the shaft to swing slightly with this point as the fulcrum. The front bearing is a hydrostatic bearing whose clearance with the journal is $R_b-R_j$ ( $R_b$ is the radius of hydrostatic bearing, and $R_j$ is the radius of shaft), which can allow the displacements of shaft in both X- and Y-directions.

Figure 1.

Schematic representation of the hydrostatic shaft system.

When the noncircular holes are machined, the shaft has two movements. The first movement is the shaft’s rotation around its axis line, and the second is the shaft’s swing (also called the revolution or precession) with the rear bearing center as fulcrum. When the shaft takes the two movements simultaneously, the expected elliptic orbit can be achieved in both the front bearing plane and the cutting point plane. When the workpiece moves along the Z-direction, the elliptic curved surface can be obtained, and by adjusting the range of the shaft’s procession, the varying elliptic curved surface can be achieved. Finally, the elliptical hole of workpiece can be machined.

For noncircular pin holes like those of the piston, the commonly used span ²⁰ of ovality is 0.02–0.06 mm, and this ovality requirement can be met by hydrostatic bearing’s clearance.

The forming principle

Figure 2 is the forming schematic of tool nose orbit in the case of synchronous forward precession at time t. The two-dimensional diagram is located in the cutting plane of tool nose. The origin O of fixed coordinate system OXY is the intersection point of the bearing center line and the cutting plane. The point $O_j$ is the intersection point of shaft center line and cutting plane. The origin O′ of moving coordinate system O′ X′ Y′ and point $O_j$ coincide. P is the tool nose and $O_{j}P$ is the rotation radius of tool nose. $R_1$ is the length of $O_{j}P$ . $S_j$ is the revolution orbit of shaft center, and S is the synthetic tool nose orbit. The fundamental principle of these movement forming methods is to obtain the expected tool nose orbit S through compositing the revolution $S_j$ of shaft center around the point O and the rotation of tool nose around the point $O_j$ . $\omega$ is the angular velocity of tool nose rotating around point $O_j$ . When $t=0$ , the angle between $O{O}_j$ and $O_{j}P$ is expressed as $\alpha$ , as seen in Figure 3. $P_0(x_0,y_0)$ is the coordinate of tool nose in the coordinate system OXY when $t=0$ , and $O^{\prime }/O_{j0}(x_{j0},0)$ is the coordinate of shaft center in the coordinate system OXY when $t=0$ . $\alpha$ is the initial phase angle of the tool nose, which is defined as the angle between $O{O}_{j0}$ ( $O{O}_{j0}$ is the direction of the link line of the shaft center and the bearing center) and $O_{j0}{P}_0$ ( $O_{j0}{P}_0$ is the direction of the tool nose) when $t=0$ .

Figure 2.

Schematic representation of the forming of tool nose orbit in the case of synchronous forward precession.

Figure 3.

Schematic representation of the initial phase angle in the case of synchronous precession.

The rotation equations of tool nose

It is assumed that the tool nose rotates around point $O_j$ in the counter-clockwise direction with the angular velocity $\omega$ . According to the geometrical relationship shown in Figure 2, the rotation of tool nose in the coordinate system O′ X′ Y′ can be expressed as

$$\{\begin{array}{l}{x}_d=R_1\cos (\omega t+\alpha )\\ y_d=R_1\sin (\omega t+\alpha )\end{array}$$

(1)

where $(x_d,y_d)$ is the coordinate of P in the coordinate system O′ X′ Y′.

The synchronous revolution equations of shaft center

For a better description, it might as well first assume that the shaft center orbit is an ellipse whose semi-major and semi-minor axis lengths are ${\Delta }_y$ and ${\Delta }_x({\Delta }_x<{\Delta }_y)$ , respectively. The angular velocity of the shaft center revolution is equal to the rotation of the tool nose. The equation of the shaft center orbit in the coordinate system OXY can be written as

$$\{\begin{array}{l}{x}_j={\Delta }_x\cos \omega t\\ y_j={\Delta }_y\sin \omega t\end{array}$$

(2)

When $\omega$ is positive, the revolution of the shaft center and the rotation of tool nose are in the same direction (both counter-clockwise), and this motion is called synchronous forward precession. When $\omega$ is negative, the revolution of the shaft center and the rotation of tool nose are in opposite directions, and the motion is called synchronous backward precession. In that case, equation (2) can be rewritten as

$$\{\begin{array}{l}{x}_j={\Delta }_x\cos \omega t\\ y_j=-{\Delta }_y\sin \omega t\end{array}$$

(3)

Combining equations (2) and (3), the general equation of the shaft center orbit can be expressed as

$$\{\begin{array}{l}{x}_j={\Delta }_x\cos \omega t\\ y_j=\epsilon {\Delta }_y\sin \omega t\end{array}$$

(4)

where $\epsilon =1$ in the case of synchronous forward precession and $\epsilon =-1$ in the case of synchronous backward precession.

The synthesis motion equations

According to the principle of motion synthesis, the coordinate of the tool nose P in coordinate system OXY can be obtained and given as follows

$$\{\begin{array}{l}x=x_j+x_d\\ y=y_j+y_d\end{array}$$

(5)

Substituting equations (1) and (2) into equation (5) yields the equations of tool nose orbit in the case of synchronous forward precession, which can be given as follows

$$\{\begin{array}{l}x={\Delta }_x\cos \omega t+R_1\cos (\omega t+\alpha )\\ y={\Delta }_y\sin \omega t+R_1\sin (\omega t+\alpha )\end{array}$$

(6)

Similarly, the equations of tool nose orbit in the case of synchronous backward precession can be written as

$$\{\begin{array}{l}x={\Delta }_x\cos \omega t+R_1\cos (\omega t+\alpha )\\ y=-{\Delta }_y\sin \omega t+R_1\sin (\omega t+\alpha )\end{array}$$

(7)

Combining equations (6) and (7), one obtains the general equation of the tool nose orbit as

$$\{\begin{array}{l}x={\Delta }_x\cos \omega t+R_1\cos (\omega t+\alpha )\\ y=\epsilon {\Delta }_y\sin \omega t+R_1\sin (\omega t+\alpha )\end{array}$$

(8)

Analysis of the movement forming methods

The type of the shaft center orbit varies with the values of ${\Delta }_x$ and ${\Delta }_y$ . When ${\Delta }_x\ne {\Delta }_y\ne 0$ , the shaft center orbit represented by equation (4) is an ellipse; when ${\Delta }_x={\Delta }_y\ne 0$ , the shaft center orbit is a circle; when ${\Delta }_x=0$ and ${\Delta }_y\ne 0$ , the shaft center orbit is a line segment in Y-axis; when ${\Delta }_y=0$ and ${\Delta }_x\ne 0$ , the shaft center orbit is a line segment in X-axis. The shaft center orbits of different types, composited with the rotation of tool nose, will form the tool nose orbits of different types. The corresponding relationship will be discussed in the following.

Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is an ellipse ( ${\Delta }_x\ne {\Delta }_y\ne 0$ )

The necessary conditions for the elliptic tool nose orbit

Defining $F\left(\omega t\right)$ to be the square of the distance between tool nose P and the origin O, $F\left(\omega t\right)$ can be written as

$$F(\omega t)=x^2+y^2$$

(9)

The necessary conditions for the elliptic tool nose orbit are as follows: (1) there exist two equal maximum values and two equal minimum values in the function $F\left(\omega t\right)$ and (2) the phase difference between two maximum points and the phase difference between two minimum points are both 180°, while the phase difference between one maximum point and the adjacent minimum point is 90°. The existence and equivalence of the extreme values and extreme point’s phase relations will be demonstrated in the following discussion.

The existence and equivalence of the extreme values

Substituting equation (8) into equation (9) and taking the first-order derivative of $F\left(\omega t\right)$ gives

$$F\prime (\omega t)=({\Delta }_y^2-{\Delta }_x^2)\sin 2\omega t+2(\epsilon {\Delta }_y{R}_1-{\Delta }_x{R}_1)\sin (2\omega t+\alpha )$$

(10)

Setting $F\prime \left(\omega t\right)$ to be zero and rearranging equation (10) yield

$$\{\begin{array}{l}\tan 2\omega t=\frac{-2\lambda \sin \alpha }{{\Delta }_y^2-{\Delta }_x^2+2\lambda \cos \alpha }\\ \lambda =\epsilon {\Delta }_y{R}_1-{\Delta }_x{R}_1\end{array}$$

(11)

The double-angle formula of tangent can be written as

$$\tan 2\omega t=\frac{2\tan \omega t}{1-{\tan }^2\omega t}$$

The solutions of equation $F\prime \left(\omega t\right)=0$ vary with the values of $\sin \alpha$ . When $\sin \alpha \ne 0$ , combining equation (11) and the double-angle formula of tangent yields the expression of $\tan \omega t$ , and taking arctangent change will get the equation’s general solutions $\omega t$ ; when $\sin \alpha =0$ , the particular solutions can be obtained by substituting $\alpha =0$ or $\alpha =\pi$ into equation (10).

• 1. When $\sin \alpha \ne 0$ , namely, $\alpha \ne 0$ and $\alpha \ne \pi$

The general solutions of $F\prime \left(\omega t\right)=0$ are

$$\{\begin{array}{l}\omega t_1=\arctan \left(\frac{\Gamma -\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\right)\\ \omega t_2=\arctan \left(\frac{\Gamma +\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\right)\\ \Gamma ={\Delta }_y^2-{\Delta }_x^2+2\lambda \cos \alpha \end{array}$$

(12)

and

$$\{\begin{array}{l}\omega t_3=\arctan \left(\frac{\Gamma -\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\right)+\pi \\ \omega t_4=\arctan \left(\frac{\Gamma +\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\right)+\pi \end{array}$$

(13)

In order to judge whether the extreme value $F\left(\omega t_1\right)$ is maximum or minimum, we can take the second-order derivative of $F\left(\omega t\right)$

$$F^{″}\left(\omega t\right)=2\Gamma \cos 2\omega t-4\lambda \sin \alpha \sin 2\omega t$$

(14)

Combining equations (12) and (14), one can obtain equation (15) as (see Appendix 1)

$$F^{″}(\omega t_1)=\frac{2{\Gamma }^2+8{\lambda }^2{\sin }^2\alpha -2\Gamma \sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }-\Gamma }$$

(15)

It can be proved that $F^{″}\left(\omega t_1\right)$ is always positive regardless of the values of $\Gamma$ in equation (15), that is, there exists the first minimum $F_{\min 1}\left(\omega t\right)=F\left(\omega t_1\right)$ . Similarly, the first maximum $F_{\max 1}\left(\omega t\right)=F\left(\omega t_2\right)$ , the second minimum $F_{\min 2}\left(\omega t\right)=F\left(\omega t_3\right)$ , and the second maximum $F_{\max 2}\left(\omega t\right)=F\left(\omega t_4\right)$ can be obtained.

From equations (12) and (13), it can be obtained that $\omega t_3=\omega t_1+\pi$ . In order to prove the equivalence of these two minimum values, we can substitute $\omega t_3=\omega t_1+\pi$ into equation (8), and after simplification we can find that $x\left(\omega t_3\right)=-x\left(\omega t_1\right)$ and $y\left(\omega t_3\right)=-y\left(\omega t_1\right)$ , which indicates these two minimum points are symmetric about the origin, so it can be proved that $F_{\min }\left(\omega t\right)=F\left(\omega t_1\right)=F\left(\omega t_3\right)$ . Similarly, the equivalence of these two maximum values can be proved, namely, $F_{\max }\left(\omega t\right)=F\left(\omega t_2\right)=F\left(\omega t_4\right)$ .

• 2. When $\sin \alpha =0$ , namely, $\alpha =0$ or $\alpha =\pi$

Since the solutions of equation $F\prime \left(\omega t\right)=0$ when $\sin \alpha =0$ are not included in equations (12) and (13), the particular solutions of equation $F\prime \left(\omega t\right)=0$ when $\alpha =0$ and $\alpha =\pi$ should be considered separately. The equation’s simplified form varies with the values of $\alpha$ and $\epsilon$ . The values of $\alpha$ and $\epsilon$ can be divided into four cases: (a) $\alpha =0$ and $\epsilon =1$ ; (b) $\alpha =0$ and $\epsilon =-1$ ; (c) $\alpha =\pi$ and $\epsilon =1$ ; and (d) $\alpha =\pi$ and $\epsilon =-1$ . In the following part, the particular solutions of equation $F\prime \left(\omega t\right)=0$ in case (a), as an example, will be calculated.

Substituting $\alpha =0$ and $\epsilon =1$ into equation (10) and simplifying it yield

$$F\prime (\omega t)=({\Delta }_y-{\Delta }_x)({\Delta }_y+{\Delta }_x+2R_1)\sin 2\omega t$$

(16)

Since ${\Delta }_y\ne {\Delta }_x$ , the solutions of $F\prime \left(\omega t\right)=0$ are

$$\{\begin{array}{l}\omega t_1=0\\ \omega t_2=\frac{\pi }{2}\\ \omega t_3=\pi \\ \omega t_4=\frac{3\pi }{2}\end{array}$$

(17)

Similarly, in order to judge whether the extreme value $F\left(\omega t_1\right)$ is maximum or minimum, we can take the second-order derivative of $F\left(\omega t\right)$

$$F^{″}(\omega t)=2({\Delta }_y-{\Delta }_x)({\Delta }_y+{\Delta }_x+2R_1)\cos 2\omega t$$

(18)

Since it has been supposed that ${\Delta }_x<{\Delta }_y$ , substituting $\omega t_1=0$ into equation (18) yields $F^{″}\left(0\right)>0$ , which indicates that $F\left(\omega t\right)$ gets the first minimum value when $\omega t=\omega t_1$ , namely, $F_{\min 1}\left(\omega t\right)=F\left(\omega t_1\right)$ . Similarly, it can be proved that the second minimum value is $F_{\min 2}\left(\omega t\right)=F\left(\omega t_3\right)$ , and $F\left(\omega t\right)$ gets the maximum values when $\omega t=\omega t_2$ and $\omega t=\omega t_4$ , namely, $F_{\max 1}\left(\omega t\right)=F\left(\omega t_2\right)$ and $F_{\max 2}\left(\omega t\right)=F\left(\omega t_4\right)$ .

Substituting $\omega t_3=\omega t_1+\pi$ into equation (8) yields $x\left(\omega t_3\right)=-x\left(\omega t_1\right)$ and $y\left(\omega t_3\right)=-y\left(\omega t_1\right)$ , so it can be proved that $F_{\min }\left(\omega t\right)=F\left(\omega t_1\right)=F\left(\omega t_3\right)$ . Similarly, it can be proved that $F_{\max }\left(\omega t\right)=F\left(\omega t_2\right)=F\left(\omega t_4\right)$ .

According to the above process, one obtains the particular solutions in the other three cases (see Appendix 2). The calculated results in four cases indicate that the particular solutions are same (i.e. equation (17)); the function $F\left(\omega t\right)$ in every case always has two same maximum values and two same minimum values.

The first necessary condition for the elliptic tool nose orbit has been proved as above. The second necessary condition will be proved in the following part.

The phase relations of the extreme points

• 1. When $\alpha \ne 0$ and $\alpha \ne \pi$

According to equations (12) and (13), it can be easily proved that the phase difference between two maximum points and the phase difference between two minimum points are both 180°. In the following part, we will prove that the phase difference between the maximum point (when $\omega t=\omega t_2$ ) and the minimum point (when $\omega t=\omega t_1$ ) is 90°.

Equation (12) can be rewritten as

$$\{\begin{array}{l}\tan \omega t_1=\frac{\Gamma -\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\\ \tan \omega t_2=\frac{\Gamma +\sqrt{{\Gamma }^2+4{\lambda }^2{\sin }^2\alpha }}{2\lambda \sin \alpha }\end{array}$$

(19)

Using the formula of triangle transformation

$$\tan (\omega t_2-\omega t_1)=\frac{\tan \omega t_2-\tan \omega t_1}{1+\tan \omega t_1\tan \omega t_2}$$

yields

$$\begin{array}{l}\tan (\omega t_2-\omega t_1)\\ =\frac{\sqrt{{\Gamma }^2+4{(\epsilon {\Delta }_y{R}_1-{\Delta }_x{R}_1)}^2{\sin }^2\alpha }}{(\epsilon {\Delta }_y{R}_1-{\Delta }_x{R}_1)(1+\tan \omega t_1\tan \omega t_2)\sin \alpha }\end{array}$$

(20)

According to equation (19), it can be easily proved that equation $1+\tan \omega t_1\tan \omega t_2=0$ . Since ${\Delta }_x<{\Delta }_y$ has been supposed, equation (20) can be calculated as

$$\tan (\omega t_2-\omega t_1)=\pm \infty$$

(21)

When $0<\alpha <\pi$ and $\epsilon =1$ , $\tan (\omega t_2-\omega t_1)=+\infty$ and $\omega t_2-\omega t_1=\pi /2$ can be obtained; when $0<\alpha <\pi$ and $\epsilon =-1$ , $\tan (\omega t_2-\omega t_1)=-\infty$ and $\omega t_2-\omega t_1=-\pi /2$ can be obtained.

The two above expressions can be unified into $\omega t_2-\omega t_1=\epsilon (\pi /2)$ .

Similarly, when $\pi <\alpha <2\pi$ , $\omega t_2-\omega t_1=-\epsilon (\pi /2)$ can be gained. When $0<\alpha <\pi$ , $\omega t_4-\omega t_3=\epsilon (\pi /2)$ can be gained; when $\pi <\alpha <2\pi$ , $\omega t_4-\omega t_3=-\epsilon (\pi /2)$ can be gained.

• 2. When $\alpha =0$ or $\alpha =\pi$

According to equation (17), it can be proved that the phase difference between two maximum points and the phase difference between two minimum points are both 180° and that the phase difference between the maximum point (when $\omega t=\omega t_2$ ) and the minimum point (when $\omega t=\omega t_1$ ) is 90°.

Thus, it has been fully proved that the necessary conditions for the elliptic tool nose orbit are tenable regardless the values of $\alpha$ .

The sufficient conditions and control equations for the elliptic tool nose orbit

Since the solutions of equation $F\prime \left(\omega t\right)=0$ vary with the values of $\alpha$ , the sufficient conditions will be investigated under different values of $\alpha$ .

The sufficient conditions and control equations when $\alpha =0$ or $\alpha =\pi$

Since $\alpha =0$ and $\alpha =\pi$ are two particular values of $\alpha$ , these two particular values could be substituted into equation (8) to investigate the sufficient conditions and control equations.

Substituting $\alpha =0$ into equation (8) gives

$$\{\begin{array}{l}x=({\Delta }_x+R_1)\cos \omega t\\ y=(\epsilon {\Delta }_y+R_1)\sin \omega t\end{array}$$

(22)

Similarly, when $\alpha =\pi$ , the parametric equations of tool nose orbit can be given as follows

$$\{\begin{array}{l}x=({\Delta }_x-R_1)\cos \omega t\\ y=(\epsilon {\Delta }_y-R_1)\sin \omega t\end{array}$$

(23)

It can be found that the curves described by equations (22) and (23) are both standard ellipses whose semi-axes are in coordinate axes. The control equations of elliptic tool nose orbit can be written as

$$\{\begin{array}{l}\frac{x^2}{{({\Delta }_x+R_1)}^2}+\frac{y^2}{{(\epsilon {\Delta }_y+R_1)}^2}=1,\text{if}\alpha =0\\ \frac{x^2}{{({\Delta }_x-R_1)}^2}+\frac{y^2}{{(\epsilon {\Delta }_y-R_1)}^2}=1,\text{if}\alpha =\pi \end{array}$$

(24)

For the synchronous forward precession ( $\epsilon =1$ ), the control equations of elliptic tool nose orbit are

$$\{\begin{array}{l}\frac{x^2}{{({\Delta }_x+R_1)}^2}+\frac{y^2}{{({\Delta }_y+R_1)}^2}=1,\text{if}\alpha =0\\ \frac{x^2}{{({\Delta }_x-R_1)}^2}+\frac{y^2}{{({\Delta }_y-R_1)}^2}=1,\text{if}\alpha =\pi \end{array}$$

(25)

For the synchronous backward precession ( $\epsilon =-1$ ), the control equations of elliptic tool nose orbit are

$$\{\begin{array}{l}\frac{x^2}{{({\Delta }_x+R_1)}^2}+\frac{y^2}{{(-{\Delta }_y+R_1)}^2}=1,\text{if}\alpha =0\\ \frac{x^2}{{({\Delta }_x-R_1)}^2}+\frac{y^2}{{({\Delta }_y+R_1)}^2}=1,\text{if}\alpha =\pi \end{array}$$

(26)

From equation (25), we can find the following conclusions.

• When $\alpha =0$ , the major axis of the elliptic tool nose orbit and that of the elliptic shaft center orbit are in the same axis, and the two minor axes are in the other one, that is to say, the elliptic tool nose orbit is parallel to the elliptic shaft center orbit.

• When $\alpha =\pi$ , the major axis of the elliptic tool nose orbit and the minor axis of the elliptic shaft center orbit are in the same axis, that is, the elliptic tool nose orbit is perpendicular to the elliptic shaft center orbit.

From equation (26), we can find the following conclusions.

• When $\alpha =0$ , the major axis of the elliptic tool nose orbit is always in X-axis and the minor axis is in Y-axis, regardless of the arrangement of the elliptic shaft center orbit’s semi-axes.

• When $\alpha =\pi$ , the major axis of the elliptic tool nose orbit is always in Y-axis and its minor axis is in X-axis, regardless of the arrangement of the elliptic shaft center orbit’s semi-axes.

The sufficient condition and control equations when $\alpha \ne 0$ and $\alpha \ne \pi$

If the curve generated by rotating clockwise the tool nose orbit around the origin by a certain degree is a standard ellipse (semi-axes are in coordinate axes), the tool nose orbit before this rotation must be an ellipse. That is the sufficient condition for the elliptic tool nose orbit when $\alpha \ne 0$ and $\alpha \ne \pi$ .

Figure 4 is the schematic representation of the tool nose orbit under the situation of elliptic shaft center orbit. $M(x_1,y_1)$ and $E(x_3,y_3)$ are the nearest points of the tool nose orbit from the origin. $N(x_2,y_2)$ and $F(x_4,y_4)$ are the farthest points of the tool nose orbit from the origin. $\beta$ is the angle between OM and X-axis, and the forward direction of $\beta$ is set to be counter-clockwise with X-axis as baseline. The proving process that curve S is an ellipse is given as follows.

Figure 4.

Schematic representation of the tool nose orbit when the shaft center orbit is an ellipse.

From the geometrical relationship represented by Figure 4, it can be obtained that

$$\sin \beta =\frac{y_1}{\sqrt{x_1^2+y_1^2}}=-\frac{x_2}{\sqrt{x_2^2+y_2^2}}$$

(27)

$$\cos \beta =\frac{x_1}{\sqrt{x_1^2+y_1^2}}=\frac{y_2}{\sqrt{x_2^2+y_2^2}}$$

(28)

If the tool nose orbit described by equation (8) is rotated clockwise around the origin by $\beta$ angle, the new parametric equations of this curve can be written as

$$\{\begin{array}{l}{x}_{\beta }=x\cos \beta +y\sin \beta \\ y_{\beta }=-x\sin \beta +y\cos \beta \end{array}$$

(29)

Substituting equations (27) and (28) into equation (29) and simplifying it yield (see Appendix 3)

$$\{\begin{array}{l}{x}_{\beta }=\sqrt{x_1^2+y_1^2}\cos \theta \\ y_{\beta }=\pm \sqrt{x_2^2+y_2^2}\sin \theta \end{array}$$

(30)

Obviously, the rotated curve described by equation (30) is a standard ellipse. So, the tool nose orbit is an oblique ellipse. Its semi-major axis $a=\sqrt{x_2^2+y_2^2}$ and semi-minor axis $b=\sqrt{x_1^2+y_1^2}$ .

Thus, the sufficient condition for the elliptic tool nose orbit has been proved. The control equations of the elliptic tool nose orbit could be expressed as follows

$$\frac{{(x\cos \beta +y\sin \beta )}^2}{b^2}+\frac{{(-x\sin \beta +y\cos \beta )}^2}{a^2}=1$$

(31)

Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is a circle ( ${\Delta }_x={\Delta }_y\ne 0$ )

The parametric equations of the shaft center orbit become

$$\{\begin{array}{l}{x}_j=\Delta \cos \omega t\\ y_j=\epsilon \Delta \sin \omega t\end{array}$$

(32)

where $\Delta$ is the radius of the circle. The parametric equations of the tool nose orbit can be gained by substituting ${\Delta }_x={\Delta }_y=\Delta$ into equation (8) as

$$\{\begin{array}{l}x=\Delta \cos \omega t+R_1\cos (\omega t+\alpha )\\ y=\epsilon \Delta \sin \omega t+R_1\sin (\omega t+\alpha )\end{array}$$

(33)

Similarly, defining $F\left(\omega t\right)$ to be the square of the distance between tool nose P and the origin O, $F\left(\omega t\right)$ is written as

$$\begin{array}{l}F\left(\omega t\right)={\Delta }^2+R_1^2+2\Delta R_1\cos \omega t\cos (\omega t+\alpha )\\ +2\epsilon \Delta R_1\sin \omega t\sin (\omega t+\alpha )\end{array}$$

(34)

The simplified results of $F\left(\omega t\right)$ vary with the values of $\epsilon$ .

Synchronous forward synthesis ( $\epsilon =1$ )

Equation (34) can be simplified as

$$F\left(\omega t\right)={\Delta }^2+R_1^2+2\Delta R_1\cos \alpha$$

(35)

Since the term of $\omega t$ is absent in equation (35), the value of $F\left(\omega t\right)$ is a constant, which indicates that the tool nose orbit is a circle instead of an ellipse.

Synchronous backward synthesis ( $\epsilon =-1$ )

Equation (34) can be simplified as

$$F(\omega t)={\Delta }^2+R_1^2+2\Delta R_1\cos (2\omega t+\alpha )$$

(36)

According to the deducing steps mentioned in section “Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is an ellipse ( ${\Delta }_x\ne {\Delta }_y\ne 0$ ),” it can be proved that the tool nose orbit is always an ellipse in the cases of synchronous backward precession and a circular shaft center orbit, regardless of the values of $\alpha$ .

• 1. When $\alpha =0$ or $\alpha =\pi$

Substituting ${\Delta }_x={\Delta }_y=\Delta$ into equation (24), the control equations of tool nose orbit can be rewritten as equation (37)

$$\{\begin{array}{l}\frac{x^2}{{(\Delta +R_1)}^2}+\frac{y^2}{{(-\Delta +R_1)}^2}=1,\text{if}\alpha =0\\ \frac{x^2}{{(\Delta -R_1)}^2}+\frac{y^2}{{(\Delta +R_1)}^2}=1,\text{if}\alpha =\pi \end{array}$$

(37)

From equation (37), we can find the following conclusions.

• When $\alpha =0$ , the major axis of the elliptic tool nose orbit is always in X-axis and its minor axis is always in Y-axis; when $\alpha =\pi$ , the major axis of the elliptic tool nose orbit is always in Y-axis and its minor axis is always in X-axis.

• 2. When $\alpha \ne 0$ and $\alpha \ne \pi$

According to the proving process of section “Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is an ellipse ( ${\Delta }_x\ne {\Delta }_y\ne 0$ ),” we can gain $\beta =(\alpha /2)-(\pi /2)$ , $a^2={(R_1+\Delta )}^2$ , and $b^2={(R_1-\Delta )}^2$ . And the tool nose orbit is an oblique ellipse whose control equations can be described by equation (38)

$$\frac{{\left(x\sin \frac{\alpha }{2}-y\cos \frac{\alpha }{2}\right)}^2}{{(R_1-\Delta )}^2}+\frac{{\left(x\cos \frac{\alpha }{2}+y\sin \frac{\alpha }{2}\right)}^2}{{(R_1+\Delta )}^2}=1$$

(38)

Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is a line segment in Y-axis ( ${\Delta }_x=0$ and ${\Delta }_y\ne 0$ )

Substituting ${\Delta }_x=0$ into equation (4) yields the parametric equations of the shaft center orbit as

$$\{\begin{array}{l}{x}_j=0\\ y_j=\epsilon {\Delta }_y\sin \omega t\end{array}$$

(39)

By substituting ${\Delta }_x=0$ into equation (8), the parametric equations of the tool nose orbit can be obtained and given as follows

$$\{\begin{array}{l}x=R_1\cos (\omega t+\alpha )\\ y=\epsilon {\Delta }_y\sin \omega t+R_1\sin (\omega t+\alpha )\end{array}$$

(40)

Since this is just a special case of the situation in which the shaft center orbit is an ellipse, the same deducing steps can be adopted while ${\Delta }_x$ is set to be zero. Therefore, when the shaft center orbit is a line segment in Y-axis, the tool nose orbit is always an ellipse regardless of the values of $\alpha$ .

• 1. When $\alpha =0$ or $\alpha =\pi$

By substituting ${\Delta }_x=0$ into equation (24), the control equations of tool nose orbit become

$$\{\begin{array}{l}\frac{x^2}{R_1^2}+\frac{y^2}{{(\epsilon {\Delta }_y+R_1)}^2}=1,\text{if}\alpha =0\\ \frac{x^2}{R_1^2}+\frac{y^2}{{(\epsilon {\Delta }_y-R_1)}^2}=1,\text{if}\alpha =\pi \end{array}$$

(41)

Obviously, when $\alpha =0$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situation of forward precession while they are mutually perpendicular under the situation of backward precession; when $\alpha =\pi$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situation of backward precession while they are mutually perpendicular under the situation of forward precession.

• 2. When $\alpha \ne 0$ and $\alpha \ne \pi$

According to the proving process of section “Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is an ellipse ( ${\Delta }_x\ne {\Delta }_y\ne 0$ ),” we can prove that tool nose orbit is an oblique ellipse whose control equation has no difference with equation (31). But we should consider ${\Delta }_x=0$ in the calculation process.

Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is a line segment in X-axis ( ${\Delta }_y=0$ and ${\Delta }_x\ne 0$ )

By substituting ${\Delta }_y=0$ into equation (4), one can write the parametric equations of the shaft center as

$$\{\begin{array}{l}{x}_j={\Delta }_x\cos \omega t\\ y_j=0\end{array}$$

(42)

The parametric equations of the tool nose orbit can be obtained by equation (43)

$$\{\begin{array}{l}x={\Delta }_x\cos \omega t+R_1\cos (\omega t+\alpha )\\ y=R_1\sin (\omega t+\alpha )\end{array}$$

(43)

Since this is just another special case of the situation in which the shaft center orbit is an ellipse, the tool nose orbit is always an ellipse regardless of the values of $\alpha$ .

1. When $\alpha =0$ or $\alpha =\pi$

By substituting ${\Delta }_y=0$ into equation (24), the control equations of the tool nose orbit can be obtained and given as follows

$$\{\begin{array}{l}\frac{x^2}{{({\Delta }_x+R_1)}^2}+\frac{y^2}{R_1^2}=1,\text{if}\alpha =0\\ \frac{x^2}{{({\Delta }_x-R_1)}^2}+\frac{y^2}{R_1^2}=1,\text{if}\alpha =\pi \end{array}$$

(44)

Obviously, when $\alpha =0$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situations of both forward precession and backward precession; when $\alpha =\pi$ , they are mutually perpendicular under the situations of both forward precession and backward precession.

1. When $\alpha \ne 0$ and $\alpha \ne \pi$

According to the proving process of section “Analysis of synthesis motion of the tool nose orbit when the shaft center orbit is an ellipse ( ${\Delta }_x\ne {\Delta }_y\ne 0$ ),” we can prove that tool nose orbit is an oblique ellipse whose control equation has no difference with equation (31). But we should consider ${\Delta }_y=0$ in the calculation process.

Conclusion

This article presents new movement forming methods of machining elliptic-shaped holes, and detailed analyses of the movement forming methods are given. Necessary conditions and sufficient conditions for forming elliptic tool nose orbit are studied under the conditions of two motion modes (synchronous forward precession and synchronous backward precession) and four shaft center orbits (ellipse, circle, line segment in X-axis, and line segment in Y-axis). The conclusions can be summarized as follows.

1. When the shaft center orbit is an ellipse or a line segment in coordinate axis (X- or Y-axis), both the synchronous forward precession and synchronous backward precession can form the elliptic tool nose orbit. When the shaft center orbit is a circle, only the synchronous backward precession can form the elliptic tool nose orbit. Therefore, the machining of elliptic holes can be realized by controlling the shaft center orbit. This article presents control equations of the tool nose orbits under multiple conditions.

2. The initial phase angle of the tool nose has a great influence on the phase of the elliptic tool nose orbit. When initial phase angle $\alpha =0$ or $\alpha =\pi$ , the tool nose orbit is a standard ellipse whose semi-axes are in coordinate axes; when $\alpha \ne 0$ and $\alpha \ne \pi$ , the orbit is an oblique ellipse, that is to say, the semi-axes are not in coordinate axes.

3. When the shaft center orbit is an ellipse, both the synchronous forward precession and synchronous backward precession can form elliptic tool nose orbit. And the features are summarized as follows.

4. When the precession mode is synchronous backward precession and $\alpha =0$ , the major axis of the elliptic tool nose orbit is always in X-axis and the minor axis is in Y-axis, regardless of the arrangement of the elliptic shaft center orbit’s semi-axes; when $\alpha =\pi$ , the major axis of the elliptic tool nose orbit is always in Y-axis and its minor axis is in X-axis, regardless of the arrangement of the elliptic shaft center orbit’s semi-axes. When $\alpha \ne 0$ and $\alpha \ne \pi$ , the orbit is an oblique ellipse.

When the shaft center orbit is a circle, only motion synthesis of backward precession can form elliptic tool nose orbit. The features are summarized as follows.

When $\alpha =0$ , the major axis of the elliptic tool nose orbit is always in X-axis and its minor axis is always in Y-axis; when $\alpha =\pi$ , the major axis of the elliptic tool nose orbit is always in Y-axis and its minor axis is always in X-axis. When $\alpha \ne 0$ and $\alpha \ne \pi$ , the orbit is an oblique ellipse.

When the shaft center orbit is a line segment in axis, both the synchronous forward precession and synchronous backward precession can form elliptic tool nose orbit. And characteristics are summarized as follows.

When the shaft center orbit is a line segment in Y-axis and $\alpha =0$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situation of forward precession while they are mutually perpendicular under the situation of backward precession; when $\alpha =\pi$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situation of backward precession while they are mutually perpendicular under the situation of forward precession. When $\alpha \ne 0$ and $\alpha \ne \pi$ , the orbit is an oblique ellipse.

When the shaft center orbit is a line segment in X-axis and $\alpha =0$ , the major axis of the tool nose orbit is parallel to the line segment of the shaft center orbit under the situations of both forward precession and backward precession; when $\alpha =\pi$ , they are mutually perpendicular under the situations of both forward precession and backward precession. When $\alpha \ne 0$ and $\alpha \ne \pi$ , the orbit is an oblique ellipse.

The matching features of shaft center orbit and the tool nose orbit mentioned above can provide multiple choices for the control realization. Their specific implementations need further study.