Feasibility of the Bi-Directional Scanning Method in Acceleration/deceleration Feedrate Scheduling for CNC Machining

Abstract

The Acceleration/Deceleration (AD) feedrate scheduling is widely used to plan the feedrate for CNC machining. Since the toolpath in CNC machining consists of enormous blocks (e.g., G01 or G02 block), the AD scheduling will result in lots of successive feedrate profiles. The feedrate at the junction of the adjacent profiles can be discontinuous, which will saturate the actuator and deteriorate the machining performance. The Bi-Directional Scanning Method (BDSM) is used to make the profiles overall continuous. To alleviate the computational burden, the BDSM is usually applied within a look-ahead buffer. When the buffer is filled with feedrate profiles, the BDSM updates the feedrates in the buffer. Conventional works believe that the BDSM with look-ahead (BDSMLA) will only increase the feedrate in each updating and is always feasible. We find that the BDSM can, however, decrease the feedrates in the buffer. The feedrate decrease will result in overall feedrate discontinuity and make the BDSM infeasible. We also propose a tweak method which can guarantee the feasibility of the BDSMLA. Simulation reveals the feedrate decrease in the BDSMLA and verifies the effectiveness of the tweak method.

A Proof of Property 2

Suppose an ACC feedrate profile $f(t),t\in \left[ 0, T \right]$ transits the feedrate from $f(0)=v_s$ to $f(T)=v_e$ with a traversal time T. Since the kinematic constraints are symmetric, the first derivative $f'(t)$ is always symmetric about $t=T/2$. Based on Dirichlet’s conditions, we can represent $f'(t)$ by the following Fourier series, where $b_i$ is the Fourier coefficients.

$$\begin{aligned} f'(t) = {{{b_0}} \over 2} + \sum \limits _{i = 1}^\infty {b_i}\cos \frac{4i\pi \left( {t - {T / 2}} \right) }{T} ,t \in \left[ {0,T} \right] \end{aligned}$$

(3)

Integrating $f'(t)$ yields,

$$\begin{aligned} f(t) = {{{b_0}t} \over 2} + \sum \limits _{i = 1}^\infty {{{{b_i}T} \over {4i\pi }}\sin \frac{4i\pi \left( {t - {T / 2}} \right) }{T}} + {c_0},t \in \left[ {0,T} \right] \end{aligned}$$

(4)

where $c_0$ is the integration constant. According to the boundary feedrates, we have

$$\begin{aligned} {c_0} = {v_s},{b_0} = {{2\left( {{v_e} - {v_s}} \right) } \over T} \end{aligned}$$

(5)

Substituting (5) into (4) and integrating f(t) obtain the distance function,

$$\begin{aligned} s(t) = {{{b_0}{t^2}} \over 4} - \sum \limits _{i = 1}^\infty {{b_i}{{\left( {{T \over {4i\pi }}} \right) }^2}\cos \frac{4i\pi \left( {t - {T / 2}} \right) }{T}} + {c_0}t + {c_1},t \in \left[ {0,T} \right] \end{aligned}$$

(6)

where $c_1$ is the integration constant. Therefore, the traversal distance can be evaluated as

$$\begin{aligned} s = s\left( T \right) - s\left( 0 \right) = {{{v_s} + {v_e}} \over 2}T \end{aligned}$$

(7)