Multi-axis Linkage Linear Interpolation and Its "S Acceleration and Deceleration" Planning Algorithm
1 Introduction The multi-axis linear interpolation and its acceleration and deceleration algorithms are the core technologies of high-end numerical control systems. Acceleration/deceleration processing is performed before and after acceleration/deceleration processing. Pre-acceleration and deceleration processing Before interpolation, the advantage is that the synthesis speed is controlled, and the position accuracy is not affected, but the deceleration point is predicted. Pre-acceleration and deceleration are usually linear acceleration and deceleration. Post-acceleration and deceleration is the acceleration/deceleration control of each interpolation axis. Since there is no coordination relationship between the axes, the synthesis position may not be accurate. The S-curve acceleration and deceleration is usually used for post-acceleration and deceleration processing. When we develop a multi-coordinated linkage fiber winding machine based on an open CNC system, we apply it to the pre-acceleration and deceleration processing, and achieved good results. 2 Multi-axis linear interpolation S acceleration/deceleration S-curve acceleration/deceleration planning means that the acceleration (DER) derivative (Jerk) da/dt is constant during acceleration/deceleration. It is minimized by controlling the Jerk value. Impact on mechanical systems. In addition, the flexible acceleration and deceleration control can be realized by setting or programming the parameters of acceleration and Jerk two physical quantities to adapt to the working conditions of different types of machine tools. In the n-dimensional linear interpolation NC program, any piece of interpolation data is | P1, P2, ..., Pn, F | where: F is the synthesis speed, P1 ~ Pn is the displacement of the current segment of each interpolation axis. According to the principle of linear interpolation, the displacement and velocity ratio of each interpolation axis is equal, and there is a sub-velocity that should be applied to each interpolation axis. Let |Pi| = P =TSEP i=1, 2, ..., n Fi F (1)
Pi2) 1â„2 represents the composite displacement; TSEP represents the time required for each axis of the linear interpolation segment to reach the end point at the same time; F1 to F2 are the composite velocity corresponding to the sub-velocity of each interpolation axis. Let Ki = Pi i=1, 2, ..., n P (2) then Fi=KiF i=1, 2, ..., n (3) In the pre-acceleration and deceleration processing, plan the given speed, as shown in Figure 1. As shown, the entire acceleration and deceleration process is divided into three sections, namely the acceleration section (1, 2, 3 zones), the uniform speed section (4 zones) and the speed section (4, 5, 6 zones). In the acceleration and deceleration sections, they include variable acceleration and deceleration zones (1, 3, 5, and 7 zones) and constant acceleration and deceleration zones (2 and 6 zones): variable acceleration and deceleration zones, |da/dt|=J, Jerk For constant value; constant acceleration/deceleration zone, |a|=A, the acceleration is constant, and the speed of the uniform speed zone (4 zone) is a constant value Vc. The motion parameters of each axis are proportional to the planned synthesis velocity v(t) at the te-point power series so that ∆t=t-tx, with v(t)=v(tx)+a(tx)∆t +1â„2J(te)∆t2 (4) Similarly, each interpolation axis corresponds. The point velocity vi(t)=vi(te)+ai(te)∆t+1â„2Ji(tx)∆t2F i=1, 2, ..., n (5) According to the linear interpolation principle, the synthesizing speed and each interpolation axis Velocity has the following proportional relationship: vi(t)=Kiv(t)F i=1, 2, ..., n (6) For the above identity, there should be vi(tx)=Kiv(tx), ai(tx)=Kia (tx), Ji(tx)=KiJ(tx) (7) Since tx is an arbitrary point, this formula shows the velocity, acceleration, and Jerk of each interpolated axis during acceleration and deceleration in the segment, respectively, with the synthesized velocity, acceleration, and Jerk. Correspondence is proportional. When the synthesis speed is planned by S-curve, each interpolation axis will perform acceleration/deceleration according to the S-curve while ensuring space trajectory, that is, S-curve acceleration/deceleration may be used for front acceleration/deceleration control. At the same time, the above relationship can be used to check the velocity, acceleration, and Jerk limit values ​​of each interpolation axis. S Acceleration/deceleration interpolation recurrence formula The interpolation cycle is set as T, then the resultant displacement Sk at the end of the k-th interpolation cycle is Sk = ∫ tk v(t)dt = ∫ tk-1 v(t)dt+ ∫ tk-1+T v(t)dt=Sk-1 ∫ t (vk-1+ak-1t+1â„2Jt2)dt=Sk-1+vk-1T+1â„2ak-1T2+(1/6)JT3 0 0 tk -1 0 (8) The compositional displacement increment in the k-th interpolation period is ∆Sk=vk-1T+(1/2)ak-1T2+(1/6)JT3=vk-1T+(1/2)(ak -1+(1/3)JT)T2=vk-1+(1/2)akT2=(vk-1+(1/2)akT)=vkT (9) ak=ak-1+(1/3 JT (10) vk = vk-1 + (1/2) akT (11) Note that the above recurrence formula is partition-adaptive, ie J = { J, T ∈ [t0, t1] ∪ t6, t7 0, T∈(t1,t2)∪(t3,t4)∪(t5,t6) -J, t∈[t2,t3]∪[t4,t5] (12) As long as the initial conditions ak-1 and vk-1 are given , you can derive the combined displacement increments for each interpolation cycle. Then, the displacement increment of each interpolation axis during the interpolation cycle is obtained. The formula is ∆Pik=Pi ∆Sk=Ki∆Sk P. (13) When the acceleration/deceleration is performed within the determination segment of the interval, the servo motor speed per block is always required. Decrease to zero and then execute the next block. Therefore, the displacements of the acceleration and deceleration sections are equal, as shown in Fig. 1. The initial speed and initial acceleration of Zone 1 (t0-t1) is 0, then the displacement Pti at time t1=(1/6)Jt13, the acceleration a1=A=Jts, and the speed Vt1=(1/2)At12=( 1/2) Ats, then ts=t1=A/J (14) It can be seen from the acceleration graph in Fig. 1 that V=(1/2)Ats+Atl+(1/2)Ats=A(ts+tl (15) Then tl=(V/A)-(A/J) (16) ta=2ts+tl=(V/A)+(A/J) (17) Can be calculated from ts, tl, tm The time from t0 to t7 is used to determine the interval. The end point discriminated end point distance ∆S and each wheelbase end point distance ∆Si
Figure 1 S schematic diagram of S-curve acceleration and deceleration
Pi2) 1â„2 represents the composite displacement; TSEP represents the time required for each axis of the linear interpolation segment to reach the end point at the same time; F1 to F2 are the composite velocity corresponding to the sub-velocity of each interpolation axis. Let Ki = Pi i=1, 2, ..., n P (2) then Fi=KiF i=1, 2, ..., n (3) In the pre-acceleration and deceleration processing, plan the given speed, as shown in Figure 1. As shown, the entire acceleration and deceleration process is divided into three sections, namely the acceleration section (1, 2, 3 zones), the uniform speed section (4 zones) and the speed section (4, 5, 6 zones). In the acceleration and deceleration sections, they include variable acceleration and deceleration zones (1, 3, 5, and 7 zones) and constant acceleration and deceleration zones (2 and 6 zones): variable acceleration and deceleration zones, |da/dt|=J, Jerk For constant value; constant acceleration/deceleration zone, |a|=A, the acceleration is constant, and the speed of the uniform speed zone (4 zone) is a constant value Vc. The motion parameters of each axis are proportional to the planned synthesis velocity v(t) at the te-point power series so that ∆t=t-tx, with v(t)=v(tx)+a(tx)∆t +1â„2J(te)∆t2 (4) Similarly, each interpolation axis corresponds. The point velocity vi(t)=vi(te)+ai(te)∆t+1â„2Ji(tx)∆t2F i=1, 2, ..., n (5) According to the linear interpolation principle, the synthesizing speed and each interpolation axis Velocity has the following proportional relationship: vi(t)=Kiv(t)F i=1, 2, ..., n (6) For the above identity, there should be vi(tx)=Kiv(tx), ai(tx)=Kia (tx), Ji(tx)=KiJ(tx) (7) Since tx is an arbitrary point, this formula shows the velocity, acceleration, and Jerk of each interpolated axis during acceleration and deceleration in the segment, respectively, with the synthesized velocity, acceleration, and Jerk. Correspondence is proportional. When the synthesis speed is planned by S-curve, each interpolation axis will perform acceleration/deceleration according to the S-curve while ensuring space trajectory, that is, S-curve acceleration/deceleration may be used for front acceleration/deceleration control. At the same time, the above relationship can be used to check the velocity, acceleration, and Jerk limit values ​​of each interpolation axis. S Acceleration/deceleration interpolation recurrence formula The interpolation cycle is set as T, then the resultant displacement Sk at the end of the k-th interpolation cycle is Sk = ∫ tk v(t)dt = ∫ tk-1 v(t)dt+ ∫ tk-1+T v(t)dt=Sk-1 ∫ t (vk-1+ak-1t+1â„2Jt2)dt=Sk-1+vk-1T+1â„2ak-1T2+(1/6)JT3 0 0 tk -1 0 (8) The compositional displacement increment in the k-th interpolation period is ∆Sk=vk-1T+(1/2)ak-1T2+(1/6)JT3=vk-1T+(1/2)(ak -1+(1/3)JT)T2=vk-1+(1/2)akT2=(vk-1+(1/2)akT)=vkT (9) ak=ak-1+(1/3 JT (10) vk = vk-1 + (1/2) akT (11) Note that the above recurrence formula is partition-adaptive, ie J = { J, T ∈ [t0, t1] ∪ t6, t7 0, T∈(t1,t2)∪(t3,t4)∪(t5,t6) -J, t∈[t2,t3]∪[t4,t5] (12) As long as the initial conditions ak-1 and vk-1 are given , you can derive the combined displacement increments for each interpolation cycle. Then, the displacement increment of each interpolation axis during the interpolation cycle is obtained. The formula is ∆Pik=Pi ∆Sk=Ki∆Sk P. (13) When the acceleration/deceleration is performed within the determination segment of the interval, the servo motor speed per block is always required. Decrease to zero and then execute the next block. Therefore, the displacements of the acceleration and deceleration sections are equal, as shown in Fig. 1. The initial speed and initial acceleration of Zone 1 (t0-t1) is 0, then the displacement Pti at time t1=(1/6)Jt13, the acceleration a1=A=Jts, and the speed Vt1=(1/2)At12=( 1/2) Ats, then ts=t1=A/J (14) It can be seen from the acceleration graph in Fig. 1 that V=(1/2)Ats+Atl+(1/2)Ats=A(ts+tl (15) Then tl=(V/A)-(A/J) (16) ta=2ts+tl=(V/A)+(A/J) (17) Can be calculated from ts, tl, tm The time from t0 to t7 is used to determine the interval. The end point discriminated end point distance ∆S and each wheelbase end point distance ∆Si 
Figure 2 Interpolation calculation flow chart
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