US 11,188,056 B2 · Feedrate scheduling method for five-axis dual-spline curve interpolation

Abstract

This invention, a feedrate scheduling method for five-axis dual-spline curve interpolation, belongs to multi-axis NC (Numerical Control) machining filed, featured a feedrate scheduling method with constant speed at feedrate-sensitive regions under axial drive constraints for five-axis dual-spline interpolation. This method first discretizes the tool-tip spline with equal arc length, thus getting the relation between the axial motion and the toolpath by computing the first, second, and third order derivatives of the axial positions with respect to the tool-tip motion arc length. After that, determine the feedrate-sensitive regions with the constraints of axial drive limitations and the objective of balanced machining quality and efficiency. Finally, determine the acceleration/deceleration-start-point curve parameters by bi-directional scanning. The invented method can effectively make a balance between the feed motion stability and efficiency in five-axis machining, and possesses a high computational efficiency and a good real-time capability.

A feedrate scheduling method for five-axis dual-spline curve interpolation;

Its characteristic is that the relationship between the axial motion and tool-tip motion is obtained by computation of the first, second and third-order derivatives of the axial positions with respect to the arc-length parameter based on tool-tip motion spline equal-arc-length discretization, and the feedrate-sensitive regions are determined with axial drive constraints and the machining quality and efficiency balancing objective;
After that, determine the speed values of federate-sensitive regions and acceleration/deceleration-start-point curve parameters by bi-direction scanning;
Finally, the constant speed values are adopted within feedrate-sensitive regions, and S-shape acceleration/deceleration mode is used to schedule the smooth speed within non-sensitive regions;
The detail procedure of the method is as follows:
Step One:
Establish the relationship between the axial motion and the tool-pose trajectory;
Denote the tool-tip motion spline equation of the dual-NURBS curve to be interpolated as CP=CP(u), and the motion spline of second-point on the tool axis except the tool tip as CQ=CQ(u);
First, discretize the tool-tip motion spline with equal arc length;
Denote the arc-length step as δs, and the ith discrete point parameter after discretization as ud,i, thus by using the second-order Taylor expansion, the (i+1)th point parameter ud,i+1 can be computed as Eq (1):
u
d
,
i
+
1
=
u
d
,
i
+
1

C
P
′
⁡
(
u
d
,
i
)

⁢
δ
S
-
C
P
′
⁡
(
u
d
,
i
)
·
C
P
″
⁡
(
u
d
,
i
)
2
⁢

C
′
⁡
(
u
d
,
i
)

4
⁢
δ
S
2
(
1
)
where ∥ ∥ stands for the Euclidean norm, additionally, C′P(ud,i) and C″P(ud,i) are the first and second order derivatives of CP(u) with respect to u at ud,i;
Denote the tool tip and the tool-axis vector as R=[Rx, Ry, Rz]T and O=[Ox, Oy, Oz]T, respectively;
After obtaining of the disc e-point parameters ud,i, i=1, 2, . . . , Nd (Nd is the total amount of discrete points), compute the i-th tool tip Ri=[Rx,i, Ry,i, Rz,i]T and tool-orientation vector Oi=[Ox,i, Oy,i, Oz,i]T corresponding to i-th discrete point as:
{
R
i
=
C
P
⁡
(
u
d
,
i
)
O
i
=
C
Q
⁡
(
u
d
,
i
)
-
C
P
⁡
(
u
d
,
i
)

C
Q
⁡
(
u
d
,
i
)
-
C
P
⁡
(
u
d
,
i
)

(
2
)
Then, calculate the five joint-axis positions according to the obtained tool tip and tool orientation of Eq. (2) using inverse kinematics of five-axis machine tools, which are denoted in one vector q with five elements corresponding to the five physical axes of the machine tool, respectively;
Denote ith joint-axis motion position vector corresponding to i-th tool tip and i-th tool vector as qi, thus, the first-to-third-order derivatives of the physical axial positions with respect to the tool-tip trajectory arc-length parameter can be calculated as:
{
q
s
,
i
=
q
i
+
1
-
q
i
-
1
2
⁢
⁢
δ
s
q
ss
,
i
=
q
s
,
i
+
1
-
q
s
,
i
-
1
2
⁢
⁢
δ
s
q
sss
,
i
=
q
ss
,
i
+
1
-
q
ss
,
i
-
1
2
⁢
⁢
δ
s
(
3
)
where qs,i, qss,i, and qsss,i are first-order, second-order, and third-order derivatives of the physical axial positions with respect to the tool-tip trajectory arc-length parameter, respectively;
The Eq. (3) expresses the relationship between the axial motion and the tool-pose motion;
Step Two:
Determination of the feedrate-sensitive regions
According to the principle of differentiation, the relationship between axial and tool-tip motion velocity, acceleration, and jerk can be derived as:
{
q
.
=
q
s
⁢
s
.
q
¨
=
q
ss
⁢
s
.
2
+
q
s
⁢
s
¨
=
q
sss
⁢
s
.
3
+
3
⁢
q
ss
⁢
s
.
⁢
s
¨
+
q
ss
⁢
(
4
)
where {dot over (q)}, {umlaut over (q)}, and are first, second, and third order derivatives of axial motion position with respect to time, i.e. axial velocity, acceleration, and jerk vectors, respectively;
in addition, {dot over (s)}, {umlaut over (s)}, and are first, second, and third order derivatives of tool-tip motion position with respect to time, i.e. tool-tip velocity;
tangential acceleration, and jerk, respectively;
Denote the physical axial velocity constraint, acceleration constraint, and jerk constraint as {dot over (q)}max, {umlaut over (q)}max, and max, respectively, and the reference tool-tip motion speed constraint as vmax;
First, let the tool-tip motion acceleration constraint αt,max be the minimum value of the translational axes acceleration constraints in {umlaut over (q)}max, and let the tool-tip motion jerk constraint jt,max be the minimum value of the translational axes jerk constraints in max;
Then, aiming at balancing the machining efficiency and motion stability, optimize αt,max and jt,max so as to determine the feedrate-sensitive regions which are defined as the areas where the axial velocity, acceleration, and jerk can be exceeded when maximum tool-tip speed, acceleration, and jerk are utilized;
According to Eq. (4), it has
{
|
q
.
|
=
|
q
s
||
s
.
|
≤
|
q
s
|
v
max
|
q
¨
|
≤
|
q
ss
|
s
.
2
+
|
q
s
||
s
¨
|
≤
|
q
ss
|
v
max
2
+
|
q
s
|
a
t
,
max
|
≤
|
q
sss
||
s
.
⁢
|
3
⁢
+
3
|
q
ss
||
s
.
||
s
¨
|
+
|
q
ss
|
|
≤
|
q
sss
|
v
max
3
+
3
|
q
ss
|
v
max
⁢
a
t
,
max
+
|
q
ss
|
j
t
,
max
(
5
)
Hereby, the discretized position where one of the inequations |qs,i|vmax>qmax, |qss,i|vmax2+|qs,i|αt,max>{umlaut over (q)}max, and |qsss,i|vmax3+3|qss,i|vmaxαt,max+|qss,i|jt,max>max can be satisfied is determined as the point inside the feedrate-sensitive regions;
Denote the total number of the sensitive regions as Nsr, and the sequence number of the discretized points corresponding to the start and end position of kth sensitive region as ns,k and ne,k, respectively, thus the feedrate-sensitive regions can be expressed by discretized-point sequence number sets as {[ns,k, ne,k]}, k=1, 2, . . . , Nsr;
Hence, the total arc length lsr of the feedrate-sensitive regions on the tool-tip motion trajectory can be calculated by:
l
sr
=
δ
d
·
∑
k
=
1
N
sr
⁢
(
n
e
,
k
-
n
s
,
k
)
(
6
)
The total arc length of the tool-tip motion spline is denoted as lt;
If lsr<lt/2, indicating that most of the areas on the toolpath belong to feedrate-sensitive regions, where the maximum speed vmax can be adopted, therefore, the motion stability and efficient machining can be simultaneously ensured, and the scheduled federate-sensitive regions are satisfactory;
Otherwise, if lsr>lt/2, indicating that most areas on the toolpath belong to feedrate-non-sensitive regions, where the minimum allowable speed values within the corresponding regions should be used, thus the efficiency will be low although the motion stability is permissible;
Therefore, the αt,max and jt,max values are optimized in this case, so as to optimize the feedrate-sensitive regions;
The optimization procedure is to find tangential acceleration and jerk constraints between zero and original αt,max and jt,max by using bisection method, so as to make the total arc length of sensitive regions lsr happens to be half of the arc length of the whole toolpath lt, i.e. lsr=lt/2;
Finally, the parameter set of the obtained feedrate-sensitive regions is expressed as {[us,k, ue,k]}, k=1, 2, . . . , Nsr;
Step Three:
Calculation of the allowable speed of each feedrate-sensitive region
First, determine the initial constant speed of each feedrate-sensitive region according to axial drive constraints;
in order to make a balance between motion stability and efficiency, constant speed is scheduled inside each feedrate-sensitive region, and in this circumstance, the tangential acceleration {umlaut over (s)} and jerk are all zeros;
According to Eq. (4), the initial constant speed value {dot over (s)}o,k for kth feedrate-sensitive region can be derived as:
s
.
o
,
k
=
min
⁢
⁢
(
min
⁢
⁢
(
q
.
max
⁡
(
j
)
q
s
,
i
⁡
(
j
)
)
,
min
⁢
⁢
(
q
¨
max
⁡
(
j
)
q
ss
,
⁢
i
⁡
(
j
)
)
,
min
⁢
⁢
(
max
⁢
(
j
)
q
sss
,
i
⁡
(
j
)
3
)
)
,
i
∈
{
n
s
,
k
,
…
⁢
,
n
e
,
k
}
,
j
∈
{
1
,
…
⁢
,
5
}
(
7
)
where min( ) means the minimum function;
Then, update the speed values of feedrate-sensitive regions with tangential acceleration and jerk constraints in S-shape acceleration/deceleration mode;
When the tangential acceleration and jerk constraints are αt,max and jt,max, respectively, the required displacement sreq({dot over (s)}start, {dot over (s)}end) from starting velocity {dot over (s)}start to ending velocity {dot over (s)}end in S-shape acceleration/deceleration mode is calculated by:
s
req
⁡
(
v
start
,
v
end
)
=
{
s
.
start
⁡
(
t
1
+
t
2
+
t
3
)
+
1
2
⁢
j
t
,
max
⁢
t
1
2
⁡
(
t
2
+
t
3
)
+
1
2
⁢
a
max
⁡
(
t
2
2
+
t
3
2
)
+
a
max
⁢
t
2
⁢
t
3
,
v
start
<
v
end
s
.
start
⁡
(
t
1
+
t
2
+
t
3
)
-
1
2
⁢
j
t
,
max
⁢
t
1
2
⁡
(
t
2
+
t
3
)
-
1
2
⁢
a
max
⁡
(
t
2
2
+
t
3
2
)
-
a
max
⁢
t
2
⁢
t
3
,
v
start
>
v
end
(
8
)
where the maximum acceleration αmax=jt,maxt1, increase acceleration/deceleration time t1, constant acceleration/deceleration time t2 and decrease acceleration/deceleration time t3 are calculated as:
t
1
=
{
a
t
,
max
j
t
,
max
|
s
.
end
-
s
.
start
|
>
a
t
,
max
2
j
t
,
max
|
s
.
end
-
s
.
start
|
j
,
t
,
max
,
|
s
.
end
-
s
.
start
|
≤
a
t
,
max
2
j
t
,
max
(
9
)
t
2
=
{
|
s
.
end
-
s
.
start
|
-
a
t
,
max
2
⁢
/
⁢
j
t
,
max
a
t
,
max
,
|
s
.
end
-
s
.
start
|
>
a
t
,
max
2
j
t
,
max
0
,
|
s
.
end
-
s
.
start
|
≤
a
t
,
max
2
j
t
,
max
t
3
=
t
1
The feedrate-sensitive region speed value {dot over (s)}o,k is updated by bi-directional scanning, and the detail procedure is given below;
At first, backward scanning from k=Nsr to k=1 for scheduling of the deceleration procedure speed;
The procedure is:
1) Let k=Nsr;
2) When k=Nsr, determine whether ne,k=Nd is satisfied or not, and if ne,k=Nd, goto step 3);
else goto step 5);
On contrary, when k≠Nsr, goto step 6);
3) In this case, the ending region of the toolpath belongs to the feedrate-sensitive region;
Calculate the required displacement sreq({dot over (s)}o,k, 0) of deceleration process from {dot over (s)}o,k to 0 using Eq. (8), and the arc length of this sensitive region which is denoted as lk:
lk=δd·(ne,k−ns,k) (10)
If sreq({dot over (s)}o,k, 0)≤lk/2, goto step 4);
else, find a speed value {dot over (s)}B,k to meet sreq({dot over (s)}B,k, 0)=lk/2 using dichotomy method between {dot over (s)}o,k and 0;
Update {dot over (s)}o,k by letting {dot over (s)}o,k={dot over (s)}B,k, and recalculate sreq({dot over (s)}o,k, 0) then goto step 4);
4) When the ending region of the toolpath belongs to the feedrate-sensitive region, the final deceleration process must be executed within the sensitive region;
Therefore, it is necessary to determine whether the axial drive constraints are exceeded when performing deceleration within the last sensitive region;
To this end, the number of discretized points of the final deceleration, denoted as nf, is first computed by
n
f
=
int
⁢
⁢
(
s
req
⁡
(
s
.
o
,
k
,
0
)
δ
d
)
(
11
)
where int( ) means round down;
According to the relationship between displacement and tangential velocity, acceleration and jerk during S-shape acceleration/deceleration process, reversely solve the discretized-point speed {dot over (s)}f,j, tangential acceleration {umlaut over (s)}f,j, and tangential jerk (j∈{nd−nf, . . . , nd}) within the deceleration process of Nsr-th federate-sensitive region;
Then solve the request axial speed {dot over (q)}f,j, the axial acceleration {umlaut over (q)}f,j, and the axial jerk f,j (j∈{nd−nf, . . . , nd}) within the deceleration process of Nsr-th federate-sensitive region by substituting {dot over (s)}f,j, {umlaut over (s)}f,j, and to Eq. (4);
Finally, calculate the minimum ratio Kf between the axial drive limitations and the request axial drive parameters in the final deceleration process by
K
f
=
min
⁢
⁢
(
min
⁢
⁢
(
|
q
.
max
⁡
(
i
)
q
.
f
,
j
⁡
(
i
)
|
)
,
min
⁢
⁢
(
|
q
¨
max
⁡
(
i
)
q
¨
f
,
j
⁡
(
i
)
|
)
,
min
⁢
⁢
(
|
f
,
j
⁢
(
i
)
⁢
|
3
)
)
,
i
∈
{
1
,
…
⁢
,
5
}
,
j
∈
{
n
s
,
k
,
…
⁢
,
n
e
,
k
}
(
12
)
By comparing Kf and 1, determine a tuning ratio for the tangential speed, acceleration, and jerk in the deceleration process of Nsr-th federate-sensitive region, denoted as Kf,t, by the following equation:
Kf,t=min(Kf,1) (13)
Let the tangential acceleration constraint of the deceleration process of Nsr-th federate-sensitive region αf,max=Kf,t2·αt,max, and the tangential jerk constraint of the deceleration process of Nsr-th federate-sensitive region jf,max=Kf,t3·jt,max;
Then update the allowable speed value corresponding to Nsr-th feedrate-sensitive region by letting {dot over (s)}o,k=Kf,t·{dot over (s)}o,k;
5) When k=Nsr and ne,k≠Nd, which indicating that there exists feedrate-non-sensitive region between the final point on the toolpath and the last sensitive region, schedule a deceleration process within this non-sensitive region;
Calculate the required displacement sreq({dot over (s)}o,k, 0) of deceleration process from {dot over (s)}o,k to 0 using Eq. (8), and the arc length lf of this final non-sensitive region by
lf=δd·(Nd−ne,k) (14)
If sreq({dot over (s)}o,k, 0)>lr,k, find a speed value {dot over (s)}B,k to meet sreq({dot over (s)}B,k, 0)=lr,k using dichotomy method between 0 and {dot over (s)}o,k;
Then, update the allowable speed value {dot over (s)}o,k of this feedrate-sensitive region by letting {dot over (s)}o,k={dot over (s)}B,k and goto step 7);
6) Evaluate whether it is a deceleration procedure;
If {dot over (s)}o,k≤{dot over (s)}o,k+1 which indicates it is an acceleration process, goto step 7);
else, calculate the required displacement sreq({dot over (s)}o,k, {dot over (s)}o,k+1) of deceleration process from {dot over (s)}o,k to, and the arc length of the kth feedrate-non-sensitive region lr,k:
lr,k=δd·(ns,k+1−ne,k) (15)
When sreq({dot over (s)}o,k, {dot over (s)}o,k+1)>lr,k+1, find a speed value {dot over (s)}B,k to meet sreq({dot over (s)}o,k, {dot over (s)}o,k+1)=lr,k+1 using dichotomy method between 0 and {dot over (s)}0,k;
Then, update the allowable feedrate value {dot over (s)}o,k of this sensitive region by letting {dot over (s)}o,k={dot over (s)}B,k and goto step 7);
7) if k=1, end the backward scanning;
else, let k=k−1 and goto step 2);
Then, forward scanning from k=1 to k=Nsr for scheduling of the feedrate-sensitive region speed values of the acceleration procedure;
1) Let k=1;
2) When k=1, determine whether ns,k=1 is satisfied;
If it is satisfied, goto step 3);
else goto step 5);
On contrary;
when k≠1, goto step 6);
3) in this case, the starting region of the toolpath belongs to the feedrate-sensitive region;
Calculate the required displacement sreq(0, {dot over (s)}o,k) of the acceleration process from 0 to {dot over (s)}o,k using Eq. (8), and the arc length of this sensitive region which is denoted as lk using Eq. (10);
If sreq(0, {dot over (s)}o,k)≤lk/2, goto step 4);
else, find a speed value {dot over (s)}F,k to meet sreq({dot over (s)}F,k, 0)=lk/2, using dichotomy method between 0 and {dot over (s)}o,k;
Then, update the allowable speed value {dot over (s)}o,k of this feedrate-sensitive region by letting {dot over (s)}o,k={dot over (s)}F,k, and recalculate sreq(0, {dot over (s)}o,k) after that, goto step 4);
4) When the starting region of the toolpath belongs to the feedrate-sensitive region, the initial acceleration process must be executed within the sensitive region;
Therefore, it is necessary to determine whether the axial drive constraints are exceeded when performing acceleration within the first sensitive region;
Calculate the number of discretized points of the initial acceleration process, denoted as nb, by
n
b
=
int
⁢
⁢
(
s
req
⁡
(
0
,
s
.
o
,
k
)
δ
d
)
(
16
)
According to the relationship between displacement and tangential speed, acceleration, and jerk of the S-shape acceleration/deceleration mode, reversely solve the tool-tip speed {dot over (s)}b,j, the tangential acceleration {umlaut over (s)}b,j, and the tangential jerk (j∈{1, . . . , nb}) within the acceleration process in the first feedrate-sensitive region;
Then solve the request axial speed {dot over (q)}b,j, the axial acceleration {umlaut over (q)}b,j, and the axial jerk (j∈{1, . . . , nb}) at each discretized point within the acceleration process of the first feedrate-sensitive region by substituting {dot over (s)}b,j, {umlaut over (s)}b,j, and to Eq. (4);
Finally, calculate the minimum ratio Kb between the axial drive limitations and the request axial drive parameters by
K
b
=
min
⁢
⁢
(
min
⁢
⁢
(
|
q
.
max
⁡
(
i
)
q
.
b
,
j
⁡
(
i
)
|
)
,
min
⁢
⁢
(
|
q
¨
max
⁡
(
i
)
q
¨
b
,
j
⁡
(
i
)
|
)
,
min
⁢
⁢
(
|
b
,
j
⁢
(
i
)
⁢
|
3
)
)
,
i
∈
{
1
,
…
⁢
,
5
}
,
j
∈
{
1
,
…
⁢
,
n
b
}
(
17
)
By comparing Kb and 1, determine a tuning ratio for the tangential speed, acceleration, and jerk in the acceleration process of the first federate-sensitive region, denoted as Kb,t by the following equation:
Kb,t=min(Kb,1) (18)
Let the tangential acceleration constraint of the acceleration process in the first federate-sensitive region αb,max=Kb,t2·αb,max, and the tangential jerk constraint of the deceleration process in the first federate-sensitive region jb,max=Kb,t3·jt,max;
Then update the allowable speed value corresponding to the first feedrate-sensitive region by letting {dot over (s)}o,k=Kb,t·{dot over (s)}o,k;
5) When k=1 and ns,k≠1 which indicating that there exists feedrate-non-sensitive region between the start point on the toolpath and the first sensitive region, schedule an acceleration process within this non-sensitive region;
Calculate the required displacement sreq(0, {dot over (s)}o,k) of the acceleration process from 0 to {dot over (s)}o,k using Eq. (8), and the arc length lr,k of this first non-sensitive region by Eq. (15);
If sreq(0, {dot over (s)}o,k)>lr,k, find a speed value {dot over (s)}F, k to meet sreq(0, {dot over (s)}o,k)=lr,k using dichotomy method between 0 and {dot over (s)}o,k;
Then, update the allowable speed value {dot over (s)}o,k of this feedrate-sensitive region by letting {dot over (s)}o,k={dot over (s)}F,k and goto step 7);
6) Evaluate whether it is an acceleration procedure;
If {dot over (s)}o,k≥{dot over (s)}o,k+1 is satisfied, indicating that it is an deceleration procedure, goto step 7);
else, calculate the required displacement sreq({dot over (s)}o,k,{dot over (s)}o,k+1) of the acceleration process from {dot over (s)}o,k to {dot over (s)}o,k+1, and the arc length lr,k of the k-th non-sensitive region;
If sreq({dot over (s)}o,k,{dot over (s)}o,k+1)>lr,k, find a speed value {dot over (s)}F,k to meet sreq({dot over (s)}o,k,{dot over (s)}o,k+1)=lr,k using dichotomy method between 0 and {dot over (s)}o,k;
Then, update the allowable speed value {dot over (s)}o,k of this feedrate-sensitive region by letting {dot over (s)}o,k={dot over (s)}F,k and goto step 7);
7) if k=Nsr−1, goto step 8);
Else, let k=k+1 and goto step 2);
8) When ne,k=Nd and Kf,t<1, judge whether the allowable speed {dot over (s)}o,k corresponding to the Nsr-th feedrate-sensitive region is updated by acceleration-procedure speed scheduling or not;
If yes, calculate the ratio Kbαbetween the speed value after and before updating, and let Kf,t=Kbα·Kf,t;
Then update the tangential acceleration constraint of the deceleration process in Nsr-th feedrate-sensitive region by αf,max=Kf,t2·αt,max, and the tangential jerk constraint of the deceleration process in Nsr-th feedrate-sensitive region by jf,max=Kf,t3·jt,max;
Thus, end the bi-directional scanning;
After above backward scanning speed scheduling for deceleration procedure and forward scanning speed scheduling for acceleration procedure, the final obtained {dot over (s)}o,k is thus the final axial-drive-constraint limited speed value corresponding to k-th feedrate-sensitive region;
Step Four:
Calculation of the acceleration/deceleration-start-point parameters and their corresponding speed values
First, determine whether the arc length lr,k of the non-sensitive region is long enough for acceleration from the allowable speed {dot over (s)}o,k of the k-th sensitive region to the programmed maximum tool-tip motion speed vmax and then deceleration to the allowable speed {dot over (s)}o,k+1 of the (k+1)-th sensitive region;
Videlicet judge if the inequation (19) is satisfied;
lr,k>sreq({dot over (s)}o,k,vmax)+sreq(vmax,{dot over (s)}o,k+1) (19)
Once Eq. (19) can be satisfied, perform acceleration from {dot over (s)}o,k to vmax and deceleration from vmax to {dot over (s)}o,k+1 processes between the kth and (k+1)th sensitive regions in sequence;
The acceleration-start-point parameter is ue,k, and the corresponding feedrate is {dot over (s)}o,k;
In addition, the deceleration-start-point parameter udc is computed as
{
n
dp
=
int
⁢
⁢
(
s
req
⁡
(
v
max
,
s
.
o
,
k
+
1
)
δ
d
)
n
dc
=
n
s
,
k
+
1
-
n
dp
l
dc
=
s
req
⁡
(
v
max
,
s
.
o
,
k
+
1
)
-
δ
s
·
n
dp
u
dc
=
u
d
,
n
dc
-
1

C
P
′
⁡
(
u
d
,
n
dc
)

⁢
l
dc
-
C
p
′
⁡
(
u
d
,
n
dc
)
·
C
P
″
⁡
(
u
d
,
n
dc
)
2
⁢

C
′
⁡
(
u
d
,
n
dc
)

4
⁢
l
dc
2
(
20
)
The feedrate corresponding to the deceleration-start point is vmax;
On contrary, if Eq. (19) cannot be satisfied, perform acceleration or deceleration merely between the kth and (k+1)th sensitive regions in order to keep a smooth and stable tool-tip motion;
If {dot over (s)}o,k<{dot over (s)}o,k+1, execute the acceleration process;
The acceleration-start-point parameter is ue,k and the corresponding feedrate is {dot over (s)}o,k;
If {dot over (s)}o,k>{dot over (s)}o,k+1, execute the deceleration process;
The deceleration-start-point parameter udc is calculated using the following equation;
{
n
dp
=
int
⁢
⁢
(
s
req
⁡
(
s
.
o
,
k
,
s
.
o
,
k
+
1
)
δ
d
)
n
dc
=
n
s
,
k
+
1
-
n
dp
l
dc
=
s
req
⁡
(
s
.
o
,
k
,
s
.
o
,
k
+
1
)
-
δ
s
·
n
dp
u
dc
=
u
d
,
n
dc
-
1

C
P
′
⁡
(
u
d
,
n
dc
)

⁢
l
dc
-
C
p
′
⁡
(
u
d
,
n
dc
)
·
C
P
″
⁡
(
u
d
,
n
dc
)
2
⁢

C
′
⁡
(
u
d
,
n
dc
)

4
⁢
l
dc
2
(
21
)
The feedrate corresponding to this deceleration-start point is {dot over (s)}o,k
After above computation, denote the total amount of the obtained acceleration/deceleration-start points as n and the set consisted by the acceleration/deceleration-start-point parameters and the corresponding feedrates as {uri, vui}, i=1, 2, . . . , n, where uri is the i-th acceleration/deceleration-start-point parameter, and vui is the i-th acceleration/deceleration-start-point speed;
Step Five:
Real-time calculation of interpolation-point parameter and feedrate
In real-time interpolation, judge which one of the parameter range [uri, uri+1] that the current interpolation-point parameter uj lays in, and then calculate t1, t2, and t3:
t
1
=
{
a
t
,
m
j
t
,
m
|
vu
i
+
1
-
vu
i
|
>
a
t
,
m
2
j
t
,
m
|
vu
i
+
1
-
vu
i
|
j
,
t
,
m
,
|
vu
i
+
1
-
vu
i
|
≤
a
t
,
m
2
j
t
,
m
(
22
)
t
2
=
{
|
vu
i
+
1
-
vu
i
|
-
a
t
,
m
2
⁢
/
⁢
j
t
,
m
a
t
,
m
,
|
vu
i
+
1
-
vu
i
|
>
a
t
,
m
2
j
t
,
m
0
,
|
vu
i
+
1
-
vu
i
|
≤
a
t
,
m
2
j
t
,
m
t
3
=
t
1
where when i=1, αt,m=αb,max and jt,m=jb,max;
when i=n, αt,m=αf,max and jt,m=jf,max;
otherwise, αt,m=αt,max and jt,m=jt,max;
When the elapsed time after the interpolation-point parameter passes into the current range t is smaller than t1+t2+t3, the feedrate vj for current interpolation point is:
v
j
=
{
vu
i
+
1
2
⁢
j
t
,
m
⁢
t
2
,
0
≤
t
<
t
1
v
1
+
a
m
⁡
(
t
-
t
1
)
,
t
1
≤
t
<
t
1
+
t
2
v
1
=
vu
i
+
1
2
⁢
j
t
,
m
⁢
t
1
2
,
v
2
+
a
m
⁡
(
t
-
t
1
-
t
2
)
-
v
2
=
v
1
+
a
m
⁢
t
2
,
t
1
+
1
2
⁢
j
t
,
m
⁡
(
t
-
t
1
-
t
2
)
2
,
t
2
≤
t
≤
t
1
+
t
2
+
t
3
(
23
)
where αm=jf,max·t1;
Otherwise, when t≥t1+t2+t3, the current point feedrate vj=vui+1;
Calculate the next interpolation-point parameter uj+1 according to the second-order Taylor's expansion method as:
u
j
+
1
=
u
j
+
v
j

C
P
′
⁡
(
u
j
)

⁢
T
-
v
j
2
⁡
(
C
P
′
⁡
(
u
j
)
,
C
P
″
⁡
(
u
j
)
)
2
⁢

C
P
′
⁡
(
u
j
)

4
⁢
T
2
(
24
)
where T is the interpolation period;
Judge whether the ending point of the curve is reached, and if yes, end the interpolation;
else, let j=j+1 and back to the beginning of Step Five;
Hence, the feedrate scheduling with axial drive constraints for five-axis dual-spline interpolation can be realized five-axis machining based on the feedrate scheduling.

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Feedrate scheduling method for five-axis dual-spline curve interpolation

Life of this patent

Feedrate scheduling method for five-axis dual-spline curve interpolation

Life of this patent

Abstract

Claims1