CN102354151A - Tangential following interpolation method applied to multilayer shoe leather numerical control cutting machine tool - Google Patents

Abstract

The invention provides a tangential following interpolation method applied to a multilayer shoe leather numerical control cutting machine tool. A multilayer shoe leather high speed cutting processing process comprises high frequency vibration of a sheet shape cutter, plane motion of a cutter, and rotation of the sheet shape cutter in a processing process. A high frequency vibration direction is defined as a Z direction, and directions of the plane motion are defined as X and Y directions. In a process of cutting a straight line, according to a feeding speed F and an equivalent interpolation period T0, wherein, T0=nT, and T represents a unit interpolation period, a contour step l of each interpolation period is calculated, and a relation of F, T0 and l is l=FT0. Since a Z axis angle of linear tangential following interpolation is not changed, a linear tangential following interpolation formula is obtained after transformation. The tangential following interpolation method applied to the multilayer shoe leather numerical control cutting machine tool has the characteristics of high cutting precision and high work efficiency.

Description

A kind of tangential following interpolating method that lathe is cut applied to multilayer shoe leather numerical control

Technical field

The present invention relates to the control method that lathe is cut applied to multilayer shoe leather numerical control.

Background technology

Numerical control cuts technology and includes state-of-the-art mechanical technique, computer and the information processing technology, systems technology, automatic control technology, sensing and detection technique, servo drive technology.With the development of Numeric Control Technology, numerical control cuts technology and progressively developed towards open, intelligent, high speed direction.External leather cuts system and has developed into integrated and intelligentized CNC integrated system, still, and external advanced numerical control leather Cutting machine is although powerful, but expensive, and takes China technical monopoly and closing policy.

For the technical method, related studies in China mechanism has also carried out certain exploration and research.Shanghai University Zhang Jianhong completes the system design to numerical control clothing tailoring machine using mechatronics integration method；University Of Ningbo's Lee's national wealth etc. have studied the Control System Design of numerical control clothing tailoring machine.The Ding Wenjie of Xi'an Communications University has inquired into the Digital Program Control technology secondary development based on lattice cypress cutting.Shen Junjia employs knife point optimization method under the Cutter based on adaptive M MAS ant group algorithms and lower knife point is optimized, the profile smooth corners transition tailoring technique based on Bezier is employed the excessive turning of curve is handled.The Zhu Nianjun of Zhejiang University has inquired into the cutter Control System Design for leather.The actual complex stressing conditions of Cutter when Shanghai University Zhao Yi peaks analyze cutter work, detail discussion has been carried out to Cutter correction problem, and using the method for fuzzy control, processing cutter is under different stressing conditions the problem of Cutter deflection angle.The great project for bidding of Zhejiang University's Chen Zi occasion and Wang Wen combinations Zhejiang Province " the quasi-flexible manufacturing technology exploitation (2003C11023) of leather and fur products; develop a set of leather cutting process automation system, and applied for national inventing patent function integrated numerically controlled automatic leather cutting method (200610155436.3)

Country's correlative technology field is mainly studied tool-path planning and speed control at present, wherein being mostly to be directed to individual layer leather to cut processing.Because individual layer leather is cut using technologies such as laser, high speed water ions, " cut constantly " for multilayer shoe leather, therefore production efficiency is extremely low.Cutting die blanking techniques are the more multilayer shoe leather process technologies of current domestic application, but are due to that automaticity is low, deposit cutting die space greatly and with certain risk, therefore are gradually eliminated by modern enterprise.Under the conditions of such technical background, multilayer shoe leather numerical control Cutting machine arises at the historic moment.Multilayer shoe leather cuts control technology and uses the sheet cutter of dither to be cut in real time to multilayer shoe leather according to planned trajectory.Due to the laminated structure of dither cutter, therefore it is essential to ensure that the state that the sheet cutter during cutting mutually is cut with cutting path to keep in real time.Otherwise two kinds of situations are directly resulted in：One is that can complete to cut in process, because the Cutter direction of motion and edge direction are inconsistent, but angle is again when being not very big, causes to cut precision and efficiency is substantially reduced,；Two be to work as the Cutter direction of motion to reach certain value with cutting edge angle, or even when vertical, it is impossible to completion cuts process, cuts cutter and fractures or twist off.

The content of the invention

Cut that precision is relatively low, ineffective deficiency applied to what multilayer shoe leather numerical control cut lathe tool-path planning technology to overcome, the present invention provides a kind of tangential following interpolating method that lathe is cut applied to multilayer shoe leather numerical control for cutting high precision, high working efficiency.

The technical solution adopted for the present invention to solve the technical problems is：

A kind of tangential following interpolating method that lathe is cut applied to multilayer shoe leather numerical control, cut process includes the rotation during the dither of sheet cutter, the plane motion for cutting cutter and sheet tool sharpening to the multilayer shoe leather at a high speed, it is Z-direction to define dither direction, and it is X, Y-direction to define plane motion direction；

During straight line is cut, Z axis is independent axes and not linked with XY axles, it is known that the starting point coordinate of straight line is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), then the initial angle angle value of straightway is expressed as (unit for °)：

Carry out drift angle amendment and obtain final cutter start angle for θ '_s=θ_s-α。

Feed speed F and equivalent interpolation cycle T in Machining Instruction₀, T₀=nT, T represent unit interpolation cycle, calculate the profile step-length l of each interpolation cycle, then the relation of three is expressed as：L=FT₀, because rectilinear tangential follows the Z axis angle of interpolation constant, so there is rectilinear tangential to follow interpolation formula as follows：

$\left\{\begin{array}{c}\mathrm{\Δx}=l\cos a=l\frac{x_e-x_s}{\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2}}\\ \mathrm{\Δy}=\mathrm{\Δ}x\tan a=\frac{\mathrm{\Δx}(y_e-y_s)}{x_e-x_s}\\ \mathrm{\Δc}=0\end{array}\right.---1-2$

For every section of straight line, c is constant, is obtained by x and y relation, and obtaining rectilinear tangential by conversion follows interpolation formula：

$\left\{\begin{array}{c}{x}_{i+1}=x_i+\mathrm{\Δx}\\ y_{i+1}=y_i+\mathrm{\Δy}\\ c_{i+1}={\mathrm{\θ}}_s^{\′}\end{array}\right.---1-3.$

Further, if known cut a section circular arcFor inverse circular arc, circular arc starting point is P_s(x_s, y_s, c_s), terminal is P_e(x_e, y_e, c_e), center of circle O is the origin of coordinates, then the Z coordinate value angle positive equivalent to the line and X-axis in current point and the center of circle, then the parametric equation for being processed curve is represented by：

$\left\{\begin{array}{c}x=R\cos \mathrm{\φ}\\ y=R\sin \mathrm{\φ}\\ c=\mathrm{\φ}+\mathrm{\π}/2\end{array}\right.---1-4$

Wherein R is arc radius, and φ is the angle of current point and line and the X-axis forward direction in the center of circle.

During circular arc is cut, Z axis, which will keep linking with XY axles, make it that the blade moment is consistent with circular arc secant direction, first has to obtain the corner of cutter starting point according to circular arc starting point, it is known that the starting point coordinate of arc section is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), central coordinate of circle is p_o(x_o, y_o), initial angle angle value is expressed as：

When inverse circular arc is cut,

When being cut along circular arc,

If P_i(x_i, y_i, c_i) it is cutter current location point, interpolation, which is calculated, to be required to reach the next position P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1), P_iPoint coordinate be：

$\left\{\begin{array}{c}{x}_i=R\cos {\mathrm{\φ}}_i\\ y_i=R\sin {\mathrm{\φ}}_i\\ c_i={\mathrm{\φ}}_i+\mathrm{\π}/2\end{array}\right.---1-5$

Once equivalent interpolation cycle T during circular arc tangential following interpolation₀Determine that its interpolation step angle α is with feeding linear velocity V：

$\mathrm{\&alpha;}=\frac{\mathrm{\&omega;T}}{n}=\mathrm{\&omega;}{\mathrm{<figure-callout\; id="0"\; label="T"\; filenames="HDA0000081116480000042.png,HDA0000081116480000061.png"\; state="\{\{state\}\}">T</figure-callout>}}_0=\frac{V}{R}{T}_0---1-6$

If in Interpolation Process, the coordinate P of current point_i(x_i, y_i, c_i), it is known that in order to obtain the feeding increment on two coordinate directions, obtaining next interpolated point P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1)：

$\left\{\begin{array}{c}{x}_{i+1}=R\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ y_{i+1}=R\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ c_{i+1}={\mathrm{\&phi;}}_i+\mathrm{\&alpha;}\end{array}\right.---1-7$

Cutter is along X, Y, C amount of feeding Δ x_i+1, Δ y_i+1, Δ c_i+1For：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=R[\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\cos {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;y}}_{i+1}=R[\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\sin {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;c}}_{i+1}=\mathrm{\&alpha;}\end{array}\right.---1-8$

Formula 1-3 abbreviations are obtained：

$\left\{\begin{array}{c}{x}_{i+1}=x_i\cos \mathrm{\&alpha;}-y_i\sin \mathrm{\&alpha;}\\ y_{i+1}=y_i\cos \mathrm{\&alpha;}+x_i\sin \mathrm{\&alpha;}\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-9$

Tan (α/2) ≈ α/2 are set, thus are had：

$\cos \mathrm{\&alpha;}=\frac{1-{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}{1+{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}=\frac{4-{\mathrm{\&alpha;}}^2}{4+{\mathrm{\&alpha;}}^2}---1-10$

Similarly have：

$\sin \mathrm{\&alpha;}=\frac{4\mathrm{\&alpha;}}{4+{\mathrm{\&alpha;}}^2}---1-11$

Make A_i=cos α, B_i=sin α, then data sampling method tangential following interpolation formula be：

$\left\{\begin{array}{c}{x}_{i+1}=A_i{x}_i-B_i{y}_i\\ y_{i+1}=A_i{y}_i+B_i{x}_i\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-12$

The feeding increment of each axle is：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=x_{i+1}-x_i\\ {\mathrm{\&Delta;y}}_{i+1}=y_{i+1}-y_i\\ {\mathrm{\&Delta;c}}_{i+1}=c_{i+1}-c_i\end{array}\right.---1-13.$

The present invention technical concept be：During being cut applied to multilayer shoe leather numerical control, keep sheet Cutter in real time and cut the interpolating method of the tangent state in path, including：For cutting for straight line path, Cutter is in real time with cutting that path is tangent, therefore the interpolation operation of tangential following motion need not be carried out.But, may different be that cutting gross thickness are different due to being cut the leather number of plies, cutting force gradually increases as cutting thickness increases；Even if same cutting thickness, cutting force gradually increases with the increase of cutting speed.Therefore to ensure that cutting force is equal during lathe work, and then ensure that machine tool motion is relatively steady, with the increase of the number of plies, correspondingly shorten interpolation cycle, interpolation rate is improved, has reached that lathe still can improve production efficiency because the cutting hides number of plies is different with smooth working；For the angle that cutting path is straight line formation, then using " griffing " mode that is processed of+rotation angle+" roll setting " mode；

To transition arc, into circular arc before using pretreatment reduction of speed mode；Depart from and accelerated before circular arc using pretreatment Mode.And accelerating degree to be adjusted according to the number of plies, the deceleration of pre-treating speed linearly reduces with the increase of the number of plies, to improve the stability of lathe.The difference between initial angle and end angle is calculated according to two circular arc end points, central coordinate of circle simultaneously, the synchronism moved according to feed motion and tangential following calculates interpolation rate and derives interpolating method；To starting circular arc, into circular arc after using pretreatment accelerated mode, depart from circular arc before using pretreatment accelerated mode, acceleration magnitude also with the linear inverse relation of the number of plies.Initial angle and end angle difference are calculated according to two circular arc end points, central coordinate of circle simultaneously, the synchronism moved according to feed motion and tangential following calculates interpolation rate and derives interpolating method；To terminate circular arc, into circular arc before using pretreatment ways of deceleration, depart from circular arc before using pretreatment reduction of speed mode, acceleration magnitude also with the linear inverse relation of the number of plies.Initial angle and end angle difference are calculated according to two circular arc end points, central coordinate of circle simultaneously, the synchronism moved according to feed motion and tangential following calculates interpolation rate and derives interpolating method.

For small arc-shaped, introduce lathe minimum process circular arc and define as " critical circular arc ", and there is processing " critical circular arc " to replace the circular arc smaller than it；For super large circular arc, " with straight Dai Qu, repeatedly griffing " mode due to its curvature very little, under the precondition for meeting machining accuracy, is used to carry out cutting processing.

Beneficial effects of the present invention are mainly manifested in：Be directed to numerical control shoe leather Cutting machine at a high speed cut during, lathe while feed motion, can make dither cutter in real time with feed motion rail keep mutually cut state --- tangential following interpolation technique.By this interpolating method, technical staff need to only cut intensity, footwear leather according to footwear leather and cut the number of plies and processing technology requirement, input and adjust correspondingly technical data, it is possible to complete multilayer shoe leather high-efficiency high-accuracy and cut work.In addition, this tangential following method can also extend to and process the apparel industries such as cloth equally to cut mode and be promoted.

Brief description of the drawings

Fig. 1 is the schematic diagram of feed motion (X, Y) and tangential following motion (Z rotations).

Fig. 2 is tangential following linear interpolation flow chart.

Fig. 3 is tangential following interpolation flow chart.

Fig. 4 is tangential following linear interpolation schematic diagram.

Fig. 5 is the tangential schematic diagram of circular arc.

Fig. 6 is all quadrants Z axis angle schematic diagram in inverse circular arc.

Fig. 7 is along all quadrants Z axis angle schematic diagram in circular arc.

Fig. 8 is no tangential following interpolation technique design sketch.

Fig. 9 is that have tangential following interpolation technique design sketch.

Embodiment

The invention will be further described below in conjunction with the accompanying drawings.

Reference picture 1, Fig. 1 is the schematic diagram that feed motion and tangential following are moved, according to the figure：Multilayer shoe leather is cut in process at a high speed, there is three movement：The dither (Z-direction) of main motion --- sheet cutter, complete multilayer shoe leather cuts motion, realizes that cutter is cut to multilayer shoe leather；Feed motion --- the plane motion (X, Y-direction) of cutter is cut, completion cuts motion of the cutter along CAD stock layouts track；Synkinesia --- the rotation (Z during sheet tool sharpening

Direction), i.e., tangential following is moved, and is completed sheet cutter during multilayer shoe leather is cut and is remained consistent with cutting track tangential direction, realizes the cutter compensation that cutter is cut to multilayer shoe leather.

Fig. 2 is multilayer shoe leather numerical control Cutting machine tangential following beeline interpolation algorithm flow chart.

During straight line is cut, because cutter blade direction keeps constant, Z axis is independent axes and not linked with XY axles.Z axis need to only rotate to straight line direction vector before taking the air line just can be with, and straight line unit number of plies interpolation cycle is multiplied by into a factor of n (n represents the leather number of plies).Therefore, the key of the tangential following control of straightway is to obtain the corner of cutter starting point.The starting point coordinate of known straight line is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), then the initial angle angle value of straightway is represented by (unit for °)：

There is alignment error in view of cutter, Cutter may be equipped with a drift angle α with returning knife raw bits, in this regard, can carry out drift angle amendment obtains final cutter start angle for θ '_s=θ_s-α。

Feed speed F and equivalent interpolation cycle T in Machining Instruction₀(T₀=nT, T represent unit interpolation cycle), the profile step-length l of each interpolation cycle is calculated, then the relation of three is represented by：L=FT₀.Because rectilinear tangential follows the Z axis angle of interpolation constant, so there is rectilinear tangential to follow interpolation formula as follows：

$\left\{\begin{array}{c}\mathrm{\&Delta;x}=l\cos a=l\frac{x_e-x_s}{\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2}}\\ \mathrm{\&Delta;y}=\mathrm{\&Delta;}x\tan a=\frac{\mathrm{\&Delta;x}(y_e-y_s)}{x_e-x_s}\\ \mathrm{\&Delta;c}=0\end{array}\right.---1-2$

For every section of straight line, c is constant, can be obtained by x and y relation.Rectilinear tangential, which can be obtained, by conversion follows interpolation formula：

$\left\{\begin{array}{c}{x}_{i+1}=x_i+\mathrm{\&Delta;x}\\ y_{i+1}=y_i+\mathrm{\&Delta;y}\\ c_{i+1}={\mathrm{\&theta;}}_s^{\&prime;}\end{array}\right.---1-3$

Fig. 6 is multilayer shoe leather numerical control Cutting machine tangential following arc interpolation flow chart.

If known procedure section circular arc

For inverse circular arc, circular arc starting point is P_s(x_s, y_s, c_s), terminal is P_e(x_e, y_e, c_e), center of circle O is the origin of coordinates, then understands the Z coordinate value angle positive equivalent to the line and X-axis in current point and the center of circle by accompanying drawing 5.The parametric equation for being then processed curve is represented by：

$\left\{\begin{array}{c}x=R\cos \mathrm{\&phi;}\\ y=R\sin \mathrm{\&phi;}\\ c=\mathrm{\&phi;}+\mathrm{\&pi;}/2\end{array}\right.---1-4$

Wherein R is arc radius, and φ is the angle of current point and line and the X-axis forward direction in the center of circle.

During circular arc is cut, Z axis, which will keep linking with XY axles, make it that the blade moment is consistent with circular arc secant direction.First have to obtain the corner of cutter starting point according to circular arc starting point.The starting point coordinate of known arc section is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), central coordinate of circle is p_o(x_o, y_o), because the corner of cutter when Machining Arc clockwise is with processing counterclockwise is different, therefore it can discuss respectively.As shown in accompanying drawing 4 and 6, its initial angle

Angle value is represented by (unit for °)：

When inverse circular arc is cut,

When being cut along circular arc,

Because the secant direction of each point on circular curve is different, therefore, when cutting circular arc, it is necessary to control cutter with the motion in XY directions synchronous rotary.According to the general principle of time-divided method, propose a kind of i.e. step pitch horn cupping such as use of Real-time interpolation algorithm of circular curve tangential following and carry out tangential following control, its method is to determine often to walk walked step angle according to sampling period of default and given feed speed, then calculates the amount of feeding of each unit time each axle respectively；In view of increasing with the number of plies, under the same conditions, moment of torsion also gradually increases angular speed, therefore, to ensure moment of torsion in the case where the number of plies is different, moment of torsion is still identical suffered by cutter, reach that cutting steadily takes into account cutting efficiency, therefore tangential following unit interpolation cycle T is multiplied by a factor

(n represents the number of plies), therefore set P in Fig. 5_i(x_i, y_i, c_i) it is cutter current location point, interpolation, which is calculated, to be required to reach the next position P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1), P_iPoint coordinate be：

$\left\{\begin{array}{c}{x}_i=R\cos {\mathrm{\&phi;}}_i\\ y_i=R\sin {\mathrm{\&phi;}}_i\\ c_i={\mathrm{\&phi;}}_i+\mathrm{\&pi;}/2\end{array}\right.---1-5$

Once equivalent interpolation cycle T during circular arc tangential following interpolation₀Determine that its interpolation step angle α is with feeding linear velocity V：

$\mathrm{\&alpha;}=\frac{\mathrm{\&omega;T}}{n}=\mathrm{\&omega;}{\mathrm{<figure-callout\; id="0"\; label="T"\; filenames="HDA0000081116480000042.png,HDA0000081116480000061.png"\; state="\{\{state\}\}">T</figure-callout>}}_0=\frac{V}{R}{T}_0---1-6$

If in Interpolation Process, the coordinate P of current point_i(x_i, y_i, c_i), it is known that in order to obtain on two coordinate directions Increment is fed, it is necessary to obtain next interpolated point P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1)：

$\left\{\begin{array}{c}{x}_{i+1}=R\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ y_{i+1}=R\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ c_{i+1}={\mathrm{\&phi;}}_i+\mathrm{\&alpha;}\end{array}\right.---1-7$

16 cutters are along X, Y, C amount of feeding Δ x_i+1, Δ y_i+1, Δ c_i+1For：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=R[\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\cos {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;y}}_{i+1}=R[\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\sin {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;c}}_{i+1}=\mathrm{\&alpha;}\end{array}\right.---1-8$

Formula (1-3) abbreviation is obtained：

$\left\{\begin{array}{c}{x}_{i+1}=x_i\cos \mathrm{\&alpha;}-y_i\sin \mathrm{\&alpha;}\\ y_{i+1}=y_i\cos \mathrm{\&alpha;}+x_i\sin \mathrm{\&alpha;}\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-9$

In order to which the interpolation for saving system CPU calculates the time, approximate calculation is used to cos α and sin α, the site error of the interpolated point caused by it will be analyzed below.Because α/2 are very small, so there are tan (α/2) ≈ α/2, thus have：

$\cos \mathrm{\&alpha;}=\frac{1-{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}{1+{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}=\frac{4-{\mathrm{\&alpha;}}^2}{4+{\mathrm{\&alpha;}}^2}---1-10$

Similarly have：

$\sin \mathrm{\&alpha;}=\frac{4\mathrm{\&alpha;}}{4+{\mathrm{\&alpha;}}^2}---1-11$

Make A_i=cos α, B_i=sin α, then data sampling method tangential following interpolation formula be：

$\left\{\begin{array}{c}{x}_{i+1}=A_i{x}_i-B_i{y}_i\\ y_{i+1}=A_i{y}_i+B_i{x}_i\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-12$

The feeding increment of each axle is：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=x_{i+1}-x_i\\ {\mathrm{\&Delta;y}}_{i+1}=y_{i+1}-y_i\\ {\mathrm{\&Delta;c}}_{i+1}=c_{i+1}-c_i\end{array}\right.---1-13$

There is formula (1-12) to understand, each increment of coordinate value of interpolation cycle is only relevant with the coordinate value and step angle of a upper interpolated point in circular arc tangential following interpolation algorithm.After step-length is determined with arc radius, step-length angle is a constant value during whole arc machining, at this moment x, y-coordinate increment size are only related to the coordinate value of a upper interpolated point, and the incremental angular of Z axis is the positive and negative values at step-length angle, and are immobilized in whole section of circular interpolation.It can be obtained shown in circular arc tangential following interpolation algorithm flow chart accompanying drawing 3 according to formula (1-13).Pair radius carries out tangential following interpolation for 40mm circular arc, and interpolation result when taking step-length for 4mm is as shown in table 3-1.

The circular arc that table 3-1 radiuses are 40.000mm uses Interpolation step-length for the result of 4.000mm progress tangential following interpolations

Walk sequence	X(mm)	Y(mm)	C(°)	△(mm)
					1	40	4	4.50	0.20
2	40	8	9.00	0.80
					3	40	12	13.50	1.76
4	36	12	18.00	2.06
					5	36	16	22.50	0.60
6	36	20	27.00	1.18
					7	32	20	31.50	2.26
8	32	24	36.00	0
					9	32	28	40.50	2.52
10	28	28	45.00	0.40
					11	24	28	49.50	3.13
12	24	32	54.00	0
					13	24	36	58.50	3.27
14	20	36	63.00	1.18
					15	16	36	67.50	0.60
16	12	36	72.00	2.06
					17	12	40	76.50	1.75
18	8	40	81.00	0.79
					19	4	40	85.50	0.20
20	0	40	90	0

The circular arc that table 3-2 radiuses are 40.000mm uses Interpolation step-length for the result of 8.000mm progress tangential following interpolations

Walk sequence	X(mm)	Y(mm)	C(°)	△(mm)
					1	40	8	9.00	0.79
2	40	16	18.00	3.08
					3	32	16	27.00	4.23
4	32	24	36.00	0
					5	32	32	45.00	5.24
6	24	32	54.00	0
					7	16	32	63.00	5.78
8	16	40	72.00	3.08
					9	8	40	81.00	0.79
10	0	40	90.00	0

Claims (2)

1. a kind of tangential following interpolating method that lathe is cut applied to multilayer shoe leather numerical control, cut process includes the rotation during the dither of sheet cutter, the plane motion for cutting cutter and sheet tool sharpening to the multilayer shoe leather at a high speed, it is characterised in that：It is Z-direction to define dither direction, and it is X, Y-direction to define plane motion direction；

During straight line is cut, Z axis is independent axes and not linked with XY axles, it is known that the starting point coordinate of straight line is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), then the initial angle angle value of straightway is expressed as：

Carry out drift angle amendment and obtain final cutter start angle for θ '_s=θ_s-α；

Feed speed F and equivalent interpolation cycle T in Machining Instruction₀, T₀=nT, T represent unit interpolation cycle, calculate the profile step-length l of each interpolation cycle, then the relation of three is expressed as：L=FT₀, because rectilinear tangential follows the Z axis angle of interpolation constant, so there is rectilinear tangential to follow interpolation formula as follows：

$\left\{\begin{array}{c}\mathrm{\&Delta;x}=l\cos a=l\frac{x_e-x_s}{\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2}}\\ \mathrm{\&Delta;y}=\mathrm{\&Delta;}x\tan a=\frac{\mathrm{\&Delta;x}(y_e-y_s)}{x_e-x_s}\\ \mathrm{\&Delta;c}=0\end{array}\right.---1-2$

For every section of straight line, c is constant, is obtained by x and y relation, and obtaining rectilinear tangential by conversion follows interpolation formula：

$\left\{\begin{array}{c}{x}_{i+1}=x_i+\mathrm{\&Delta;x}\\ y_{i+1}=y_i+\mathrm{\&Delta;y}\\ c_{i+1}={\mathrm{\&theta;}}_s^{\&prime;}\end{array}\right.---1-3.$

2. a kind of tangential following interpolating method that lathe is cut applied to multilayer shoe leather numerical control as claimed in claim 1, it is characterised in that：If known cut a section circular arcFor inverse circular arc, circular arc starting point is P_s(x_s, y_s, c_s), terminal is P_e(x_e, y_e, c_e), center of circle O is the origin of coordinates, then the Z coordinate value angle positive equivalent to the line and X-axis in current point and the center of circle, then the parametric equation for being processed curve is expressed as：

$\left\{\begin{array}{c}x=R\cos \mathrm{\&phi;}\\ y=R\sin \mathrm{\&phi;}\\ c=\mathrm{\&phi;}+\mathrm{\&pi;}/2\end{array}\right.---1-4$

Wherein R is arc radius, and φ is the angle of current point and line and the X-axis forward direction in the center of circle.

During circular arc is cut, Z axis, which will keep linking with XY axles, make it that the blade moment is consistent with circular arc secant direction, first has to obtain the corner of cutter starting point according to circular arc starting point, it is known that the starting point coordinate of arc section is p_s(x_s, y_s), terminal point coordinate is p_e(x_e, y_e), central coordinate of circle is p_o(x_o, y_o), initial angle angle value is expressed as：

When inverse circular arc is cut,

When being cut along circular arc,

If P_i(x_i, y_i, c_i) it is cutter current location point, interpolation, which is calculated, to be required to reach the next position P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1), P_iPoint coordinate be：

$\left\{\begin{array}{c}{x}_i=R\cos {\mathrm{\&phi;}}_i\\ y_i=R\sin {\mathrm{\&phi;}}_i\\ c_i={\mathrm{\&phi;}}_i+\mathrm{\&pi;}/2\end{array}\right.---1-5$

Once equivalent interpolation cycle T during circular arc tangential following interpolation₀Determine that its interpolation step angle α is with feeding linear velocity V：

$\mathrm{\&alpha;}=\frac{\mathrm{\&omega;T}}{n}=\mathrm{\&omega;}{T}_0=\frac{V}{R}{T}_0---1-6$

If in Interpolation Process, the coordinate P of current point_i(x_i, y_i, c_i), it is known that in order to obtain the feeding increment on two coordinate directions, obtaining next interpolated point P after an interpolation cycle_i+1(x_i+1, y_i+1, c_i+1)：

$\left\{\begin{array}{c}{x}_{i+1}=R\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ y_{i+1}=R\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})\\ c_{i+1}={\mathrm{\&phi;}}_i+\mathrm{\&alpha;}\end{array}\right.---1-7$

Cutter is along X, Y, C amount of feeding Δ x_i+1, Δ y_i+1, Δ c_i+1For：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=R[\cos ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\cos {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;y}}_{i+1}=R[\sin ({\mathrm{\&phi;}}_i+\mathrm{\&alpha;})-\sin {\mathrm{\&phi;}}_i]\\ {\mathrm{\&Delta;c}}_{i+1}=\mathrm{\&alpha;}\end{array}\right.---1-8$

Formula 1-3 abbreviations are obtained：

$\left\{\begin{array}{c}{x}_{i+1}=x_i\cos \mathrm{\&alpha;}-y_i\sin \mathrm{\&alpha;}\\ y_{i+1}=y_i\cos \mathrm{\&alpha;}+x_i\sin \mathrm{\&alpha;}\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-9$

Tan (α/2) ≈ α/2 are set, thus are had：

$\cos \mathrm{\&alpha;}=\frac{1-{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}{1+{\left(\frac{\mathrm{\&alpha;}}{2}\right)}^2}=\frac{4-{\mathrm{\&alpha;}}^2}{4+{\mathrm{\&alpha;}}^2}---1-10$

Similarly have：

$\sin \mathrm{\&alpha;}=\frac{4\mathrm{\&alpha;}}{4+{\mathrm{\&alpha;}}^2}---1-11$

Make A_i=cos α, B_i=sin α, then data sampling method tangential following interpolation formula be：

$\left\{\begin{array}{c}{x}_{i+1}=A_i{x}_i-B_i{y}_i\\ y_{i+1}=A_i{y}_i+B_i{x}_i\\ c_{i+1}=c_i+\mathrm{\&alpha;}\end{array}\right.---1-12$

The feeding increment of each axle is：

$\left\{\begin{array}{c}{\mathrm{\&Delta;x}}_{i+1}=x_{i+1}-x_i\\ {\mathrm{\&Delta;y}}_{i+1}=y_{i+1}-y_i\\ {\mathrm{\&Delta;c}}_{i+1}=c_{i+1}-c_i\end{array}\right.---1-13.$

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