CN104155916B - A kind of high accuracy quickly space circular arc interpolation method - Google Patents

Abstract

A kind of high accuracy quickly space circular arc interpolation method, the method has six big steps: step 1, definition space circular interpolation code format；Step 2, decoding obtains space circular arc interpolation data message；Step 3, calculates the maximum adjustment of feed-rate value under circular interpolation radius, central angle, arc length and interpolation bow high level error constraint；Step 4, calculates interpolation parameters increment；Step 5, calculates interpolated point；Step 6, interpolation endpoint.This interpolating method can vow the circular interpolation in any space plane according to given circular arc starting point, terminal and circular arc planar process.

Description

A kind of high accuracy quickly space circular arc interpolation method

Technical field

The present invention relates to a kind of high accuracy quickly space circular arc interpolation method, belong to Computerized Numerical Control processing technology field.

Background technology

Existing numerical control coefficient the most only possesses G01 space line interpolation and G02/03 two dimension circular interpolation, interpolation circular arc Shi Bixu specifies circular interpolation plane simultaneously, i.e. XZ plane that X/Y plane that G17 code is specified, G18 code are specified or G19 code The YZ plane specified.Therefore, space free curve is typically all separated into small space line section before processing, is formed big Amount continuous print G01 code, digital control system uses space line interpolating method to complete adding of space free curve according to G01 code Work.It practice, use space circular arc approximate spatial free curve can not only reduce the quantity of code in a large number, raising curve approaches essence Degree, simultaneously can make machine tool motion more steady, it is to avoid the turning sudden change of straightway junction.Accordingly, it would be desirable to exploitation one Interpolation precision is high, calculate fireballing space circular arc interpolation method approaches the processing mode of free curve to realize space circular arc.

Summary of the invention

The present invention is directed to problem above, it is provided that a kind of high accuracy quickly space circular arc interpolation method.This interpolating method energy The circular interpolation in any space plane has been vowed according to given circular arc starting point, terminal and circular arc planar process.The present invention is led to Crossing techniques below scheme to realize, Fig. 1 show interpolating method flow chart, specifically comprises the following steps that

Step 1, definition space circular interpolation code format:

Definition space circular interpolation code format of the present invention is: G02.1X (U) _ Y (V) _ Z (W) _ I_J_K_NX_NY_NZ_F

X, Y, Z are circular arc terminal absolute coordinate, and U, V, W are circular arc terminal relative coordinate (relative to starting point)；

I, J, K are circular arc center of circle relative coordinate (relative to starting point)；

NX, NY, NZ are that the method for circular arc plane is vowed, and specify that winding vows counterclockwise interpolation circular arc；

F is interpolation feed speed.

Step 2, decoding acquisition space circular arc interpolation data message:

The starting point making circular interpolation is (x_s,y_s,z_s), i.e. the terminal of code the last period interpolation, for known；

If the circular interpolation code read is: G02.1Xx_e Yy_e Zz_e Ii_c Jj_c Kk_c NXn_x NYn_y NZn_zFf, then The terminal of circular interpolation is (x_e,y_e,z_e), the center of circle is (x_c,y_c,z_c)=(x_s+i_c,y_s+j_c,z_s+k_c), circular arc planar process is vowed and is (n_x,n_y,n_z), maximum feed speed is f；

Step 3, calculates the maximum feeding speed under circular interpolation radius, central angle, arc length and interpolation bow high level error constraint Degree correction value: circular interpolation radius: $R=\sqrt{{(x_e-x_c)}^2+{(y_e-y_c)}^2+{(z_e-z_c)}^2};$

Circular interpolation central angle: zequin vector is V_s=(x_s-x_c,y_s-y_c,z_s-z_c), terminal vector is V_e=(x_e-x_c, y_e-y_c,z_e-z_c), circular arc planar process is vowed as N=(n_x,n_y,n_z), zequin vector, terminal vector are mixed with what circular arc planar process was vowed Conjunction is amassed as (V_s,V_e, N) and=(V_s×V_e) N, if (V_s,V_e, N) and ＞ 0, then circular interpolation central angle $\mathrm{\θ}=\arccos \left(\frac{V_s\·V_e}{R^2}\right).$ If (Vs, Ve, N)≤0, then the circular interpolation center of circle $\mathrm{\θ}=2\mathrm{\π}-\arccos \left(\frac{V_s\·V_e}{R^2}\right);$

Circular interpolation arc length: L=R θ；

Maximum adjustment of feed-rate value: if the maximum interpolation bow high level error of regulation is δ, interpolation cycle is T_c, then maximum is entered It is modified to speed $f=\min (f,\frac{2}{T_c}\sqrt{R^2-{(R-\mathrm{\δ})}^2});$

Step 4, calculating interpolation parameters increment:

If according to circular arc arc length and maximum feed speed, carrying out the feeding of the i-th interpolation cycle after feed speed planning Speed is V_i, acceleration is A_i, acceleration is J_i, then Interpolation step-length isInterpolation parameters increases Amount is calculated asThe three-dimensional euclidean mould that wherein | | N | | vows for circular arc planar process is long.

Step 5, calculating interpolated point:

Making current interpolated point is (x_i,y_i,z_i), next interpolated point is (x_i+1,y_i+1,z_i+1), it is calculated as follows:

$\begin{array}{c}{x}_{i+1}=x_i+(2h(n_z{y}_c-n_z{y}_i-n_y{z}_c+n_y{z}_i+h{n}_y^2{x}_c+h{n}_z^2{x}_c-h{n}_y^2{x}_i-h{n}_z^2{x}_i)\\ -2h{n}_x(h{n}_y{y}_c-h{n}_y{y}_i+h{n}_z{z}_c-h{n}_z{z}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{y}_{i+1}=y_i+(2h(n_z{x}_c-n_z{x}_i-n_x{z}_c+n_x{z}_i+h{n}_x^2{x}_c+h{n}_z^2{y}_c-h{n}_x^2{y}_i-h{n}_z^2{y}_i)\\ -2h{n}_y(h{n}_x{x}_c-h{n}_x{x}_i+h{n}_z{z}_c-h{n}_z{z}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{z}_{i+1}=z_i+(2h(n_y{x}_c-n_y{x}_i-n_x{y}_c+n_x{y}_i+h{n}_x^2{x}_c+h{n}_y^2{z}_c-h{n}_x^2{z}_i-h{n}_y^2{z}_i)\\ -h{n}_x{n}_z{x}_c+h{n}_x{n}_z{x}_i+h{n}_y{n}_z{y}_c+h{n}_y{n_{z}y}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

Calculated (x_i+1,y_i+1,z_i+1) it is interpolated point, control space circular arc interpolation by servo position.

Step 6, interpolation endpoint:

Update the residue a length of L=L-L of interpolation_iIf, L ＞ 0, then return step 4 and continue interpolation；Otherwise, interpolation terminates.

Wherein, the interpolation bow high level error related in described step 3 is that circular interpolation process interpolation track approaches circular arc and causes Geometric error, as in figure 2 it is shown, figure cathetus section is that (actual interpolation track is the least straightway to interpolation track, is herein Amplify result), interpolation track goes to approach when being interpolated circular arc, and the maximum deviation between interpolation track and circular arc is bow high level error, I.e. bow high level error δ shown in figure.

Wherein, described step 4 relates to feed speed planning, for the feed speed planned according to circular interpolation arc length Information, belongs to feed speed planning aspect, obtains the feed speed of each interpolation cycle, acceleration and acceleration etc. after planning Information is for calculating Interpolation step-length and the interpolation parameters increment of each interpolation cycle.

Wherein, described step 5 relates to current interpolated point be known point, initial value be circular interpolation rise Point, each interpolation cycle calculates an interpolated point.

A kind of high accuracy quickly space circular arc interpolation method that the present invention provides has the advantage that

1. Interpolation Spaces.Vow with circular arc planar process and describe circular arc place plane, the circle of space arbitrary plane can be completed Arc interpolation；

2. interpolation uniformity.Specify to vow counterclockwise interpolation around circular arc planar process, unified the forward and reverse interpolation of circular arc, by Circular arc plane direction of normal determines circular interpolation direction；

3. interpolation precision is high.Through theoretical validation, the interpolated point calculated by step 5 can guarantee accurate original On circle, therefore the radial error of interpolated point is 0；

4. interpolation rate is fast.Interpolated point only has simple four arithmetic operation in calculating, and calculates in interpolation parameters increment and the most only relates to And square root, it is to avoid the calculating of trigonometric function, therefore computational efficiency is high, and speed is fast；

5. interpolation flow process is simple.The quadrant avoiding circular interpolation judges, algorithm flow is simple, it is easy to accomplish.

Accompanying drawing explanation

Fig. 1 is interpolating method flow chart of the present invention；

Fig. 2 is that circular interpolation bends high level error schematic diagram；

Fig. 3 is to implement example interpolation result.

Symbol description:

In Fig. 2, δ represents bow high level error；

In Fig. 3, N represents that circular arc planar process is vowed；

In Fig. 3, C represents the circular arc center of circle.

Detailed description of the invention

Below in conjunction with the accompanying drawings the present invention is implemented example to elaborate.See with a full circle for interpolation object, flow chart Fig. 1, one of the present invention high accuracy quickly space circular arc interpolation method, there is a step in detail below:

Step 1, definition space circular interpolation code format:

With a radius as 50mm, the center of circle is in zero (0,0,0), and circular arc planar process vows that the full circle for (4 ,-3,1) is Interpolation object, definition circular interpolation Origin And Destination is (30,40,0), and interpolation feed speed is 100mm/s, G code form For:

G02.1X30Y40Z0I-30J-40K0NX4NY-3NZ1F100

Step 2, decoding acquisition space circular arc interpolation data message:

Being obtained circular interpolation data message by G code decoding is:

Circular arc terminal: (30,40,0)

The circular arc center of circle: (0,0,0)

Circular arc planar process is vowed: (4 ,-3,1)

Interpolation maximum feed speed: 100

Step 3, calculates the maximum feeding speed under circular interpolation radius, central angle, arc length and interpolation bow high level error constraint Degree correction value: circular interpolation radius: $R=\sqrt{{(x_e-x_c)}^2+{(y_e-y_c)}^2+{(z_e-z_c)}^2}=\sqrt{{30}^2+{40}^2}=50;$

Circular interpolation central angle: starting point vector V_s=(x_s-x_c,y_s-y_c,z_s-z_c)=(30,40,0), terminal vector V_e= (x_e-x_c,y_e-y_c,z_e-z_c)=(30,40,0), circular arc planar process is vowed as N=(n_x,n_y,n_z)=(4 ,-3,1), zequin to Mixed product (the V that amount, terminal vector is vowed with circular arc planar process_s,V_e, N) and=(V_s×V_e) N=0, then the circular interpolation center of circle $\mathrm{\θ}=2\mathrm{\π}-\arccos \left(\frac{V_s\·V_e}{R^2}\right)=2\mathrm{\π};$

Circular interpolation arc length: L=R θ=2 π R；

Maximum adjustment of feed-rate value: if the maximum interpolation bow high level error of regulation is δ=0.001mm, interpolation cycle is T_c =1ms, then maximum adjustment of feed-rate is:

$f=\min (f,\frac{2}{T_c}\sqrt{R^2-{(R-\mathrm{\δ})}^2})=\min \left(\mathrm{100,632,4524}\right)=100\mathrm{mm}/s.$

Step 4, calculating interpolation parameters increment:

If during constant speed interpolation, interpolation rate V_i=100mm/s, acceleration and acceleration are 0, then Interpolation step-length is L_i =V_iT_c=0.1mm.Interpolation parameters increment is $h=\frac{L_i}{\left|\right|N\left|\right|\sqrt{4R^2-L_i^2}}{1.961162332\×10}^{-4},$

Step 5, calculating interpolated point:

Making current interpolated point is (x_i,y_i,z_i), next interpolated point is (x_i+1,y_i+1,z_i+1), it is calculated as follows:

$\begin{array}{c}{x}_{i+1}=x_i+(2h(n_z{y}_c-n_z{y}_i-n_y{z}_c+n_y{z}_i+h{n}_y^2{x}_c+h{n}_z^2{x}_c-h{n}_y^2{x}_i-h{n}_z^2{x}_i)\\ -2h{n}_x(h{n}_y{y}_c-h{n}_y{y}_i+h{n}_z{z}_c-h{n}_z{z}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{y}_{i+1}=y_i+(2h(n_z{x}_c-n_z{x}_i-n_x{z}_c+n_x{z}_i+h{n}_x^2{x}_c+h{n}_z^2{y}_c-h{n}_x^2{y}_i-h{n}_z^2{y}_i)\\ -2h{n}_y(h{n}_x{x}_c-h{n}_x{x}_i+h{n}_z{z}_c-h{n}_z{z}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{z}_{i+1}=z_i+(2h(n_y{x}_c-n_y{x}_i-n_x{y}_c+n_x{y}_i+h{n}_x^2{x}_c+h{n}_y^2{z}_c-h{n}_x^2{z}_i-h{n}_y^2{z}_i)\\ -h{n}_x{n}_z{x}_c+h{n}_x{n}_z{x}_i+h{n}_y{n}_z{y}_c+h{n}_y{n_{z}y}_i))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

When making i=0, interpolated point is circular arc starting point (30,40,0), and the interpolated point obtained by the recurrence calculation of above formula is such as Under:

In upper table, i represents i-th interpolation cycle；(x, y z) represent the interpolated point that each interpolation cycle generates；R represents every The interpolated point of generation of individual cycle and the distance in the center of circle；L represents the distance of interpolated point that each cycle generates and a upper interpolated point, I.e. Interpolation step-length.As can be seen from the above table, calculated each interpolated point all on original circle, the radial error of interpolated point It is 0 (because the distance in each interpolated point and the center of circle is 50, equal to arc radius).And the interpolated point generated is in strict accordance with regulation Interpolation step-length generate, the distance between adjacent two interpolated points is the most equal with Interpolation step-length, the most there is not velocity error (Interpolation step-length of regulation and the relative error of actual Interpolation step-length).Therefore, the radial error of 0 represents with the velocity error of 0 The interpolation precision of the present invention is the highest.

Step 6, interpolation endpoint:

Update the residue a length of L=L-L of interpolation_iIf, L ＞ 0, then return step 4 and continue interpolation；Otherwise, interpolation terminates.Insert The interpolation track formed after having mended is as shown in Figure 2；Fig. 3 is to implement example interpolation result, is a space circle, and some C is the center of circle, Vector N is that the method for this circle place plane is vowed, interpolation direction is direction shown in arrow in figure, around vector N interpolation counterclockwise.

Claims (4)

1. a high accuracy quick space circular arc interpolation method, it is characterised in that: the method specifically comprises the following steps that

Step 1, definition space circular interpolation code format:

Definition space circular interpolation code format is: G02.1X (U) _ Y (V) _ Z (W) _ I_J_K_NX_NY_NZ_FX, and Y, Z are circle Arc terminal absolute coordinate, U, V, W are the circular arc terminal relative coordinate relative to starting point；

I, J, K are the circular arc center of circle relative coordinate relative to starting point；

NX, NY, NZ are that the method for circular arc plane is vowed, and specify that winding vows counterclockwise interpolation circular arc；

F is interpolation feed speed；

Step 2, decoding acquisition space circular arc interpolation data message:

The starting point making circular interpolation is (x_s,y_s,z_s), i.e. the terminal of code the last period interpolation, for known；

If the circular interpolation code read is: G02.1 Xx_e Yy_e Zz_e Ii_c Jj_c Kk_c NXn_x NYn_y NZn_zFf, then justify The terminal of arc interpolation is (x_e,y_e,z_e), the center of circle is (x_c,y_c,z_c)=(x_s+i_c,y_s+j_c,z_s+k_c), circular arc planar process is vowed as (n_x, n_y,n_z), maximum feed speed is f；

Step 3, the maximum feed speed calculated under circular interpolation radius, central angle, arc length and interpolation bow high level error constraint is repaiied On the occasion of:

Circular interpolation radius:

Circular interpolation central angle: zequin vector is V_s=(x_s-x_c,y_s-y_c,z_s-z_c), terminal vector is V_e=(x_e-x_c,y_e- y_c,z_e-z_c), circular arc planar process is vowed as N=(n_x,n_y,n_z), the mixing that zequin vector, terminal vector are vowed with circular arc planar process Amass as (V_s,V_e, N) and=(V_s×V_e) N, if (V_s,V_e, N) and ＞ 0, then circular interpolation central angleIf (V_s,V_e, N)≤0, then the circular interpolation center of circle

Circular interpolation arc length: L=R θ；

Maximum adjustment of feed-rate value: if the maximum interpolation bow high level error of regulation is δ, interpolation cycle is T_c, then maximum feed speed It is modified to

Step 4, calculating interpolation parameters increment:

According to circular arc arc length and maximum feed speed, the feed speed carrying out the i-th interpolation cycle after feed speed planning is V_i, acceleration is A_i, acceleration is J_i, then Interpolation step-length isInterpolation parameters incremental computations ForThe three-dimensional euclidean mould that wherein | | N | | vows for circular arc planar process is long；

Step 5, calculating interpolated point:

Making current interpolated point is (x_i,y_i,z_i), next interpolated point is (x_i+1,y_i+1,z_i+1), it is calculated as follows:

$\begin{array}{c}{x}_{i+1}=x_i+(2h\left(n_z{y}_c-n_z{y}_i-n_y{z}_c+n_y{z}_i+{\mathrm{hn}}_y^2{x}_c+{\mathrm{hn}}_z^2{x}_c-{\mathrm{hn}}_y^2{x}_i-{\mathrm{hn}}_z^2{x}_i\right)\\ -2{\mathrm{hn}}_x\left({\mathrm{hn}}_y{y}_c-{\mathrm{hn}}_y{y}_i+{\mathrm{hn}}_z{z}_c-{\mathrm{hn}}_z{z}_i\right))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{y}_{i+1}=y_i-(2h\left(n_z{x}_c-n_z{x}_i-n_x{z}_c+n_x{z}_i-{\mathrm{hn}}_x^2{y}_c-{\mathrm{hn}}_z^2{y}_c+{\mathrm{hn}}_x^2{y}_i+{\mathrm{hn}}_z^2{y}_i\right)\\ +2{\mathrm{hn}}_y\left({\mathrm{hn}}_x{x}_c-{\mathrm{hn}}_x{x}_i+{\mathrm{hn}}_z{z}_c-{\mathrm{hn}}_z{z}_i\right))/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

$\begin{array}{c}{z}_{i+1}=z_i+\left(2h\right(n_y{x}_c-n_y{x}_i-n_x{y}_c+n_x{y}_i+{\mathrm{hn}}_x^2{z}_c+{\mathrm{hn}}_y^2{z}_c-{\mathrm{hn}}_x^2{z}_i-{\mathrm{hn}}_y^2{z}_i\\ -{\mathrm{hn}}_x{n}_z{x}_c+{\mathrm{hn}}_x{n}_z{x}_i-{\mathrm{hn}}_y{n}_z{y}_c+{\mathrm{hn}}_y{n}_z{y}_i\left)\right)/(h^2{n}_x^2+h^2{n}_y^2+h^2{n}_z^2+1)\end{array}$

Calculated (x_i+1,y_i+1,z_i+1) it is interpolated point, control space circular arc interpolation by servo position；

Step 6, interpolation endpoint:

Update the residue a length of L=L-L of interpolation_iIf, L ＞ 0, then return step 4 and continue interpolation；Otherwise, interpolation terminates.

A kind of high accuracy quickly space circular arc interpolation method the most according to claim 1, it is characterised in that: described step 3 In relate to interpolation bow high level error be that circular interpolation process interpolation track approaches the geometric error that circular arc causes, straightway is interpolation Track, interpolation track goes to approach when being interpolated circular arc, and the maximum deviation between interpolation track and circular arc is bow high level error δ.

A kind of high accuracy quickly space circular arc interpolation method the most according to claim 1, it is characterised in that: described step 4 In relate to feed speed planning, for the feed speed information planned according to circular interpolation arc length, belong to feed speed planning side Face, obtains the feed speed of each interpolation cycle, acceleration with acceleration information for calculating each interpolation cycle after planning Interpolation step-length and interpolation parameters increment.

A kind of high accuracy quickly space circular arc interpolation method the most according to claim 1, it is characterised in that: described step 5 In the current interpolated point that relates to be known point, initial value is the starting point of circular interpolation, and each interpolation cycle calculates once Interpolated point.