CN103268081A - Precision error control based orbit segment transfer processing algorithm for numerical control machine tool - Google Patents

Abstract

The invention provides a precision error control based orbit segment transfer processing algorithm for numerical control machine tool; elaboration and analysis are performed aiming at the current two transfer processing methods of connecting the orbit segment with a straight line and connecting the orbit segment with an arc and the error situation; for the error of the transfer processing of the orbit segment and the straight line, the change relation of the generated error and an interpolation surplus and an included angle is analyzed according to the difference of the velocity vector included angle; the transfer processing method and the error calculation of the orbit segment and the arc are elaborated based on two aspects of whether the orbit segment and the arc are on the same plane and the included angle of the orbit segment and the tangent line of the arc. Compared with the prior art, the analysis for the transfer processing method and the error are more complete; furthermore, the analysis for various transfer situations is more detailed and more thorough, and therefore, the higher transfer precision and the higher transfer velocity can be obtained.

Description

A kind of orbit segment switching Processing Algorithm based on trueness error control for numerically-controlled machine

Technical field

The present invention relates to a kind of orbit segment switching Processing Algorithm based on trueness error control for numerically-controlled machine, belong to the computer numerical control technology field.

Background technology

Because what moulding no matter any workpiece be, finally all is made up of a section orbit segment (circular arc, straight line), just has transit point between the orbit segment, produce the problem of switching transition method and switching precision thus.For guaranteeing workpiece accuracy, having the scholar to propose will carry out acceleration and deceleration between each orbit segment handles, be every section all experience accelerate, at the uniform velocity, these three processes of slowing down, so just cause the frequent start-stop of lathe, speed of feed slow excessively, problem such as desired speed can't guarantee, crudy is undesirable.The adjacent smooth-path section that has the scholar will not have flex point is done as a wholely to carry out interpolation, but in fact, it is a kind of special circumstances that the no flex point between the adjacent track section connects, and angle therebetween can be 0 °～180 °.There is the scholar to propose under the situation that guarantees the tie point precision, the realization track is processed continuously, it mainly is that the mini line segment during for big obtuse angle connects at switching place, and for the connection situation between circular arc, the straight line, how the speed when being connected to acute angle between straight line is handled does not relate to.

Because restrictions such as cycle of interpolation operation, precision, often there is trueness error in the junction between orbit segment, and namely orbit segment remains surplus at last and can't satisfy the required distance of last interpolation cycle.There is the scholar to pass through to change interpolation cycle according to last surplus, thereby change step-length and make it to arrive last final on trajectory, reach the purpose of evading site error, but owing to need carry out the modification of interpolation cycle, and it is often as system's reference clock, after changing, can cause the change of system time benchmark, have certain immesurable factor.There is the scholar then to connect according to different orbit segments: straight line and straight line, straight line and circular arc, on according to existing forwarding method basis, thereby it is carried out the constraint of error analysis acquisition speed, but be primarily aimed at the error analysis that connects between straight line, too conservative for the error analysis of acute angle connection simultaneously, cause the speed of asking for too small.

Above-mentioned analysis mainly is to guarantee to carry out finding the solution according to the switching speed of accuracy constraint under the immovable situation of original movement locus for the processing of switching place as can be known.It is the connection processing method that mainly adopts at present, but mainly be to handle at the connection between the straight-line segment up to now, and for handling straight line and being connected of circular arc shorter mention, while is also comparatively rare for the correlative study that straight line is connected to acute angle, can not carry out deep analysis and research, and often at be that little line segment connects and handles, handle then not necessarily very suitable for the connection of big orbit segment.

Current CNC processing technology, the final movement locus of carrying out mainly is to be made of straight line and circular arc, and a transit point must be arranged between orbit segment, because the factor of operational precision and velocity process, tend to occur the phenomenon that the remaining surplus of orbit segment is not equal to last interpolation cycle required separation distance, the processing of so just need transferring on the basis that guarantees profile errors.By analyzing as can be known, finding the solution with the line style of current orbit segment of transit point is irrelevant, but determined by the line style of next orbit segment.

Summary of the invention

The invention provides a kind of orbit segment switching Processing Algorithm based on trueness error control, purpose be at current mainly be the research handled of the continuous smooth transition switching of mini line segment and for the present situation of the less consideration of other situation, the present invention has elaborated the forward processing method of various orbit segment switching types, and the obtaining of constraint speed that be solved to of profile errors provide foundation, and carried out simulating, verifying.Be more complete with respect to existing algorithm for the analysis of its forward processing method and error.

For achieving the above object, first kind of technical scheme that the present invention adopts is: a kind of orbit segment switching Processing Algorithm based on trueness error control for numerically-controlled machine, comprise workpiece to be processed is placed on the workbench of described numerically-controlled machine, the operate portions of described numerically-controlled machine is carried out process operation to described processing work

The first step: ask transit point

In the interpolation operation of described numerically-controlled machine, the publicity of step pitch is:

ΔL＝VT；

In the formula, Δ L represents step pitch; T represents interpolation cycle, and V represents the movement velocity of the operate portions of described numerically-controlled machine;

If arc P _SeP _EeSome P _SeTangent line be P _qP _SeP _sCoordinate be (x _s, y _s, z _s), P _SeCoordinate be (x _Se, y _Se, z _Se), P _eCoordinate be (x _e, y _e, z _e), P _EeCoordinate be (x _Eey _Ee, z _Ee), arc P _SeP _EeThe coordinate of center of circle O be (x _o, y _o, z _o), arc P _SeP _EeRadius be R; P wherein _s, P _Se, P _EeCoordinate figure be known, need ask for P now _eCoordinate figure, P wherein _sP _eDistance be Δ L;

One, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are in same space plane;

Expression formula by face can be learnt: when

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _s-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _s-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (201)；

The time, represent that namely described current orbit segment and arc track section are on same plane;

(1) works as P _sP _SeWith P _SeThe angle of O is during greater than 90 °; When the last interpolation cycle of described current orbit segment, the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, need the final position of described last interpolation cycle is transformed in next coupled arc track section, namely with a P _sBeing the center of circle, serves as that distance is drawn circle with Δ L, with described arc track section arc P _SeP _EeThe intersection point that intersects is described transit point P _eThen can obtain formula (202) according to geometric relationship:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (202)；

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _O) ²＝R ²

And some P _eIn plane P _SeOP _EeIn, then get formula (203):

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (203)：

Can obtain two groups of solution of equations by finding the solution, again because of described transit point Pe at described arc track section P _SeP _EeBetween, then also have constraint condition: (x _Se-x _e) (x _e-x _Ee)＜0, thus unique solution obtained, and then can determine described transit point P _eCoordinate figure;

(2) work as P _sP _SeWith P _SeThe angle of O is during less than 90 °, when the last interpolation cycle of described current orbit segment, and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, along track P _sP _SeExtend to P _eMake P _sP _eLength equal Δ L, described some P _eBe transit point; Then can obtain formula (204) according to geometric relationship:

$\frac{x_e-x_{\mathrm{se}}}{x_s-x_{\mathrm{se}}}=\frac{y_e-y_{\mathrm{se}}}{y_s-y_{\mathrm{se}}}=\frac{z_e-z_{\mathrm{se}}}{z_s-z_{\mathrm{se}}}$

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (204)；

Can obtain two groups of solution of equations by finding the solution, satisfy constraint condition simultaneously again: (x _e-x _o) ²+ (y _e-y _o) ²+ (z _e-z _o) ²＞R ²Thereby, obtain unique solution, and then can determine described transit point P _eCoordinate figure;

Two, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are not in same space plane; Define described current orbit segment P _sP _SeProjection line at the face at described arc track section place is P _{S '}P _Se

Expression formula by face can be learnt: when

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _s-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (205)；

+(z _s-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]≠0

The time, represent that namely described current orbit segment and arc track section are on same plane;

(1) works as P _{S '}P _SeWith P _SeThe angle of O is during greater than 90 °, the last interpolation cycle of described current orbit segment, the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, need the final position of described last interpolation cycle is transformed in next coupled arc track section, namely with a P _sBeing the center of circle, serves as that distance is drawn circle with Δ L, with described arc track section arc P _SeP _EeThe intersection point that intersects is described transit point P _eThen can obtain formula (206) according to geometric relationship:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ²

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _O) ²＝R ² (206)；

And some P _eIn plane P _SeOP _EeIn, then get formula (207):

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (207)；

Can obtain two groups of solution of equations by finding the solution, again because of described transit point P _eAt described arc track section P _SeP _EeBetween, then also have constraint condition: (x _Se-x _e) (x _e-x _Ee)＜0, thus unique solution obtained, and then can determine described transit point P _eCoordinate figure;

(2) work as P _{S '}P _SeWith P _SeThe angle of O is during less than 90 °, when the last interpolation cycle of described current orbit segment, and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, along P _{S '}P _SeExtend to P _eMake P _{S '}P _eLength equal Δ L, described some P _eBe transit point;

Straight line P _sP _SeVector be:

n＝(x _s-x _se，y _s-y _se，z _s-z _se)＝(M，N，P) (208)；

The equation on the plane at described arc track section place is:

(x-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (209)；

Abbreviation is:

AX+By+Cz+D＝0 (210)；

The normal line vector on the plane at described arc track section place is:

m＝(A，B，C) (211)；

Described straight line P _sP _SeAngle with the plane at described arc track section place

For:

If through a P _s, P _Se, P _eThe plane, and it is perpendicular to the plane at described arc track section place, then its plane equation is:

(x-x _se)(BP-NC)+(y-y _se)(-AP+MC)

+(z-z _se)(AN-MB)＝0 (213)；

Described through a P _s, P _Se, P _eThe normal on plane be:

P _n＝(U，V，W) (214)；

Abbreviation is:

Ux+Vy+Wz+E＝0 (215)；

Can obtain according to geometric relationship:

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (216)；

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (217)；

Ux _e+Vy _e+Wz _e+E＝0 (218)；

(x _e-x _se)(BP-NC)-(y _e-y _se)(-AP+MC)

+(z _e-z _se)(AN-MB)＝0 (219)；

Can obtain two groups of solution of equations by solving simultaneous equation, satisfy constraint condition: min[(x simultaneously again _e-x _Se) ²+ (y _e-y _Se) ²+ (z _e-z _Se) ²], thereby obtain unique solution, and then can determine described transit point P _eCoordinate figure;

Second step: switching error analysis

One, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are in same space plane;

(1) ∠ P _sP _Se0＞90 °

Arc P _SeP _eWith respect to arc P _SeP _Ee, its length is short, and it is equal to line segment P _SeP _eLength

(1) as ∠ P _sP _SeP _q∈ [0 ° 90 °] namely is equivalent to described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween velocity angle α when being acute angle,

The last interpolated point P of described current orbit segment _sWith described transit point P _eBetween distance, delta L be:

$\mathrm{\ΔL}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(220\right);$

The last interpolated point P of described current orbit segment _sAnd P _SeBetween distance, delta L ₁For:

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(221\right);$

The cosine of described angle α:

$\cos \mathrm{\&alpha;}=\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(222\right);$

According to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{{2\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(223\right):$

And then obtain:

${{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2=\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {(\mathrm{\&pi;}-\mathrm{\&alpha;})}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-\mathrm{\&Delta;}{L}^2)}---\left(224\right):$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(225\right):$

Bring the calculated value of formula (220), formula (221), formula (222), formula (223) and formula (224) into described error formula (225) simultaneously; According to described error formula (225) as can be known, error only with Δ L ₁, Δ L is relevant with angle α; That is: when

The time, along with Δ L ₁Increase and error increases, when The time, then along with Δ L ₁The increase error reduce;

(2) as ∠ P _sP _SeP _q∈ [90 ° 180 °],

1, as described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(226\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(227\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(228\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{2{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(229\right);$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2={\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(230\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(234\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(235\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

2, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (236)；

3, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(237\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(238\right);$

Error formula is:

e＝ΔL ₁ (239)；

(2) ∠ P _sP _Se0＜90 °

As ∠ P _sOP _Ee〉=∠ P _SeOP _EeThe time, ∠ P then _sP _SeP _q∈ [0 ° 90 °]

Described transit point P _eTo described arc track section P _SeP _EeThe distance and described arc P of center of circle O _SeP _Ee'sThe difference of radius is namely:

$e=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---\left(240\right)$

As ∠ P _sOP _Ee＜∠ P _SeOP _EeThe time, ∠ P then _sP _SeP _q∈ [90 ° 180 °], error formula is:

e＝ΔL-ΔL ₁ (241)；

Two, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are not in same space plane;

(1) ∠ P _s' P _Se0＞90 °

Arc P _SeP _eWith respect to arc P _SeP _Ee, its length is short, and it is equal to line segment P _SeP _eLength

(1) as ∠ P _s' P _SeP _q∈ [0 ° 90 °], namely be equivalent to described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween velocity angle α when being acute angle,

The last interpolated point P of described current orbit segment _sWith described transit point P _eBetween distance, delta L be:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(242\right);$

The last interpolated point P of described current orbit segment _sAnd P _SeBetween distance, delta L ₁For:

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(243\right);$

The cosine of described angle α:

$\cos \mathrm{\&alpha;}=\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(244\right);$

According to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{{2\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(245\right):$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2=\mathrm{\&Delta;}{\mathrm{<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {(\mathrm{\&pi;}-\mathrm{\&alpha;})}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-\mathrm{\&Delta;}{L}^2)}---\left(246\right):$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(247\right):$

Bring the calculated value of formula (242), formula (243), formula (244), formula (245) and formula (246) into described error formula (247) simultaneously; According to described error formula (247) as can be known, error only with Δ L ₁, Δ L is relevant with angle α; That is: when

The time, along with Δ L ₁Increase and error increases, when

The time, then along with Δ L ₁The increase error reduce;

(2) as ∠ P _s' P _SeP _q∈ [90 ° 180 °],

1, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(248\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(249\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(250\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{2{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(251\right);$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2={\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(252\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(253\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(254\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

2, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (255)；

3, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(256\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(257\right);$

Error formula is:

e＝ΔL ₁ (258)；

(2) ∠ P _s' P _Se0＜90 °

(1) as ∠ P _s' P _Sep _q∈ [0 ° 90 °], there are two kinds in error:

$e_1=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---\left(259\right)$

e ₂As follows:

1, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(260\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(261\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(262\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{2{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(263\right);$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2={\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(264\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(265\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(266\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

2, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (267)；

3, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(268\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(269\right);$

Error formula is:

e＝ΔL ₁ (270)；

Then obtain max (e ₁e ₂);

(1) as ∠ P _s' P _SeP _q∈ [90 ° 180 °], there are two kinds in error:

$e_1=\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}---\left(271\right)$

e ₂As follows:

1, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ l]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(272\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(273\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(274\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{2{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---\left(275\right);$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2={\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(276\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(277\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(278\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

2, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is: e=Δ L ₁(279);

3, as described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(280\right);$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(281\right);$

Error formula is: e=Δ L ₁(282);

Then obtain max (e ₁e ₂);

(3) when the plane perpendicular to described arc track section place, the plane at described current orbit segment place, then: Δ L ₁=Δ L (283).

Description of drawings

Accompanying drawing 1 is that current orbit segment is connected with straight line;

Accompanying drawing 2 is that current orbit segment is connected (on a plane) and angle greater than 90 ° with circular arc;

Accompanying drawing 3 is that current orbit segment is connected (on a plane) and angle less than 90 ° with circular arc;

Accompanying drawing 4 is that current orbit segment is connected (not on a plane) and angle greater than 90 ° with circular arc;

Accompanying drawing 5 is that current orbit segment is connected (not on a plane) and angle greater than 90 ° with circular arc;

Accompanying drawing 6 is the three-dimensional switching method synoptic diagram that prolongs;

Accompanying drawing 7 is connected with straight-line segment and the error analysis of velocity angle when being acute angle;

Graph of errors when accompanying drawing 8 different residue interpolation surpluses and angle;

When accompanying drawing 9 velocity angles are 60 °, the error curve diagram of different residue interpolation surpluses;

Accompanying drawing 10 is connected with straight-line segment and the error analysis of velocity angle when being the obtuse angle;

Graph of errors when accompanying drawing 11 different residue interpolation surpluses and angle;

Accompanying drawing 12 orbit segment angles are [45 ° 90 °], and surplus is less than Δ L ₁Error analysis during ctg α;

Error analysis when accompanying drawing 13 orbit segment angles are [0 ° 45 °];

Accompanying drawing 14 ∠ p _sP _SeSwitching error analysis in the time of 0＞90 °;

Accompanying drawing 15 ∠ P _sP _SeSwitching error analysis in the time of 0＜90 °;

Accompanying drawing 16 current orbit segments be connected with circular arc and angle greater than 90 ° error analysis;

Accompanying drawing 17 current orbit segments be connected with circular arc and angle less than 90 ° error analysis;

Accompanying drawing 18 forwarding method plane errors are analyzed;

Accompanying drawing 19 processing work X-Y schemes;

Accompanying drawing 20B point transit point track.

Embodiment

Below in conjunction with drawings and Examples the present invention is further described:

Embodiment: a kind of orbit segment switching Processing Algorithm based on trueness error control for numerically-controlled machine

Current orbit segment can be divided into two classes with being connected of next orbit segment: 1, link to each other with straight-line segment; 2, link to each other with arc section.

When namely linking to each other with straight line for first kind of situation, as the current orbit segment of Fig. 1 with shown in straight line is connected, in interpolation operation, thus because interpolation cycle has determined that with the movement velocity combination distance of walking is step pitch Δ L=VT at every turn, T represents interpolation cycle, and V represents movement velocity.Because it is the phenomenon of its integral multiple that interpolation cycle and the actual distance that will walk of present segment often might occur non-, so can appear at last interpolation cycle of current orbit segment the time, P _sP _SeDistance situation less than Δ L might appear.Under these circumstances, if do not change direction but carry out moving interpolation according to the working direction of theory, then can arrive P _E1Thereby, produced and cut, and this situation is absolute prohibition.Its disposal route is to guarantee under the prerequisite of machining precision, and the final position of this interpolation cycle is transformed in the coupled next line segment, namely with P _sBe the center of circle, Δ L is distance, draws circular arc, with line segment P _SeP _EeJoining P _eFluctuation can not occur with regard to assurance speed like this, make it smooth variation, improve actual travelling speed and the efficient of processing.But but can be owing to not arriving theoretical terminal point P _SeThereby generation error.

Obtaining of transit point:

Current orbit segment links to each other with circular arc

1, two orbit segments are in the same space plane

Expression formula by face can be learnt: when

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _s-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (2-3)

+(z _s-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0

The time, expression circular arc and current orbit segment are on the same plane, because the singularity of circular arc, when two orbit segments are in meeting in the same plane, according to working as orbit segment P shown in Figure 2 _sP _SeWith P _SeThere are following two kinds of connected modes in the angle relation of O, and corresponding forward processing method also is different.

Situation 1: as shown in Figure 2: ∠ OP _SeP _q=90 ° (2-4)

Be line segment P _qP _SeTangent with circular arc, then as orbit segment P _sP _SeWith P _SeThe O angle is during greater than 90 ° (concrete determination methods see below 3.2.2 trifle described), as orbit segment P _sP _SeThe remaining distance of last interpolation cycle is Δ L ₁, if carry out moving interpolation along its former direction, then reach P _Le, can produce and cut, therefore with p _sBeing the center of circle, is radius with Δ L, intersects at P with circular arc _e

Obtaining of transit point position: P sets up an office _s, P _Se, P _e, P _Ee, the coordinate of O is as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), (x _Ee, y _Ee, z _Ee), (x _o, y _o, z _o), arc radius is R, according to relevant geometric relationship as can be known:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (2-5)

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _O) ²＝R ²

And some P _eIn plane P _SeOP _EeIn, then:

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _xe-z _o)] (2-6)

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0

Find the solution this quadratic equation group and can obtain P _eCoordinate figure x _E1, y _E1, z _E1, x _E2, y _E2, z _E2, simultaneously again owing to will satisfy:

(x _se-x _e)(x _e-x _ee)＜0 (2-7)

So can obtain unique solution.

Situation 2: as shown in Figure 3, ∠ P _qP _SeO=90 ° is line segment P _xP _SeTangent with circular arc, then as line segment P _sP _SeWith P _SeThe O angle is during less than 90 °, as orbit segment P _sP _SeThe remaining distance of last interpolation cycle is Δ L ₁, if according to the method described above, with P _sBeing the center of circle, is radius with Δ L, intersects at P with circular arc _Le, then over-cutting can occur, and this is to allow anything but to occur, so should take in this case to carry out moving interpolation along former trajectory direction.

Obtaining of transit point position: P sets up an office _s, P _Se, P _e, P _Ee, the coordinate of O is as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), (x _Ee, y _Ee, z _Ee), (x _o, y _o, z _o), arc radius is R, known P _s, P _Se, P _Ee, the O coordinate figure need be asked for P now _e, P wherein _sP _eDistance known, be set at Δ L, then can learn according to geometric formula

$\frac{x_e-x_{\mathrm{se}}}{x_s-x_{\mathrm{se}}}=\frac{y_e-y_{\mathrm{se}}}{y_s-y_{\mathrm{se}}}=\frac{z_e-z_{\mathrm{se}}}{z_s-z_{\mathrm{se}}}---(2-8)$

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ²

Separate this quadratic equation group and can obtain P _eCoordinate figure x _E1, y _E1, z _E1, x _E2, y _E2, z _E2, satisfy formula (2-9) again simultaneously, so can obtain unique solution.

(x _e-x _o) ²+(y _e-y _o) ²+(z _e-z _o) ²＞R ² (2-9)

2, two orbit segments are not in the same space plane

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

Namely-(y _s-y _Se) [(x _Se-xo) (z _Se-z _Ee)-(x _Se-x _Ee) (z _Se-z _o)] (2-10)

+(z _s-z _se)[(z _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]≠0

The time, expression present segment and circular arc be not on same plane, and be similar on same plane to two orbit segments, and it is according to being P according to the projection of present segment on the circular arc plane shown in figure below _s' P _SeWith P _SeO angle relation is divided into two kinds of connected modes.

Situation 1: as shown in Figure 4: ∠ p _qP _SeO=90 °, i.e. line segment P _qP _SeTangent with circular arc, the projection of present segment on the circular arc plane is P _s' P _SeWith p _SeThe O angle is during greater than 90 °, current orbit segment P _sp _SeThe remaining distance of last interpolation cycle is Δ L1, if carry out moving interpolation along rectilinear direction, then reaches P _Le, can produce and cut, therefore with p _sBeing the center of circle, is radius with Δ L, intersects at P with circular arc _eP sets up an office _s, p _Se, P _e, P _Ee, the coordinate of O is as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), (x _Ee, y _Ee, z _Ee), (x _o, y _o, z _o), arc radius is R, known P _s, p _Se, P _Ee, the O coordinate figure need be asked for P now _e, P wherein _sP _eDistance known, be set at

According to relevant geometric relationship as can be known:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (2-11)

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _o) ²＝R ²

And some P _e, in plane P _SeOP _EeIn, then

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (2-12)

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0

Above-mentioned quadratic equation group found the solution to obtain P _eCoordinate figure x _E1, y _E1, z _E1, x _E2, y _E2, z _E2, simultaneously again owing to need to satisfy: (x _Se-x _e) (x _e-x _Ee)＜0 (2-13);

So can obtain unique solution.

Situation 2: as shown in Figure 5, ∠ P _qP _SeO=90 °, i.e. line segment p _qP _SeTangent with circular arc, when the projection P of current orbit segment at arc surface _s' P _SeWith p _SeThe O angle is during less than 90 °, current orbit segment P _sP _SeThe remaining distance of last interpolation cycle is Δ L ₁, if according to the method described above, with P _sBeing the center of circle, is radius with Δ L, intersects at P with circular arc _Le2, or move along former orbit segment direction, then arrive P _Le1, this dual mode all over-cutting can occur, and this allows to occur anything but.Therefore ought should not take above-mentioned switching method in this case.

As shown in Figure 6, this forwarding method be according to current orbit segment with and at circular arc plane projection straight line p _s' P _SeIn the plane that constitutes, change original direction of motion, make it to arrive p _s' P _SeExtended line on p _SeThe point.

Obtaining of transit point position: P sets up an office _s, P _Se, P _e, P _Ee, the coordinate of O is as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), (x _Ee, y _Ee, z _Ee), (x _o, y _o, z _o), arc radius is R, known P _s, P _Se, P _Ee, the O coordinate figure.

Straight line P _sP _SeVector:

n＝(x _s-x _se，y _s-y _se，z _s-z _se)＝(M，N，P) (2-14)

The equation on circular arc plane is:

(x-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (2-15)

+(z-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0

Can be reduced to: Ax+By+Cz+D=0 (2-16)

The normal line vector on circular arc plane: m=(A, B, C) (2-17)

Then straight line and circular arc plane included angle:

If through a P _s, P _Se, P _eThe plane, and it is perpendicular to the circular arc plane, then its plane equation is:

(x-x _se)(BP-NC)+(y-y _se)(-AP+MC) (2-19)

+(z-z _se)(AN-MB)＝0

And its normal line vector: P _n=(U, V, W) (2-20)

And then be reduced to: Ux+Vy+Wz+E=0 (2-21)

Then as shown in Figure 6:

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (2-22)

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (2-23)

Ux _e+Vy _e+Wz _e+E＝0 (2-24)

(x _e-x _se)(BP-NC)-(y _e-y _se)(-AP+MC) (2-25)

+(z _e-z _se)(AN-MB)＝0

Solving simultaneous equation can obtain P _eCoordinate figure x _E1, y _E1, z _E1, x _E2, y _E2, z _E2, simultaneously again owing to require: min[(x _e-x _Se) ²+ (y _e-y _Se) ²+ (z _e-z _Se) ²], so can obtain unique solution.

The analysis of switching Algorithm Error

The switching error analysis that current orbit segment links to each other with straight-line segment

According to the velocity angle of two orbit segments can acute angle, the two kinds of situations in obtuse angle.

The velocity angle is acute angle, namely angle α ∈ as shown in Figure 7 (0 ° 90 °], P _s, P _Se, P _eCoordinate as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), then

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---(2-26)$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---(2-27)$

Article two, included angle of straight line cosine:

$\cos \mathrm{\&alpha;}=\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---(2-28)$

According to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{{2\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---(2-29)$

And then obtain:

${{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2=\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {(\mathrm{\&pi;}-\mathrm{\&alpha;})}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-\mathrm{\&Delta;}{L}^2)}---(2-30)$

As shown in Figure 7, its error is:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---(2-31)$

Bring following formula into error formula simultaneously.Have this to learn, error only with Δ L ₁, Δ L, relevant with two sections path velocity vector angle α.

As can be seen from Figure 8, along with the increase of angle, error increases, simultaneously Δ L ₁With the ratio of Δ L not simultaneously, its amplification curve also is different.By analyzing as can be known, when

The time, along with Δ L ₁Increase and error increases, when

The time, then along with Δ L ₁The increase error reduce, specifically as shown in Figure 9

Angle α ∈ between two orbit segments [45 ° 90 °], and Δ L ₁∈ [ctg α Δ L], P _s, P _Se, P _eCoordinate as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e),

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---(2-32)$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---(2-33)$

Article two, included angle of straight line cosine:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---(2-34)$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2+{{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2}^2-{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}^2}{2{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;L}}_2}---(2-35)$

And then obtain:

${\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2={\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1}^2-{\mathrm{\&Delta;L}}^2)}---(2-36)$

Error is as shown in Figure 10:

$e=\frac{{\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---(2-37)$

Can be learnt that by Figure 11 along with the increase of orbit segment angle, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---(2-38)$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing.As 50 ° the time,

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\mathrm{<figure-callout\; id="50"\; label="sin"\; filenames="DEST\_PATH\_HSB00001105275500032.png"\; state="\{\{state\}\}">sin</figure-callout>}\frac{50}{2}}=1.18\mathrm{\&Delta;L}---(2-39)$

So, have monotone increasing.And during for 70/80 °, because

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\mathrm{<figure-callout\; id="70"\; label="sin"\; filenames="HSA00000868392400012.png,HSA00000868392400021.png"\; state="\{\{state\}\}">sin</figure-callout>}\frac{70}{2}}=0.8717\mathrm{\&Delta;L}>\mathrm{\&Delta;Lctg\&alpha;}=0.3640\mathrm{\&Delta;L}---(2-40)$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\sin \frac{\mathrm{\&alpha;}}{2}}=\frac{\mathrm{\&Delta;<figure-callout\; id="2"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png"\; state="\{\{state\}\}">L</figure-callout>}}{2\mathrm{<figure-callout\; id="80"\; label="sin"\; filenames="DEST\_PATH\_HSB00001105275500032.png,DEST\_PATH\_HSB00001105275500033.png"\; state="\{\{state\}\}">sin</figure-callout>}\frac{80}{2}}=0.7779\mathrm{\&Delta;L}>\mathrm{\&Delta;Lctg\&alpha;}=0.1763\mathrm{\&Delta;L}---(2-41)$

Has maximum value.

Angle α ∈ between 2 o'clock [45 ° 90 °], and Δ l ₁∈ [0ctg α],

If P _s, P _Se, P _eCoordinate as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e), error is as shown in Figure 12:

e＝ΔL ₁ (2-42)

Angle α ∈ between 2 o'clock (0 ° 45 °]

If P _s, P _Se, P _eCoordinate as follows: (x _s, y _s, z _s), (x _Se, y _Se, z _Se), (x _e, y _e, z _e),

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---(2-43)$

${\mathrm{\&Delta;<figure-callout\; id="1"\; label="L"\; filenames="DEST\_PATH\_HSB00001105275500072.png,HSA00000868392400012.png"\; state="\{\{state\}\}">L</figure-callout>}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---(2-44)$

Error is as shown in Figure 13:

e＝ΔL ₁ (2-45)

For the acute angle switching, change owing to have bigger vector acceleration, should guarantee that motor is in its acceleration range, otherwise can exceed the ability of motor self, cause bigger trajectory error.Simultaneously because the existence of its switching line segment, make not so little, relatively large of acute angle, so guarantee the workpiece of wedge angle for needs, this algorithm is inapplicable.

The switching error analysis that current orbit segment links to each other with circular arc

1, present segment and circular arc are on same plane

∠ P _sP _Se0＞90 ° is the left side of tangent line

It is consistent that error analysis and line segment directly connect, its angular range [0 ° 180 °].

As shown in Figure 14: because the value of its line segment is smaller for circle, can wait be all with circular arc general.So, under these circumstances, as ∠ P _sp _SeP _qDuring ∈ [0 ° 90 °], it can be equal to next orbit segment is straight-line segment, and the angle of velocity is the situation of acute angle.Then find the solution can be identical with it for error.

As ∠ P _sP _SeP _qDuring ∈ [90 ° 180 °], its error is found the solution, and can be equal to next orbit segment be straight-line segment, and the angle of its velocity is the situation at obtuse angle.Specific algorithm can be with reference to above.

∠ P _sP _SeP _qThe judgement of corner dimension can be according to ∠ P _sOP _EeWith ∠ P _SeOP _EeSize draw, as ∠ P _sOP _Ee≤ ∠ P _SeOP _EeThe time, then represent ∠ P _sP _SeP _q∈ [90 ° 180 °], otherwise ∠ P then _sP _SeP _q∈ [0 ° 90 °].

∠ P _sP _Se0＜90 ° is the right side of tangent line

Angular range [0 ° 180 °].As ∠ P _sOP _Ee〉=∠ P _SeOP _EeThe time, then represent ∠ P _sP _SeP _q∈ [0 ° 90 °]

Can know its error by inference by Figure 15 is: actual interpolated point to the difference of the distance in the center of circle and arc radius namely:

$e=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---(2-46)$

As ∠ P _sOP _Ee＜∠ P _SeOP _EeThe time, then represent ∠ P _sP _SeP _q∈ [90 ° 180 °]

Can know its error by inference by figure is distance between actual interpolated point and the present segment theoretical end point, that is:

e＝ΔL-ΔL ₁ (2-47)

2, current orbit segment and circular arc be not in same plane

Current orbit segment is connected with circular arc and angle is ∠ P greater than 90 ° _s' P _Se0＞90 °

Then connect between error analysis and the line segment consistent, angular range [0 ° 180 °].

Since its line segment with respect to the circle for, very little, can wait be all with circular arc general, so, under these circumstances, ∠ P _s' P _SeP _qThe angular range determination methods is with identical when the same plane, namely according to ∠ P _s' OP _EeWith ∠ P _SeOP _EeMagnitude relationship judge and to obtain.

As ∠ P _s' P _SeP _q∈ [0 ° 90 °], then can be equal to next orbit segment is that straight-line segment and both velocity angles are the situation of acute angle, its error can be obtained according to the computing method of above introducing.

As ∠ P _s' P _SeP _q∈ [90 ° 180 °], then can be equal to next orbit segment is that straight-line segment and both velocity angles are the situation at obtuse angle, its error can be obtained according to the computing method of above introducing.

Current orbit segment is connected with circular arc and angle is ∠ P less than 90 ° _s' P _Se0＜90 °, angular range [0 ° 180 °].When its projection line and tangent line angle are that [0 ° 90 °] are ∠ P _s' P _SeP _q∈ [0 ° 90 °].

As ∠ P _s' P _SeP _q∈ [0 ° 90 °], as Figure 17, Figure 18 as can be known, this forwarding method produces error and has two kinds, a, on the circular arc plane, actual interpolated point and the center of circle directly apart from the difference of arc radius.B, owing to be the forwarding method that the solid that adopts prolongs, rather than directly along the direction of present segment, certain included angle is arranged, its value size and Δ L, Δ L ₁, Relevant, and the variation of this angle should be in the allowed band of motor self.

Finding the solution for situation a of error is

$e_1=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---(2-48)$

And for situation b, then ask for e according to the content of trifle above ₂, being equivalent to next orbit segment is straight line, and the situation of angle when being the obtuse angle, then obtains max (e ₁e ₂).

When its projection line and tangent line angle are that [90 ° 180 °] are ∠ P _s' P _SeP _q∈ [90 ° 180 °].

As above two figure as can be known, this forwarding method produces error and has two kinds, a, on the circular arc plane, error between actual interpolated point and the mathematical point is the distance between Pe and the Pse, b, owing to be that the solid that adopts prolongs, rather than directly along the direction of present segment, certain included angle is arranged, value size and Δ L, Δ L ₁, Relevant, and the variation of this angle should be in the allowed band of motor self.

Finding the solution for situation a of error is

$e_1=\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}---(2-49)$

And for b, then ask for e according to the content of trifle above ₂, then obtain max (e ₁e ₂).

∠ P _s' P _SeP _qThe judgement of corner dimension can be according to ∠ P _s' OP _EeWith ∠ P _SeOP _EeSize draw, as ∠ P _s' OP _Ee＜∠ P _SeOP _EeThe time, then represent ∠ P _s' P _SeP _q∈ [90 ° 180 °], otherwise ∠ P then _s' P _SeP _q∈ [0 ° 90 °].

Also has a limiting case simultaneously: when the plane of present segment perpendicular to the circular arc place, then in this case, can only carry out forced deceleration, thereby make Δ L ₁=Δ L (2-50)

The processing checking

Following Figure 19 is equipped with the workpiece drawing of processing above the 8-axes 5-linkage digital control system TDNC-H8 of independent research for the five-axis machine tool TDNC-W200 in seminar, it is to be made of jointly circular arc, straight line, the algorithm that article is set forth is processed checking in the TDNC-H8 of independent research Five Axis CNC System, its interpolation cycle is 2ms, and permissible error is 3 μ m.Material Processing is: the 6082-T651 aluminium sheet.Process tool is: ADL aluminum end milling cutter, 55 ° of helix angles, 3 swords, the place of production: Taiwan.Specification: φ 3.0X9CX10DX80L-3F.The correlation parameter of processing: speed of feed: 25mm/s, the speed of mainshaft: 1500r/min, cutting depth: 1mm.Shown in following two forms of each key point coordinate and radius parameter:

Table 2-1 processing work key point parameter

Table 2-2 processing work correlation parameter

Its theoretical error at each point, substantial measurement errors (error is the numerical value of the influence acquisition that the removal tool radius causes when measuring herein, the point of a knife locus of points error when namely carrying out dry running by the measurement cutter) and switching speed are shown in following table 2-3.Wherein A is initial and terminating point.

Table 2-3 transit point error

Be example with the B point, its concrete synoptic diagram is: its angle is less than 45 °, so error: e=L1=2.8 μ m, L=5.96 μ m, switching key point B1, B, shown in the following form of B2 three point coordinate values:

Table 2-4P _BPoint transit point coordinate

Figure 20 is B point transit point track

As can be known from the results, the application of this algorithm can obtain than higher in the past switching precision and switching speed.

Beneficial effect: the two kinds of forward processing methods and the error condition that connect, be connected with circular arc at current orbit segment and straight line elaborate and analyze: for the error of the switching processing of the straight-line segment different error of its generation and the variation relations of interpolation surplus and angle analyzed according to its velocity angle.For whether calculating then proposition foundation on same plane with forward processing method and the error of circular arc, set forth with circular arc tangential line angle two aspects.More complete for the analysis of its forward processing method and error with respect to existing document.And because detailed more and deep for various transit case analyses, thereby can obtain than higher in the past switching precision and switching speed.

Above-described embodiment only is explanation technical conceive of the present invention and characteristics, and its purpose is to allow the personage who is familiar with this technology can understand content of the present invention and enforcement according to this, can not limit protection scope of the present invention with this.All equivalences that spirit essence is done according to the present invention change or modify, and all should be encompassed within protection scope of the present invention.

Claims (1)

1. orbit segment switching Processing Algorithm based on trueness error control that is used for numerically-controlled machine, comprise workpiece to be processed is placed on the workbench of described numerically-controlled machine, the operate portions of described numerically-controlled machine is carried out process operation to described processing work, it is characterized in that:

The first step: ask transit point

In the interpolation operation of described numerically-controlled machine, the publicity of step pitch is:

ΔL＝VT；

In the formula, Δ L represents step pitch; T represents interpolation cycle, and V represents the movement velocity of the operate portions of described numerically-controlled machine;

If arc P _SeP _EeSome P _SeTangent line be P _qP _SeThe coordinate of Ps is (x _s, y _s, z _s), P _SeCoordinate be (x _Se, y _Se, z _Se), P _eCoordinate be (x _e, y _e, z _e), P _EeCoordinate be (x _Ee, y _Ee, z _Ee), arc P _SeP _EeThe coordinate of center of circle O be (x _o, y _o, z _o), arc P _SeP _EeRadius be R; P wherein _s, P _Se, P _EeCoordinate figure be known, need ask for P now _eCoordinate figure, P wherein _sP _eDistance be Δ L;

One, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are in same space plane;

Expression formula by face can be learnt: when

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _s-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _s-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (201)；

The time, represent that namely described current orbit segment and arc track section are on same plane;

(1) works as P _sP _SeWith P _SeThe angle of O is during greater than 90 °; When the last interpolation cycle of described current orbit segment, the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, need the final position of described last interpolation cycle is transformed in next coupled arc track section, namely with a P _sBeing the center of circle, serves as that distance is drawn circle with Δ L, with described arc track section arc P _SeP _EeThe intersection point that intersects is described transit point Pe; Then can obtain formula (202) according to geometric relationship:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (202)；

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _O) ²＝R ²

And some Pe is in plane P _SeOP _SeIn, then get formula (203):

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (203)：

Can obtain two groups of solution of equations by finding the solution, again because of described transit point P _eAt described arc track section P _SeP _EeBetween, then also have constraint condition: (x _Se-x _e) (x _e-x _Ee)＜0, thus unique solution obtained, and then can determine the coordinate figure of described transit point Pe;

(2) work as P _sP _SeWith P _SeThe angle of O is during less than 90 °, when the last interpolation cycle of described current orbit segment, and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, along track P _sP _SeExtend to P _eMake P _sP _eLength equal Δ L, described some P _eBe transit point; Then can obtain formula (204) according to geometric relationship:

$\frac{x_e-x_{\mathrm{se}}}{x_s-x_{\mathrm{se}}}=\frac{y_e-y_{\mathrm{se}}}{y_s-y_{\mathrm{se}}}=\frac{z_e-z_{\mathrm{se}}}{z_s-z_{\mathrm{se}}}$

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (204)；

Can obtain two groups of solution of equations by finding the solution, satisfy constraint condition simultaneously again: (x _e-x _o) ²+ (y _e-y _o) ²+ (z _e-z _o) ²＞R ²Thereby, obtain unique solution, and then can determine described transit point P _eCoordinate figure;

Two, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are not in same space plane; Define described current orbit segment P _sP _SeProjection line at the face at described arc track section place is P _{S '}P _Se

Expression formula by face can be learnt: when

(x _s-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _s-y _se)[(z _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)] (205)；

+(z _s-z _se)[(x _se-x _o)(y _se-y _ee)-(z _se-x _ee)(y _se-y _o)]≠0

The time, represent that namely described current orbit segment and arc track section are on same plane;

(1) works as P _{S '}P _SeWith P _SeThe angle of O is during greater than 90 °, the last interpolation cycle of described current orbit segment, the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, need the final position of described last interpolation cycle is transformed in next coupled arc track section, namely with a P _sBeing the center of circle, serves as that distance is drawn circle with Δ L, with described arc track section arc P _SeP _EeThe intersection point that intersects is described transit point P _eThen can obtain formula (206) according to geometric relationship:

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ²

(x _e-x _O) ²+(y _e-y _O) ²+(z _e-z _O) ²＝R ² (206)；

And some P _eIn plane P _SeOP _EeIn, then get formula (207):

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (207)；

Can obtain two groups of solution of equations by finding the solution, again because of described transit point P _eAt described arc track section P _SeP _EeBetween, then also have constraint condition: (x _Se-x _e) (x _e-x _Ee)＜0, thus unique solution obtained, and then can determine described transit point P _eCoordinate figure;

(2) work as P _{S '}P _SeWith P _SeThe angle of O is during less than 90 °, when the last interpolation cycle of described current orbit segment, and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁Less than Δ L, along P _{S '}P _SeExtend to P _eMake P _{S '}P _eLength equal Δ L, described some P _eBe transit point;

Straight line P _sP _SeVector be:

n＝(x _s-x _se，y _s-y _se，z _s-z _se)＝(M，N，P) (208)；

The equation on the plane at described arc track section place is:

(x-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z-z _se)[(x _se-x _o)(y _se-y _ee)-(z _se-x _ee)(y _se-y _o)]＝0 (209)；

Abbreviation is:

Ax+By+Cz+D＝0 (210)；

The normal line vector on the plane at described arc track section place is:

m＝(A，B，C) (211)；

Described straight line P _sP _SeAngle with the plane at described arc track section place

For:

If through a P _s, P _Se, P _eThe plane, and it is perpendicular to the plane at described arc track section place, then its plane equation is:

(x-x _se)(BP-NC)+(y-y _se)(-AP+MC)

+(z-z _se)(AN-MB)＝0 (213)；

Described through a P _s, P _Se, P _eThe normal on plane be:

P _n＝(U，V，W) (214)；

Abbreviation is:

Ux+Vy+Wz+E＝0 (215)；

Can obtain according to geometric relationship:

(x _e-x _se)[(y _se-y _o)(z _se-z _ee)-(y _se-y _ee)(z _se-z _o)]

-(y _e-y _se)[(x _se-x _o)(z _se-z _ee)-(x _se-x _ee)(z _se-z _o)]

+(z _e-z _se)[(x _se-x _o)(y _se-y _ee)-(x _se-x _ee)(y _se-y _o)]＝0 (216)；

(x _e-x _s) ²+(y _e-y _s) ²+(z _e-z _s) ²＝ΔL ² (217)；

Ux _e+Vy _e+Wz _e+E＝0 (218)；

(x _e-x _se)(BP-NC)-(y _e-y _se)(-AP+MC)

+(z _e-z _se)(AN-MB)＝0 (219)；

Can obtain two groups of solution of equations by solving simultaneous equation, satisfy constraint condition: min[(x simultaneously again _e-x _Se) ²+ (y _e-y _Se) ²+ (z _e-z _Se) ²], thereby obtain unique solution, and then can determine described transit point P _eCoordinate figure;

Second step: switching error analysis

One, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are in same space plane;

(1) ∠ P _sP _Se0＞90 °

Arc P _SeP _eWith respect to arc P _SeP _Ee, its length is short, and it is equal to line segment P _SeP _eLength

(1) as ∠ P _sP _SeP _q∈ [0 ° 90 °] namely is equivalent to described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween velocity angle α when being acute angle,

The last interpolated point P of described current orbit segment _sWith described transit point P _eBetween distance, delta L be:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(220\right);$

The last interpolated point P of described current orbit segment _sAnd P _SeBetween distance, delta L ₁For:

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(221\right);$

The cosine of described angle α:

$\cos \mathrm{\&alpha;}=\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(222\right);$

According to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{{2\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(223\right):$

And then obtain:

${{\mathrm{\&Delta;L}}_2=\mathrm{\&Delta;L}}_1\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {(\mathrm{\&pi;}-\mathrm{\&alpha;})}^2-({{\mathrm{\&Delta;L}}_1}^2-\mathrm{\&Delta;}{L}^2)}---\left(224\right):$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1\mathrm{\&Delta;}{L}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(225\right):$

Bring the calculated value of formula (220), formula (221), formula (222), formula (223) and formula (224) into described error formula (225) simultaneously; According to described error formula (225) as can be known, error only with Δ L ₁, Δ L is relevant with angle α; That is: when The time, along with Δ L ₁Increase and error increases, when

The time, then along with Δ L ₁The increase error reduce;

(2) as ∠ P _sP _SeP _q∈ [90 ° 180 °],

As described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween angle a ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(226\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(227\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(228\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{2{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(229\right);$

And then obtain:

${\mathrm{\&Delta;L}}_2={\mathrm{\&Delta;L}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;L}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(230\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(234\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;L}}_1=\frac{\mathrm{\&Delta;L}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(235\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (236)；

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(237\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(238\right);$

Error formula is:

e＝ΔL ₁ (239)；

(2) ∠ P _sP _Se0＜90 °

As ∠ P _sOp _Ee〉=∠ P _SeOP _EeThe time, ∠ P then _sP _SeP _q∈ [0 ° 90 °]

Described transit point P _eTo described arc track section P _SeP _EeThe distance and described arc P of center of circle O _SeP _Ee'sThe difference of radius is namely:

$e=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---\left(240\right)$

As ∠ P _sOP _Ee＜∠ P _SeOP _EeThe time, ∠ P then _sP _SeP _q∈ [90 ° 180 °], error formula is:

e＝ΔL-ΔL ₁ (241)；

Two, be connected with an arc track section when described current orbit segment, and described current orbit segment and arc track section are not in same space plane;

(1) ∠ P _{S '}P _Se0＞90 °

Arc P _SeP _eWith respect to arc P _SeP _Ee, its length is short, and it is equal to line segment P _SeP _eLength

(1) as ∠ P _s' P _SeP _q∈ [0 ° 90 °], namely be equivalent to described current orbit segment P _sP _SeWith described line segment P _SeP _eBetween velocity angle α when being acute angle,

The last interpolated point P of described current orbit segment _sWith described transit point P _eBetween distance, delta L be:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(242\right);$

The last interpolated point P of described current orbit segment _sAnd P _SeBetween distance, delta l ₁For:

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(243\right);$

The cosine of described angle α:

$\cos \mathrm{\&alpha;}=\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(244\right);$

According to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{{2\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(245\right):$

And then obtain:

${\mathrm{\&Delta;L}}_2=\mathrm{\&Delta;}{L}_1\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {(\mathrm{\&pi;}-\mathrm{\&alpha;})}^2-({{\mathrm{\&Delta;L}}_1}^2-\mathrm{\&Delta;}{L}^2)}---\left(246\right):$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(247\right):$

Bring the calculated value of formula (242), formula (243), formula (244), formula (245) and formula (246) into described error formula (247) simultaneously; According to described error formula (247) as can be known, error only with Δ l ₁, Δ L is relevant with angle α; That is: when

The time, along with Δ L ₁Increase and error increases, when

The time, then along with Δ L ₁The increase error reduce;

(2) as ∠ P _{S '}P _SeP _q∈ [90 ° 180 °],

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(248\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(249\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(250\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{2{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(251\right);$

And then obtain:

${\mathrm{\&Delta;L}}_2={\mathrm{\&Delta;L}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;L}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(252\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(253\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;L}}_1=\frac{\mathrm{\&Delta;L}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(254\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (255)；

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(256\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(257\right);$

Error formula is:

e＝ΔL ₁ (258)；

(2) ∠ P _{S '}P _Se0＜90 °

(1) as ∠ P _{S '}P _SeP _q∈ [0 ° 90 °], there are two kinds in error:

$e_1=\sqrt{{(x_e-x_O)}^2+{(y_e-y_O)}^2+{(z_e-z_O)}^2}-R---\left(259\right)$

e ₂As follows:

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(260\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(261\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(262\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{2{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(263\right);$

And then obtain:

${\mathrm{\&Delta;L}}_2={\mathrm{\&Delta;L}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;L}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(264\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(265\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;L}}_1=\frac{\mathrm{\&Delta;L}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(266\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point p _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (267)；

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(268\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(269\right);$

Error formula is:

e＝ΔL ₁ (270)；

Then obtain max (e ₁e ₂);

(1) as ∠ P _{S '}P _SeP _q∈ [90 ° 180 °], there are two kinds in error:

$e_1=\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}---\left(271\right)$

e ₂As follows:

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [ctg α Δ L]:

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(272\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(273\right);$

The cosine of described angle α:

$\cos (\mathrm{\&pi;}-\mathrm{\&alpha;})=$

$\frac{(x_{\mathrm{se}}-x_s)\&CenterDot;(x_e-x_{\mathrm{se}})+(y_{\mathrm{se}}-y_s)\&CenterDot;(y_e-y_{\mathrm{se}})+(z_{\mathrm{se}}-z_s)\&CenterDot;(z_e-z_{\mathrm{se}})}{\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}+\sqrt{{(x_e-x_{\mathrm{se}})}^2+{(y_e-y_{\mathrm{se}})}^2+{(z_e-z_{\mathrm{se}})}^2}}---\left(274\right);$

Then according to the cosine law as can be known:

$\cos \mathrm{\&alpha;}=-\frac{{{\mathrm{\&Delta;L}}_1}^2+{{\mathrm{\&Delta;L}}_2}^2-{\mathrm{\&Delta;L}}^2}{2{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2}---\left(275\right);$

And then obtain:

${\mathrm{\&Delta;L}}_2={\mathrm{\&Delta;L}}_1\cos \left(\mathrm{\&alpha;}\right)+\sqrt{{{\mathrm{\&Delta;L}}_1}^2\cos {\left(\mathrm{\&alpha;}\right)}^2-({{\mathrm{\&Delta;L}}_1}^2-{\mathrm{\&Delta;L}}^2)}---\left(276\right);$

Error formula is:

$e=\frac{{\mathrm{\&Delta;L}}_1{\mathrm{\&Delta;L}}_2\sin \mathrm{\&alpha;}}{\mathrm{\&Delta;L}}---\left(277\right);$

Along with the increase of angle α, error is reducing, for Δ L ₁,

${\mathrm{\&Delta;L}}_1=\frac{\mathrm{\&Delta;L}}{2\sin \frac{\mathrm{\&alpha;}}{2}}\&Element;\left[\mathrm{ctg\&alpha;\&Delta;L}\right]---\left(278\right);$

When its not in this scope, error is along with Δ L ₁Increase and have the trend of monotone increasing;

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ [45 ° 90 °], and the last interpolated point P of described current orbit segment _sWith distal point P _SeBetween distance, delta L ₁During ∈ [0ctg α],

Error formula is:

e＝ΔL ₁ (279)；

As described current orbit segment and described line segment P _SeP _eBetween angle α ∈ (0 ° 45 °],

$\mathrm{\&Delta;L}=\sqrt{{(x_e-x_s)}^2+{(y_e-y_s)}^2+{(z_e-z_s)}^2}---\left(280\right);$

${\mathrm{\&Delta;L}}_1=\sqrt{{(x_{\mathrm{se}}-x_s)}^2+{(y_{\mathrm{se}}-y_s)}^2+{(z_{\mathrm{se}}-z_s)}^2}---\left(281\right);$

Error formula is:

e＝ΔL ₁ (282)；

Then obtain max (e ₁e ₂);

(3) when the plane perpendicular to described arc track section place, the plane at described current orbit segment place, then:

ΔL ₁＝ΔL (283)。

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