CN103646150B - Continuous transition curve constitution method based on quintic Bezier curve - Google Patents

Abstract

The invention discloses a G3 continuous transition curve constitution method based on a quintic Bezier curve. The method comprises the following steps: step 1, working out the first-order steering vector and the second-order steering vector at the end point of a primitive curve; step 2, enabling the primitive curve and a transition curve to satisfy a G2 continuity condition at the connection point; step 3, on the basis of the step 2, enabling the primitive curve and the transition curve to satisfy G3 continuity at the connection point; step 4, inversely calculating the reference mark of a quintic Bezier transition curve; step 5, normally calculating the quintic Bezier transition curve according to the control peak in the step 4. According to the invention, a Bezier curve is adopt to construct the transition curve, so that the advantages of controlling the shape of the curve easily and visually and providing convenience for interactive design can be achieved; besides, the few the parameters are, the few the orders are, so that the calculation frequency is low, and the memory space occupied in a computer is relatively small; the invention can be applied to construction of transition curves in CAD modeling.

Description

A kind of g3 continuous transition curve constructing method based on 5 bezier curves

Technical field

The present invention relates to a kind of g3 continuous transition curve constructing method based on 5 bezier curves, belong to cad technology neck Domain.

Background technology

G3 is continuously sometimes very useful in cad modeling, such as: the fillet curve of blade front and rear edge, iphone phone housing, vapour Car design etc..G3 continuously belongs to curvature variation continuously, and this continuous rank not only has g0, g1, g2 continuous level another characteristic Outside, at junction point, curvature variation is also continuous, and this makes Curvature varying more smooth, the surface of this continuous rank There is the visual effect more smooth than g2.Some cad softwares have the construction ability of g3 full curve curved surface, such as autodesk at present Alias2011essentials constructs g3 full curve with 7 order polynomials.But 7 order polynomial degree of curve are high, curve easily occurs " vibration " is so difficult model is interacted with formula design, and curve exponent number is higher it is desirable to the number of times calculating is more, and storage accounts for Also bigger with space.For above-mentioned problem present invention utilizes three order derivatives of two ends restrictive curve are it is proposed that use 5 Secondary bezier curve construction g3 continuous transition curve method.

Content of the invention

1st, purpose: it is an object of the present invention to provide a kind of g3 continuous transition curve constructing method based on 5 bezier curves, Improve the quality of easement curve moulding in cad modeling process.

2nd, technical scheme: the purpose of the present invention is achieved through the following technical solutions.

A kind of g3 continuous transition curve constructing method based on 5 bezier curves, it comprises the following steps:

Step one, obtain single order at end points for the primitive curve and lead resultant second order and lead arrow；

Step 2, primitive curve and easement curve is made to meet the g2 condition of continuity at junction point；

Step 3, on the basis of step 2, make primitive curve and easement curve meet g3 at junction point continuous；

Step 4, according to 5 bezier easement curve control point of step 3 inverse；

Step 5, just calculating 5 bezier easement curves according to the control vertex in step 4.

Wherein, " primitive curve " described in step one refers to need to carry out two parameter curves of transition connection, " single order Lead resultant second order and lead arrow " refer to the derivation to general parameterss for the parameter curve；

Wherein, " the g2 condition of continuity " described in step 2 refers to that two groups of curves meet g0, g1 continuous strip at junction point Under part, public curvature is had to swear (size and Orientation)；G0 continuously refers to an end points of a curve and the one of another curve End points contacts；G1 continuously refer to two curves junction meet g0 continuous under the conditions of, cut arrow direction identical.

Wherein, " g3 is continuous " described in step 3 refers to 5 bezier curves and primitive curve in two junction points The derivative of the curvature at place is continuous；

Wherein, " 5 bezier easement curve control point of inverse " described in step 4 refer to drawing 5 bezier mistakes First control vertex is calculated, 5 times bezier easement curve control point has 6 before crossing curve；

Wherein, " just calculating 5 bezier easement curves " described in step 5 refers to according to the control top obtaining in step 4 Point draws the expression formula of 5 bezier easement curves.

3rd, advantage and effect

The invention provides with the continuous easement curve method of 5 bezier curve construction g3, using bezier curve structure Make easement curve and there is the shape being easy to intuitively controlling curve, the advantage facilitating interaction design, and number of parameters is fewer, rank Number is lower, and calculation times are also fewer, and memory space is less in a computer, and the present invention can be used for easement curve in cad modeling Construction.

Brief description

Fig. 1 combs schematic diagram for curvature of curve

Fig. 2 is for the present invention with 5 bezier curve construction g3 continuous transition curve synoptic diagrams

Fig. 3 is easement curve one, second dervative schematic diagram at junction point

Fig. 4 is FB(flow block) of the present invention.

In figure code name, symbol description are as follows:

G0, g1, g2, g3 curve g0, g1, g2, g3 are continuous

p₀, p₁, p₂, p₃, p₄, p₅5 bezier curve controlled summits

r₀Curve 0 and the junction point of easement curve

r₁Curve 1 and the junction point of easement curve

R ' easement curve first derivative

R ' ' easement curve second dervative

Specific embodiment

See Fig. 1 Fig. 4, a kind of g3 continuous transition curve constructing method based on 5 bezier curves of the present invention, the method Specific implementation step is as follows:

Step one: obtain the single order in junction for the primitive curve and lead resultant second order and lead arrow,Represent respectively The one of one, two derivatives of curve 0 and easement curve junction point and curve 1 and easement curve junction point, second dervative, " point " represents To primitive curve derivation, " slash " represents to easement curve derivation；

Step 2: curve 0,1 and easement curve meet g2 continuously, and the expression formula meeting this condition is:r′′=λv₁+ω²v₂, wherein ω, λ are arbitrary coefficient；Can make in this stepWherein, α, β are unit vector, and have α ⊥ β, v₂For?Vertical Projection vector on direction；

Step 3: according to curvature formulationsIf easily proving curve r (t) at end points, second dervative It is changed to r ' and r ' ', then curvature at end points for the r (t) is constant.By adjust ω make the mould length of first derivative take reasonable value (if Parameter field is [0,1], then | r ' | takes curve arc long), suitably choose λ value and make curve reach curvature derivative and be continuouslyAt this Can set in step $f\left(t\right)={\left(k^2\right)}^{\′}={\left(k^2\right)}^{\·}=\frac{(r^{\′}\×r^{\′},r^{\′}\×r^{\′})}{{(r^{\′},r^{\′})}^3},$ And set $d_0=\left|\stackrel{\·}{r}\right|,d_2=\left|v_2\right|$ Can get following formula:

$\begin{array}{c}{f}^{\′}\left(t\right)=\frac{{(r^{\′}\×r^{\′},r^{\′}\×r^{\′})}^{\′}{(r^{\′},r^{\′})}^3-6(r^{\′}\×r^{\′},r^{\′}\×r^{\′}){(r^{\′},r^{\′})}^2(r^{\′},r^{\′})}{{(r^{\′},r^{\′})}^6}\\ =\frac{2{(r^{\′}\×r^{\′},r^{\′}\×r^{\′\′})}^{\′}(r^{\′},r^{\′})-6(r^{\′}\×r^{\′},r^{\′}\×r^{\′})(r^{\′},r^{\′})}{{(r^{\′},r^{\′})}^4}\\ =f^{\′}\left(t\right)=\frac{2(r^{\′\′},\mathrm{\β})d_0{d}_2-{6\mathrm{\λ\ωd}}_2^2}{{\mathrm{\ω}}^2{d}_0^5}\end{array}$

Step 4: 5 times bezier curve representation formula is:b_j,nT () is bernstein Base,

So, have at endpoint curve: p (0)=p₀, p (1)=p₅, according to even Continuous condition has r₀=p₀, r₁=p₅.One, two, three ranks at 5 bezier curve two-end-points lead arrow:

p₀′=5(p₁-p₀) (2)

p₅′=5(p₅-p₄) (3)

p₀′′=20(p₂-2p₁+p₀) (4)

p₅′′=20(p₅-2p₄+p₃) (5)

p₀′′′=60(p₃-3p₂+3p₁-p₀) (6)

p′₀、p′₅、p″₀、p″₅、p″′₀、p″′₅(7)

Can be obtained by equation (1), (2), (4), (6):

$-\frac{{\mathrm{\ω}}_0{d}_{02}}{d_{00}}{\mathrm{\λ}}_0+(v_{11},{\mathrm{\β}}_0){\mathrm{\λ}}_1={3\mathrm{\ω}}_0^2{d}_{02}-{\mathrm{\ω}}_1^2(v_{12},{\mathrm{\β}}_0)-20(-p_0+{2p}_4-p_5,{\mathrm{\β}}_0)+\frac{{\mathrm{\ω}}_0^1{d}_{00}^4{k}_0{\stackrel{\·}{k}}_0}{{3d}_{02}}---\left(8\right)$

Can be obtained by equation (1), (3), (5), (7):

$(v_{01},{\mathrm{\β}}_1){\mathrm{\λ}}_0+\frac{{\mathrm{\ω}}_1{d}_{12}}{d_{10}}{\mathrm{\λ}}_1={-\mathrm{\ω}}_0^2(v_{02},{\mathrm{\β}}_1)+{3\mathrm{\ω}}_1^2{d}_{12}+20(p_0-{2p}_1+p_5,{\mathrm{\β}}_1)-\frac{{\mathrm{\ω}}_1^2{d}_{10}^4{k}_1{\stackrel{\·}{k}}_1}{{3d}_{12}}---\left(9\right)$

Wherein, ω_i、λ_i、k_i、β_i, v_i1、v_i2、d_i0、d_i2Represent respectively curve i ω when being connected with easement curve, λ, k,β、v₁、v₂、d₀、d₂Value (i=0,1), λ can be solved by equation (8), (9)₀、λ₁；

Step 5: the λ being drawn by step 4₀Obtain p with equation (2), (4)₂, λ₁Obtain p with equation (3), (5)₃, All of control point all obtains afterwards, and then obtains 5 bezier easement curves of two groups of curves.

Present invention utilizes three order derivatives of two ends restrictive curve, only need to cutting along junction point in cad modeling process Control point p is moved in arrow direction₁And p₄, thus determine ω value, this movement can be designed to Interactive control, facilitate user Intuitively change the shape of easement curve, after ω determines, two other control point is obtained by step of the present invention, thus 5 times Bezier easement curve also determines that.

Claims (1)

1. a kind of g3 continuous transition curve constructing method based on 5 bezier curves it is characterised in that: it includes following step Rapid:

Step one: obtain the single order in junction for the primitive curve and lead resultant second order and lead arrow,Represent curve 0 respectively With the one of one, two derivatives of easement curve junction point and curve 1 and easement curve junction point, second dervative, " point " represents to former Beginning curve derivation, " slash " represents to easement curve derivation；

Step 2: curve 0,1 and easement curve meet g2 continuously, and the expression formula meeting this condition is:R "=λ v₁+ω²v₂, wherein ω, λ are arbitrary coefficient；Make in this step Wherein, α, β are unit vector, and have α ⊥ β, v₂For?Vertical direction on projection Vector；

Step 3: according to curvature formulationsIf easily proving that curve r (t) at end points, second dervative are changed to R ' and r ", then curvature at end points for the r (t) is constant；The mould length of first derivative is made to take reasonable value by adjusting ω, if parameter field For [0,1], then | r ' | takes curve arc long, chooses λ value and makes curve reach curvature derivative and is continuouslyThis step 3 setsAnd setObtain following formula:

$\begin{array}{c}{f}^{\′}\left(t\right)=\frac{{(r^{\′}\×r^{\′\′},r^{\′}\×r^{\′\′})}^{\′}{(r^{\′},r^{\′})}^3-6{(r^{\′}\×r^{\′\′},r^{\′}\×r^{\′\′})}^{\′}{(r^{\′},r^{\′})}^2(r^{\′},r^{\′\′})}{{(r^{\′},r^{\′})}^6}\\ =\frac{2{(r^{\′}\×r^{\′\′},r^{\′}\×r^{\′\′\′})}^{\′}(r^{\′},r^{\′})-6(r^{\′}\×r^{\′\′},r^{\′}\×r^{\′\′})(r^{\′},r^{\′\′})}{{(r^{\′},r^{\′})}^4}\\ =f^{\′}\left(t\right)=\frac{2(r^{\′\′\′},\mathrm{\β})d_0{d}_2-6{\mathrm{\λ\ωd}}_2^2}{{\mathrm{\ω}}^2{d}_0^5}\end{array}---\left(1\right)$

Step 4: 5 times bezier curve representation formula is:b_j,nT () is bernstein base,So, have at endpoint curve: p (0)=p₀, p (1)=p₅, according to the condition of continuity There is r₀=p₀, r₁=p₅；One, two, three ranks at 5 bezier curve two-end-points lead arrow:

p₀'=5 (p₁-p₀) (2)

p₅'=5 (p₅-p₄) (3)

p₀"=20 (p₂-2p₁+p₀) (4)

p₅"=20 (p₅-2p₄+p₃) (5)

p₀" '=60 (p₃-3p₂+3p₁-p₀) (6)

p′₀、p′₅、p″₀、p″₅、p″′₀、p″′₅(7)

Obtained by equation (1), (2), (4), (6):

$-\frac{{\mathrm{\ω}}_0{d}_{02}}{d_{00}}{\mathrm{\λ}}_0+(v_{11},{\mathrm{\β}}_0){\mathrm{\λ}}_1=3{\mathrm{\ω}}_0^2{d}_{02}-{\mathrm{\ω}}_1^2(v_{12},{\mathrm{\β}}_0)-20(-p_0+2p_4-p_5,{\mathrm{\β}}_0)+\frac{{\mathrm{\ω}}_0^2{d}_{00}^4{k}_0{\stackrel{\·}{k}}_0}{3d_{02}}---\left(8\right)$

Obtained by equation (1), (3), (5), (7):

$(v_{01},{\mathrm{\β}}_1){\mathrm{\λ}}_0+\frac{{\mathrm{\ω}}_1{d}_{12}}{d_{10}}{\mathrm{\λ}}_1=-{\mathrm{\ω}}_0^2(v_{02},{\mathrm{\β}}_1)+3{\mathrm{\ω}}_1^2{d}_{12}+20(p_0-2p_1+p_5,{\mathrm{\β}}_1)-\frac{{\mathrm{\ω}}_1^2{d}_{10}^4{k}_1{\stackrel{\·}{k}}_1}{3d_{12}}---\left(9\right)$

Wherein, ω_i、λ_i、k_i、β_i, v_i1、v_i2、d_i0、d_i2Represent respectively curve i ω when being connected with easement curve, λ, k,β、 v₁、v₂、d₀、d₂Value, i=0,1, solve λ by equation (8), (9)₀、λ₁；

Step 5: the λ being drawn by step 4₀Obtain p with equation (2), (4)₂, λ₁Obtain p with equation (3), (5)₃, finally own Control point all obtain, and then obtain 5 bezier easement curves of two groups of curves.