The singular properties of the Hermit curve on Unit Circle

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1 The singular properties of the Hermit curve on Unit Circle Ching-Shoei Chiang 1, Ching-Tsorng Tsai 1 Computer Science and Information Management, Soochow University, Taipei, Taiwan Computer Science, Tunghai University Taichung, Taiwan Abstract - The design of curves, surface, and solid are important in computer aided geometric design (CAGD). Images, surround by boundary curves, are also investigated by many researchers. One way to describe an image is using the medial axis transform. Under this consideration, the properties of the boundary curves tangent to circles become important to design for D images. In this paper, we define a special class of curves, which is the curves whose end points is on a circle, and endpoints tangent vectors are parallel to the tangent line of circle at the end points. There are 4 parameters to design this curve, two are the length of the tangent vectors 0 and 1 on the boundary curve, one is the angle, associated with the circle, between two endpoints, and the final parameter is the variable t for the curves. Different curves design for singular point can be achieved by given two values or constraints among these 4 parameters. We can use the singular point design in calligraphic strokes with cusp to describe the edge contour. Using these properties, with more constraints on the singular point or maximum curvature at specified parameter, we survey the couture of the curves, so that the design of the curve may be simplify by giving constraint. Definition and Theorem Let introduce the Hermit curve first. Given two points P 0 and P 1, with two associated tangent vectors V 0 and V 1, the Hermit curve C(t) defined as (See figure 1): 1 1 C(t) = [t, t, t, 1 ] 1 P [ ] [ 1 ] (1) V V 1 P 0 Keywords: Computer Graphics, computer-aided geometric design, Hermit Curve, geometric modeling 1 Introduction The D image can be stored by many different ways, including its boundary curves, a medial axis curves with radius function, union of many primitive figures, and so on. There are many researchers use different curves to simulate different images. For example, Cinque, Levialdi and Malizia [1] uses cubic Bezier curve to do the shape description. Yang, Lu and Lee[] use Bezier curve to approach the shape description for Chinese calligraphy characters. Chang and Yan[] derived an algorithm to approach the hand-drawn image by using cubic Bezier curve. Cao and Kot[4] derived an algorithm to do data embedding in electronic inks without losing data. The boundary curves can be also used to derived the offset curves and medial axis of an images[5], and also to simulate the nature objects, such as flowers[6]. In this paper, we would like to investigate the cubic Hermit curves, where the end points are on a unit circle, and its end points tangent line is parallel to the tangent line of circles. Figure 1. Hermite Curve To simplify our problem, we assume that the end points P 0 and P 1 are on the unit circle, the first point P 0 =(-1,0), and the second point P 1 =(-cosθ, sinθ). It s associated tangent vectors parallel to (0,1) and (sinθ, cosθ). We have the following definition for this curve. Definition1:Unit Circle Hermit Curve Two points P 0 =(-1, 0) P 1 =(-cosθ, sinθ) on the unit circle with its associated tangent vector V 0 =α 0 (0, 1) V 1 =α 1 (sinθ, cosθ) produced a Hermite curve, we call it unit circle hermit curve, denoted H 1 (t;θ,α 0, α 1 ),where α 0 >0 α 1 >0 0<t<1 0<θ<π

2 Notice that for the general case, we can always convert the design problem into the design problem for H 1 (t;θ,α 0, α 1 ). We can always translate the center to the origin, scale the radius into one and rotate the circle so that the first point is on (-1,0). We call this process the standardization of the problem. After we solve the problem, produce the curve we want, we can always inverse the process, which is rotate, scale, and translate to see the designed curves for the original design problem. Let us consider the cusp on the curve. Theorm1: Singular Point Let C(t)=H 1 (t;θ,α 0, α 1 ),then C (t)=(x (t),y (t))=(0, 0) α 0 B(t)=α 1 D(t)=A(t)tan θ,where A(t)=(t-1) B(t)=(t-1) (t-1) D(t)=t(t-) Proof: We separate our proof into two parts, including if ( ) and only if ( ). (1) C (t)=(x (t), y (t))=(0, 0) α 0 B(t)=α 1 D(t)=A(t)tan θ After the first derivative of the curve, we have ( 1,0) C (t) = [A(t), A(t), B(t), D(t)] ( cosθ, sinθ) α 0 (0,1) () [ α 1 (sinθ, cosθ)] Where A(t)=(t-1) B(t)=(t-1)(t-1) D(t)=t(t-) From C (t)=(0, 0),the following equations holds: { A(t) + cosθa(t) + α 1sinθD(t) = 0 () A(t)sinθ + α 0 B(t) + α 1 cosθd(t) = 0 (4) From Cramer s rule to solve sinθ and cosθ, with the properties sin θ +cos θ =1, we have: or { A (t) + α 1 D (t) = 0 (5) α 0 B (t) α 1 D (t) = 0 (6) Notice that equation (5) never holds under the assumption that α 0 >0 α 1 >0 0<t<1 We can simplify the equation (6) into α 0 B(t)=±α 1 D(t): or{ α 0B(t) = α 1 D(t) (6. a) α 0 B(t) = α 1 D(t) (6. b) Solve the system of equation () (4) and(6.a),we have α 0 B(t)=α 1 D(t)=A(t)tan θ (0<θ<π) On the other hand, solve the system of equation () (4) and (6.b),we have α 1 D(t)=A(t)tan θ =-A(t)cotθ,conclude sec θ =0,since we cannot find any θ so that sec θ =0,so α 0B(t)=-α 1 D(t) will never happened () C (t)=(x (t), y (t))=(0, 0) α 0 B(t)=α 1 D(t)=A(t)tan θ Via equation (), we have: (x (t),y (t))=(-a(t)+a(t)cosθ+α 1 D(t)sinθ,-A(t)sinθ+α 0 B(t)+α 1 D( t)cosθ)= (A(t)sin θ α 1D(t)cos θ ) (sin θ, cos θ ) = cosθ (A(t)tan θ α 1D(t)) (sin θ, cos θ ) Since α 0 B(t)=α 1 D(t)=A(t)tan θ,we got A(t)tan θ α 1D(t)=0, so C (t)=(0, 0) Observation: Notice that there are some values of t degenerate the case. Assume t can be 0 (A(t)=D(t)=0) or 1(A(t)=B(t)=0), then there are no cusp in the defined curve. Now we have 4 variables (t θ α 0 α 1 ) with two constraints (x (t)=0 y (t)=0),in general case, we have solutions by given the values of two variables or two constraints. These two variables can be (θ, t) (θ, α 0 ) (θ, α 1 ) (t, α 0 ) (t, α 1 ) (α 0, α 1 ) Constraints can be something like α 0 =α 1. So, we have the following corollaries:.1 Corollary1:θ t are given Let C(t)=H 1 (t ; θ, α 0,α 1 ),and gives the value of θ t,then α 0 = tan t 1 (θ) α 1 = 6(t 1) tan t (θ),where t 1 t From Corollary1, we can know whether α 0 and α 0 are positive, negative, or zero by given the range of θ and t,as shown in Table 1. Let 0<t<1, 0<θ<π, t 1, t, θ π, we have 4 different cases concerning the positive/negative values of α 0 and α 1, that is, (α 0 >0 α 1 >0) (α 0 >0 α 1 <0) (α 0 <0 α 1 >0) and (α 0 <0 α 1 <0). These cases, associated with the range of θ and t, are list in Table.. Corollary :θ α 0 are given Let C(t)=H 1 (t;θ,α 0, α 1 ),if we know the value of θ and α 0, then α 1 = 4 α 0 tan θ tan θ t = α 0 where θ 0 α 0 4tan θ (α 0 tan θ )

3 . Corollary :θ α 1 are given Let C(t)=H 1 (t;θ,α 0, α 1 ),if we know the value of θ and α 1, then α 0 = 4 α 1 tan θ tan θ t = α 1 4tan θ α 1 tan θ α 1 tan θ,where θ 0 Notice that the similar equation for α 1 in corollary and α 0 in corollary..5 Corollary5:given t α 1 Let C(t)=H 1 (t;θ,α 0,α 1 ),if we know the value of t α 1, then α 0 = α 1t(t ) (t 1)(t 1) θ = tan 1 ( α 1(t ) 6(t 1) ),where t 1 t 1.4 Corollary4:Given t α 0 Let C(t)=H 1 (t;θ,α 0,α 1 ), if we know the value of t α 0,then α 1 = α 0(t 1)(t 1) θ = tan 1 ( α 0(t 1) ),where t 0 t t(t ) tan θ t 1 6(t 1) t α 0 α 1 Table 1 Positive/Negative of α 0, α 1 >0 =0 <0 0<θ<π θ=0 π<θ<π t> 1 or t<0 t=0 0<t<1 t>1 or t< (t> 1 ) (0<θ<π) (t<0) (0<θ<π) (0<t< 1 ) (π<θ<π) (t>1) (0<θ<π) (t< ) (0<θ<π) ( <t<1) (π<θ<π) t=1 t=0 θ=0 t=1 θ=0 <t<1 (t> 1 ) (π<θ<π) (t<0) (π<θ<π) (0<t< 1 ) (0<θ<π) (t>1) (π<θ<π) (t< ) (π<θ<π) ( <t<1) (0<θ<π) Table Range of α 0, α 1 from range of θ and t Case A ( 1 <t< ) (0<θ<π) (α 0>0 α 1 >0) Case B Case C ( <t<1) (0<θ<π) or (0 <t< 1 ) (π<θ<π) (α 0 >0 α 1 <0) (0 <t< 1 ) (0<θ<π) or ( <t<1) (π<θ<π) (α 0 <0 α 1 >0) Case D ( 1 <t< ) (π<θ<π) (α 0<0 α 1 <0).6 Corollary6:Given α 0 α 1 Let C(t)=H 1 (t;θ,α 0,α 1 ),if we know the value of α 0 α 1,then t = α 1+α 0 ± α 1 α 0 α 1 +α 0 (α 0 α 1 ) θ = tan 1 ( α 0(t 1) ),where α 0 α 1 t 0,or t= 1 θ = tan 1 ( α 6 ),where α 0 = α 1 = α.7 Corollary7:given θ and α 0 =α 1 Let C(t)=H 1 (t;θ,α 0, α 1 ),assume α 0 =α 1,and give the value of θ, then α 0 = α 1 = an θ t = 1 Experimental Result From the theorem and corollaries describes in the last section, we give examples in this section..1 Example 1:θ t are given In corollary 1, given θ t, we can find α 0 and α 0, the result is shown in Table and Figure. The last column of Table lists the different cases shown in Table. Table Findingα 0 and α 0 by giving θ, t # Given value Obtained value Cases θ t α 0 α 1 (a) C (b) A (c) A (d) A (e) B (f) D

4 Figure Fixed θ, vary t Table 4 and Figure shows the cases that we fixed the value t, and vary the value θ.. Example : Figure Fixed t, θ vary Let θ=80 and α 0 =10,we find t , α 1.780,as shown in Figure 4(a). Let θ=110, α 1 =8,we find t α ,as shown in Figure 4(b). Table 4 Fixed t, θ vary # Given value Obtained value Cases θ T α 0 α 1 (a) A (b) A (c) A (d) D (e) D (f) D (a). Example : (b) Figrue 4 (θ, α 0 ) or (θ, α 1 ) are given Let t=0.5 α 0 =5 we find α 1 =5 θ as shown in Figure 5(a). Let t=0.57 α 1 =5,we have α 0 = θ ,as shown in Figure 5(b). (a) (b) Figure 5 (t, α 0 ) or (t, α 1 ) are given

5 .4 Example 4:α 0 α 1 are given Let C(t)=H 1 (t;θ,α 0,α 1 ). If α 0 =8 α 1 =0, we find t ,when t=0.5954, we find θ , as shown in Figure 6(a). If α 0 =5 α 1 =5,we find t= 1 θ , as shown in Figure 6(b) (a) (b) Figure 6 (α 0, α 1 ) or (α 0, α 1 ) are given We consider the case that α 0 =α 1 and the value of θ are given.5 Example 5: Let C(t)=H 1 (t;θ,α 0, α 1 ), when the value of θ are given, we can find the value of α 0 α 1 t,we list the result in Table 5 and Figure 7. For example, when θ=50,we find α 0 =α 1 = t=1/, as shown in Figure 7(a). # Given value Table 5 Given θ with α 0 =α 1 Obtained value Θ α 0 α 1 t (a) (b) (c) (d) (e) (f) Figure 7 Given θ with α 0 =α 1 4 Conclusion and future research The unit circle hermit curve, H 1 (t;θ,α 0,α 1 ), can be used to design curves and images. On many cases, the singular point may be needed to be considered. For example, the Chinese characters may have cusp on the boundary. Before the design of the character, the relationship between the parameter value and the contour of the boundary curve is important. When we give two values of the parameters for H 1 (t;θ,α 0,α 1 ) curve, the can find the other two values with simple computation. When we define a curve, the only memory we need is 7 locations to store these 4 parameters, and other parameters for the standardization process. So, the design process saved not only the time, but also the memory There are more constraints we can used to design the curves. For example, the maximum curvature happened at t 0, the curve passing through a point p 0, the curve tangent to a line L 0, and so on. We believe the result will as simple as the case we introduced here. Acknowledge This work was supported in part by the National Science Council in Taiwan under Grants 10-1-E

6 Reference [1] Cinque, L., Levialdi, S. and Malizia, A. Shape description using cubic polynomial Bezier curves ; Pattern Recognition Letters, Vol. 19., pp.81-88, [] Yang, H.-M., Lu, J.-J. and Lee, H.-J.. A Bezier curve-based approach to shape description for Chinese calligraphy characters. ; In "Proceedings of the Sixth International Conference on Document Analysis and Recognition", pp (001) [] Chang, H.-H. and Yan, H. Vectorization of hand-drawn image using piecewise cubic Bezier curves fitting. ; Pattern Recognition, Vol. 1, No. 11, pp (1998) [4] Cao, H., Kot, A.C. Lossless Data Embedding in Electronic Inks ; IEEE Transactions on Information Forensics and Security, Vol. 5, No., pp.14-. (010) [5] Cao, L., Jia, Z., Liu, J. Computation of medial axis and offset curves of curved boundaries in planar domains based on the Cesaro s approach. ; Computer Aided Geometric Design, Vol. 6, No.4, pp (009) [6] Qin, Peiyu and Chen, Chuanbo Simulation Model of Flower Using the Integration of L-systems with Bezier Surfaces ; International Journal of Computer Science and Network Security, Vol.6, No., pp (006)