Mustafa et al. SpringerPlus (2016) 5:1493 DOI 10.1186/s40064-016-3018-3
Open Access
RESEARCH
Engineering images designed by fractal subdivision scheme Ghulam Mustafa1* , Mehwish Bari1 and Saba Jamil2 *Correspondence: ghulam.
[email protected] 1 Department of Mathematics, The Islamia University of Bahawalpur, Baghdad Campus, Bahawalpur 63100, Punjab, Pakistan Full list of author information is available at the end of the article
Abstract This paper is concerned with the modeling of engineering images by the fractal properties of 6-point binary interpolating scheme. Association between the fractal behavior of the limit curve/surface and the parameter is obtained. The relationship between the subdivision parameter and the fractal dimension of the limit fractal curve of subdivision fractal is also presented. Numerical examples and visual demonstrations show that 6-point scheme is good choice for the generation of fractals for the modeling of fractal antennas, bearings, garari’s and rock etc. Keywords: Interpolating subdivision scheme, Engineering images, Fractal antennas, Bearings Mathematics Subject Classification: 28A80, 65D05, 65D10
Background Fractals are infinitely complex patterns that are self-similar across different scales. They are made by rehashing a basic procedure again and again in an ongoing feedback loop. Its uses in various areas of the study of materials and of other areas of engineering are examples of practical prose. Its uses in physical theory, especially in conjunction with the basic equations of mathematical physics. Let us, instead, give a few typical examples: to rock base, an essential of engineering, furthermore a proceeding with objective of science, is to depict nature quantitatively. Another illustration is viscus flow through porous media, i.e. the stream of water pushing oil, has demonstrated as of late to admit to a few administrations, one of which is a ‘front‘, which is a compelling and attractive arrangement and another is ‘fractal fingering‘ which is undesirable. A fractal antenna (i.e. Fig. 1) is an antenna that uses a fractal, self-similar design to maximize the length, or increase the perimeter of material that can receive or transmit electromagnetic radiation within a given total surface area or volume. Another mechanical parts in engineering such as gears, chain, grari etc can be constructed based on ideas of fractal geometry. Owing to the rapid emergence and growth of techniques in the engineering application of fractals, it has become necessary to gather the most recent advances on a regular basis. This study starts from the question, can we design mechanical parts in engineering by simple iterative techniques? Firstly, Mandelbrot (1982) found fractal geometry that deals with geometric shape which is self-similar, irregular and has detailed structure at arbitrarily small scales. © 2016 The Author(s). This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
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Fig. 1 Fractal antenna
Nowadays, many techniques to generate fractals have been devised, such as IFS (iterated function systems) method Barnsley and Demko (1985), L-system method Prusinkiewicz and Lindenmayer (1990) and few others. Recently it has been shown that subdivision technique is not only an important tool for the fast generation of smooth engineering objects, but also an efficient tool for the fast generation of fractals. Zheng et al. (2007a, b) analyzed fractal properties of 4-point binary and three point ternary interpolatory subdivision schemes. Wang et al. (2011) discussed the fractal properties of the generalized chaikin corner-cutting subdivision scheme with two tension parameters. They gave the fractal range of scheme on the basis of the discussion of limit points on the limit curve. Li et al. (2013) designed the fractal curves by using the normal vector based subdivision scheme. Sarfraz et al. (2015) designed some engineering images by using rational spline interpolation. In this paper, we explore the properties of Weissman (1989) fractal subdivision scheme in different areas including engineering images. We conclude that 6-point scheme of Weissman can create engineering images for curves and surfaces with true fractal allotting and can provide some ways of shape control. The paper is organized as follows. In "Fractal properties of the scheme" section, fractal range of 6-point subdivision scheme is being discussed. In "Numerical examples and demonstrations" section, some numerical examples are presented to confirm the correctness and effectiveness of the engineering images in the form of curve and surface. Finally, we give some concluding remarks in "Conclusions" section.
Fractal properties of the scheme The well known 6-point binary interpolating scheme Weissman (1989) is k+2 = pik+1 p2i k+2 p2i+1
1 9 9 k+1 k+1 k+1 + 3µ pi−1 + + 2µ pi + + 2µ pi+1 = − 16 16 16 1 k+1 k+1 + 3µ pi+2 + µpi+3 . − 16 k+1 µpi−2
(1)
√
2 3 C 0- and C 1- continuous curves at − 16 < µ < −1+3 The √ scheme (1) produces and 16 √ −6+3 2 3 −1+ 19 < µ < µ = µ = 0 respectively. By substituting and , we get 4-point 32 32 256 and 6-point schemes as given in Deslauriers and Dubuc (1989) respectively.
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According to interpolatory property p0k ≡ p00, k ≥ 0. Suppose pim and pjm are two fixed control points after m subdivision steps, ∀ m ∈ Z, m ≥ 0. The role of parameter µ is required to be evaluated on the sum of all small edges among the two points after another k iterations. First we discuss and analyze the effect of µ among the two initial control points p00 and p10. k+1 of scheme (1), we have For i = −2 in the odd rule p2i+1 9 1 k+2 k+1 k+1 k+1 k+1 + 3µ p−3 + 2µ p−2 p−3 = µ p−4 + p1k+1 − + p0k+1 + + p−1 . 16 16 (2) k+1 k+1 k k = p−2 = p−1 Substituting p−4 and p−2 in (2), we get 1 9 k+2 k+1 k+1 k k + 3µ p−3 + 2µ p−1 p−3 = µ p−2 + p1k+1 − + p00 + + p−1 . 16 16 k+1 k+1 k = p−1 of scheme (1) by using p−2 and p2k+1 = p1k , we Putting i = −1, 0 in odd rule p2i+1 have 1 9 k+2 k+1 k+1 k+1 k k + 3µ p−1 + p1 + 2µ p−1 p−1 = µ p−3 + p1 − + p00 , + 16 16
and
p1k+2
1 9 k+1 k+1 k k + 3µ p−1 + p1 + + 2µ p1k+1 + p00 . = µ p−1 + p3 − 16 16
Here, we define two edge vectors between the points p00 and p10 after k steps defined by 1 = p1k−1 and vk = p1k − p0k and Rk = p2k − p1k . Since pik = 2ik so we have p2k = 22k = 2k−1 we can write as
Rk = p1k−1 − p0k−1 + p0k − p1k = vk−1 − vk .
(3)
k k k k − p0k + p−1 , Wk = p0k − p−1 , we can write it as Uk − Wk = p1k − p−1 Let Uk = p1k − p−1 = p1k − p0k = vk . Further Uk = vk + Wk . This implies 1 k+1 k k + 3µ p1k − p−1 Uk+2 = −µ p1k − p−1 + µ p3k+1 − p−3 − 16 9 1 k+1 k+1 k+1 + 3µ p1 − p−1 + + 2µ p1k+1 − p−1 . + 16 16 k+1 i Since by dyadic parametrization pik+1 = 2k+1 then the term µ p3k+1 − p−3 can be written as 2 −2 −1 1 −1 1 −1 1 µ k+1 + k+1 − k+1 − k+1 = µ k+1 − k+1 + k − k . 2 2 2 2 2 2 2 2
Therefore 1 k+1 k k + µ p1k+1 − p−1 − Uk+2 = −µ p1k − p−1 + p1k − p−1 + 3µ 16 9 1 k+1 k+1 k+1 k k . + 3µ p1 − p−1 + + 2µ p1k+1 − p−1 × p1 − p−1 + 16 16
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This implies
Uk+2 = µUk+1 −
1 1 9 + 3µ Uk + + 3µ Uk+1 + + 2µ Uk+1 . 16 16 16
So we have
Uk+2 =
1 10 + 6µ Uk+1 − + 3µ Uk . 16 16
(4)
Equation (4) is the second order linear difference equation, such that 1 10 + 6µ Uk+1 + + 3µ Uk = 0. Uk+2 − 16 16 The characteristic equation is 1 10 2 + 6µ Q + + 3µ = 0. Q − 16 16
(5)
1 . By solving (5), we get Q1 = 18 + 6µ and Q2 = 12, (Q1 � = Q2 ) when µ � = 16 0 0 1 1 Let U0 = p1 − p−1 and U1 = p1 − p−1. For k + 1 = 0 and i = 0 in (1), we get
p11
9 9 1 0 0 + 3µ p−1 + + 2µ p0 + + 2µ p10 − = 16 16 16 1 − + 3µ p20 + µp30 . 16 0 µp−2
This implies
p11
1 9 0 0 0 0 + 3µ p−1 + p2 + + 2µ p00 + p10 . = µ p−2 + p3 − 16 16
(6)
Substitute k + 1 = 0 and i = −1 in (1), we have 1 p−1
1 9 9 0 0 + 3µ p−2 + + 2µ p−1 + + 2µ p00 = − 16 16 16 1 + 3µ p10 + µp20 . − 16 0 µp−3
This implies 1 p−1
1 9 0 0 0 0 0 + 3µ p−2 + p1 + + 2µ p−1 = µ p−3 + p2 − + p00 . 16 16
(7)
Subtracting (7) from (6), we have
1 5 5 1 0 0 0 + 4µ p−2 + 5µ p−1 + 5µ p10 p11 − p−1 = − µp−3 + − + 16 8 8 1 + 4µ p20 + µp30 . − 16
(8)
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The solution of Eq. (8) is k k 1 1 + 6µ c1 + Uk = c2 , 8 2 where
c1 =
and
8 1 1 1 0 0 0 + 4µ p−2 + 5µ p−1 + 5µ p10 µp−3 − + − 3 − 48µ 16 8 8 1 + 4µ p20 − µp30 + 16
1 1 1 8 0 0 0 + 4µ p−2 + − µ p−1 − − µ p10 c2 = − µp−3 − 3 − 48µ 16 2 2 1 0 0 + 4µ p2 − µp3 . + 16
When µ =
1 16,
the solution of Eq. (4) will be
k k 1 1 Uk = c´1 + k c´2 2 2 where 0 c´1 = p10 − p−1
and
c´2 = −
1 0 0 0 p−3 − 5p−2 + 7p−1 − 7p10 + 5p20 − p30 . 8
Since
vk+2
9 8 k+1 k+1 k+1 − + 2µ + µ p−1 + p2k+1 = µ p−2 + p3 − 16 16 9 + + 2µ p0k+1 + p1k+1 − p0k+2 . 16
k+1 and vk = p1k − p0k , we get Then by taking Uk+1 = p1k+1 − p−1 9 9 k+1 k+1 + 2µ Uk+1 − + 2µ vk − µ p2k+1 − p−2 vk+2 = + µ p3k+1 − p−1 16 16 1 k+1 1 k+1 + p−1 + p2 − p0k+2 . 2 2 k+1 k+1 + µ p3k+1 − p−1 = 0 then Since by dyadic parametrization −µ p2k+1 − p−2 1 9 1 k 9 k+1 + 2µ Uk+1 − + 2µ vk + vk+2 = p1 − p0k + p−1 − p0k+1 . 16 16 2 2 k . Then Since vk = p1k − p0k and Wk+1 = p0k − p−1 9 1 1 9 + 2µ Uk+1 − + 2µ vk + vk − Wk+1 . vk+2 = 16 16 2 2
(9)
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Substitute Wk+1 = Uk+1 − vk+1 in (9), we get 9 1 1 9 + 2µ Uk+1 − + 2µ vk + vk − Uk+1 − vk+1 . vk+2 = 16 16 2 2 Simplifying, we have
1 vk+2 − vk+1 + 2
1 1 + 2µ vk = + 2µ Uk+1 . 16 16
Now we will find the solution of Eq. (10), first consider 1 1 + 2µ = 0. γ2 − γ + 2 16
(10)
(11)
√ √ 2 3 < µ < −6+3 , the solution of Eq. (11) is γ1 = 41 + −2µ and Case 1. When − 16 32 √ γ2 = 41 − −2µ, and γ1 �= γ2, γ1 , γ2 �= 18 + 6µ, γ1 , γ2 �= 12. Equation (10) can be expressed as
1 vk+2 − vk+1 + 2
k+1 k+1 1 1 1 1 . + 2µ vk = + 2µ + 6µ c1 + c2 16 16 8 2
The solution of above equation is
vk = β1 γ1k + β2 γ2k +
k k 65 1 1 1 + 6µ + c2 c1 , 144 8 2 2
(12)
where β1 and β2 are given as β1 =
√ √ √ 1 √ −3568 −2 wp−1 + 6912 p0 −2 w − 56 −2 wp2 13824 −2 w(−1 + 16 w) √ √ √ √ + 56 −2 wp−2 − 3344 p1 −2 w + 896 −2 wwp3 + 59776 p1 −2 ww √ √ √ √ − 110592 p0 −2 ww + 50816 −2 wp−1 w − 896 −2 wwp−3 − 3584 −2 wp2 w √ + 3584 −2 wwp−2 + 1296 p0 − 878 p1 + 209 p−2 + 223 p2 − 850 p−1 − 3344 wp−3 − 3568 wp3 + 3344 wp−2 − 28416 p1 w2 + 3568 p2 w + 221184 p0 w2 + 60672 w2 p3
− 89088 w2 p−2 − 34560 p0 w + 17392 p1 w + 49920 w2 p−3 + 17168 p−1 w − 132096 p2 w2 − 82176 p−1 w2 , √ √ 1 √ −209 p−2 − 3344 p1 −2 w − 56 −2 wp2 β2 = 13824 −2 w(−1 + 16 w) √ + 6912 p0 −2 w − 1296 p0 + 878 p1 − 223 p2 − 17392 p1 w − 49920 w2 p−3
+ 89088 w2 p−2 + 82176 p−1 w2 + 28416 p1 w2 + 132096 p2 w2 − 60672 w2 p3 √ √ √ − 221184 p0 w2 − 896 −2 wwp−3 − 3568 −2 wp−1 + 56 −2 wp−2 √ √ √ − 110592 p0 −2 ww + 50816 −2 wp−1 w + 59776 p1 −2 ww √ √ √ + 896 −2 wwp3 − 3584 −2 wp2 w + 3584 −2 wwp−2 + 34560 p0 w
− 3344 wp−2 + 3568 wp3 + 3344 wp−3 − 17168 p−1 w − 3568 p2 w + 850 p−1 .
From Eqs. (3) and (12), we have
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k 1 1 7 − 48µ 65 1 k k c1 + 6µ − 1 β1 γ1 + − 1 β2 γ2 + Rk = γ1 γ2 1 + 48µ 144 8 k 1 1 + c2 . 2 2
(13)
√ √ √ 2 Case 2. When −1+32 19 < µ < −1+3 , the solution of Eq. (11) is ρ1 = 41 + i 2µ and 16 √ ρ2 = 41 − i 2µ, ρ1 �= ρ2, ρ1 , ρ2 �= 18 + 6µ, ρ1 , ρ2 �= 12. The solution of Eq. (10) can be expressed as k k 65 1 1 1 + 6µ + c2 vk = χ1 ρ1k + χ2 ρ2k + c1 , (14) 144 8 2 2
where χ1 and χ2 are given as χ1 =
√ √ √ 1 √ −3568i 2 wp−1 + 6912i p0 2 w − 56i 2 wp2 13824i 2 w(−1 + 16 w) √ √ √ √ + 56i −2 wp2 − 3344i p1 2 w + 896i 2 wwp3 + 59776i p1 2 ww √ √ √ √ − 110592i p0 2 ww + 50816i 2 wp−1 w − 896i 2 wwp−3 − 3584i 2 wp2 w √ + 3584i −2 wwp−2 + 1296 p0 − 878 p1 + 209 p−2 + 223 p2 − 850 p−1 − 3344 wp−3 − 3568 wp3 + 3344 wp−2 − 28416 p1 w2 + 3568 p2 w + 221184 p0 w2 + 60672 w2 p3
− 89088 w2 p−2 − 34560 p0 w + 17392 p1 w + 49920 w2 p−3 + 17168 p−1 w − 132096 p2 w2 − 82176 p−1 w2 , √ √ 1 √ −209 p−2 − 3344i p1 2 w − 56i 2 wp2 χ2 = 13824i 2 w(−1 + 16 w) √ + 6912i p0 2 w − 1296 p0 + 878 p1 − 223 p2 − 17392 p1 w − 49920 w2 p−3
+ 89088 w2 p−2 + 82176 p−1 w2 + 28416 p1 w2 + 132096 p2 w2 − 60672 w2 p3 √ √ √ − 221184 p0 w2 − 896i 2 wwp−3 − 3568i 2 wp−1 + 56i 2 wp−2 √ √ √ − 110592i p0 2 ww + 50816i 2 wp−1 w + 59776i p1 2 ww √ √ √ + 896i 2 wwp3 − 3584i 2 wp2 w + 3584i 2 wwp−2 + 34560 p0 w
− 3344 wp−2 + 3568 wp3 + 3344 wp−3 − 17168 p−1 w − 3568 p2 w + 850 p−1 .
From Eqs. (3) and (14), we have
Rk =
k 1 1 7 − 48µ 65 1 c1 + 6µ − 1 χ1 ρ1k + − 1 χ2 ρ2k + ρ1 ρ2 1 + 48µ 144 8 k 1 1 + c2 . 2 2
(15)
From the Eqs. (12)–(15), we have the following theorems. √ 19
Theorem 1 For −1+32 tal curve.
