• MATLAB | 艺术就是画圈圈


    什么是艺术??艺术就是画圈圈!!

    这两天刷到了Hamid Naderi Yeganeh大佬的一系列线条艺术,感觉非常惊艳,顺手就用MATLAB实现了一下。大佬的大部分作品都在以下网站,大家有兴趣可以去瞅瞅去试试实现其他的作品:

    https://www.ams.org/publicoutreach/math-imagery/yeganeh

    飞鸟一

    This image shows 9,830 circles. For k = 1 , 2 , 3 , … , 9830 k=1,2,3, \ldots, 9830 k=1,2,3,,9830 , the center of the circle is ( X ( k ) , Y ( k ) ) (X(k), Y(k)) (X(k),Y(k)) and the radius of the k-th circle is R ( k ) R(k) R(k) , where:
    X ( k ) = ( sin ⁡ ( π k 20000 ) ) 12 ( 1 2 ( cos ⁡ ( 31 π k 1000 ) ) 16 sin ⁡ ( 6 π k 10000 ) + 1 6 ( sin ⁡ ( 31 π k 1000 ) ) 20 ) + 3 k 20000 + ( cos ⁡ ( 31 π k 10000 ) ) 6 sin ⁡ ( π 2 ( k − 10000 10000 ) 7 − π 5 ) Y ( k ) = − 9 4 ( cos ⁡ ( 31 π k 10000 ) ) 6 cos ⁡ ( π 2 ( k − 10000 10000 ) 7 − π 5 ) ( 2 3 + ( sin ⁡ ( π k 20000 ) sin ⁡ ( 3 π k 2000 ) ) 6 ) + 3 4 ( cos ⁡ ( 3 π k − 10000 10000 ) ) 10 ( cos ⁡ ( 9 π k − 10000 10000 ) ) 10 ( cos ⁡ ( 36 π k − 10000 100000 ) ) 14 + 7 10 ( k − 10000 10000 ) 2 R ( k ) = ( sin ⁡ ( π k 20000 ) ) 10 ( 1 4 ( cos ⁡ ( 31 π k 10000 + 25 π 32 ) ) 20 + 1 20 ( cos ⁡ ( 31 π k 10000 ) ) 2 ) + 1 30 ( 3 2 − ( cos ⁡ ( 62 π k 10000 ) ) 2 ) \begin{aligned} &X(k)=\left(\sin \left(\frac{\pi k}{20000}\right)\right)^{12}\left(\frac{1}{2}\left(\cos \left(\frac{31 \pi k}{1000}\right)\right)^{16} \sin \left(\frac{6 \pi k}{10000}\right)+\frac{1}{6}\left(\sin \left(\frac{31 \pi k}{1000}\right)\right)^{20}\right)+\frac{3 k}{20000}+\left(\cos \left(\frac{31 \pi k}{10000}\right)\right)^6 \sin \left(\frac{\pi}{2}\left(\frac{k-10000}{10000}\right)^7-\frac{\pi}{5}\right) \\ &Y(k)=\frac{-9}{4}\left(\cos \left(\frac{31 \pi k}{10000}\right)\right)^6 \cos \left(\frac{\pi}{2}\left(\frac{k-10000}{10000}\right)^7-\frac{\pi}{5}\right)\left(\frac{2}{3}+\left(\sin \left(\frac{\pi k}{20000}\right) \sin \left(\frac{3 \pi k}{2000}\right)\right)^6\right)+\frac{3}{4}\left(\cos \left(3 \pi \frac{k-10000}{10000}\right)\right)^{10}\left(\cos \left(9 \pi \frac{k-10000}{10000}\right)\right)^{10}\left(\cos \left(36 \pi \frac{k-10000}{100000}\right)\right)^{14}+\frac{7}{10}\left(\frac{k-10000}{10000}\right)^2 \\ &R(k)=\left(\sin \left(\frac{\pi k}{20000}\right)\right)^{10}\left(\frac{1}{4}\left(\cos \left(\frac{31 \pi k}{10000}+\frac{25 \pi}{32}\right)\right)^{20}+\frac{1}{20}\left(\cos \left(\frac{31 \pi k}{10000}\right)\right)^2\right)+\frac{1}{30}\left(\frac{3}{2}-\left(\cos \left(\frac{62 \pi k}{10000}\right)\right)^2\right) \end{aligned} X(k)=(sin(20000πk))12(21(cos(100031πk))16sin(100006πk)+61(sin(100031πk))20)+200003k+(cos(1000031πk))6sin(2π(10000k10000)75π)Y(k)=49(cos(1000031πk))6cos(2π(10000k10000)75π)(32+(sin(20000πk)sin(20003πk))6)+43(cos(3π10000k10000))10(cos(9π10000k10000))10(cos(36π100000k10000))14+107(10000k10000)2R(k)=(sin(20000πk))10(41(cos(1000031πk+3225π))20+201(cos(1000031πk))2)+301(23(cos(1000062πk))2)

    完整代码

    % bird 1
    K=1:9830;
    t=linspace(0,2*pi,200);
    
    X=@(k) (sin(pi.*k./2e4)).^12.*(cos(31.*pi.*k./1e4).^16.*sin(6.*pi.*k./1e4)./2+sin(31.*pi.*k./1e4).^20./6)...
            +3.*k./2e4+cos(31.*pi.*k./1e4).^6.*sin(pi./2.*(k./1e4-1).^7-pi./5);
    Y=@(k) -9./4.*cos(31.*pi.*k./1e4).^6.*cos(pi./2.*(k./1e4-1).^7-pi./5).*(2./3+(sin(pi.*k./2e4).*sin(3.*pi.*k./2e4)).^6)...
            +3./4.*cos(3.*pi.*(k-1e4)./1e5).^10.*cos(9.*pi.*(k-1e4)./1e5).^10.*cos(36.*pi.*(k-1e4)./1e5).^14+7./10.*((k-1e4)./1e4).^2;
    R=@(k) sin(pi.*k./2e4).^10.*(1./4.*cos(31.*pi.*k./1e4+25.*pi./32).^20+1./20.*cos(31.*pi.*k./1e4).^2)+1./30.*(3./2-cos(62.*pi.*k./1e4).^2);
    
    CX=[X(K')+cos(t).*R(K'),K'.*nan]';
    CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
    plot(CX(:),CY(:),'Color',[0,0,0,.2]);
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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    飞鸟二

    This image shows all circles of form ( x − A ( k ) ) 2 + ( y − B ( k ) ) 2 = R ( k ) 2 (x-A(k))^2+(y-B(k))^2=R(k)^2 (xA(k))2+(yB(k))2=R(k)2 for k = − 20000 , − 19999 , … , 20000 k=-20000,-19999,\dots,20000 k=20000,19999,,20000, where:

    A ( k ) = 3 k 45000 + sin ⁡ ( 17 π 20 ( k 20000 ) 5 ) cos ⁡ 6 ( 41 π k 20000 ) + ( 1 3 cos ⁡ 16 ( 41 π k 20000 ) + 1 3 cos ⁡ 80 ( 41 π k 20000 ) ) cos ⁡ 12 ( π k 40000 ) sin ⁡ ( 6 π k 20000 ) B ( k ) = 15 30 ( k 20000 ) 4 − cos ⁡ ( 17 π 20 ( k 20000 ) 5 ) ( 11 10 + 45 20 cos ⁡ 8 ( π k 40000 ) cos ⁡ 6 ( 3 π k 40000 ) ) cos ⁡ 6 ( 41 π k 20000 ) + 12 20 cos ⁡ 10 ( 3 π k 200000 ) cos ⁡ 10 ( 9 π k 200000 ) cos ⁡ 10 ( 18 π k 200000 ) R ( k ) = 1 50 + 1 40 sin ⁡ 2 ( 41 π k 20000 ) sin ⁡ 2 ( 9 π k 200000 ) + ( 1 17 ) cos ⁡ 2 ( 41 π k 20000 ) cos ⁡ 10 ( π k 40000 ) \begin{aligned} &A(k)=\frac{3 k}{45000}+\sin \left(\frac{17 \pi}{20}\left(\frac{k}{20000}\right)^5\right) \cos ^6\left(\frac{41 \pi k}{20000}\right)+\left(\frac{1}{3} \cos ^{16}\left(\frac{41 \pi k}{20000}\right)+\frac{1}{3} \cos ^{80}\left(\frac{41 \pi k}{20000}\right)\right) \cos ^{12}\left(\frac{\pi k}{40000}\right) \sin \left(\frac{6 \pi k}{20000}\right)\\ &B(k)=\frac{15}{30}\left(\frac{k}{20000}\right)^4-\cos \left(\frac{17 \pi}{20}\left(\frac{k}{20000}\right)^5\right)\left(\frac{11}{10}+\frac{45}{20} \cos ^8\left(\frac{\pi k}{40000}\right) \cos ^6\left(\frac{3 \pi k}{40000}\right)\right) \cos ^6\left(\frac{41 \pi k}{20000}\right)+\frac{12}{20} \cos ^{10}\left(\frac{3 \pi k}{200000}\right) \cos ^{10}\left(\frac{9 \pi k}{200000}\right) \cos ^{10}\left(\frac{18 \pi k}{200000}\right)\\ &R(k)=\frac{1}{50}+\frac{1}{40} \sin ^2\left(\frac{41 \pi k}{20000}\right) \sin ^2\left(\frac{9 \pi k}{200000}\right)+\left(\frac{1}{17}\right)\cos ^2\left(\frac{41 \pi k}{20000}\right) \cos ^{10} \left(\frac{\pi k}{40000}\right) \end{aligned} A(k)=450003k+sin(2017π(20000k)5)cos6(2000041πk)+(31cos16(2000041πk)+31cos80(2000041πk))cos12(40000πk)sin(200006πk)B(k)=3015(20000k)4cos(2017π(20000k)5)(1011+2045cos8(40000πk)cos6(400003πk))cos6(2000041πk)+2012cos10(2000003πk)cos10(2000009πk)cos10(20000018πk)R(k)=501+401sin2(2000041πk)sin2(2000009πk)+(171)cos2(2000041πk)cos10(40000πk)

    完整代码

    % bird 2
    K=-2e4:2e4;
    t=linspace(0,2*pi,200);
    
    X=@(k) k./15e3+sin(17.*pi./20.*(k./2e4).^5).*cos(41.*pi.*k./2e4).^6+...
            (1./3.*cos(41.*pi.*k./2e4).^16+1./3.*cos(41.*pi.*k./2e4).^80).*cos(pi.*k./4e4).^12.*sin(6.*pi.*k./2e4);
    Y=@(k) 1./2.*(k./2e4).^4-cos(17.*pi./20.*(k./2e4).^5).*(11./10+45./20.*cos(pi.*k./4e4).^8.*cos(3.*pi.*k./4e4).^6).*cos(41.*pi.*k./2e4).^6+...
            12./20.*cos(3.*pi.*k./2e5).^10.*cos(9.*pi.*k./2e5).^10.*cos(8.*pi.*k./2e5).^10;
    R=@(k) 1./50+1./40.*sin(41.*pi.*k./2e4).^2.*sin(9.*pi.*k./2e5).^2+1./17.*cos(41.*pi.*k./2e4).^2.*cos(pi.*k./4e4).^10;
    
    CX=[X(K')+cos(t).*R(K'),K'.*nan]';
    CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
    plot(CX(:),CY(:),'Color',[0,0,0,.2]);
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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    蝴蝶一

    This image shows 40000 circles. For k = 1 , 2 , 3 , … , 40000 k=1,2,3, \ldots, 40000 k=1,2,3,,40000 , the center of the circle is ( X ( k ) , Y ( k ) ) (X(k), Y(k)) (X(k),Y(k)) and the radius of the k-th circle is R ( k ) R(k) R(k) , where:

    X ( k ) = ( 3 / 2 ) ( ( cos ⁡ ( 141 π k / 40 , 000 ) ) 9 ) ∗ ( 1 − ( 1 / 2 ) sin ⁡ ( π k / 40 , 000 ) ) ∗ ( 1 − ( 1 / 4 ) ( ( cos ⁡ ( 2 π k / 40 , 000 ) ) 30 ) ∗ ( 1 + ( cos ⁡ ( 32 π k / 40 , 000 ) ) 20 ) ) ∗ ( 1 − ( 1 / 2 ) ( ( sin ⁡ ( 2 π k / 40 , 000 ) ) 30 ) ∗ ( ( sin ⁡ ( 6 π k / 40 , 000 ) ) 10 ) ∗ ( ( 1 / 2 ) + ( 1 / 2 ) ( sin ⁡ ( 18 π k / 40 , 000 ) ) 20 ) ) Y ( k ) = cos ⁡ ( 2 π k / 40 , 000 ) ∗ ( ( cos ⁡ ( 141 π k / 40 , 000 ) ) 2 ) ∗ ( 1 + ( 1 / 4 ) ( ( cos ⁡ ( π k / 40 , 000 ) ) 24 ) ∗ ( ( cos ⁡ ( 3 π k / 40 , 000 ) ) 24 ) ∗ ( cos ⁡ ( 21 π k / 40 , 000 ) ) 24 ) R ( k ) = ( 1 / 100 ) + ( 1 / 40 ) ( ( ( cos ⁡ ( 141 π k / 40 , 000 ) ) 14 ) + ( sin ⁡ ( 141 π k / 40 , 000 ) ) 6 ) ∗ ( 1 − ( ( cos ⁡ ( π k / 40 , 000 ) ) 16 ) ( ( cos ⁡ ( 3 π k / 40 , 000 ) ) 16 ) ∗ ( cos ⁡ ( 12 π k / 40 , 000 ) ) 16 ) . \begin{aligned} X(k)=&(3 / 2)\left((\cos (141 \pi k / 40,000))^9\right) * \\ &(1-(1 / 2) \sin (\pi k / 40,000)) * \\ &\left(1-(1 / 4)\left((\cos (2 \pi k / 40,000))^{30}\right) *\right.\\ &\left.\left(1+(\cos (32 \pi k / 40,000))^{20}\right)\right) * \\ &\left(1-(1 / 2)\left((\sin (2 \pi k / 40,000))^{30}\right) *\right.\\ &\left((\sin (6 \pi k / 40,000))^{10}\right) * \\ &\left.\left((1 / 2)+(1 / 2)(\sin (18 \pi k / 40,000))^{20}\right)\right)\\ Y(k)=& \cos (2 \pi k / 40,000) * \\ &\left((\cos (141 \pi k / 40,000))^2\right) * \\ &\left(1+(1 / 4)\left((\cos (\pi k / 40,000))^{24}\right) *\right.\\ &\left((\cos (3 \pi k / 40,000))^{24}\right) * \\ &\left.(\cos (21 \pi k / 40,000))^{24}\right)\\ R(k)=&(1 / 100)+(1 / 40)\left(\left((\cos (141 \pi k / 40,000))^{14}\right)+(\sin (141 \pi k / 40,000))^6\right) * \\ &\left(1-\left((\cos (\pi k / 40,000))^{16}\right)\left((\cos (3 \pi k / 40,000))^{16}\right) *\right.\\ &\left.(\cos (12 \pi k / 40,000))^{16}\right) . \end{aligned} X(k)=Y(k)=R(k)=(3/2)((cos(141πk/40,000))9)(1(1/2)sin(πk/40,000))(1(1/4)((cos(2πk/40,000))30)(1+(cos(32πk/40,000))20))(1(1/2)((sin(2πk/40,000))30)((sin(6πk/40,000))10)((1/2)+(1/2)(sin(18πk/40,000))20))cos(2πk/40,000)((cos(141πk/40,000))2)(1+(1/4)((cos(πk/40,000))24)((cos(3πk/40,000))24)(cos(21πk/40,000))24)(1/100)+(1/40)(((cos(141πk/40,000))14)+(sin(141πk/40,000))6)(1((cos(πk/40,000))16)((cos(3πk/40,000))16)(cos(12πk/40,000))16).

    完整代码

    % butterfly 1
    K=1:4e4;
    t=linspace(0,2*pi,200);
    
    X=@(k) 3./2.*cos(141.*pi.*k./4e4).^9.*(1-1./2.*sin(pi.*k./4e4)).*(1-1./4.*cos(2.*pi.*k./4e4).^30.*(1+cos(32.*pi.*k./4e4).^20)).*...
            (1-1./2.*sin(2.*pi.*k./4e4).^30.*sin(6.*pi.*k./4e4).^10.*(1./2+1./2.*sin(18.*pi.*k./4e4).^20));
    Y=@(k) cos(2.*pi.*k./4e4).*cos(141.*pi.*k./4e4).^2.*(1+1./4.*cos(pi.*k./4e4).^24.*cos(3.*pi.*k./4e4).^24.*cos(21.*pi.*k./4e4).^24);
    R=@(k) 1./100+1./40.*(cos(141.*pi.*k./4e4).^14+sin(141.*pi.*k./4e4).^6).*(1-cos(pi.*k./4e4).^16.*cos(3.*pi.*k./4e4).^16.*cos(12.*pi.*k./4e4).^16);
    
    CX=[X(K')+cos(t).*R(K'),K'.*nan]';
    CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
    plot(CX(:),CY(:),'Color',[0,0,0,.2]);
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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    蝴蝶二

    This image shows 40000 circles. For k = 1 , 2 , 3 , … , 40000 k=1,2,3, \ldots, 40000 k=1,2,3,,40000 , the center of the circle is ( X ( k ) , Y ( k ) ) (X(k), Y(k)) (X(k),Y(k)) and the radius of the k-th circle is R ( k ) R(k) R(k) , where:

    X ( k ) = ( 6 / 5 ) ( ( cos ⁡ ( 141 π k / 40 , 000 ) ) 9 ) ( 1 − ( 1 / 2 ) ( sin ⁡ ( π k / 40 , 000 ) ) 3 ) ∗ ( 1 − ( 1 / 4 ) ( ( cos ⁡ ( 2 π k / 40 , 000 ) ) 30 ) ( 1 + ( 2 / 3 ) ( cos ⁡ ( 30 π k / 40 , 000 ) ) 20 ) − ( ( sin ⁡ ( 2 π k / 40 , 000 ) ) 10 ) ( ( sin ⁡ ( 6 π k / 40 , 000 ) ) 10 ) ∗ ( ( 1 / 5 ) + ( 4 / 5 ) ( cos ⁡ ( 24 π k / 40 , 000 ) ) 20 ) ) Y ( k ) = cos ⁡ ( 2 π k / 40 , 000 ) ( ( cos ⁡ ( 141 π k / 40 , 000 ) ) 2 ) ( 1 + ( 1 / 4 ) ( ( cos ⁡ ( π k / 40 , 000 ) ) 24 ) ∗ ( ( cos ⁡ ( 3 π k / 40 , 000 ) ) 24 ) ( cos ⁡ ( 19 π k / 40 , 000 ) ) 24 ) R ( k ) = ( 1 / 100 ) + ( 1 / 40 ) ( ( ( cos ⁡ ( 2820 π k / 40 , 000 ) ) 6 ) + ( sin ⁡ ( 141 π k / 40 , 000 ) ) 2 ) ( 1 − ( ( cos ⁡ ( π k / 40 , 000 ) ) 16 ) ∗ ( ( cos ⁡ ( 3 π k / 40 , 000 ) ) 16 ) ( cos ⁡ ( 12 π k / 40 , 000 ) ) 16 ) . \begin{aligned} X(k)=&(6 / 5)\left((\cos (141 \pi k / 40,000))^9\right)\left(1-(1 / 2)(\sin (\pi k / 40,000))^3\right) * \\ &\left(1-(1 / 4)\left((\cos (2 \pi k / 40,000))^{30}\right)\left(1+(2 / 3)(\cos (30 \pi k / 40,000))^{20}\right)-\right.\\ &\left((\sin (2 \pi k / 40,000))^{10}\right)\left((\sin (6 \pi k / 40,000))^{10}\right) * \\ &\left.\left((1 / 5)+(4 / 5)(\cos (24 \pi k / 40,000))^{20}\right)\right) \\ Y(k)=& \cos (2 \pi k / 40,000)\left((\cos (141 \pi k / 40,000))^2\right)\left(1+(1 / 4)\left((\cos (\pi k / 40,000))^{24}\right) *\right.\\ &\left.\left((\cos (3 \pi k / 40,000))^{24}\right)(\cos (19 \pi k / 40,000))^{24}\right) \\ R(k)=&(1 / 100)+(1 / 40)\left(\left((\cos (2820 \pi k / 40,000))^6\right)+\right.\\ &\left.(\sin (141 \pi k / 40,000))^2\right)\left(1-\left((\cos (\pi k / 40,000))^{16}\right) *\right.\\ &\left.\left((\cos (3 \pi k / 40,000))^{16}\right)(\cos (12 \pi k / 40,000))^{16}\right) . \end{aligned} X(k)=Y(k)=R(k)=(6/5)((cos(141πk/40,000))9)(1(1/2)(sin(πk/40,000))3)(1(1/4)((cos(2πk/40,000))30)(1+(2/3)(cos(30πk/40,000))20)((sin(2πk/40,000))10)((sin(6πk/40,000))10)((1/5)+(4/5)(cos(24πk/40,000))20))cos(2πk/40,000)((cos(141πk/40,000))2)(1+(1/4)((cos(πk/40,000))24)((cos(3πk/40,000))24)(cos(19πk/40,000))24)(1/100)+(1/40)(((cos(2820πk/40,000))6)+(sin(141πk/40,000))2)(1((cos(πk/40,000))16)((cos(3πk/40,000))16)(cos(12πk/40,000))16).

    完整代码

    % butterfly 2
    K=1:4e4;
    t=linspace(0,2*pi,200);
    
    X=@(k) 6./5.*cos(141.*pi.*k./4e4).^9.*(1-1./2.*sin(pi.*k./4e4).^3).*(1-1./4.*cos(2.*pi.*k./4e4).^30.*(1+2./3.*cos(30.*pi.*k./4e4).^20)-...
            sin(2.*pi.*k./4e4).^10.*sin(6.*pi.*k./4e4).^10.*(1./5+4./5.*cos(24.*pi.*k./4e4).^20));
    Y=@(k) cos(2.*pi.*k./4e4).*cos(141.*pi.*k./4e4).^2.*(1+1./4.*cos(pi.*k./4e4).^24.*cos(3.*pi.*k./4e4).^24.*cos(19.*pi.*k./4e4).^24);
    R=@(k) 1./100+1./40.*(cos(2820.*pi.*k./4e4).^6+sin(141.*pi.*k./4e4).^2).*(1-cos(pi.*k./4e4).^16.*cos(3.*pi.*k./4e4).^16.*cos(12.*pi.*k./4e4).^16);
    
    CX=[X(K')+cos(t).*R(K'),K'.*nan]';
    CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
    plot(CX(:),CY(:),'Color',[0,0,0,.2]);
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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    橄榄枝

    This image shows 4000 circles. For k = 1 , 2 , 3 , … , 4000 k=1,2,3, \ldots, 4000 k=1,2,3,,4000 , the center of the circle is ( X ( k ) , Y ( k ) ) (X(k), Y(k)) (X(k),Y(k)) and the radius of the k-th circle is R ( k ) R(k) R(k) , where:

    X ( k ) = ( 2 k / 4000 ) + ( 1 / 28 ) sin ⁡ ( 42 π k / 4000 ) + ( 1 / 9 ) ( ( sin ⁡ ( 21 π k / 4000 ) ) 8 ) + ( 1 / 4 ) ( ( sin ⁡ ( 21 π k / 4000 ) ) 6 ) ∗ sin ⁡ ( ( 2 π / 5 ) ( k / 4000 ) 12 ) Y ( k ) = ( 1 / 4 ) ( k / 4000 ) 2 + ( 1 / 4 ) ( ( ( sin ⁡ ( 21 π k / 4000 ) ) 5 ) + ( 1 / 28 ) sin ⁡ ( 42 π k / 4000 ) ) ∗ ( cos ⁡ ( ( π / 2 ) ( k / 4000 ) 12 ) ) R ( k ) = ( 1 / 170 ) + ( 1 / 67 ) ( ( sin ⁡ ( 42 π k / 4000 ) ) 2 ) ∗ ( 1 − ( ( cos ⁡ ( 21 π k / 4000 ) ) 4 ) ) . \begin{aligned} X(k)=&(2 k / 4000)+(1 / 28) \sin (42 \pi k / 4000)\\ &+(1 / 9)\left((\sin (21 \pi k / 4000))^8\right)\\ &+(1 / 4)\left((\sin (21 \pi k / 4000))^6\right) * \\ &\sin \left((2 \pi / 5)(k / 4000)^{12}\right)\\ Y(k)=&(1 / 4)(k / 4000)^2\\ &+(1 / 4)\left(\left((\sin (21 \pi k / 4000))^5\right)\right.\\ &+(1 / 28) \sin (42 \pi k / 4000)) * \\ &\left(\cos \left((\pi / 2)(k / 4000)^{12}\right)\right)\\ R(k)=&(1 / 170)+(1 / 67)\left((\sin (42 \pi k / 4000))^2\right) * \\ &\left(1-\left((\cos (21 \pi k / 4000))^4\right)\right) . \end{aligned} X(k)=Y(k)=R(k)=(2k/4000)+(1/28)sin(42πk/4000)+(1/9)((sin(21πk/4000))8)+(1/4)((sin(21πk/4000))6)sin((2π/5)(k/4000)12)(1/4)(k/4000)2+(1/4)(((sin(21πk/4000))5)+(1/28)sin(42πk/4000))(cos((π/2)(k/4000)12))(1/170)+(1/67)((sin(42πk/4000))2)(1((cos(21πk/4000))4)).

    完整代码

    % olive
    K=1:4e3;
    t=linspace(0,2*pi,200);
    
    X=@(k) 2.*k./4e3+1./28.*sin(42.*pi.*k./4e3)+1./9.*sin(21.*pi.*k./4e3).^8+1./4.*sin(21.*pi.*k./4e3).^6.*sin(2.*pi./5.*(k./4e3).^12);
    Y=@(k) 1./4.*(k./4e3).^2+1./4.*(sin(21.*pi.*k./4e3).^5+1./28.*sin(41.*pi.*k./4e3)).*cos(pi./2.*(k./4e3).^12);
    R=@(k) 1./170+1./67.*sin(42.*pi.*k./4e3).^2.*(1-cos(21.*pi.*k./4e3).^4);
    
    CX=[X(K')+cos(t).*R(K'),K'.*nan]';
    CY=[Y(K')+sin(t).*R(K'),K'.*nan]';
    plot(CX(:),CY(:),'Color',[0,0,0,.2]);
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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    当然除了画圈圈,该网站上还有一些其他有趣的图形,可以自行去查看,最后,给大家比个心~

    心形

    This image shows the following curve:

    X ( t ) = 4 9 sin ⁡ ( 2 t ) + 1 3 ( sin ⁡ ( t ) ) 8 cos ⁡ ( 3 t ) + 1 8 sin ⁡ ( 2 t ) ( cos ⁡ ( 247 t ) ) 4 Y ( t ) = sin ⁡ ( t ) + 1 3 ( sin ⁡ ( t ) ) 8 sin ⁡ ( 3 t ) + 1 8 sin ⁡ ( 2 t ) ( sin ⁡ ( 247 t ) ) 4 , 0 ≤ t ≤ π \begin{aligned} &X(t)=\frac{4}{9} \sin (2 t)+\frac{1}{3}(\sin (t))^8 \cos (3 t)+\frac{1}{8} \sin (2 t)(\cos (247 t))^4 \\ &Y(t)=\sin (t)+\frac{1}{3}(\sin (t))^8 \sin (3 t)+\frac{1}{8} \sin (2 t)(\sin (247 t))^4, 0 \leq t \leq \pi \end{aligned} X(t)=94sin(2t)+31(sin(t))8cos(3t)+81sin(2t)(cos(247t))4Y(t)=sin(t)+31(sin(t))8sin(3t)+81sin(2t)(sin(247t))4,0tπ

    完整代码

    % heart
    t=linspace(0,pi,5e3);
    X=@(t) 4./9.*sin(2.*t)+1./3.*sin(t).^8.*cos(3.*t)+1./8.*sin(2.*t).*cos(247.*t).^4;
    Y=@(t) sin(t)+1./3.*sin(t).^8.*sin(3.*t)+1./8.*sin(2.*t).*sin(247.*t).^4;
    
    hLine=plot(X(t),Y(t),'-','LineWidth',1);
    colorNum=length(hLine.XData);
    colorData=uint8([(pink(colorNum).*255)';255+t.*0]);
    pause(1e-16) 
    set(hLine.Edge,'ColorBinding','interpolated', 'ColorData',colorData)
    set(gca,'DataAspectRatio',[1,1,1],'XColor','none','YColor','none');
    
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  • 原文地址:https://blog.csdn.net/slandarer/article/details/127702433