How can I createa new flower from this script??

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Leonardo Toglia
Leonardo Toglia el 31 de Mzo. de 2021
Comentada: darova el 1 de Abr. de 2021
n=800;
r=linspace(0,1,n);
theta=linspace(0,2*pi,n);
[R,THETA]=ndgrid(r,theta);
% define the number of petals we want per cycle. Violets have 5!
petalNum=5;
x = 1 - (1/2)*((5/4)*(1 - mod(petalNum*THETA, 2*pi)/pi).^2 - 1/4).^2;
% update the formula for phi to stop the spiralling effect
phi = (pi/2)*exp(-2*pi/(8*pi));
y = 1.95653*(R.^2).*(1.27689*R - 1).^2.*sin(phi);
R2 = x.*(R.*sin(phi) + y.*cos(phi));
X=R2.*sin(THETA);
Y=R2.*cos(THETA);
Z=x.*(R.*cos(phi)-y.*sin(phi));
% color code for blueviolet RGB 138,43,226
% Indigo rgb value (75,0,130)
mapSize=20;
blue_map=linspace(138,75,mapSize)';
blue_map(:,2)=linspace(43,0,mapSize)';
blue_map(:,3)=linspace(226,130,mapSize)';
% and a pretty bit in the middle
gold_map=[255 215 0; 250 210 0];
% combine to form a full flower map
violet_map=[gold_map; blue_map];
% Now lets plot it and try and pick an attractive angle
surf(X,Y,Z,'LineStyle','none')
colormap(violet_map/255)
view([-12.700 81.200])
  2 comentarios
darova
darova el 31 de Mzo. de 2021
I've just run your code
Leonardo Toglia
Leonardo Toglia el 31 de Mzo. de 2021
Hi @darova but how can I modify to obtain a tulip flower like this?? thanks for replying soon

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darova
darova el 31 de Mzo. de 2021
See this modification
R2 = x.*(R.*sin(phi) + y.*cos(phi));
RR = (R2-min(R2(:)))./(max(R2(:))-min(R2(:))); % scale inbetween [0 .. 1]
R3 = sin(RR*pi*0.9); % create new R
X=R3.*sin(THETA);
Y=R3.*cos(THETA);
  3 comentarios
Leonardo Toglia
Leonardo Toglia el 31 de Mzo. de 2021
for a tighter version of the petals I add another R? @darova
darova
darova el 1 de Abr. de 2021
Yes, experiment with RR variable

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