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程序如下
gama = 2*pi/9;
h = 300;
Rr = 400;
<p无奈的爆米花;
R1 = h/(2*sin(gama/2));
theta1 = linspace(0,gama/8,10);
theta2 = linspace(gama/8,gama/4,10);
theta3 = linspace(gama/4,3*gama/8,10);
theta4 = linspace(3*gama/8,gama/2,10);
theta5 = linspace(gama/2,5*gama/8,10);
theta6 = linspace(5*gama/8,3*gama/4,10);
theta7 = linspace(3*gama/4,7*gama/8,10);
theta8 = linspace(7*gama/8,gama,10);
theta = [theta1 theta2 theta3 theta4 theta5 theta6 theta7 theta8];
<pdxz = -80.9186*theta1.^3+279.4697*theta1.^2-153.9023*theta1+474.3163;
<pcqdwd = -80.9186*(theta2-gama/8).^3+258.2853*(theta2-gama/8).^2-106.7218*(theta2-gama/8)+462.9603;
<p欢喜的背包 = -407.0496*(theta3-gama/4).^3+237.1009*(theta3-gama/4).^2-63.7437*(theta3-gama/4)+455.5382;
<pdrdzc = 388.6254*(theta4-3*gama/8).^3+130.5355*(theta4-3*gama/8).^2-31.6614*(theta4-3*gama/8)+451.5106;
<p自觉的故事 = -388.6254*(theta5-gama/2).^3+232.2774*(theta5-gama/2).^2+450.0000;
<pttddd = 407.0496*(theta6-5*gama/8).^3+130.5355*(theta6-5*gama/8).^2+31.6614*(theta6-5*gama/8)+451.5106;
<p优美的发箍 = 80.9186*(theta7-3*gama/4).^3+237.1009*(theta7-3*gama/4).^2+63.7437*(theta7-3*gama/4)+455.5382;
<p内向的大米 = 80.9186*(theta8-7*gama/8).^3+258.2853*(theta8-7*gama/8).^2+106.9743*(theta8-7*gama/8)+462.9603;
c1 = -R1.*(sin(gama-theta)+sin(theta));
c2 = R1.*(cos(gama-theta)-cos(theta));
c3 = -h^2*sin(theta)./(sin(gama-theta)+sin(theta))+h^2/2-…
R1^2*sin(gama).*(cos(gama-theta)-cos(theta))./(sin(gama-theta)+sin(theta));
c5 = h.*sin(theta)./(sin(gama-theta)+sin(theta));
YY4 = [Y4(1) Y4(2) Y4(3) Y4(4) Y4(5) Y4(6) Y4(7) Y4(8) Y4(9) ceil(Y4(10))];
<pdxz1 = [Y1 Y2 Y3 YY4];
<pcqdwd2 = [Y5 Y6 Y7 Y8];
Y =[Y11 Y22];
<pdfdbl = -sqrt(Rr^2-(Y11-Dr).^2);
<pcqdjd = sqrt(Rr^2-(Y22-Dr).^2);
Z = [Z1 Z2];
X = -c3./c1-h./c1.*Z-c2./c1.*Y;
R1 = h/(2*sin(gama/2));
Xc = 0;
Yc = R1*sin(gama)./(sin(gama-theta)+sin(theta));
Zc = h*sin(theta)./(sin(gama-theta)+sin(theta))-h/2;
r = sqrt((X-Xc).^2+(Y-Yc).^2+(Z-Zc).^2);
plot(theta,r)
谢谢各位大哥，想要实现的就是将r绕theta轴进行旋转，成为空间曲面
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