用空心曲面表示山体的方程

Question

我想用函数，其中x和y是坐标形成的圆，z(x，y)从边缘的0增长到R/2的最大值，在中心再增长到0，没有剧烈的变化。我和你在一起

t = -pi:pi/180:pi;
R = 5;
x = R*cos(t);
y = R*sin(t);

for i = 1:361
  for j = 1:361
     z(i,j) = exp( sqrt((x(i)).^2 + (y(j)).^2));
  end
end
[u, v] = meshgrid(x, y);
mesh(u, v, z), grid on;

我应该如何意识到z把这个滴加到中间呢？如有任何建议请见谅！

Answer 1

也许用不同的标准差减去两个二维高斯？

% the area
x = linspace(-5,5,1E2);
y = linspace(-5,5,1E2);

sig1=1;
sig2=2;

%2D gaussian
efac = 1/(2*sig1);
X   = exp(-efac*x.^2);
Y   = exp(-efac*y.^2)';
z1    = Y*X;
z1=z1./max(z1(:));

%2D gaussian
efac = 1/(2*sig2);
X   = exp(-efac*x.^2);
Y   = exp(-efac*y.^2)';
z2    = Y*X;
z2=z2./max(z2(:));

[u, v] = meshgrid(x, y);
mesh(u, v, z2-z1), grid on;

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在R之外有一个零的替代方案，它更符合您自己的代码：

x = linspace(-2*pi,2*pi,1E2);
y = linspace(-2*pi,2*pi,1E2);
[u, v] = meshgrid(x, y);
r = sqrt(u.^2+v.^2);
z = sin(r);
z(r>pi)=0;

mesh(u, v, z), grid on;

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或者让它在底部少一点锋利：

z = sin(r).^2;

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