同學問的,查了下資料。
%需要擬合的點的坐標為(0,-174.802,990.048),(0.472,-171.284,995.463),(0.413,-168.639,1003.55),(0.064,-167.862,1019.55),%(0,-170.357,1035.44),(0,-172.142,1044.78),(0.215,-174.759,1047.84),(0.171,-176.586,1048.13),(0,-179.832,1043.34),(0,181.589,1040.11),(0,-182.76,1032.62),(0,-184.13,1017.55),(0.113,-183.445,1003.17)function my_fit_new()% 日期:2011年12月29日% 作者:半人馬alpha% 適用於你說的情況% 你的數據擬合結果是一個旋轉雙曲面(a,c均為虛數,即 a^2<0,c^2<0)% 我按擬合出的參數給你把圖畫了一下,是旋轉雙曲面的一支% step0:生成擬合數據(例)x = [0,0,0,0,0,0,0,0.064,0.113,0.171,0.215,0.413,0.472]';y = [-174.802,-170.357,-172.142,-179.832,181.589,-182.760,-184.130,-167.862,-183.445,-176.586,-174.759,-168.639,-171.284]';z = [990.048,1035.44,1044.78,1043.34,1040.11,1032.62,1017.55,1019.55,1003.17,1048.13,1047.84,1003.55,995.463]';% step1:擬合,k表示系數,行向量% 待擬合方程:F = z^2 = (-c^2/a^2*x^2) + (c^2/a^2*2*x1*x) + (- c^2/b^2*y^2) +% (c^2/b^2*2*y1*y) + (2*z1*z) +% (-c^2/a^2*x1^2 - c^2/b^2*y1^2 - z1^2 + c^2)% x,y,z 均要先轉化為列向量% k(1) = -c^2/a^2 由k值就可求出橢圓所有參數!!!% k(2) = c^2/a^2*2*x1% k(3) = - c^2/b^2% k(4) = c^2/b^2*2*y1% k(5) = 2*z1% k(6) = -c^2/a^2*x1^2 - c^2/b^2*y1^2 - z1^2 + c^2xdata = [x,y,z];ydata = z.^2; %% 先把 z 值平方,再進行擬合k0 = ones(1,6); %% k 的運行初值,不會影響最終結果F = @(k,xdata) k(1)*xdata(:,1).^2 + k(2)*xdata(:,1) + k(3)*xdata(:,2).^2 + k(4)*xdata(:,2) + k(5)*xdata(:,3) + k(6);[k,resnorm]=lsqcurvefit(F,k0,xdata,ydata);% step2:橢圓參數求解x1 = -k(2)/k(1)/2;y1 = -k(4)/k(3)/2;z1 = k(5)/2;c = sqrt(z1^2 + k(6) - k(1)*x1^2 - k(3)*y1^2);a = c/sqrt(-k(1));b = c/sqrt(-k(3));disp('x1:');disp(x1);disp('y1:');disp(y1);disp('z1:');disp(z1);disp('a軸:');disp(a);disp('b軸:');disp(b);disp('c軸:');disp(c);end