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这是一条镜像帖。来源:北邮人论坛 / matlab / #5167同步于 2009/4/16
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Matlab机器人发帖

【新人求助】如何能在图像中弄出最大值?

jymingyue
2009/4/16镜像同步1 回复
程序如下 [a1,b1]=meshgrid(0.21:0.0305:0.82,0.205:0.02425:0.69); for i=1:21 for j=1:21 ZJ3(i,j)=FJ3fun(a1(i,j),b1(i,j)); end end mesh(a1,b1,ZJ3); function FJ3=Fj3fun(Lhn,Lkm) G1=1700; G2=1450; L1=0.205; Lbc=0.24; Lhj=0.4; Ljn=0.5045; Lde=0.24; Lfd=0.24; Ljm=0.443; Ljk=0.32; Ldb=2.4; Ldo1=0.945; Ldg=0.4; x3=50.6*pi/180; x6=pi*61.7/180; x7=pi/2; x9=pi/3; cosx3_x1=(Lhj*Lhj+Ljn*Ljn-Lhn*Lhn)/(2*Lhj*Ljn); x1=x3-acos(cosx3_x1); cosx6_x2_x7=(Ljm*Ljm+Ljk*Ljk-Lkm*Lkm)/(2*Ljk*Ljm); x2=x7-x6+acos(cosx6_x2_x7); F1=(G1*L1)/(Lbc*sin(x9-x1)); FX1=F1*cos(x1); FY1=F1*sin(x1)+G1; F2=(F1*Lde*sin(x9-x1))/(Lfd*sin(pi-x2)); Fx21=F2*cos(x2)-F1*cos(x1); Fy21=F2*sin(x2)-F1*sin(x1); F3=(FY1*Ldb*cos(x1)+G2*Ldo1*cos(x1)-FX1*Ldb*sin(x1))/(Ldg*sin(x2-x1)); Fx22=FX1+F3*cos(x2); Fy22=F3*sin(x2)+G2+FY1; FX2=Fx21+Fx22; FY2=Fy21+Fy22; sin_x4=(Lhj*sin(x3-x1))/Lhn; FJ3=(F3*Lhj*sin(x2-x1))/(Ljn*sin_x4); 但运行后只出个图形,我想知道最大值,如何让MATLAB再生成的图像中给出最大值。 就是让上图中有入下图的数字显示
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hyades机器人#1 · 2009/4/21
FIGURE TOOLS里面有个选项是取点值的