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比例導引三維仿真軌跡 ,matlab程序 ,仿真效果不錯,對彈道仿真很有幫助

代碼片段和文件信息
%給出了比例導引法的差分方程?采用Matlab語言?對比例導引法的導彈理想彈道進行了三維數據仿真?繪制出三維理想彈道。隨導彈與目標參數變化情況?對理想彈道的特性進行了分析
clear?all
clc
clear
tt=0.1;
sm=0.6*tt;
st=0.42*tt;
x(1)=0;y(1)=0;z(1)=0;
pmr(:1)=[x(1);y(1);z(1)];%導彈起始位置
ptr(:1)=[25;5;10];%目標起始位置
m=3;%比例導引系數
q(1)=0;
o(1)=0;
a(1)=0;
for(k=2:600)
ptr(:k)=[25-0.42*cos(pi/6)*tt*k;5;10+0.42*sin(pi/6)*k*tt];
r(k-1)=sqrt((ptr(1k-1)-pmr(1k-1))^2+(ptr(2k-1)-pmr(2k-1))^2+(ptr(3k-1)-pmr(3k-1))^2);?%目標和導彈相對距離
c=sqrt((ptr(1k)-pmr(1k-1))^2+(ptr(2k)-pmr(2k-1))^2+(ptr(3k)-pmr(3k-1))^2);
b=acos((r(k-1)^2+st^2-c^2)/(2*r(k-1)*st));
dq=acos((r(k-1)^2-st^2+c^2)/(2*r(k-1)*c));
if?abs(imag(b))>0
b=0.0000001;
end?
if?abs(imag(dq))>0
dq=0.0000001;
end?
q(k)=q(k-1)+dq;
o(k)=o(k-1)+m*dq;
a(k)=o(k)-q(k);
c1=r(k-1)*sin(b)/sin(a(k)+b);
c2=r(k-1)*sin(a(k))/sin(a(k)+b);
c3=sqrt((c1-sm)^2+(c2-st)^2+2*(c1-sm)*(c2-st)*cos(a(k)+b));
dq=a(k)-acos(((c1-sm)^2+c3^2-(c2-st)^2)/(2*(c1-sm)*c3));
if?abs(imag(dq))>0
dq=0.0000001;
end?
q(k)=q(k-1)+dq;
o(k)=o(k-1)+m*dq;
a(k)=o(k)-q(k);
c1=r(k-1)*sin(b)/sin(a(k)+b);
c2=r(k-1)*sin(a(k))/sin(a(k)+b);
c3=sqrt((c1-sm)^2+(c2-st)^2+2*(c1-sm)*(c2-st)*cos(a(k)+b));
dq=a(k)-acos(((c1-sm)^2+c3^2-(c2-st)^2)/(2*(c1-sm)*c3));
if?abs(imag(dq))>0
dq=0.0000001;
end
q(k)=q(k-1)+dq;
o(k)=o(k-1)+m*dq;
a(k)=o(k)-q(k);
c1=r(k-1)*sin(b)/sin(a(k)+b);
c2=r(k-1)*sin(a(k))/sin(a(k)+b);
c3=sqrt((c1-sm)^2+(c2-st)^2+2*(c1-sm)*(c2-st)*cos(a(k)+b));
x1(k)=ptr(1k-1)+c2/st*(ptr(1k)-ptr(1k-1));
y1(k)=ptr(2k-1)+c2/st*(ptr(2k)-ptr(2k-1));
z1(k)=ptr(3k-1)+c2/st*(ptr(3k)-ptr(3k-1));
x(k)=pmr(1k-1)+sm/c1*(x1(k)-pmr(1k-1));
y(k)=pmr(2k-1)+sm/c1*(y1(k)-pmr(2k-1));
z(k)=pmr(3k-1)+sm/c1*(z1(k)-pmr(3k-1));
pmr(:k)=[x(k);y(k);z(k)];
r(k)=sqrt((ptr(1k)-pmr(1k))^2+(ptr(2k)-pmr(2k))^2+(ptr(3k)-pmr(3k))^2);
if?r(k)<0.06;
break;
end;
end
sprintf(‘遭遇時間:%3.1f‘0.1*k);
figure(1);
plot3(pmr(11:k)pmr(21:k)pmr(31:k)‘g‘ptr(1:)ptr(2:)ptr(3:)‘r‘);
axis([0?25?0?5?0?25]);
text(x(80)y(80)z(80)‘\leftarrow?比例導引‘);
grid?on
?屬性????????????大小?????日期????時間???名稱
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?????文件???????2152??2011-06-24?17:37??bilidaoyin.m
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?????????????????2152????????????????????1
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