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資源簡介

這個程序主要是用matlab編寫代碼實現慣性導航的仿真,對其工作原理進行仿真,大家可以借鑒

資源截圖

代碼片段和文件信息

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%?????????指北方位捷聯式慣性導航系統?????????%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%初始化
clear

load(‘jlfw‘);

%定義常量
Re=6378245;??????????????????????????????????????????????????????????
e=1/298.3;
wie_scalar=7.292115147e-5;
g0=9.7803267714;
gk1=0.00193185138639;
gk2=0.00669437999013;
%變量賦初值
longitude=[116.344695283*pi/180?zeros(147999)];?
latitude=[39.975172*pi/180?zeros(147999)];
h=30;
f=zeros(348000);
wnb=zeros(348000);
theta=[0.120992605?zeros(147999)]*pi/180;
gamma=[0.010445947?zeros(147999)]*pi/180;
psi=[360-91.637207?zeros(147999)]*pi/180;
Q=[cos(psi(1)/2)*cos(theta(1)/2)*cos(gamma(1)/2)+sin(psi(1)/2)*sin(theta(1)/2)*sin(gamma(1)/2);
???cos(psi(1)/2)*sin(theta(1)/2)*cos(gamma(1)/2)+sin(psi(1)/2)*cos(theta(1)/2)*sin(gamma(1)/2);
???cos(psi(1)/2)*cos(theta(1)/2)*sin(gamma(1)/2)-sin(psi(1)/2)*sin(theta(1)/2)*cos(gamma(1)/2);
???cos(psi(1)/2)*sin(theta(1)/2)*sin(gamma(1)/2)-sin(psi(1)/2)*cos(theta(1)/2)*cos(gamma(1)/2)];
velocity=[0.000048637?0.000206947?0.007106781];
v=[zeros(148000)];

for?i=2:48000
%計算所在地參數
??%Rx為卯酉圈的曲率半徑
??Rx=Re/(1-e*sin(latitude(i-1))^2);
??%Ry為子午圈的曲率半徑
??Ry=Re/(1+2*e-3*e*sin(latitude(i-1))^2);
??%使用WGS-84模型計算該位置的重力加速度
??g=g0*(1+gk1*sin(latitude(i-1))^2)*(1-2*h/Re)/sqrt(1-gk2*sin(latitude(i-1))^2);
??
??%計算姿態矩陣
??Cnb=[Q(1)^2+Q(2)^2-Q(3)^2-Q(4)^2?2*(Q(2)*Q(3)-Q(1)*Q(4))?2*(Q(2)*Q(4)+Q(1)*Q(3));
???????2*(Q(2)*Q(3)+Q(1)*Q(4))?Q(1)^2-Q(2)^2+Q(3)^2-Q(4)^2?2*(Q(3)*Q(4)-Q(1)*Q(2));
???????2*(Q(2)*Q(4)-Q(1)*Q(3))?2*(Q(3)*Q(4)+Q(1)*Q(2))?Q(1)^2-Q(2)^2-Q(3)^2+Q(4)^2];
???
??Cbn=[Q(1)^2+Q(2)^2-Q(3)^2-Q(4)^2?2*(Q(2)*Q(3)+Q(1)*Q(4))?2*(Q(2)*Q(4)-Q(1)*Q(3));
???????2*(Q(2)*Q(3)-Q(1)*Q(4))?Q(1)^2-Q(2)^2+Q(3)^2-Q(4)^2?2*(Q(3)*Q(4)+Q(1)*Q(2));
???????2*(Q(2)*Q(4)+Q(1)*Q(3))?2*(Q(3)*Q(4)-Q(1)*Q(2))?Q(1)^2-Q(2)^2-Q(3)^2+Q(4)^2];
??
??%使用畢卡法(角增量法)求解四元數微分方程
??win=[-velocity(2)/Ry;wie_scalar*cos(latitude(i-1))+velocity(1)/Rx;
??????wie_scalar*sin(latitude(i-1))+velocity(1)/Rx*tan(latitude(i-1))];
??wnb(:i)=wib_INSc(:i)-Cbn*win;
??delta_big_theta=[0?-wnb(1i)?-wnb(2i)?-wnb(3i);
?????????????wnb(1i)?0?wnb(3i)?-wnb(2i);
?????????????wnb(2i)?-wnb(3i)?0?wnb(1i);
?????????????wnb(3i)?wnb(2i)?-wnb(1i)?0]/80;
??delta_theta=sqrt((wnb(1i)/80)^2+(wnb(2i)/80)^2+(wnb(3i)/80)^2);
??Q=(eye(4)*(1-delta_theta^2/8+delta_theta^4/384)+(1/2-delta_theta^2/48)*delta_big_theta)*Q;
??
??%計算姿態角
??theta(i)=asin(Cbn(23));
??gamma(i)=atan(-Cbn(13)/Cbn(33));
??psi(i)=atan(Cbn(21)/Cbn(22));
????if?Cbn(22)<0
????????psi(i)=psi(i)+pi;
????else?if?Cbn(21)<0
????????psi(i)=psi(i)+2*pi;
????????end
????end
????
????if?Cbn(33)<0
????????if?gamma(i)>0?
????????????gamma(i)=gamma(i)-pi;
????????else?gamma(i)=gamma(i)+pi;
????????end
????end
??
??%計算速度與經緯度
??f(:i)=Cnb*f_INSc(:i);
??ax=f(1i)+2*wie_scalar*sin(latitude(i-1))*velocity(2)+velocity(1)*velocity(2)*tan(latitude(i-1))/Rx;
??ay=f(2i)-2*wie_scalar*sin(latitude(i-1

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????目錄???????????0??2014-05-06?09:31??慣性導航\
?????文件????????4672??2009-07-27?14:40??慣性導航\jielian.m
?????文件?????2304256??2003-12-12?21:33??慣性導航\jlfw.mat
?????文件???????53248??2007-11-19?13:36??慣性導航\文檔.doc

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