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  • 大小: 7KB
    文件類型: .cpp
    金幣: 2
    下載: 1 次
    發布日期: 2021-06-17
  • 語言: C/C++
  • 標簽: 相對定向??

資源簡介

利用連續像對相對定向的方法進行立體像對相對定向,得到像對方位元素

資源截圖

代碼片段和文件信息

#include?
#include??
#include?
const?int?N=8;??
void?Mult(double?R[8][8]?double?b[8][8]double?c[8][8]?int?m?int?t?int?p);??
void?inverse(double?c[N][N]int?t);?
void?main()?
{?double??R[N][N]lx[N]ly[N]rx[N]ry[N]A[N][N]l[N][N]ATA[N][N];?
??double?f=0.024Q=0W=0K=0x=0y=0X[8][8]bubvbw;?
??int?ijn;?
??int?q=0k;??
??cout<<“請輸入同名像點的對數“<>n;???????????
??lx[1]=0.001983lx[2]=0.000924lx[3]=0.001068lx[4]=0.001208lx[5]=-0.000514lx[6]=0.001293??????????
??ly[1]=-0.006091ly[2]=0.007098ly[3]=0.004538ly[4]=0.006858ly[5]=-0.010050ly[6]=-0.008089???????????
??rx[1]=-0.003202rx[2]=-0.002830rx[3]=-0.002878rx[4]=-0.002578rx[5]=-0.005642rx[6]=-0.003981???????
??ry[1]=-0.005564ry[2]=0.007694ry[3]=0.005098ry[4]=0.007429ry[5]=-0.009152ry[6]=-0.007441??????//輸入像點的坐標
??bu=rx[1]-lx[1];???
??do?
????{?q++;??????
??????bv=bu*x;bw=bu*y;????
??????double??R[8][8]b[8][8]l[8][8]c[8][8]A[8][8]AT[8][8]Z[8][8]ATA[8][8]d[8][8];???
??????R[0][0]=cos(Q)*cos(K)-sin(Q)*sin(W)*sin(K);?????
??????R[0][1]=-cos(Q)*sin(K)-sin(Q)*sin(W)*cos(K);?????
??????R[0][2]=-sin(Q)*cos(W);??
??R[1][0]=cos(W)*sin(K);??
??????R[1][1]=cos(W)*cos(K);??
??R[1][2]=-sin(W);???
??????R[2][0]=sin(Q)*cos(K)+cos(Q)*sin(W)*sin(K);??
??????R[2][1]=-sin(Q)*sin(K)+cos(Q)*sin(W)*cos(K);
??????R[2][2]=cos(Q)*cos(W);??????//把相對方位元素帶入公式,計算得到9個方向余弦??
??????b[0][0]=lx[1];
??????b[1][0]=ly[1];
??????b[2][0]=-f;
??????b[0][2]=lx[3];
??????b[1][2]=ly[3];
??????b[2][2]=-f;
??????b[0][4]=lx[5];
??????b[1][4]=ly[5];
??????b[2][4]=-f;?
??????b[0][1]=lx[2];
??????b[1][1]=ly[2];
??????b[2][1]=-f;
??????b[0][3]=lx[4];
??????b[1][3]?=ly[4];
??????b[2][3]=-f;
??????b[0][5]=lx[6];
??????b[1][5]=ly[6];
??????b[2][5]=-f;?????
??????d[0][0]=rx[1];
??????d[1][0]=ry[1];
??????d[2][0]=-f;
??????d[0][2]=rx[3];
??????d[1][2]=ry[3];
??????d[2][2]=-f;
??????d[0][4]=rx[5];
??????d[1][4]=ry[5];
??????d[2][4]=-f;???
??????d[0][1]=rx[2];
??????d[1][1]=ry[2];
??????d[2][1]=-f;
??????d[0][3]=rx[4];
??????d[1][3]?=ry[4];
??????d[2][3]=-f;
??????d[0][5]=rx[6];
??????d[1][5]=ry[6];
??????d[2][5]=-f;???//將坐標值賦值給矩陣
??????Mult(Rdc336);??????
???????for(j=0;j<=5;j++)???
???????{??l[j][0]=b[1][j]*(bu*c[2][j]-bw*c[0][j])/(c[2][j]*b[0][j]-c[0][j?]*b[2][j])-c[1][j]*(bu*b[2][j]-bw*b[0][j])/(c[2][j]*b[0][j]-c[0][j]*b[2][j])-bv;????
??????????A[j][0]=-c[0][j]*c[1][j]*(bu*b[2][j]-bw*b[0][j])/((c[2][j]*b[0][j]-c[0][j]*b[2][j])*c[2][j]);??
??????????A[j][1]=-(c[2][j]+c[1][j]*c[1][j]/c[2][j])*(bu*b[2][j]-bw*b[0][j])/(c[2][j]*b[0][j]-c[0][j]*b[2][j]);???
??????????A[j][2]=(bu*b[2][j]-bw*b[0][j])/(c[2][j]*b[0][j]-c[0][j]*b[2][j])*c[0][j];???
??????????A[j][3]=bu;?????
??????????A[j][4]=-bu*c[1][j]/c[2][j];?
???????}??????
???????for(i=0;i<=5;i++)???????
?????????for(k=0;k<=4;k++)??????
?????????{??
???????????AT[k][i]=A[i][k];//右片像空間坐標
???????????}?????????
??????????Mult(ATAATA565);????
?

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