贝塞尔大地主题解算分析 PPT

大地主题解算一、实验目的：1.提高运用计算机语言编程开发的能力；2.加深对大地主题解算计算公式及辅助参数的理解并掌握计算步骤；3.通过编程语言实现大地主题解算。

二、工具：Windows XP Mode 环境下的Microsoft Visual C++ 6.0三、注意事项：1.计算所需变量多，容易混淆；2.正反算函数的编写；3.函数调用；4.弧度与角度之间的转化。

四、实验要求：1.提交报告，实验总结，编写代码；2.独立编程，调试运行；3.上交成果：编写思想，编写过程，问题分析，源代码，计算结果；五、编程过程实现：1.对白塞尔法大地主题解算有一定的了解，并参考教材P148-P150；2.由于参数较多，而在C语言环境下很多符号无法定义，需要符合要求的定义符号替代书本上那些无法直接在C语言环境下定义的符号来达到实现实验的目的；3.程序中采用弧度与度分秒之间转换的函数定义与调用，减轻一定的实验麻烦；4.在C语言环境下，数学函数fabs代替abs起绝对值作用，atan代替arctan起反函数作用；5.程序中尤其注意弧度与角度之间转换，在C语言环境下电脑默认为弧度。

六、源程序代码：#include<stdio.h>#include<math.h>double hudu(double,double,double); /*度分秒转换为弧度*/double du(double); /*弧度转换为度*/double fen(double); /*弧度转换为分*/double miao(double); /*弧度转换为秒*/#define PI 3.1415926void main (void){int k;printf("请选择执行正算或者反算，若执行正算，请输入1；若执行反算，请输入2。

\n");scanf("%d",&k);/*正算*/if(k==1){double bz,lz,az,S,bz2,lz2,az2,B1,L1,A1,B2,L2,A2,bx,by,lx,ly,ax,ay;int bx2,by2,lx2,ly2,ax2,ay2;doublee2,W1,sinu1,cosu1,sinA0,coto1,sin2o1,cos2o1,sin2o,cos2o,A,B,C,r,t,o0,o,g,si nu2,q;/*以度分秒顺序输入数据*/printf("请输入大地线起点纬度度分秒\n");scanf("%lf%lf%lf",&bx,&by,&bz);printf("请输入大地线起点经度度分秒\n");scanf("%lf%lf%lf",&lx,&ly,&lz);printf("请输入大地方位角度分秒\n");scanf("%lf%lf%lf",&ax,&ay,&az);printf("请输入大地线长度\n");scanf("%lf",&S);/*调用函数*/B1=hudu(bx,by,bz);L1=hudu(lx,ly,lz);A1=hudu(ax,ay,az);/*白塞尔大地主题解算*/e2=0.006693421622966;W1=sqrt(1-e2*sin(B1)*sin(B1));sinu1=sin(B1)*(sqrt(1-e2))/W1;cosu1=cos(B1)/W1;sinA0=cosu1*sin(A1);coto1=cosu1*cos(A1)/sinu1;sin2o1=2*coto1/(coto1*coto1+1);cos2o1=(coto1*coto1-1)/(coto1*coto1+1);A=6356863.020+(10718.949-13.474*(1-sinA0*sinA0))*(1-sinA0*sinA0);B=(5354.469-8.978*(1-sinA0*sinA0))*(1-sinA0*sinA0);C=(2.238*(1-sinA0*sinA0))*(1-sinA0*sinA0)+0.006;r=691.46768-(0.58143-0.00144*(1-sinA0*sinA0))*(1-sinA0*sinA0);t=(0.2907-0.0010*(1-sinA0*sinA0))*(1-sinA0*sinA0);o0=(S-(B+C*cos2o1)*sin2o1)/A;sin2o=sin2o1*cos(2*o0)+cos2o1*sin(2*o0);cos2o=cos2o1*cos(2*o0)-sin2o1*sin(2*o0);o=o0+(B+5*C*cos2o)*sin2o/A;g=(r*o+t*(sin2o-sin2o1))*sinA0;/*求B2*/sinu2=sinu1*cos(o)+cosu1*cos(A1)*sin(o);B2=atan(sinu2/(sqrt(1-e2)*sqrt(1-sinu2*sinu2)));/*求L2*/q=atan(sin(A1)*sin(o)/(cosu1*cos(o)-sinu1*sin(o)*cos(A1)));/*判断q*/if(sin(A1)>0 && tan(q)>0)q=fabs(q);else if(sin(A1)>0 && tan(q)<0)q=PI-fabs(q);else if(sin(A1)<0 && tan(q)<0)q=-fabs(q);elseq=fabs(q)-PI;L2=L1+q-g/3600/180*PI;/*求A2*/A2=atan(cosu1*sin(A1)/(cosu1*cos(o)*cos(A1)-sinu1*sin(o))); /*判断A2*/if(sin(A1)<0 && tan(A2)>0)A2=fabs(A2);else if(sin(A1)<0 && tan(A2)<0)A2=PI-fabs(A2);else if(sin(A1)>0 && tan(A2)>0)A2=PI+fabs(A2);elseA2=2*PI-fabs(A2);/*调用函数*/bx2=(int)(du(B2));by2=(int)(fen(B2));bz2=miao(B2);lx2=(int)(du(L2));ly2=(int)(fen(L2));lz2=miao(L2);ax2=(int)(du(A2));ay2=(int)(fen(A2));az2=miao(A2);printf("大地线终点纬度度分秒分别为：\n%d\n%d\n%lf\n",bx2,by2,bz2);printf("大地线终点经度度分秒分别为：\n%d\n%d\n%lf\n",lx2,ly2,lz2);printf("终点大地方位角度分秒分别为：\n%d\n%d\n%lf\n",ax2,ay2,az2);}/*反算*/else if(k==2){doublebz,lz,bz2,lz2,az,az2,B1,L1,B2,L2,S,A1,A2,bx,by,lx,ly,bx2,by2,lx2,ly2;int ax,ay,ax2,ay2;doublee2,W1,W2,sinu1,sinu2,cosu1,cosu2,L,a1,a2,b1,b2,g,g2,g0,r,p,q,sino,coso,o,si nA0,x,t1,t2,A,B,C,y;/*以度分秒顺序输入数据*/printf("请输入大地线起点纬度度分秒\n");scanf("%lf%lf%lf",&bx,&by,&bz);printf("请输入大地线起点经度度分秒\n"); scanf("%lf%lf%lf",&lx,&ly,&lz);printf("请输入大地线终点纬度度分秒\n"); scanf("%lf%lf%lf",&bx2,&by2,&bz2); printf("请输入大地线终点经度度分秒\n"); scanf("%lf%lf%lf",&lx2,&ly2,&lz2);/*调用函数*/B1=hudu(bx,by,bz);L1=hudu(lx,ly,lz);B2=hudu(bx2,by2,bz2);L2=hudu(lx2,ly2,lz2);/*白塞尔大地主题解算*/e2=0.006693421622966;W1=sqrt(1-e2*sin(B1)*sin(B1));W2=sqrt(1-e2*sin(B2)*sin(B2));sinu1=sin(B1)*sqrt(1-e2)/W1;sinu2=sin(B2)*sqrt(1-e2)/W2;cosu1=cos(B1)/W1;cosu2=cos(B2)/W2;L=L2-L1;a1=sinu1*sinu2;a2=cosu1*cosu2;b1=cosu1*sinu2;b2=sinu1*cosu2;/*逐次趋近法求解A1*/g0=-662.904266/3600*PI/180; g=0;r=L;while(1){p=cosu2*sin(r);q=b1-b2*cos(r);A1=atan(p/q);/*判断A1*/if(p>0 && q>0)A1=fabs(A1);else if(p>0 && q<0)A1=PI-fabs(A1);else if(p<0 && q<0)A1=PI+fabs(A1);elseA1=2*PI-fabs(A1);sino=p*sin(A1)+q*cos(A1);coso=a1+a2*cos(r);o=atan(sino/coso);/*判断o*/if(coso>0)o=fabs(o);elseo=PI-fabs(o);sinA0=cosu1*sin(A1);x=2*a1-(1-sinA0*sinA0)*coso;t1=(33523299-(28189-70*(1-sinA0*sinA0))*(1-sinA0*sinA0))*0.0000000001;t2=(28189-94*(1-sinA0*sinA0))*0.0000000001;g2=(t1*o-t2*x*sino)*sinA0;/*检验循环次数*/printf("\ng2=%lf\ng0=%lf\n",g2,g0);if(g2<=g0)break;elser=L+g2;}/*求解S*/A=6356863.020+(10708.949-13.474*(1-sinA0*sinA0))*(1-sinA0*sinA0);B=10708.938-17.956*(1-sinA0*sinA0);C=4.487;y=((1-sinA0*sinA0)*(1-sinA0*sinA0)-2*x*x)*coso; S=A*o+(B*x+C*y)*sino;/*求解A2*/A2=atan(cosu1*sin(r)/(b1*cos(r)-b2));/*判断A2*/if(p<0 && q<0)A2=fabs(A2);else if(p<0 && q>0)A2=PI-fabs(A2);else if(p>0 && q>0)A2=PI+fabs(A2);elseA2=2*PI-fabs(A2);/*调用函数*/ax=(int)(du(A1));ay=(int)(fen(A1));az=miao(A1);ax2=(int)(du(A2));ay2=(int)(fen(A2));az2=miao(A2);printf("起点大地方位角度分秒分别为：\n%d\n%d\n%lf\n",ax,ay,az);printf("终点大地方位角度分秒分别为：\n%d\n%d\n%lf\n",ax2,ay2,az2);printf("大地线长度为：%lf\n",S);}/*数据错误*/elseprintf("数据错误，请重新输入\n");}/*度分秒转换为弧度*/double hudu(double a0,double b0,double c0){double A0;A0=(a0+b0/60+c0/3600)*PI/180;return A0;}/*弧度转换为度*/double du(double B0){double x0;x0=(int)(B0*180/PI);return x0;}/*弧度转换为分*/double fen(double C0){double _y,y0;_y=(int)(C0*180/PI);y0=(fabs)((int)((C0*180/PI-_y)*60));return y0;}/*弧度转换为秒*/double miao(double D0){double _z1,_z2,z0;_z1=(int)(D0*180/PI);_z2=(int)((D0*180/PI-_z1)*60);z0=(fabs)((double)(((D0*180/PI-_z1)*60-_z2)*60));return z0;}大地主题解算正算：大地主题解算反算：以上检验数据来自书本P151，P152。

大地主题解算(正算)代码与白塞尔大地主题解算大地主题解算(正算)代码：根据经纬度和方向角以及距离计算另外一点坐标新建模块->拷贝下面的大地主题(正算)代码，调用方法示例：起点经度：116.235(度)终点纬度：37.435(度)方向角：50（度）长度：500（米）终点经纬度（"经度,纬度"）=Computation(37.435,116.235,50,500)Const Pi = 3.14Private a, b, c, alpha, e, e2, W, V As Double'a 长轴半径'b 短轴'c 极曲率半径'alpha 扁率'e 第一偏心率'e2 第二偏心率'W 第一基本纬度函数'V 第二基本纬度函数Private B1, L1, B2, L2 As Double'B1 点1的纬度'L1 点1的经度'B2 点1的纬度'L2 点2的经度Private S As Double '''''大地线长度Private A1, A2 As Double'A1 点1到点2的方位角'A2 点2到点1的方位角Function Computation(STARTLAT, STARTLONG, ANGLE1, DISTANCE As Double) As StringB1 = STARTLATL1 = STARTLONGA1 = ANGLE1S = DISTANCEa = 6378245b = 6356752.3142c = a ^ 2 / balpha = (a - b) / ae = Sqr(a ^ 2 - b ^ 2) / ae2 = Sqr(a ^ 2 - b ^ 2) / bB1 = rad(B1)L1 = rad(L1)A1 = rad(A1)W = Sqr(1 - e ^ 2 * (Sin(B1) ^ 2))V = W * (a / b)Dim W1 As DoubleE1 = e ''''第一偏心率'// 计算起点的归化纬度W1 = W ''Sqr(1 - e1 * e1 * Sin(B1 ) * Sin(B1 )) sinu1 = Sin(B1) * Sqr(1 - E1 * E1) / W1cosu1 = Cos(B1) / W1'// 计算辅助函数值sinA0 = cosu1 * Sin(A1)cotq1 = cosu1 * Cos(A1)sin2q1 = 2 * cotq1 / (cotq1 ^ 2 + 1)cos2q1 = (cotq1 ^ 2 - 1) / (cotq1 ^ 2 + 1)'// 计算系数AA,BB,CC及AAlpha, BBeta的值。

大地主题的数值解法范业明1,王解先1,2,刘慧芹1(1 同济大学测量与国土信息工程系,上海 200092;2 现代工程测量国家测绘局重点实验室,上海 200092)摘要:本文基于椭球面上大地线的微分方程,将法截弧方位角与大地线方位角之间的关系作为初始条件,通过用数值方法求解大地线的微分方程,进行大地主题的正反解。

并以实际数据验证了其正确性与可行性。

本法均采用封闭公式计算,精度高,公式简单,特别适用于计算机解算。

关键词:大地线;微分方程;数值解法中图分类号:P226+1文献标识码:BAbstract :Based on differential equation of geodesic on the surface of ellipsoid,the problem of direct and inverse solution of geodetic is solved through numerical method.The relationship of normal arc and geodesic is deduced and introduced as the initial condition for the differential equation.Numerical experiments are given,and the validity and feasibility of the method proposed in this paper have been proved.Besides,all the formulas are close,simple and easy to be realized by computer.Key words :geodesic;differential equation;numerical method 收稿日期:2006 02 15;修订日期:2006 05 10作者简介:范业明(1982-),男(汉族),辽宁沈阳人,硕士研究生.0 引言一直以来由于计算工具的限制,大地主题解算一般都采用级数展开形式,如短距离的高斯平均引数法,长距离的贝塞尔-赫尔默特方法等等。

贝塞尔大地坐标解算（正算)#include<stdio.h>#include<math.h>#define PI 3.1415926535897323int main(){int q,p;double h,i,j,L,B,A1,S;double a,g,c,d,e,f;void n(double b,double l,double a1,double s,double a,double d,double e,double f);double z(double d,double m,double s);void r(double dd);printf("请选择：\n1.我国1954北京坐标系\n2.我国1980西安坐标系\n3.我国2000国家大地坐标系\n");scanf("%d",&p);if(p==1){a=6378245.0;g=6356863.0188;c=6399698.9018;d=0.00335232986926;e=0.00669342162297;f=0.00673852541468;}else if(p==2){a=6378140.0;g=6356755.2882;c=6399596.6520;d=0.0033528131779;e=0.00669438499959;f=0.00673950181947;}else if(p==3){a=6378137.0;g=6356752.3141;c=6399593.6259;d=1/298.257222101;e=0.0066943800229;f=0.00673949677547;}printf("请输入注：输入格式为角度(例如：30'40'50')\nL1=");scanf("%lf'%lf'%lf'",&h,&i,&j);L=z(h,i,j);printf("请输入注：输入格式为角度(例如：30'40'50')\nB1=");scanf("%lf'%lf'%lf'",&h,&i,&j);B=z(h,i,j);printf("请输入注：输入格式为角度(例如：30'40'50')\nA1=");scanf("%lf'%lf'%lf'",&h,&i,&j);A1=z(h,i,j);printf("请输入\nS=");scanf("%lf",&S);n(B,L,A1,S,a,d,e,f);scanf("%d",&q);}double z(double d,double m,double s){double e;double sign=(d<0.0)?-1.0:1.0;if(d==0){sign=(m<0.0)?-1.0:1.0;if(m==0){sign=(s<0.0)?-1.0:1.0;}}if(d<0)d=d*(-1.0);if(m<0)m=m*(-1.0);if(s<0)s=s*(-1.0);e=sign*(d*3600+m*60+s)*PI/(3600*180);return e;}void r(double dd){int de;int d,m;double s;double sign=(dd<0.0)?-1.0:1.0;if(dd<0)dd=fabs(dd);dd=dd*3600*180/PI;de=int(dd/3600);d=sign*de;dd=dd-de*3600;m=int(dd/60);s=dd-m*60;printf("%d'%d'%lf'\n",d,m,s);}void n(double b,double l,double a1,double s,double a,double d,double e,double f) {double u1,m,M,A2,u2,ss,ss1,k,aa,bb,cc,rr,rr1,rr2,B2,L2;double aa2,bb2,cc2,k2,ll;u1=atan(sqrt(1-e)*tan(b));m=asin(cos(u1)*sin(a1));if(m<0.0)m=m+2*PI;M=atan(tan(u1)/cos(a1));if(M<0.0)M=M+PI;k=f*cos(m)*cos(m);aa=sqrt(1+f)*(1-k/4+7*k*k/64-15*k*k*k/256)/a;bb=k/4-k*k/8+37*k*k*k/512;cc=k*k/128-k*k*k/128;ss1=aa*s;ss=aa*s+bb*sin(ss1)*cos(2*M+ss1)+cc*sin(2*ss1)*cos(4*M+2*ss1);for(;fabs(ss-ss1)>(2.8*PI/180*pow(10.0,-7.0));){ss1=ss;ss=aa*s+bb*sin(ss1)*cos(2*M+ss1)+cc*sin(2*ss1)*cos(4*M+2*ss1);}A2=atan(tan(m)/cos(M+ss));if(A2<0.0)A2=A2+PI;if(a1<=PI)A2=A2+PI;u2=atan(-cos(A2)*tan(M+ss));rr1=atan(sin(u1)*tan(a1));if(rr1<0)rr1=rr1+PI;if(m>=PI)rr1=rr1+PI;rr2=atan(sin(u2)*tan(A2));if(rr1<0)rr2=rr2+PI;if(m<PI){if((M+ss)>=PI)rr2=rr2+PI;}elseif((M+ss)<=PI)rr2=rr2+PI;rr=rr2-rr1;B2=atan(sqrt(1+f)*tan(u2));k2=e*cos(m)*cos(m);aa2=(e/2+e*e/8+e*e*e/16)-e*(1+e)*k2/16+3*e*k2*k2/128;bb2=e*(1+e)*k2/16-e*k2*k2/32;cc2=e*k2*k2/256;ll=rr-sin(m)*(aa2*ss+bb2*sin(ss)*cos(2*M+ss)+cc2*sin(2*ss)*cos(4*M+2*ss));L2=l+ll;printf("L2=");r(L2);printf("\nB2=");r(B2);printf("\nA2=");r(A2);}反算#include<stdio.h>#include<math.h>#define PI 3.1415926535897323int main(){int q,p;double h,i,j,L,B,L2,B2;double a,g,c,d,e,f;void n(double b,double l,double l2,double b2,double a,double d,double e,double f);double z(double d,double m,double s);void r(double dd);printf("请选择：\n1.我国1954北京坐标系\n2.我国1980西安坐标系\n3.我国2000国家大地坐标系\n");scanf("%d",&p);if(p==1){a=6378245.0;g=6356863.0188;c=6399698.9018;d=0.00335232986926;e=0.00669342162297;f=0.00673852541468;}else if(p==2){a=6378140.0;g=6356755.2882;c=6399596.6520;d=0.0033528131779;e=0.00669438499959;f=0.00673950181947;}else if(p==3){a=6378137.0;g=6356752.3141;c=6399593.6259;d=1/298.257222101;e=0.0066943800229;f=0.00673949677547;}printf("请输入注：输入格式为角度(例如：30'40'50')\nL1=");scanf("%lf'%lf'%lf'",&h,&i,&j);L=z(h,i,j);printf("请输入注：输入格式为角度(例如：30'40'50')\nB1=");scanf("%lf'%lf'%lf'",&h,&i,&j);B=z(h,i,j);printf("请输入注：输入格式为角度(例如：30'40'50')\nL2=");scanf("%lf'%lf'%lf'",&h,&i,&j);L2=z(h,i,j);printf("请输入注：输入格式为角度(例如：30'40'50')\nB2=");scanf("%lf'%lf'%lf'",&h,&i,&j);B2=z(h,i,j);n(B,L,L2,B2,a,d,e,f);scanf("%s",&q);}double z(double d,double m,double s){double e;double sign=(d<0.0)?-1.0:1.0;if(d==0){sign=(m<0.0)?-1.0:1.0;if(m==0){sign=(s<0.0)?-1.0:1.0;}}if(d<0)d=d*(-1.0);if(m<0)m=m*(-1.0);if(s<0)s=s*(-1.0);e=sign*(d*3600+m*60+s)*PI/(3600*180);return e;}void r(double dd){int de;int d,m;double s;double sign=(dd<0.0)?-1.0:1.0;if(dd<0)dd=fabs(dd);dd=dd*3600*180/PI;de=int(dd/3600);d=sign*de;dd=dd-de*3600;m=int(dd/60);s=dd-m*60;printf("%d'%d'%lf'\n",d,m,s);}void n(double b,double l,double l2,double b2,double a,double d,double e,double f) {double u1,m,M,A2,u2,ss,ss1,k,aa,bb,cc,rr,rr0,B2,L2,m0,A10,A1,S;double aa2,bb2,cc2,k2,ll;u1=atan(sqrt(1-e)*tan(b));u2=atan(sqrt(1-e)*tan(b2));ss=acos(sin(u1)*sin(u2)+cos(u1)*cos(u2)*cos(l2-l));m0=asin(cos(u1)*cos(u2)*sin(l2-l)/sin(ss));rr0=l2-l+0.003351*ss*sin(m0);k2=e*cos(m0)*cos(m0);aa2=(e/2+e*e/8+e*e*e/16)-e*(1+e)*k2/16+3*e*k2*k2/128;ss1=ss+sin(m0)*aa2*ss*sin(m0);m=asin(cos(u1)*cos(u2)*sin(rr0)/sin(ss1));A10=atan(sin(rr0)/(cos(u1)*tan(u2)-sin(u1)*cos(rr0)));if(A10<=0.0)A10=A10+PI;if(m<=0.0)A10=A10+PI;M=atan(sin(u1)*tan(A10)/sin(m));if(M<=0.0)M=M+PI;k2=e*cos(m)*cos(m);aa2=(e/2+e*e/8+e*e*e/16)-e*(1+e)*k2/16+3*e*k2*k2/128;bb2=e*(1+e)*k2/16-e*k2*k2/32;rr=l2-l+sin(m)*(aa2*ss1+bb2*sin(ss1)*cos(2*M+ss1));ss=acos(sin(u1)*sin(u2)+cos(u1)*cos(u2)*cos(rr));A1=atan(sin(rr)/(cos(u1)*tan(u2)-sin(u1)*cos(rr)));if(A1<=0.0)A1=A1+PI;if(m<=0.0)A1=A1+PI;A2=atan(sin(rr)/(-cos(u2)*tan(u1)+sin(u2)*cos(rr)));if(A2<=0.0)A2=A2+PI;if(m>=0.0)A2=A2+PI;k=f*cos(m)*cos(m);aa=sqrt(1+f)*(1-k/4+7*k*k/64-15*k*k*k/256)/a;bb=k/4-k*k/8+37*k*k*k/512;cc=k*k/128-k*k*k/128;S=(ss-bb*sin(ss)*cos(2*M+ss)-cc*sin(2*ss)*cos(4*M+2*ss))/aa;printf("A1=");r(A1);printf("A2=");r(A2);printf("S=%lf",S);}。

存档日期：存档编号：江苏师范大学科文学院本科生毕业设计（论文）论文题目：贝塞尔大地主题正反算及程序设计*名：**系别：环境与测绘系专业：测绘工程年级、学号：08测绘、************师：***江苏师范大学科文学院教务部印制摘要在大地测量计算过程中，大地主题解算计算繁琐复杂，手工计算易于出错，而且费时费力。

随着计算机技术的高速发展，计算机计算的速度快、准确度高、计算机语言的丰富、编程可视化等优点为我们将复杂烦琐的计算过程简单、简洁、高效化带来了契机。

为了便于工程计算，本课题着眼于研究借助计算机及其编程语言MATLAB来实现大地主题解算问题。

大地主题解算方法，主要有高斯平均引数法、勒让德级数法、贝塞尔法。

前两种方法受到大地线长度的制约，随着大地线两端点的距离加大，其解算精度明显降低。

而贝塞尔法具有不受大地线长度制约的优点，解算精度最大不超过5毫米，是大地主题解算方法中解算精度最高的一种。

因此，本文就以贝塞尔法为研究对象，开发贝塞尔大地主题解算小程序。

关键词：贝塞尔大地主题正反算，程序设计A b s t r a c tIn Geodetic computation process, the solution of geodetic problem computational complexity of manual calculation, error prone, and took the time and trouble. With the rapid development of computer technology, computational speed, high accuracy, computer language, the advantages of rich programming visualization for we will complex complicated calculating process is simple, concise, efficient change brings opportunity. For the convenience of engineering calculation, this paper focus on the research of have the aid of computer and programming language MATLAB to realize the geodetic problem solving.Solution of geodetic problem method, mainly Gauss average argument method, Legendre series expansion method, Bessel method. The former two methods by geodesic length constraints, along with the line ends point distance increase, the calculation precision significantly reduced. Bessel law is not affected by the advantages of geodesic length restriction, calculation accuracy of less than 5mm, is the theme of the earth solution method of calculating precision is highest kind. Therefore, this article on Bessel law as the object of study, the development of Bessel solution of geodetic problem of small procedures.Key words：Direct and inverse solution of geodetic problem，The designing of program目录摘要 .................................................... I Abstract (II)1. 椭球面和球面上对应元素间的关系 (1)1.1 贝塞尔法解算大地问题的基本思想 (1)1.2 对应元素关系式 (1)2. 在球面上进行大地主题解算 (5)2.1 球面上大地主题正解方法 (6)2.2 球面上大地主题反解方法 (6)3. 贝塞尔微分方程的积分 (8)3.1 用于大地主题反算时的大地线长度公式 (8)3.2 用于大地主题正算时的大地线长度公式 (10)3.3 椭球面大地线端点经差与球面经差的关系式 (11)3.4 反解时，大地线长度和球面长度关系式的简化 (13)4. 贝塞尔大地主题正解算步骤 (15)4.1 计算起点的归化纬度 (15)4.2 计算辅助函数值 (15)4.3 计算系数 (15)4.4 计算球面长度 (15)4.5 计算经差改正数 (16)4.6 计算终点大地坐标及大地方位角 (16)5. 贝塞尔大地主题正解算MATLAB 程序设计 (17)5.1 正算流程 (17)5.2 界面设计及功能模块编写 (18)6. 贝赛尔大地主题反解算步骤 (23)6.1 辅助计算 (23)6.2 用逐次趋近法同时计算起点大地方位角、球面长度及经差δλ+=l ：236.3 计算系数及大地线长度S (24)6.4 计算反方位角 (24)7. 贝塞尔大地主题反解算MATLAB 程序设计 (25)7.1 反算流程 (25)7.2 界面设计及功能模块编写 (26)8 总结 (29)参考文献 (31)致谢 (29)1. 椭球面和球面上对应元素间的关系1.1 贝塞尔法解算大地问题的基本思想基本思想：将椭球面上的大地元素按照贝塞尔投影条件投影到辅助球面上，继而在球面上进行大地问题解算，最后再将球面上的计算结果换算到椭球面上。