胡思华1208010210

大地测量学课程设计设计题目：高斯平均引数法大地主题解算学院：＿矿业学院＿＿＿专业：＿测绘工程＿＿＿班级：＿测绘121＿＿＿学号：＿1280010210＿＿学生姓名：＿胡思华＿＿＿＿指导教师：＿张俊＿＿＿2014年12月26日-1-目录1.高斯平均引数大地主题正解........................-2-2.高斯平均引数大地主题反解........................-3-3.同一平行圈弧长、子午线弧长与大地线比较大小......-4-4.VB程序代码.....................................-6-5.计算案例........................................-8-6.参考文献........................................-9-7.心得体会.......................................-10-8.教师评语.......................................-11--2-高斯平均引数大地法大地主题解算摘要：高斯平均引数正算的基本思想是：首先把勒让德级数在端点P1处展开改在大地线中点M展开，以使级数公式项数减少，收敛快，精度高；其次，考虑到求中点M的复杂性，将M点用大地线两端点平均纬度及平均方位角相对应的m点来代替，并借助迭代运算，便可顺利地实现大地主题解算。大地主题反解是已知两端点的经纬度L1，B1，L2，B2，反求两点间的的大地线长度S及正反大地方位角A12和A21。1.高斯平均引数大地主题正解(1)建立级数展开式:同理可得:(2)33213()()(4202)24MMdBdBBBBSSdSdSMMmmBABA,,21,22SSMPMP223322311()()()(4200)22468MMdBSdBSdBSBBdSdSdS223312311()()()(4201)22468MMMMdBSdBSdBSBBdSdSdS3321324MMdLdLLLLSSdSdS()()332112324MMdAdAAAASSdSdS()()2121121118022mmBBBAAA(),()mMmMBBAA,MMMmMmmMmdBfBAFBBBAAAdS()(,)(,)+MmmMmMmmmdBfffBABBAAdSBA()(,)()()()()+MmmMmMmmmdBdBdBdSdSfBABBAAdSBA()()()(,)()()()()[phone]MmMmSdBBBdSSdBdS()()22222288MmMmSdAAAdSSdAdS()()-3-同理可得：2.高斯平均引数大地主题反解高斯平均引数反算公式可以依正算公式导出:32mmmmmmmmAVVdBAAdSMcNcos()coscos323mmmmmmmVdBAdSctABBN()(cos)()cos222222222388mMmmmmmmmmSVSdBBBtAtAdSN()(sincos)22222221288MmmmmmmmSdASAAAAtdSN()sincos()222222223223333812mmMmmmmmmmmm22mmmmmmVVdBSSAAtAAS+dSNNVAAt+S+58N()coscos(sincos)sincos()次23322222332222313924243155mMmmmmmmm22mmmmmmVSdBAAtt+dSNAt+tS+()cos[sin()cos()次22222212222221232243195mmmmmmmmmmmmVSBBBSAAtNNAt()cos{[sin()cos()]}次22222222124195mmmmmmmmmmSLSBAAtNNAtsecsin{[sincos()]}次22222242221279245225mmmmmmmmmmmmSASAtAtNNAtsin{[cos()sin()]}次21212112,,180BBBLLLAAA21211111()222mBBBBBBBB1212mAAA-4-上述两式的主式为:3.同一平行圈弧长、子午线弧长与大地线比较大小子午线弧长计算公式：BaBaBaBaM8cos6cos4cos2cosa86420式中：128a1632a3271638a167321522a12835165832a88866864486422864200mmmmmmmmmmmmmmm平行圈弧长公式：''1''''cosBlblNS（cosBb''1N）222222222sinsincos[sin24cos(19)]mmmmmmmmmmmSALSANBStANSAt2222222222222coscos[sin(232)243cos(14)]mmmmmmmmmmmmmmNSABSASAtVNSAtt2sincos,cosmmmmmmNLBSANBSAV2301210323101230mmSArLrBLrLSAsBsBLsBsincos23012103AtLtBLtL3222201210333192424mmmmmmmmmmNNBNBrBrtrtcoscoscos,(),2222222101230233233248mmmmmmmmmmmNNBNssttstVcos,(),()2432012103221132212412mmmmmmmmmttBtBttBtcos,cos(),cos()sintancosmmmSAASAsinsinmmSASA122111,18022mmAAAAAA-5-B子午线弧长平行圈弧长△B=1°1′1″l=1°1′1″0°110576m1842.94m30.716m111321m1855.36m30.923m15°1106561844.2630.7381075521792.5429.87630°1108631847.7130.795964881608.1326.80245°1111431852.3930.873788481314.1421.90260°1114231857.0430.95155801930.0215.575°1116251860.4231.00728902481.718.02890°1116961861.6031.02700.000.000用高斯平均引数反解求大地线长纬度为60°，纬度差为1°，子午线弧长为111423m，大地线长为111422.494m纬度为60°，经度差为1°，平行圈弧长为55801m，大地线长为55800.438m由上述两例可知，单位经差的子午线弧长与大地线长基本相等，单位纬度差的平行圈弧长也与大地线长基本相等。这个验证了大地线是两点间的最短连线这个概念。-6-4.VB程序代码正算：eps=e2/(1#-e2)t1=Sin(B1)t2=Cos(B1)NB=a/Sqr(1#-e2*t1*t1)MB=a*(1-e2)/Sqr((1#-e2*t1*t1)^3)dB=S12*Cos(A1)/MBdL=S12*Sin(A1)/NB/Cos(B1)dA=dL*Sin(B1)DodB0=dBdL0=dLdA0=dABm=B1+dB0/2LM=L1+dL0/2Am=A1+dA0/2t1=Sin(Am)t2=Cos(Am)t=Tan(Bm)q=eps*Cos(Bm)*Cos(Bm)n=a/Sqr(1#-e2*Sin(Bm)*Sin(Bm))MB=a*(1-e2)/Sqr((1#-e2*Sin(Bm)*Sin(Bm))^3)db1=t1*t1*(2+3*t*t+2*q)db2=3*t2*t2*q*(t*t-1-(1+4*t*t)*q)dB=S12*t2*(1+S12*S12*(db1+db2)/n/n/24)/MBxt=t*txt1=t1*t1xt2=t2*t2dL1=S12*S12*(xt*xt1-xt2*(1+(1-9*xt)*q))/n/n/24dL=S12*t1*(1+dL1)/n/Cos(Bm)da1=xt2*(2+(7+9*xt+5*q)*q)da2=xt1*(2+xt+2*q)dA=S12*t*t1*(1+S12*S12*(da1+da2)/NB/NB/24)/NBXdB=Abs(dB-dB0)*p'常数p=206265XdL=Abs(dL-dL0)*pXdA=Abs(dA-dA0)*pLoopWhileXdB0.0001AndXdL0.0001AndXdA0.0001B2=B1+dBL2=L1+dLA2=A1+dAIfA1piThenA2=A2-pi-7-IfA1piThenA2=A2+pi反算：eps=e2/(1-e2)Bm=(B1+B2)/2dB=B2-B1dL=L2-L1CosBm=Cos(Bm)NB=a/Sqr(1#-e2*Sin(Bm)*Sin(Bm))t=Tan(Bm)q=eps*CosBm*CosBmv2=1+qv4=v2*v2v6=v4*v2r10=NB*CosBmr12=NB*CosBm*(1+q-9*q*t*t+q*q)/24/v4r30=-NB*CosBm^3*t^2/24s01=NB/v2s21=-NB*CosBm^2*(2+3*t*t+2*q)/24/v2s03=NB*q*(1-t*t)/8/v6't10=CosBm*t't12=CosBm*t*(3+2*q-2*q*q)/24't30=CosBm^3*t*(1+q)/12t10=CosBm*tt12=CosBm*t*(2+7*q+9*t*t*q)/24/v4t30=CosBm^3*t*(2+t*t+2*q)/24SsinAm=r10*dL+r12*dL*dB^2+r30*dL^3ScosAm=s01*dB+s21*dB*dL^2+s03*dB^3dA=t10*dL+t12*dL*dB^2+t30*dL^3'计算大地正反方位角IfdB=0ThenAm=pi/2IfdB0OrdB0ThenAm=Atn(SsinAm/ScosAm)IfAm0ThenAm=Am+piIfSsinAm0ThenAm=Am+piA1=Am-dA/2A2=Am+dA/2IfA1piThenA2=A2-piIfA1piThenA2=A2+piIfdL=0ThenS12=ScosAm/Cos(Am)Else:S12=SsinAm/Sin(Am)EndIf-8-5.计算案例克拉索夫斯基椭球1975国际椭球-9-正算过程B1=300000L1=462500S=100000mA12=750000dB=4.07495399E-03dL=1.7472216E-02dA=8.7361010E-03计算值趋近次数1234sinAm0.9670468294550.9670533918394330.96705331383970.967053313245289cosAm0.25459856566850.2545736383285760.25457393462720.254573936885357tanBm0.580