《地图制图学》课程教学课件（PPT讲稿）第四章 常用地图投影及其应用

二、等角圆锥投影等角条件m= n或a=b或w=0MdpdpapdoN cosMdppraeNM (1 -e’ sin ?)%(1 - e’ sin ?p)1-e2dpMdo? sin?N cos (@cOsOO

二、等角圆锥投影 等角条件 r a Md d    − =     N cos Md a d − =     − − − = =         1 sin cos 1 cos 2 2 2 d e e N Md a d ( ) ( ) 2 3 2 2 2 1 sin 1 e  a e M − − = ( ) 2 1 2 2 1 e sin  a N − =

二、等角圆锥投影doIn Ksincos1- @2sin-e?cos?Cos2&4sin82e"sinjcosCOsgCOsQsind sin Φ = d(esin βsin = esin p两边取微分1-sin =1-e? sin ’ = cos? cos ddΦ = ecos pdpecos pdgddcos?Φcos dsin1

( )  • − − − − = K d e e a ln 1 sin cos 1 ln 2 2 2     二、等角圆锥投影 令 sin  = esin  两边取微分 d sin  = d(esin )    2 2 2 2 1− sin =1− e sin = cos    = = −         cos cos cos 1 sin cos 2 2 2 2 d ae e d a e e d a cosd = e cosd

d1-edx =-cos dcOs@sincos@I045°tg2x=aln+c2d4592@dd2daalnUIne? sin? p)dcosdcoscos@tg2KKK=Uaaln UInOUapp

( )   • − − =   1 sin  cos 1 2 2 2 d e e dx a ( ) a U t g t g a d ae d a d e e a e ln 2 45 2 45 ln 1 sin cos cos cos 1 2 2 2 =       +       + • = − = − −                  = −     cos cos d ae d dx a a U K ln = ln  a U K =  a U K  = c t g t g x a t g ae t g c a e +       +       +  + =       − +      = + 2 45 2 45 ln 2 ln 45 2 ln 45        

正轴等角圆锥投影的公式K8=a0:ax=p.-pcos y=psin8akapm=n-rUarakP=mrUaW=0+

正轴等角圆锥投影的公式 a rU aK r a m = n = =  2 2 2       = = = a rU aK P m n

1、单标准纬线等角圆锥投影NCntgp.S'Cp.P。= N.ctgpakap-n=rUarakap.n.rUar.N.cosp。r.N.coso.U" = N.ctgp.U.rU.= sin ΦKa:N.ctg。sin。P。a

1、单标准纬线等角圆锥投影      N S C Cn tg =  =    = N ctga rU aK r a n = =  = = =1 a rU aK r a n                  sin cos = = = N ctg r N a a a a N ctg U N U a r U K             = = = sin cos