某三级公路,交点桩号为K0+518.66,转角α=18°18′36″,圆曲线半径R=300m,缓和曲线长Ls=50m,试计算平曲线测设元素和主点里程桩号。

解:解:q=Ls 2 -L 3s 240R 2=24.99m p=L 2s 24R -L 4s 2384R 3=2.471mβ0=18002πRLs=4°46'29'' T H =(R+P)tg α2 +q=73.39mL H =R π180°(α-2β0)+2Ls=145.87m E H =(R+P)sec α2-R=4.22m D H =2T H -L H =0.91mZH=JD-T H =K 0+445.27HY=ZH+L S =K 0+495.27QZ=ZH+L H 2=K 0+518.21 HZ=ZH+L H =K 0+591.14YH=HZ-L S =K 0+541.14