45.87m 180°Р αEH=(R+P)sec -R=4.22m 2Р DH=2TH-LH=0.91mР ZH=JD-TH=K0+445.27Р HY=ZH+LS=K0+495.27Р LHQZ=ZH+=K0+518.21 2Р HZ=ZH+LH=K0+591.14Р YH=HZ-LS=K0+541.14Р 2、某三岭区一般二级公路,边坡点桩号为K5+030.00,高程为427.68m,i1=+5%, i2=-4%,竖曲线半径R=2000m。试计算竖曲线诸要素以及桩号为K5+010.00和K5+080.00处的设计高程。Р 解:(1)曲线要素Р W=i1-i2=0.09 所以该竖曲线为凸形竖曲线Р 曲线长:L=Rw=2000×0.09=180mР L切线长:T= =90m 2Р T2 902外距:E===2.03m 2R2×2000Р (2)竖曲线起终点桩号Р 起点桩号=(k5+030.00)-90=k4+940.00Р 终点桩号=(k5+030.00)+90=k5+120.00Р (3)k5+010.00和k5+080.00切线标高和改正值Р k5+010.00的切线标高=427.68-(k5+030.00- k5+010.00)×Р5%=426.68mР (k+010.00-k+940.00)2 k5+010.00改正值==1.225m 2×2000Р k5+080.00切线标高=427.68-(k5+080.00- k5+030.00)×4%=425.68mР (k5+120.00-k5+080.00)2Р k5+080.00改正值= =0.40m 2×2000Р (4) k5+010.00和k5+080.00处设计高程Р k5+010.00设计高程=426.68-1.225m=425.45mР k5+080.00处设计高程=425.68-0.40=425.28m
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