题目

平原区某公路有两个交点间距为407.54m,JD1=K7+231.38,偏角α1=12°24′20″(左偏),半径R1=1200m; JD2为右偏,α2=15°32′50″,R2=1000m,试按S型曲线计算LS1、LS2长度,并计算两曲线主点里程桩号。

题目解答

答案

解:令两曲线的切线长相当,则取T1≈407.54/2=203.77m按1:1:1计算LS1:LS1=αR/2=12.2420×PI/180×1200/2=129.91取LS1=130m则经计算得,T1=195.48m203.77-195.48=8.29,即T1计算值偏短。因切线长度与缓和曲线长度的增减有近似1/2的关系,则另LS1=130+2×8.29=146.58,取LS1=140m。则,T1= 200.486mT2=407.54-T1=407.54-200.49=207.05按1:1:1计算LS2:LS2=αR/2=15.3250×PI/180×1000/2=135.68按Ls=135.68计算切线长T2得,T2=204.45m207.05-204.45=2.60取LS2=135.68+2×2.60=140.88则经计算得,T2=207.055m207.05-207.055=-0.005取LS2=140.88-2×0.005=140.87则经计算得,T2=207.05mJD1曲线要素及主点里程桩号计算JD1= K7+231.38 T1=200.49 L1=399.82 E1=7.75 J1=1.15ZH1=K7+030.89 HY1=K7+170.89 QZ1=K7+230.80YH1=K7+290.71 HZ1=K7+430.71JD2=JD1+407.54-J1=K7+231.38+407.54-1.15=K7+637.77JD2曲线要素及主点里程桩号计算JD2=K7+637.77 T2=207.05 L2=412.22 E2=10.11 J2=1.88ZH2=K7+430.72 HY2=K7+571.59 QZ2=K7+636.83YH2=K7+702.07 HZ2=K7+842.94

解析

步骤 1：计算LS1
- 令两曲线的切线长相当，取T1≈407.54/2=203.77m
- 按1:1:1计算LS1: LS1=αR/2=12.2420×PI/180×1200/2=129.91
- 取LS1=130m
- 则经计算得，T1=195.48m，203.77-195.48=8.29，即T1计算值偏短
- 因切线长度与缓和曲线长度的增减有近似1/2的关系，另LS1=130+2×8.29=146.58，取LS1=140m
- 则，T1=200.486m，T2=407.54-T1=407.54-200.49=207.05

步骤 2：计算LS2
- 按1:1:1计算LS2: LS2=αR/2=15.3250×PI/180×1000/2=135.68
- 按Ls=135.68计算切线长T2得，T2=204.45m，207.05-204.45=2.60
- 取LS2=135.68+2×2.60=140.88
- 则经计算得，T2=207.055m，207.05-207.055=-0.005
- 取LS2=140.88-2×0.005=140.87
- 则经计算得，T2=207.05m

步骤 3：计算JD1曲线要素及主点里程桩号
- JD1=K7+231.38，T1=200.49，L1=399.82，E1=7.75，J1=1.15
- ZH1=K7+030.89，HY1=K7+170.89，QZ1=K7+230.80，YH1=K7+290.71，HZ1=K7+430.71

步骤 4：计算JD2曲线要素及主点里程桩号
- JD2=JD1+407.54-J1=K7+231.38+407.54-1.15=K7+637.77
- JD2=K7+637.77，T2=207.05，L2=412.22，E2=10.11，J2=1.88
- ZH2=K7+430.72，HY2=K7+571.59，QZ2=K7+636.83，YH2=K7+702.07，HZ2=K7+842.94