
路线测量方案圆曲线主点的测设圆曲线测设元素0L=RanE=R(sec~D=2T-LT=Rig2180°J2QZ.LR二、主点测设1.里程计算交点的里程实测,主点的里程根据交点的里程推算ZY=JD-TYZ=ZY+LQZ=YZ-L/2(校核)JD=QZ+D/22.主点测设一、切线支距法1.计算公式x,=Rsin g,y=R(1-cOsO,)1.180°y3OQZR元二、偏角法1.原理根据曲线起点或终点至曲线上任一点的弦线与切线之间的弦切角Ai和弦Ci来确定点位。134--41800C=2Rsm0=1--24R22R元2
路线测量方案———圆曲线主点的测设 一、圆曲线测设元素 二、主点测设 1.里程计算 交点的里程实测,主点的里程根据交点的里程推算 ZY = JD - T YZ = ZY + L QZ = YZ - L/2 JD = QZ + D/2 (校核) 2.主点测设 一、切线支距法 1.计算公式 二、偏角法 1. 原理 根据曲线起点或终点至曲线上任一点的弦线与切线之间的弦切角Δi 和弦 Ci 来确定点位。 2 T = Rtg 180 L = R 1) 2 = (sec − E R D = 2T − L 180 (1 cos ) sin R l y R x R i i i i i i = = − = 180 2 2R l i i i = = 2 3 2 24 2 sin R l c R l i i i i = = −

JDJDA3y3QZC303A1中C2p3P2YZ?@3RR@120PL=AA,=2JC9+00=A+0A=A32p+200C3p32=A+2A1a203A,I=A,+na。Az=A,+nA,+ABR0例题:已知交点的里程为K3+182.76,测得转角α=25°48,圆曲线半径R=300m求曲线测设元素及主点里程。1.曲线测设元素:E=7.77T=68.71L135.09D=2.332.主点里程ZY=JD-T=182.76-68.71=114.05YZ=ZY+L=114.05+135.09=249.14QZ=YZ L/2=249.14 - 67.54=181.60JD=QZ+D/2=181.60+1.16=182.76(检核)圈曲线详纫设计算表盛长口信免试数aN每名彩计测长桩号第长e000000n000000U0ZYK3+114.05595034.05034055.950.065.955.95+12020.00228411.121543525.922025.95+14020.00423163.511543545.77204595+1606175220.007.231543565.422065.95+180627001.607570090867.3566.981.50QZ18.407.57353.3300#452767.546698IG+181.6018.4020.0035518234.021543548.9249.14+20020357130220,001411543529.0929.1420+2209.1435907380.52229.140.149.149.14+240YZ00DD0a00000oK3+149.14
例题:已知交点的里程为 K3+182.76,测得转角 =2548,圆曲线半径 R=300m 求曲线测设元素及主点里程。 1. 曲线测设元素: T=68.71 L=135.09 E=7.77 D=2.33 2. 主点里程 ZY = JD – T = 182.76 – 68.71 = 114.05 YZ = ZY + L = 114.05 + 135.09 = 249.14 QZ = YZ –L/2 = 249.14 – 67.54 = 181.60 JD = QZ +D/2 = 181.60 + 1.16 = 182.76 (检核) 1 0 0 3 1 0 1 0 2 1 1 2 2 2 2 2 = + + = = + + = = = A n+1 = 1 + n0 YZ = 1 + nn + B

例题:如图,设某公路的交点桩号为K0+518.66,右转角ay=180018'36",圆曲线半径R=100m,缓和曲线长Is=10m,试测设主点桩。解:(一)计算测设元素p=0.04m; q=5.00m1,180°=2°51'53βo =2R元=10.00mXo=ls40R212=0.17myo=6RTh=(R+p)tg+q=21.12mL =R(α-2β)+2l,=41.96m180°En =(R+p)sec号-R=1,33m(二)里程计算ZH=K0+497.54;HY=K0+507.54;QZ=K0+518.52;HZ=K0+539.50YH=K0+529.50(三)主点测设1.架仪JDi,后视JDi-1,量取TH,得ZH点;后视JDi+1,量取TH,得HZ点:在分角线上量取EH,得QZ点。2.分别在ZH、HZ点架仪,后视JDi方向,量取x0,再在此方向垂直方向上量取yO,得HY和YH点
例题:如图,设某公路的交点桩号为K0+518.66,右转角αy=180018'36",圆曲线半径R=100m, 缓和曲线长 ls=10m,试测设主点桩。 解:(一)计算测设元素 p=0.04m;q=5.00m (二)里程计算 ZH=K0+497.54;HY=K0+507.54;QZ=K0+518.52;HZ=K0+539.50;YH=K0+529.50 (三)主点测设 1.架仪 JDi,后视 JDi-1,量取 TH,得 ZH 点;后视 JDi+1,量取 TH,得 HZ 点;在分角 线上量取 EH,得 QZ 点。 2.分别在 ZH、HZ 点架仪,后视 JDi 方向,量取 x0,再在此方向垂直方向上量取 y0,得 HY 和 YH 点。 2 51 53 180 2 0 0 0 = = R l s = = = − = m R l y m R l x l s s s 0.17 6 10.00 40 2 0 2 3 0 TH R p tg q 21.12m 2 = ( + ) + = LH R 2l s 41.96m 180 ( 2 ) = − 0 + = EH R p R 1.33m 2 = ( + )sec − =