带有缓和曲线的圆曲线逐桩坐标计算例题

1、带有缓和曲线的圆曲线逐桩坐标计算例题：某山岭区二级公路，已知交点的坐标分别为JD1 （40961.914,91066.103 ）、JD2（40433.528,91250.097 ）、JD3（40547.416,91810.392 ） , JD2里程为 K2+200.000,R=150m缓和曲线长度为40m,计算带有缓和曲线的圆曲线的逐桩坐标。（工程测量 第202页36题）解：（1）转角、缓和曲线角、曲线常数、曲线要素、主点里程、主点坐标计算1坐标方位角12160=48ga 2378 3072转角 （左转曲线）82 17 26” 一 3缓和曲线角B。-优-L1802R兀7 誼224曲线内移值2L

2、sP 一 24R0.444 m5切线增长值q LsLsq 22240R219.988m6曲线 要素i（切线长Th =(R + p) tan 竺 +q2151.438m曲线长Lh =R(a2九)士+2Ls180Lh = 255.436mLy = 175.436m外距aEh =( R + p)seR249.781m切曲差Dh = 2Th - Lh47.440m7主点 里程ZH桩号ZH=JD （桩号）ThK2+048.562HY桩号HY=ZH （桩号）+ LsK2+088.562YH桩号YH=HY （桩号）+ LyK2+263.998HZ桩号HZ=YH （桩号）十LsK2+303.998QZ桩号QZ

3、=HZ （桩号）-Lh/2K2+176.280JD桩号JD=QZ （桩号）+ Dh/2K2+200.0008主点 坐标ZHXzH =X2 +ThCOSd2140576.543mYzh U+ThSin2191200.296mHZXhZ =X2 +ThCOSd2340463.693mYhz =丫2 *Th sin 2391398.500mQZ1ttEa23-(18)29 3920Xqz = X2 + Eh COSCt Eh40476.788mYqz = 丫2 + E h sin a Ehh91274.728m方法一：偏角法(坐标正算)(2)第一缓和段坐标计算 飞=7 38 22= 12二160 4

4、8 03桩号弧长li =桩点里程-Z丫里程偏:角方位角% =% Ai(左转)弦长ii5i i-90R2L|XiXi =Xzh +眷叫。YiY =Yzh +G singZH:K2+048.5620160480340576.54391200.296+06011.43801230160353311.43840565.75491204.097+08031.43813423159134031.43840547.14992211.446HYK2+088.56240I123247158151639.968ifZ40539.41991215.104I .(3)圆曲线段坐标计算丄切线二邛。=153 0941“桩

5、号弧长li 桩点里程-H丫里程偏角吕耳方位角(左转)Otq =O(ZY_JD - ft) - 弦长C TRsinAXXi =Xhy 七 coSXiyYhy * singHY:K2+088.562C切线 zzOZyjd Q153094140539.41991215.104+ 10011.43821104150583711.43540529.42091220.652+ 12031.43860015147092631.38040513.05591232.122+ 14051.438949261432015+ 16071.4381338371393104QZ:K2+176.28087.71816451

6、0136243186.47340476.78991274.728+ 18091.438+200111.438+220131.438+240151.438+260171.438YH:K2+263.998175.4363330211193920165.60640457.48091359.018(4)第二缓和段坐标计算-0 = 7 38 22桩号弧长h =HZ里程-桩点里程偏角3 LS 丿方位角Ctq =0(32 +(左转)弦长-90FTL；XZ1 3 “*f、 T/Xi HZ +c CO怨iYYi =YhzsinotcYH:K2+263.9984023247261032439.96840457.4

7、8091359.018+28023.99805500259253723.99640459.29091374.911+3003.9980013225830373.99840462.89791394.582HZK2+303.998v-_ 1JTI 4 Ika1.-X I1 _ 1/ 厂 i .、132258303740463.69391398.500精心整理方法二：切线支距法(坐标系转换)(2)第一缓和段坐标计算：12 =160 48 03Xi 二 Xzh xco 12 y sin :伐丫 二Yzh xsin ：伐 - ycos伐(本题为左转曲线)桩号h =桩点里程-ZY里程x1 一 l5l3l7

8、Xy40R2L：y 3 36RLS 336R3LSZH:K2+048.5621 040576.54391200.296+06011.43811.4380.04240565.75591204.096+08031.43831.4170.86340547.15692211.446HYK2+088.5624039.9291.77840539.41991215.104(3)圆曲线段坐标计算0 = 7 38 22 p=0.444mq=19.988m：12 =160 48 03 ZH( 40576.543,91200.296 )X i = XZH xcos：12 ysi n：12Y =YZH xsi n：1

9、2_ycos：12桩号li 桩点里程-HY里程店Xp =Rsin 申+qyp =Rf _cos 申弹XyHY:K2+088.56240539.41991215.104+ 10011.43812003051.1963.72640529.41991220.651+ 12031.43819385270.4249.17740513.05291232.122+ 14051.438271714+ 16071.438345536QZ:K2+176.28087.718410843118.68437.48740476.78891274.728+ 18091.4384233581.J-+200111.438+220131.438+240151.438+260171.438YH:K2+263.998175.436743904164.638110.74040457.47991359.018(4)第二缓和段坐标计算：12 =78 30 37Xi =Xhz -xcos： 23 ysin： 23Y =Yhz - xsin 23 - ycos 23 (本题为左转曲线)桩号h =桩点里程-ZY里程x-1i5|3|7Xy40R2L：y 3 36RLs 336R