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  • 【百度地图】获取某个覆盖物的总面积

      1 //参数是一个百度类型的多边形覆盖物
      2 function getArea(polygon){
      3 
      4   // 检查类型:既不是百度类型的范围又不是数组类型的数据,直接返回0
      5   if (!(polygon instanceof BMap.Polygon) && !(polygon instanceof Array)) {
      6           return 0;
      7   }
      8 
      9    //如果是百度类型的,得到点集合,不是的话,另说。
     10    var pts = new Array();
     11    if (polygon instanceof BMap.Polygon) {
     12         pts = polygon.getPath();
     13    }
     14 
     15    //判断数组的长度,如果是小于3的话,不构成面,无法计算面积
     16    if (pts.length < 3) {
     17           return 0;
     18    }
     19 
     20    var totalArea = 0;// 初始化总面积
     21    var LowX = 0.0; 
     22    var LowY = 0.0; 
     23    var MiddleX = 0.0;
     24    var MiddleY = 0.0;
     25    var HighX = 0.0;
     26    var HighY = 0.0;
     27    var AM = 0.0;
     28    var BM = 0.0;
     29    var CM = 0.0;
     30    var AL = 0.0;
     31    var BL = 0.0;
     32    var CL = 0.0;
     33    var AH = 0.0;
     34    var BH = 0.0;
     35    var CH = 0.0;
     36    var CoefficientL = 0.0;
     37    var CoefficientH = 0.0;
     38    var ALtangent = 0.0;
     39    var BLtangent = 0.0;
     40    var CLtangent = 0.0;
     41    var AHtangent = 0.0;
     42    var BHtangent = 0.0;
     43    var CHtangent = 0.0;
     44    var ANormalLine = 0.0;
     45    var BNormalLine = 0.0;
     46    var CNormalLine = 0.0;
     47    var OrientationValue = 0.0;
     48    var AngleCos = 0.0;
     49    var Sum1 = 0.0;
     50    var Sum2 = 0.0;
     51    var Count2 = 0;
     52    var Count1 = 0;
     53    var Sum = 0.0;
     54    var Radius = 6378137.0;// ,WGS84椭球半径
     55    var Count = pts.length;
     56     for ( var i = 0; i < Count; i++) {
     57           if (i == 0) {
     58               LowX = pts[Count - 1].lng * Math.PI / 180;
     59               LowY = pts[Count - 1].lat * Math.PI / 180;
     60               MiddleX = pts[0].lng * Math.PI / 180;
     61               MiddleY = pts[0].lat * Math.PI / 180;
     62               HighX = pts[1].lng * Math.PI / 180;
     63               HighY = pts[1].lat * Math.PI / 180;
     64           } else if (i == Count - 1) {
     65               LowX = pts[Count - 2].lng * Math.PI / 180;
     66               LowY = pts[Count - 2].lat * Math.PI / 180;
     67               MiddleX = pts[Count - 1].lng * Math.PI / 180;
     68               MiddleY = pts[Count - 1].lat * Math.PI / 180;
     69               HighX = pts[0].lng * Math.PI / 180;
     70               HighY = pts[0].lat * Math.PI / 180;
     71           } else {
     72               LowX = pts[i - 1].lng * Math.PI / 180;
     73               LowY = pts[i - 1].lat * Math.PI / 180;
     74               MiddleX = pts[i].lng * Math.PI / 180;
     75               MiddleY = pts[i].lat * Math.PI / 180;
     76               HighX = pts[i + 1].lng * Math.PI / 180;
     77               HighY = pts[i + 1].lat * Math.PI / 180;
     78           }
     79           AM = Math.cos(MiddleY) * Math.cos(MiddleX);
     80           BM = Math.cos(MiddleY) * Math.sin(MiddleX);
     81           CM = Math.sin(MiddleY);
     82           AL = Math.cos(LowY) * Math.cos(LowX);
     83           BL = Math.cos(LowY) * Math.sin(LowX);
     84           CL = Math.sin(LowY);
     85           AH = Math.cos(HighY) * Math.cos(HighX);
     86           BH = Math.cos(HighY) * Math.sin(HighX);
     87           CH = Math.sin(HighY);
     88           CoefficientL = (AM * AM + BM * BM + CM * CM)/ (AM * AL + BM * BL + CM * CL);
     89           CoefficientH = (AM * AM + BM * BM + CM * CM)/ (AM * AH + BM * BH + CM * CH);
     90           ALtangent = CoefficientL * AL - AM;
     91           BLtangent = CoefficientL * BL - BM;
     92           CLtangent = CoefficientL * CL - CM;
     93           AHtangent = CoefficientH * AH - AM;
     94           BHtangent = CoefficientH * BH - BM;
     95           CHtangent = CoefficientH * CH - CM;
     96           AngleCos = (AHtangent * ALtangent + BHtangent * BLtangent + CHtangent* CLtangent)/ (Math.sqrt(AHtangent * AHtangent + BHtangent* BHtangent + CHtangent * CHtangent) * Math.sqrt(ALtangent * ALtangent + BLtangent* BLtangent + CLtangent * CLtangent));
     97           AngleCos = Math.acos(AngleCos);
     98           ANormalLine = BHtangent * CLtangent - CHtangent * BLtangent;
     99           BNormalLine = 0 - (AHtangent * CLtangent - CHtangent* ALtangent);
    100           CNormalLine = AHtangent * BLtangent - BHtangent * ALtangent;
    101           if (AM != 0)
    102               OrientationValue = ANormalLine / AM;
    103           else if (BM != 0)
    104               OrientationValue = BNormalLine / BM;
    105           else
    106               OrientationValue = CNormalLine / CM;
    107           if (OrientationValue > 0) {
    108               Sum1 += AngleCos;
    109               Count1++;
    110           } else {
    111               Sum2 += AngleCos;
    112               Count2++;
    113           }
    114       }
    115 
    116       var tempSum1, tempSum2;
    117       tempSum1 = Sum1 + (2 * Math.PI * Count2 - Sum2);
    118       tempSum2 = (2 * Math.PI * Count1 - Sum1) + Sum2;
    119       if (Sum1 > Sum2) {
    120           if ((tempSum1 - (Count - 2) * Math.PI) < 1)
    121               Sum = tempSum1;
    122           else
    123               Sum = tempSum2;
    124       } else {
    125           if ((tempSum2 - (Count - 2) * Math.PI) < 1)
    126               Sum = tempSum2;
    127           else
    128               Sum = tempSum1;
    129       }
    130       totalArea = (Sum - (Count - 2) * Math.PI) * Radius * Radius;
    131 
    132       return totalArea; // 返回总面积
    133 }
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  • 原文地址:https://www.cnblogs.com/zwqh/p/7737133.html
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