Question

I tried to use haversine i couldn t succeed Basically my problem is i have 3 map locations with their Lat and Long values. What i trying to do is getting this triangle s middle point.

照片是这样的

""https://i.stack.imgur.com/xkPEV.jpg" alt="在这里输入图像描述"/ >

I Have the locations of H, F and I points. i need to know question mark s location. 41.040035,28.984026 41.040868,28.985807 41.039136,28.984981

PHP, MYSQL or Objective C any of these languages are ok as an answer. Even suggestions are appreciated. Thanks.

Answer 1

我检查了这个例子:http://geomidpoint.com/example.html

并在PHP中写下一个函数,希望这有帮助...

[edit] 我忘了转换成弧度进行计算, 所以它给出了不同的输出, 所以现在应该是正常的...

   <?php

function middlepoint($lat1,$lon1,$lat2,$lon2,$lat3,$lon3){
    $w1=1095.75;$w2=730.5;$w3=365.25;$tw=$w1+$w2+$w3;   //weighting factors

    $x1=cos(floatval(deg2rad($lat1)))*cos(floatval(deg2rad($lon1)));$y1=cos(floatval(deg2rad($lat1)))*sin(floatval(deg2rad($lon1)));$z1=sin(floatval(deg2rad($lat1)));$x2=cos(floatval(deg2rad($lat2)))*cos(floatval(deg2rad($lon2)));$y2=cos(floatval(deg2rad($lat2)))*sin(floatval(deg2rad($lon2)));$z2=sin(floatval(deg2rad($lat2)));$x3=cos(floatval(deg2rad($lat3)))*cos(floatval(deg2rad($lon3)));$y3=cos(floatval(deg2rad($lat3)))*sin(floatval(deg2rad($lon3)));$z3=sin(floatval(deg2rad($lat3)));  //convert lat/long to cartesian (x,y,z) coordinates

    $x = ($x1*$w1+$x2*$w2+$x3*$w3)/$tw;$y=($y1*$w1+$y2*$w2+$y3*$w3)/$tw;$z=($z1*$w1+$z2*$w2+$z3*$w3)/$tw;   //Compute combined weighted cartesian coordinates

    $lon=atan2($y,$x);$hyp=sqrt(pow($x,2)+pow($y,2));$lat=atan2($z,$hyp);   //Convert cartesian coordinate to latitude and longitude for the midpoint

    $midpoint[0] = $lon * (180/pi());$midpoint[1] = $lat * (180/pi());  //Convert from radians to degrees

    return $midpoint;   //return an array(lat,lon);

}

$test = middlepoint(41.040035,28.984026,41.040868,28.985807,41.039136,28.984981);
print_r($test);

?>

Answer 2

在你工作的规模上,将拉特/长坐标作为飞机坐标处理不会造成任何伤害,并使用欧几里德几何学的基本操作来确定三角形的中点等位置。

您也可能对< a href=>" barycentic坐标。

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