我正在研究一种移动设备的“指南针”。我有以下几点意见:
point 1 (current location): Latitude = 47.2246, Longitude = 8.8257
point 2 (target location): Latitude = 50.9246, Longitude = 10.2257
另外,我还有以下信息(来自我的android手机):
The compass-direction in degree, which bears to the north.
For example, when I direct my phone to north, I get 0°
我如何创建一个“指南针”箭头来显示指向该点的方向?
这有一个数学问题吗?
编辑:好的,我找到了一个解决方案,看起来是这样的:
/**
* Params: lat1, long1 => Latitude and Longitude of current point
* lat2, long2 => Latitude and Longitude of target point
*
* headX => x-Value of built-in phone-compass
*
* Returns the degree of a direction from current point to target point
*
*/
function getDegrees(lat1, long1, lat2, long2, headX) {
var dLat = toRad(lat2-lat1);
var dLon = toRad(lon2-lon1);
lat1 = toRad(lat1);
lat2 = toRad(lat2);
var y = Math.sin(dLon) * Math.cos(lat2);
var x = Math.cos(lat1)*Math.sin(lat2) -
Math.sin(lat1)*Math.cos(lat2)*Math.cos(dLon);
var brng = toDeg(Math.atan2(y, x));
// fix negative degrees
if(brng<0) {
brng=360-Math.abs(brng);
}
return brng - headX;
}
发布于 2014-03-30 12:00:56
别忘了说我最终找到了答案。该应用程序用于确定公交车辆的指南针方向及其目的地。本质上,花哨的数学获取地球的曲率,找到一个角度/罗盘读数,然后将这个角度与一个通用的指南针数值进行匹配。当然,您可以只保留compassReading并将其应用为图像的旋转量。请注意,这是到终点(汽车站)的车辆方向的平均确定,这意味着它不能知道道路正在做什么(因此,这可能最适用于飞机或轮滑比赛)。
//example obj data containing lat and lng points
//stop location - the radii end point
endpoint.lat = 44.9631;
endpoint.lng = -93.2492;
//bus location from the southeast - the circle center
startpoint.lat = 44.95517;
startpoint.lng = -93.2427;
function vehicleBearing(endpoint, startpoint) {
endpoint.lat = x1;
endpoint.lng = y1;
startpoint.lat = x2;
startpoint.lng = y2;
var radians = getAtan2((y1 - y2), (x1 - x2));
function getAtan2(y, x) {
return Math.atan2(y, x);
};
var compassReading = radians * (180 / Math.PI);
var coordNames = ["N", "NE", "E", "SE", "S", "SW", "W", "NW", "N"];
var coordIndex = Math.round(compassReading / 45);
if (coordIndex < 0) {
coordIndex = coordIndex + 8
};
return coordNames[coordIndex]; // returns the coordinate value
}
例如: vehicleBearing(mybus,busstation)可能会返回"NW“,意思是它正向西北方向移动
发布于 2017-08-29 10:05:27
我在math here中找到了一些有用的gps坐标公式。对于这种情况,我的解决方案如下
private double getDirection(double lat1, double lng1, double lat2, double lng2) {
double PI = Math.PI;
double dTeta = Math.log(Math.tan((lat2/2)+(PI/4))/Math.tan((lat1/2)+(PI/4)));
double dLon = Math.abs(lng1-lng2);
double teta = Math.atan2(dLon,dTeta);
double direction = Math.round(Math.toDegrees(teta));
return direction; //direction in degree
}
发布于 2018-01-31 19:29:03
我不能很好地理解你的解决方案,计算斜率对我很有效。修改efwjames的答案和你的答案。这应该是-
import math
def getDegrees(lat1, lon1, lat2, lon2,head):
dLat = math.radians(lat2-lat1)
dLon = math.radians(lon2-lon1)
bearing = math.degrees(math.atan2(dLon, dLat))
return head-bearing
https://stackoverflow.com/questions/8502795
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