我想把GPS的位置(纬度,经度)转换成x,y坐标。我找到了许多关于这个主题的链接,并应用了它,但它没有给我正确的答案!
我遵循以下步骤来测试答案:(1)首先,我采取两个位置,并使用地图计算它们之间的距离。(2)然后将两个位置转换为x,y坐标。(3)然后再次计算x,y坐标中两点之间的距离,看看它是否在点(1)中给出了相同的结果。
我找到了下面的解决方案之一,但它没有给我正确的答案!
latitude = Math.PI * latitude / 180;
longitude = Math.PI * longitude / 180;
// adjust position by radians
latitude -= 1.570795765134; // subtract 90 degrees (in radians)
// and switch z and y
xPos = (app.radius) * Math.sin(latitude) * Math.cos(longitude);
zPos = (app.radius) * Math.sin(latitude) * Math.sin(longitude);
yPos = (app.radius) * Math.cos(latitude);
我也尝试了这个link,但仍然不能很好地与我一起工作!
如何将(纬度,经度)转换为(x,y)?
谢谢,
发布于 2018-12-18 14:15:23
我想和你分享我是如何处理这个问题的。我使用了等角投影,就像@MvG说的那样,但这种方法给出了与地球(或整个地图)相关的X和Y位置,这意味着你得到了全局位置。在我的例子中,我想转换一小块区域(大约500米见方)的坐标,所以我将投影点关联到另外两个点,获得全局位置并与本地(屏幕上)位置相关,就像这样:
首先,我在想要投影的区域周围选择两个点(左上角和右下角),就像下面的图片:
一旦我有了lat和lng格式的全局引用区域,我就可以对屏幕位置执行相同的操作。包含此数据的对象如下所示。
//top-left reference point
var p0 = {
scrX: 23.69, // Minimum X position on screen
scrY: -0.5, // Minimum Y position on screen
lat: -22.814895, // Latitude
lng: -47.072892 // Longitude
}
//bottom-right reference point
var p1 = {
scrX: 276, // Maximum X position on screen
scrY: 178.9, // Maximum Y position on screen
lat: -22.816419, // Latitude
lng: -47.070563 // Longitude
}
var radius = 6371; //Earth Radius in Km
//## Now I can calculate the global X and Y for each reference point ##\\
// This function converts lat and lng coordinates to GLOBAL X and Y positions
function latlngToGlobalXY(lat, lng){
//Calculates x based on cos of average of the latitudes
let x = radius*lng*Math.cos((p0.lat + p1.lat)/2);
//Calculates y based on latitude
let y = radius*lat;
return {x: x, y: y}
}
// Calculate global X and Y for top-left reference point
p0.pos = latlngToGlobalXY(p0.lat, p0.lng);
// Calculate global X and Y for bottom-right reference point
p1.pos = latlngToGlobalXY(p1.lat, p1.lng);
/*
* This gives me the X and Y in relation to map for the 2 reference points.
* Now we have the global AND screen areas and then we can relate both for the projection point.
*/
// This function converts lat and lng coordinates to SCREEN X and Y positions
function latlngToScreenXY(lat, lng){
//Calculate global X and Y for projection point
let pos = latlngToGlobalXY(lat, lng);
//Calculate the percentage of Global X position in relation to total global width
pos.perX = ((pos.x-p0.pos.x)/(p1.pos.x - p0.pos.x));
//Calculate the percentage of Global Y position in relation to total global height
pos.perY = ((pos.y-p0.pos.y)/(p1.pos.y - p0.pos.y));
//Returns the screen position based on reference points
return {
x: p0.scrX + (p1.scrX - p0.scrX)*pos.perX,
y: p0.scrY + (p1.scrY - p0.scrY)*pos.perY
}
}
//# The usage is like this #\\
var pos = latlngToScreenXY(-22.815319, -47.071718);
$point = $("#point-to-project");
$point.css("left", pos.x+"em");
$point.css("top", pos.y+"em");
如你所见,我是用javascript写的,但是计算结果可以翻译成任何语言。
附注:我将转换后的位置应用于id为"point- to -project“的HTML元素。要在您的项目中使用这段代码,您应该创建此元素(样式为position absolute)或更改"usage“块。
发布于 2018-08-08 19:12:36
由于当我搜索同样的问题时,这个页面显示在google的顶部,所以我想提供一个更实用的答案。MVG的答案是正确的,但却是理论上的。
我用javascript为fitbit ionic做了一个轨迹绘制应用程序。下面的代码是我如何解决这个问题的。
//LOCATION PROVIDER
index.js
var gpsFix = false;
var circumferenceAtLat = 0;
function locationSuccess(pos){
if(!gpsFix){
gpsFix = true;
circumferenceAtLat = Math.cos(pos.coords.latitude*0.01745329251)*111305;
}
pos.x:Math.round(pos.coords.longitude*circumferenceAtLat),
pos.y:Math.round(pos.coords.latitude*110919),
plotTrack(pos);
}
plotting.js
plotTrack(position){
let x = Math.round((this.segments[i].start.x - this.bounds.minX)*this.scale);
let y = Math.round(this.bounds.maxY - this.segments[i].start.y)*this.scale; //heights needs to be inverted
//redraw?
let redraw = false;
//x or y bounds?
if(position.x>this.bounds.maxX){
this.bounds.maxX = (position.x-this.bounds.minX)*1.1+this.bounds.minX; //increase by 10%
redraw = true;
}
if(position.x<this.bounds.minX){
this.bounds.minX = this.bounds.maxX-(this.bounds.maxX-position.x)*1.1;
redraw = true;
};
if(position.y>this.bounds.maxY){
this.bounds.maxY = (position.y-this.bounds.minY)*1.1+this.bounds.minY; //increase by 10%
redraw = true;
}
if(position.y<this.bounds.minY){
this.bounds.minY = this.bounds.maxY-(this.bounds.maxY-position.y)*1.1;
redraw = true;
}
if(redraw){
reDraw();
}
}
function reDraw(){
let xScale = device.screen.width / (this.bounds.maxX-this.bounds.minX);
let yScale = device.screen.height / (this.bounds.maxY-this.bounds.minY);
if(xScale<yScale) this.scale = xScale;
else this.scale = yScale;
//Loop trough your object to redraw all of them
}
发布于 2019-11-12 21:04:36
为了完整起见,我喜欢添加我的python改编的@allexrm代码,它工作得非常好。再次感谢!
radius = 6371 #Earth Radius in KM
class referencePoint:
def __init__(self, scrX, scrY, lat, lng):
self.scrX = scrX
self.scrY = scrY
self.lat = lat
self.lng = lng
# Calculate global X and Y for top-left reference point
p0 = referencePoint(0, 0, 52.526470, 13.403215)
# Calculate global X and Y for bottom-right reference point
p1 = referencePoint(2244, 2060, 52.525035, 13.405809)
# This function converts lat and lng coordinates to GLOBAL X and Y positions
def latlngToGlobalXY(lat, lng):
# Calculates x based on cos of average of the latitudes
x = radius*lng*math.cos((p0.lat + p1.lat)/2)
# Calculates y based on latitude
y = radius*lat
return {'x': x, 'y': y}
# This function converts lat and lng coordinates to SCREEN X and Y positions
def latlngToScreenXY(lat, lng):
# Calculate global X and Y for projection point
pos = latlngToGlobalXY(lat, lng)
# Calculate the percentage of Global X position in relation to total global width
perX = ((pos['x']-p0.pos['x'])/(p1.pos['x'] - p0.pos['x']))
# Calculate the percentage of Global Y position in relation to total global height
perY = ((pos['y']-p0.pos['y'])/(p1.pos['y'] - p0.pos['y']))
# Returns the screen position based on reference points
return {
'x': p0.scrX + (p1.scrX - p0.scrX)*perX,
'y': p0.scrY + (p1.scrY - p0.scrY)*perY
}
pos = latlngToScreenXY(52.525607, 13.404572);
位置x和位置y包含经纬度的转换x&y坐标(52.525607,13.404572)
我希望这对像我这样的人有帮助,以便正确地解决将lat lng转换为本地参考坐标系的问题。
最好的
https://stackoverflow.com/questions/16266809
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