我需要根据方位 Angular 和半径在坐标系上定位一个点。
使用以下代码,我可以将一个点定位在距中心正确的距离处,但这仅适用于水平方向。
我需要的是基于方位和半径的点的顶部和左侧位置
setTimeout(function() {
var km2show = 100;
var testDistance = 50;
var width = $('#radar').width(); // Width & height of radar
var center = width/2; // Radar center
var px2km = center/km2show; // Pixels per km
var radius = radius2coor((px2km*testDistance),center);
var bearing = 45;
// Set height of radar based on the width
$('#radar').height(width);
// Set dot in center of radar
$('#centerDot').css({top: center, left: center});
$('#container').append("<div style='position:absolute;top:0px;left:"+radius+"px;color:white;'>*</div>");
},100);
function radius2coor(radius,center) {
var res = radius-center;
return res;
}
请参阅 jsFiddle
那么我该如何获取机器人点的顶部和左侧位置?
最佳答案
您遇到的主要问题是 Angular 不是弧度,所以我们要做的第一件事是将 45 度转换为 pi/4。
此外,当从常规 Angular 坐标转换为 x,y 坐标时,您可以将半径乘以 Angular 正弦来找到 y 坐标,然后将半径乘以 Angular 余弦来获得 x 坐标。只要考虑一下单位圆就可以理解了。
var bearing = parseInt(prompt("enter angle in degrees", "0"));
if(!isNaN(bearing)){
setTimeout(function() {
var km2show = 100;
var testDistance = 50;
var width = $('#radar').width(); // Width & height of radar
var center = width/2; // Radar center
var px2km = center/km2show; // Pixels per k
//not sure what this is doing so I set a random radius
//(called it distanceFromCenter). If you need this to be
//the distance between two cartesian points then you can
//just implement the distance formula.
//var radius = radius2coor((px2km*testDistance),center);
var radius = 100;
var radianBearing = (bearing/180)*Math.PI
// Set height of radar based on the width
$('#radar').height(width);
// Set dot in center of radar
$('#centerDot').css({top: center, left: center});
//the main issue you were encountering was that the angle wasn't in radians so I converted it.
positionDot(radius, radianBearing);
},100);
}
function positionDot(distanceFromCenter, bearing, width)
{
//when going from regular angular coordinates to x,y coordinates you multiply the radius by sine of the angle to find y coordinate and you multiply the radius by the cosine of the angle to get the x coordinate.
$('#container').append("<div style='position:absolute;top:"+(-distanceFromCenter*Math.sin(bearing)).toString()+"px;left:"+distanceFromCenter*Math.cos(bearing)+"px;color:white;'>*</div>");
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.0/jquery.min.js"></script>
<div id='radar' style='width:100%;max-width:400px;height:400px;border:1px black solid;border-radius:400px;background-color:#3c3c3c;position:relative;'>
<div id='centerDot' style='position:absolute;color:white;'>
<div style='position:relative;' id='container'></div>
<b>*</b>
</div>
</div>
关于javascript - 从方位获取像素位置,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41345343/