Joh*_*rto 8 c++ algorithm math matlab angle
我正在开展一个非常酷的项目,我正在收集有关光标移动的数据,但我遇到了一个问题,我认为我可以使用一些帮助.我不断阅读有关光标x和y位置的数据(以及其他相关数据),一旦光标超过y维度的某个阈值,我需要计算运动方向(角度).让我用一个图来说明我画的:
往往会发生的是光标以略微的直线移动,但随后朝向运动的末端弯曲.我需要计算θ,即蓝色矢量相对于正x轴的角度.我想出的想法是使用最后2个样本来很大程度上确定移动方向是什么,否则如果我使用太多样本,我会扭曲实际角度.为了给出一个极端的案例,让我跟进另一张图片:
这里每个点代表一个样本.请注意,如果我使用BOTH点,我想要的真实角度将是错误的(同样,我需要找到光标最后移动的方向,这是在行尾绘制的矢量).我不希望这个案子出现太多,但想知道是否有办法解决它,如果它.
最后,请注意,这些运动可以在第一或第二象限中发生,如果这有所不同.
我真的很感激这里的任何帮助.我用C++编写代码,但我想我可以翻译任何答案.谢谢.
这应该可以帮助您开始 http://jsfiddle.net/0ao9oa7a/
代码
function process(points) {
if(points.length === 0) {
txt = "Not enough points\n" + txt;
return null;
}
// compress points, removing those that are too close together
var newPoints = [];
newPoints.push(points[0]);
for(var i = 1; i < points.length; i++) {
if(Math.sqrt(Math.pow(points[i].x - newPoints[newPoints.length - 1].x, 2) + Math.pow(points[i].y - newPoints[newPoints.length - 1].y, 2)) > 5) {
newPoints.push(points[i]);
}
}
points = newPoints;
if(points.length < 2) {
txt = "Not enough points\n" + txt;
return null;
}
// get all of the angles
var angles = [];
for(var i=0; i < points.length - 1; i++) {
var rad = Math.atan2(points[i + 1].y - points[i].y, points[i + 1].x - points[i].x);
angles[i] = rad;
txt += "x: " + (points[i].x|0) + " y: " + (points[i].y|0) + " x: " + (points[i+1].x|0) + " y: " + (points[i+1].y|0) + " [" + ((rad * 180 / Math.PI)|0) + "]" + "\n";
}
txt += "\n";
// get all of the diffs between angles
// save the index of the max diff
var absDiffs = [];
var maxDiff = -1;
var maxDiffAngleIndex = -1;
for(var i=0; i < points.length - 1; i++) {
var delta = Math.abs(angles[i] - angles[i + 1]);
if(delta >= maxDiff) {
maxDiff = delta;
maxDiffAngleIndex = i + 1;
}
}
if(maxDiffAngleIndex == -1) {
txt = "Angle: " + angles[0] + " : " + (angles[0] * 180 / Math.PI) + "\n" + txt;
return angles[0];
} else if(maxDiffAngleIndex == angles.length - 1) {
txt = "Angle: " + angles[angle.length - 1] + " : " + (angles[angles.length - 1] * 180 / Math.PI) + "\n" + txt;
return angles[angles.length - 1];
} else {
// find the average angle from the index to the end
var sumX = 0;
var sumY = 0;
for(var i = maxDiffAngleIndex; i < angles.length; i++) {
sumX += Math.cos(angles[i]);
sumY += Math.sin(angles[i]);
}
var avgX = sumX / (angles.length - maxDiffAngleIndex);
var avgY = sumY / (angles.length - maxDiffAngleIndex);
//
var avgAngle = Math.atan2(avgY, avgX);
txt = "Angle: " + avgAngle + " : " + (avgAngle * 180 / Math.PI) + "\n" + txt;
return avgAngle;
}
}
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