如何计算两个受约束线段的旋转角度？
如何计算两个受约束线段的旋转角度？
提问于 2018-05-30 23:43:51
我有两个向量，Y向对齐是固定的，因此X向对齐可以旋转。这些向量通过两个固定长度的线段连接在一起。给定两个向量之间的角度(82.74)和所有线段的长度，如何获得两个连接线段的角度(24.62和22.61)？
给定的内容:矢量的大小，以及X轴和OG之间的角度：
var magOG = 3,
magOE = 4,
magGH = 3,
magEH = 2,
angleGamma = 90;这是我的起点：angleGamma = 90 -然后，我将拥有以下向量：
var vOG = new vec2(-3,0),
vOE = new vec2(0,-4);从现在开始，我将尝试获取angleGamma值小于90度的angleAlpha和angleBeta。
受约束线段的幅值：
段HG和HE必须满足以下条件：
/
| OG*OG+ OE*OE = (HG + HE)*(HG + HE)
>
| OG - HG = OE - HE
\这将导致以下两个解决方案(如公认的答案-双边处理中所指出的)：
Solution 1:
========================================================
HG = 0.5*(-Math.sqrt(OG*OG + OE*OE) + OG - OE)
HE = 0.5*(-Math.sqrt(OG*OG + OE*OE) - OG + OE)
Solution 2:
========================================================
HG = 0.5*(Math.sqrt(OG*OG + OE*OE) + OG - OE)
HE = 0.5*(Math.sqrt(OG*OG + OE*OE) - OG + OE)SCRATCHPAD:
这是一个包含完整解决方案的游乐场。这里使用的可视化库是很棒的JSXGraph。感谢拜罗伊特大学数字技术移动学习中心。
圆相交函数的积分:在这个问题的接受答案中：01AutoMonkey：A JavaScript function that returns the x,y points of intersection between two circles?
function deg2rad(deg) {
return deg * Math.PI / 180;
}
function rad2deg(rad) {
return rad * 180 / Math.PI;
}
function lessThanEpsilon(x) {
return (Math.abs(x) < 0.00000000001);
}
function angleBetween(point1, point2) {
var x1 = point1.X(), y1 = point1.Y(), x2 = point2.X(), y2 = point2.Y();
var dy = y2 - y1, dx = x2 - x1;
var t = -Math.atan2(dx, dy); /* range (PI, -PI] */
return rad2deg(t); /* range (180, -180] */
}
function circleIntersection(circle1, circle2) {
var r1 = circle1.radius, cx1 = circle1.center.X(), cy1 = circle1.center.Y();
var r2 = circle2.radius, cx2 = circle2.center.X(), cy2 = circle2.center.Y();
var a, dx, dy, d, h, h2, rx, ry, x2, y2;
/* dx and dy are the vertical and horizontal distances between the circle centers. */
dx = cx2 - cx1;
dy = cy2 - cy1;
/* angle between circle centers */
var theta = Math.atan2(dy,dx);
/* vertical and horizontal components of the line connecting the circle centers */
var xs1 = r1*Math.cos(theta), ys1 = r1*Math.sin(theta), xs2 = r2*Math.cos(theta), ys2 = r2*Math.sin(theta);
/* intersection points of the line connecting the circle centers */
var sxA = cx1 + xs1, syA = cy1 + ys1, sxL = cx2 - xs2, syL = cy2 - ys2;
/* Determine the straight-line distance between the centers. */
d = Math.sqrt((dy*dy) + (dx*dx));
/* Check for solvability. */
if (d > (r1 + r2)) {
/* no solution. circles do not intersect. */
return [[sxA,syA], [sxL,syL]];
}
thetaA = -Math.PI - Math.atan2(cx1,cy1); /* Swap X-Y and re-orient to -Y */
xA = +r1*Math.sin(thetaA);
yA = -r1*Math.cos(thetaA);
ixA = cx1 - xA;
iyA = cy1 - yA;
thetaL = Math.atan(cx2/cy2);
xL = -r2*Math.sin(thetaL);
yL = -r2*Math.cos(thetaL);
ixL = cx2 - xL;
iyL = cy2 - yL;
if(d === 0 && r1 === r2) {
/* infinite solutions. circles are overlapping */
return [[ixA,iyA], [ixL,iyL]];
}
if (d < Math.abs(r1 - r2)) {
/* no solution. one circle is contained in the other */
return [[ixA,iyA], [ixL,iyL]];
}
/* 'point 2' is the point where the line through the circle intersection points crosses the line between the circle centers. */
/* Determine the distance from point 0 to point 2. */
a = ((r1*r1) - (r2*r2) + (d*d)) / (2.0 * d);
/* Determine the coordinates of point 2. */
x2 = cx1 + (dx * a/d);
y2 = cy1 + (dy * a/d);
/* Determine the distance from point 2 to either of the intersection points. */
h2 = r1*r1 - a*a;
h = lessThanEpsilon(h2) ? 0 : Math.sqrt(h2);
/* Now determine the offsets of the intersection points from point 2. */
rx = -dy * (h/d);
ry = +dx * (h/d);
/* Determine the absolute intersection points. */
var xi = x2 + rx, yi = y2 + ry;
var xi_prime = x2 - rx, yi_prime = y2 - ry;
return [[xi, yi], [xi_prime, yi_prime]];
}
function plot() {
var cases = [
{a: 1.1, l: 1.9, f: 0.3073},
{a: 1.0, l: 1.7, f: 0.3229}
];
var testCase = 1;
var magA = cases[testCase].a, magL = cases[testCase].l;
var maxS = Math.sqrt(magA*magA+magL*magL), magS1 = maxS * cases[testCase].f, magS2 = maxS - magS1;
var origin = [0,0], board = JXG.JSXGraph.initBoard('jxgbox', {boundingbox: [-5.0, 5.0, 5.0, -5.0], axis: true});
var drawAs = {dashed: {dash: 3, strokeWidth: 0.5, strokeColor: '#888888'} };
board.suspendUpdate();
var leftArm = board.create('slider', [[-4.5, 3], [-1.5, 3], [0, -64, -180]]);
var leftLeg = board.create('slider', [[-4.5, 2], [-1.5, 2], [0, -12, -30]]);
var rightArm = board.create('slider', [[0.5, 3], [3.5, 3], [0, 64, 180]]);
var rightLeg = board.create('slider', [[0.5, 2], [3.5, 2], [0, 12, 30]]);
var lh = board.create('point', [
function() { return +magA * Math.sin(deg2rad(leftArm.Value())); },
function() { return -magA * Math.cos(deg2rad(leftArm.Value())); }
], {size: 3, name: 'lh'});
var LA = board.create('line', [origin, lh], {straightFirst: false, straightLast: false, lastArrow: true});
var cLS1 = board.create('circle', [function() { return [lh.X(), lh.Y()]; }, function() { return magS1; }], drawAs.dashed);
var lf = board.create('point', [
function() { return +magL * Math.sin(deg2rad(leftLeg.Value())); },
function() { return -magL * Math.cos(deg2rad(leftLeg.Value())); }
], {size: 3, name: 'lf'});
var LL = board.create('line', [origin, lf], {straightFirst: false, straightLast: false, lastArrow: true});
var cLS2 = board.create('circle', [function() { return [lf.X(), lf.Y()]; }, function() { return magS2; }], drawAs.dashed);
var lx1 = board.create('point', [
function() { return circleIntersection(cLS1, cLS2)[0][0]; },
function() { return circleIntersection(cLS1, cLS2)[0][1]; }
], {size: 3, face:'x', name: 'lx1'});
var lx2 = board.create('point', [
function() { return circleIntersection(cLS1, cLS2)[1][0]; },
function() { return circleIntersection(cLS1, cLS2)[1][1]; }
], {size: 3, face:'x', name: 'lx2'});
/* Angle between lh, lx1 shall be between 0 and -180 */
var angleLAJ = board.create('text', [-3.7, 0.5, function(){ return angleBetween(lh, lx1).toFixed(2); }]);
/* Angle between lf, lx1 shall be between 0 and 180 */
var angleLLJ = board.create('text', [-2.7, 0.5, function(){ return angleBetween(lf, lx1).toFixed(2); }]);
var rh = board.create('point', [
function() { return +magA * Math.sin(deg2rad(rightArm.Value())); },
function() { return -magA * Math.cos(deg2rad(rightArm.Value())); }
], {size: 3, name: 'rh'});
var RA = board.create('line', [origin, rh], {straightFirst: false, straightLast: false, lastArrow: true});
var cRS1 = board.create('circle', [function() { return [rh.X(), rh.Y()]; }, function() { return magS1; }], drawAs.dashed);
var rf = board.create('point', [
function() { return +magL * Math.sin(deg2rad(rightLeg.Value())); },
function() { return -magL * Math.cos(deg2rad(rightLeg.Value())); }
], {size: 3, name: 'rf'});
var RL = board.create('line', [origin, rf], {straightFirst: false, straightLast: false, lastArrow: true});
var cRS2 = board.create('circle', [function() { return [rf.X(), rf.Y()]; }, function() { return magS2; }], drawAs.dashed);
var rx1 = board.create('point', [
function() { return circleIntersection(cRS1, cRS2)[1][0]; },
function() { return circleIntersection(cRS1, cRS2)[1][1]; }
], {size: 3, face:'x', name: 'rx1'});
var rx2 = board.create('point', [
function() { return circleIntersection(cRS1, cRS2)[0][0]; },
function() { return circleIntersection(cRS1, cRS2)[0][1]; }
], {size: 3, face:'x', name: 'rx2'});
var angleRAJ = board.create('text', [+1.3, 0.5, function(){ return angleBetween(rh, rx1).toFixed(2); }]);
var angleRLJ = board.create('text', [+2.3, 0.5, function(){ return angleBetween(rf, rx1).toFixed(2); }]);
board.unsuspendUpdate();
}
plot();<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="//cdnjs.cloudflare.com/ajax/libs/jsxgraph/0.99.7/jsxgraph.css" />
<link rel="stylesheet" href="style.css">
<script type="text/javascript" charset="UTF-8" src="//cdnjs.cloudflare.com/ajax/libs/jsxgraph/0.99.7/jsxgraphcore.js"></script>
</head>
<body>
<div id="jxgbox" class="jxgbox" style="width:580px; height:580px;"></div>
</body>
</html>
回答 1
Stack Overflow用户
回答已采纳
发布于 2018-05-31 15:28:47
根据你的草图，E和G的坐标是：
E = (0, -magOE)
G = magOG * ( -sin(gamma), -cos(gamma) )然后，计算H的位置是一个trilateration问题。实际上，这只是双边测量，因为你错过了第三个距离。因此，您将获得H的两个可能的位置。
首先，让我们定义一个新的坐标系，其中E位于原点，G位于x轴。在我们的原始坐标系中的x轴方向是：
x = (G - E) / ||G - E||Y轴是：
y = ( x.y, -x.x )E和G在这个新坐标系中的坐标是：
E* = (0, 0)
G* = (0, ||G - E||)现在，我们可以很容易地在这个坐标系中找到H的坐标，直到前面提到的模糊性。我将缩写||G - E|| = d，就像维基百科文章中使用的符号一样：
H.x* = (magGH * magGH - magEH * magEH + d * d) / (2 * d)
H.y* = +- sqrt(magGH * magGH - H.x* * H.x*)因此，我们对H.y有两种解决方案，一种是积极的，另一种是消极的。
最后，我们只需要将H转换回原来的坐标系：
H = x * H.x* + y * H.y* - (0, magOE)给定H的坐标，计算角度非常简单：
alpha = arccos((H.x - G.x) / ||H - G||)
beta = arccos((H.y - E.y) / ||H - E||)示例
从您的示例中获取值
magOG = 3
magOE = 4
magGH = 3
magEH = 2
angleGamma = 82.74°我们首先得到：
E = (0, -4)
G = 3 * ( -sin(82.74°), -cos(82.74°) )
= (-2.976, -0.379)我们的坐标系：
X= (-0.635，0.773) y=( 0.773，0.635)
在此坐标系中：
E* = (0, 0)
G* = (0, 4.687)那么，H在我们的辅助坐标系中的坐标是：
H* = (2.877, +- 0.851)我将只关注H*.y的正值，因为这是您在草图中标记的点。
变换回原始坐标系：
H = (-1.169, -1.237)最后计算角度：
alpha = 25.41°
beta = 22.94°值的细微差异可能是由舍入误差引起的(在我的计算中或在您的计算中)。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50608788
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