说我有两个向量:
V1 = {x:3.296372727813439,y:-14.497928014719344,z:12.004105246875968}
V2 = {x:2.3652551657790695,y:-16.732085083053185,z:8.945905454164146}
如何确定v1需要旋转的角度才能直接观察v2?
说英语:说我确切地知道我在太空中的确切位置,以及另一个人在太空其他地方的确切位置......数学上,我怎样才能弄清楚用手指指向它们的角度?
Here's what my axis looks like
我当前(不正确)公式
v2.x -= v1.x; // move v2 (vector 2) relative to the new origin
v2.y -= v1.y;
v2.z -= v1.z;
v1.x = 0; // set v1 (vector 1) as the origin
v1.y = 0;
v1.z = 0;
var r = Math.sqrt(Math.pow(v2.x,2) + Math.pow(v2.y,2) + Math.pow(v2.z,2));
var θ = Math.acos((Math.pow(v2.x,2) + Math.pow(v2.z,2))/(r*Math.sqrt(Math.pow(v2.x,2) + Math.pow(v2.z,2))));
var ϕ = Math.acos(v2.x/Math.sqrt(Math.pow(v2.x,2) + Math.pow(v2.z,2)));
然后我用theta和phi旋转v1。
v1.rotation.y = θ;
v2.rotation.x = ϕ;
但是这给了我错误的轮换。
θ= 0.6099683401012933
φ= 1.8663452274936656
但是如果我使用THREE.js并使用lookAt函数,它会反而吐出这些旋转,它可以很好地工作:
y/θ:-0.24106818240525682
x/φ:2.5106584861123644
提前感谢所有帮助!我不能在最终结果中使用THREE.js,我正在尝试使用纯粹的香草JS,以便将其移植到另一种语言。
答案 0 :(得分:2)
正确的公式是:
var dx = v2.x-v1.x; //-0.93
var dy = v2.y-v1.y; //-31.22
var dz = v2.z-v1.z;
var rxy = Math.sqrt( Math.pow(dx,2) + Math.pow(dy,2) );
var lambda = Math.atan(dy/dx);
var phi = Math.atan(dz/rxy)
phi
和lambda
的上述公式需要根据您的向量所在的四分之一进行调整。我为您提供了便利:
//if you do the calculations in degrees, you need to add 180 instead of PI
if (dx < 0) phi = phi + Math.PI;
if (dz < 0) lambda = -1 * lambda;
答案 1 :(得分:0)
使用矩阵怎么样?我认为v1是你的观点,并期待v2。矩阵是显示方向的好方法。欧拉角是对方向的另一种解释。
我的想法是构建从对象空间到世界空间的转换矩阵,您想要做的事情可以分三步进行翻译:
(0,0,0)
世界空间与物体空间相同。 v1'(0,0,0)
。v1(3.296372727813439,-14.497928014719344,12.004105246875968)
,物体空间与世界空间有偏移,但物体空间轴与世界空间轴平行,相机旋转仍为(0,0,0)。v2
,如您所见,相机旋转会发生变化。如果我可以构建转换矩阵表示上面的所有操作,我可以获得方向。
我们使用基轴获得旋转矩阵。
您可能需要设置相机向上矢量。默认值为(0,1,0)
。在对象空间中,基础 z 轴可以由v1-v2
计算。
z = (v1.x-v2.x,v1.y-v2.y,v1.z-v2.z).normalize()
基础 x 向量:我们知道基向量是垂直于 z-up 平面的向量,我们通过叉积得到 x 向量up和 z 。
x = up.crossproduct(z)
基础 y 向量, y 垂直于 z-x 平面。
y = z.product(x)
我们可以将旋转矩阵构建为3 x 3矩阵:
然后,我们终于得到了转换矩阵:
我们可以使用矩阵表示相机方向。如果你需要欧拉角或四元数。有一些方法在他们之间转换。你可以在这本书中找到:3D Math Primer for Graphics and Game Developmen
Three.js实现与我的方式相同的LookAt()
功能。
这是three.js源代码,我添加了一些注释:
function lookAt( eye, target, up ) //eye : your camera position; target : which point you want to look at; up : camera up vector
{
if ( x === undefined ) {
x = new Vector3();
y = new Vector3();
z = new Vector3();
}
var te = this.elements; //this.elements is a 4 x 4 matrix stored in a list.
z.subVectors( eye, target ).normalize(); // set z vector with the direction from your camera to target point.
if ( z.lengthSq() === 0 ) {
z.z = 1;
}
x.crossVectors( up, z ).normalize(); // set the x vector by cross product up and z vector, you know cross product would get a //vector which perpendicular with these two vectors.
if ( x.lengthSq() === 0 ) {
z.z += 0.0001; // if z is ZERO vector, then, make a little addition to z.z
x.crossVectors( up, z ).normalize();
}
y.crossVectors( z, x ); // set y by cross product z and x.
// using basic axises to set the matrix.
te[ 0 ] = x.x; te[ 4 ] = y.x; te[ 8 ] = z.x;
te[ 1 ] = x.y; te[ 5 ] = y.y; te[ 9 ] = z.y;
te[ 2 ] = x.z; te[ 6 ] = y.z; te[ 10 ] = z.z;
return this;
};
// now you get the transformation matrix, you can set the rotation or orientation with this matrix.
您可以使用像three.js这样的列表实现矩阵。
我还有另一个想法 - 球面极坐标系。球面坐标总是被标记为(r,θ,Φ),θ是前进角,Φ是俯仰角。你需要做的是将v1和v2笛卡尔坐标转换为球坐标。因为球形的第二和第三个元素是角度,我们可以计算v1和v2之间的角位移。然后,将此位移作为相机旋转的补充。
这是我的代码(假设您的相机位于世界原点(0,0,0)):
//convert v1 and v2 Cartesian coordinates to Spherical coordinates;
var radiusV1 = Math.sqrt( Math.pow(v1.x) + Math.pow(v1.y) + Math.pow(v1.z));
var headingV1 = Math.atan2(v1.x , v1.z);
var pitchV1 = Math.asin(-(v1.y) / radiusV1);
var radiusV2 = Math.sqrt( Math.pow(v2.x) + Math.pow(v2.y) + Math.pow(v2.z));
var headingV2 = Math.atan2(v2.x , v2.z);
var pitchV2 = Math.asin(-(v2.y) / radiusV2);
//calculate angular displacement.
var displacementHeading = headingV2 - headingV1;
var displacementPitch = pitchV2 - pitchV1;
//make this displacement as an addition to camera rotation.
camera.rotation.x += displacementPitch;
camera.rotation.y += displacementHeading;
顺便说一下,3D数学非常有用且值得学习,我参考的所有公式或概念都可以在书中找到。
希望它可以帮到你。
答案 2 :(得分:0)
x /φ围绕x轴旋转,因此它等于y,z之间的角度,对于y /θ,我们必须找到x,z之间的角度。
V1 = { x: 3.296372727813439, y: -14.497928014719344, z: 12.004105246875968 }
V2 = { x: 2.3652551657790695, y: -16.732085083053185, z: 8.945905454164146 }
var v={dx:V2.x-V1.x, dy:V2.y-V1.y, dz:V2.z-V1.z}
testVector(v);
function testVector(vec){
console.log();
var angles=calcAngles(vec);
console.log("phi:"+angles.phi+" theta:"+angles.theta);
}
function calcAngles(vec){
return {
theta:(Math.PI/2)+Math.atan2(vec.dz, vec.dx),
phi:(3*Math.PI/2)+Math.atan2(vec.dz, vec.dy)
};
}
答案 3 :(得分:0)
我从最新版本的THREE.js(r84)中提取了相关代码。 我认为这是获得结果的最佳方式。
// Unless otherwise noted by comments, all functions originate from the latest version of THREE.js (r84)
// https://github.com/mrdoob/three.js/tree/master
// THREE.js is licensed under MIT (Copyright © 2010-2017 three.js authors)
//
// Some functions have been changed by K Scandrett to work within this setting,
// but not the calculations.
// Any mistakes are considered mine and not the authors of THREE.js.
// I provide no guarantees that I haven't created any bugs in reworking the original code
// so use at your own risk. Enjoy the pizza.
var v1 = {x: 3.296372727813439, y: -14.497928014719344, z: 12.004105246875968};
var v2 = {x: 2.3652551657790695, y: -16.732085083053185,z: 8.945905454164146};
var startVec = {x: v1.x, y: v1.y, z: v1.z, w: 0};
var endVec = {x: v2.x, y: v2.y, z: v2.z, w: 0};
var upVec = {x: 0, y: 1, z: 0}; // y up
var quat = lookAt(startVec, endVec, upVec);
var angles = eulerSetFromQuaternion(quat);
console.log(angles.x + " " + angles.y + " " + angles.z);
/* KS function */
function magnitude(v) {
return Math.sqrt(v.x * v.x + v.y * v.y + v.z * v.z);
}
/* KS function */
function normalize(v) {
var mag = magnitude(v);
return {
x: v.x / mag,
y: v.y / mag,
z: v.z / mag
};
}
function subVectors(a, b) {
return {
x: a.x - b.x,
y: a.y - b.y,
z: a.z - b.z
};
}
function crossVectors(a, b) {
var ax = a.x,
ay = a.y,
az = a.z;
var bx = b.x,
by = b.y,
bz = b.z;
return {
x: ay * bz - az * by,
y: az * bx - ax * bz,
z: ax * by - ay * bx
};
}
function lengthSq(v) {
return v.x * v.x + v.y * v.y + v.z * v.z;
}
function makeRotationFromQuaternion(q) {
var matrix = new Float32Array([
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1
]);
var te = matrix;
var x = q.x,
y = q.y,
z = q.z,
w = q.w;
var x2 = x + x,
y2 = y + y,
z2 = z + z;
var xx = x * x2,
xy = x * y2,
xz = x * z2;
var yy = y * y2,
yz = y * z2,
zz = z * z2;
var wx = w * x2,
wy = w * y2,
wz = w * z2;
te[0] = 1 - (yy + zz);
te[4] = xy - wz;
te[8] = xz + wy;
te[1] = xy + wz;
te[5] = 1 - (xx + zz);
te[9] = yz - wx;
te[2] = xz - wy;
te[6] = yz + wx;
te[10] = 1 - (xx + yy);
// last column
te[3] = 0;
te[7] = 0;
te[11] = 0;
// bottom row
te[12] = 0;
te[13] = 0;
te[14] = 0;
te[15] = 1;
return te;
}
function RotationMatrix(m) {
// http://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToQuaternion/index.htm
// assumes the upper 3x3 of m is a pure rotation matrix (i.e, unscaled)
var _w, _x, _y, _z;
var te = m,
m11 = te[0],
m12 = te[4],
m13 = te[8],
m21 = te[1],
m22 = te[5],
m23 = te[9],
m31 = te[2],
m32 = te[6],
m33 = te[10],
trace = m11 + m22 + m33,
s;
if (trace > 0) {
s = 0.5 / Math.sqrt(trace + 1.0);
_w = 0.25 / s;
_x = (m32 - m23) * s;
_y = (m13 - m31) * s;
_z = (m21 - m12) * s;
} else if (m11 > m22 && m11 > m33) {
s = 2.0 * Math.sqrt(1.0 + m11 - m22 - m33);
_w = (m32 - m23) / s;
_x = 0.25 * s;
_y = (m12 + m21) / s;
_z = (m13 + m31) / s;
} else if (m22 > m33) {
s = 2.0 * Math.sqrt(1.0 + m22 - m11 - m33);
_w = (m13 - m31) / s;
_x = (m12 + m21) / s;
_y = 0.25 * s;
_z = (m23 + m32) / s;
} else {
s = 2.0 * Math.sqrt(1.0 + m33 - m11 - m22);
_w = (m21 - m12) / s;
_x = (m13 + m31) / s;
_y = (m23 + m32) / s;
_z = 0.25 * s;
}
return {
w: _w,
x: _x,
y: _y,
z: _z
};
}
function eulerSetFromQuaternion(q, order, update) {
var matrix;
matrix = makeRotationFromQuaternion(q);
return eulerSetFromRotationMatrix(matrix, order);
}
function eulerSetFromRotationMatrix(m, order, update) {
var _x, _y, _z;
var clamp = function(value, min, max) {
return Math.max(min, Math.min(max, value));
};
// assumes the upper 3x3 of m is a pure rotation matrix (i.e, unscaled)
var te = m;
var m11 = te[0],
m12 = te[4],
m13 = te[8];
var m21 = te[1],
m22 = te[5],
m23 = te[9];
var m31 = te[2],
m32 = te[6],
m33 = te[10];
//order = order || this._order;
order = order || 'XYZ'; // KS added. Other code sets the rotation order default
if (order === 'XYZ') {
_y = Math.asin(clamp(m13, -1, 1));
if (Math.abs(m13) < 0.99999) {
_x = Math.atan2(-m23, m33);
_z = Math.atan2(-m12, m11);
} else {
_x = Math.atan2(m32, m22);
_z = 0;
}
} else if (order === 'YXZ') {
_x = Math.asin(-clamp(m23, -1, 1));
if (Math.abs(m23) < 0.99999) {
_y = Math.atan2(m13, m33);
_z = Math.atan2(m21, m22);
} else {
_y = Math.atan2(-m31, m11);
_z = 0;
}
} else if (order === 'ZXY') {
_x = Math.asin(clamp(m32, -1, 1));
if (Math.abs(m32) < 0.99999) {
_y = Math.atan2(-m31, m33);
_z = Math.atan2(-m12, m22);
} else {
_y = 0;
_z = Math.atan2(m21, m11);
}
} else if (order === 'ZYX') {
_y = Math.asin(-clamp(m31, -1, 1));
if (Math.abs(m31) < 0.99999) {
_x = Math.atan2(m32, m33);
_z = Math.atan2(m21, m11);
} else {
_x = 0;
_z = Math.atan2(-m12, m22);
}
} else if (order === 'YZX') {
_z = Math.asin(clamp(m21, -1, 1));
if (Math.abs(m21) < 0.99999) {
_x = Math.atan2(-m23, m22);
_y = Math.atan2(-m31, m11);
} else {
_x = 0;
_y = Math.atan2(m13, m33);
}
} else if (order === 'XZY') {
_z = Math.asin(-clamp(m12, -1, 1));
if (Math.abs(m12) < 0.99999) {
_x = Math.atan2(m32, m22);
_y = Math.atan2(m13, m11);
} else {
_x = Math.atan2(-m23, m33);
_y = 0;
}
} else {
console.warn('THREE.Euler: .setFromRotationMatrix() given unsupported order: ' + order);
}
//_order = order;
//if ( update !== false ) this.onChangeCallback();
return {
x: _x,
y: _y,
z: _z
};
}
function setFromQuaternion(q, order, update) {
var matrix = makeRotationFromQuaternion(q);
return setFromRotationMatrix(matrix, order, update);
}
function setFromRotationMatrix(m) {
// http://www.euclideanspace.com/maths/geometry/rotations/conversions/matrixToQuaternion/index.htm
// assumes the upper 3x3 of m is a pure rotation matrix (i.e, unscaled)
var _w, _x, _y, _z;
var te = m,
m11 = te[0],
m12 = te[4],
m13 = te[8],
m21 = te[1],
m22 = te[5],
m23 = te[9],
m31 = te[2],
m32 = te[6],
m33 = te[10],
trace = m11 + m22 + m33,
s;
if (trace > 0) {
s = 0.5 / Math.sqrt(trace + 1.0);
_w = 0.25 / s;
_x = (m32 - m23) * s;
_y = (m13 - m31) * s;
_z = (m21 - m12) * s;
} else if (m11 > m22 && m11 > m33) {
s = 2.0 * Math.sqrt(1.0 + m11 - m22 - m33);
_w = (m32 - m23) / s;
_x = 0.25 * s;
_y = (m12 + m21) / s;
_z = (m13 + m31) / s;
} else if (m22 > m33) {
s = 2.0 * Math.sqrt(1.0 + m22 - m11 - m33);
_w = (m13 - m31) / s;
_x = (m12 + m21) / s;
_y = 0.25 * s;
_z = (m23 + m32) / s;
} else {
s = 2.0 * Math.sqrt(1.0 + m33 - m11 - m22);
_w = (m21 - m12) / s;
_x = (m13 + m31) / s;
_y = (m23 + m32) / s;
_z = 0.25 * s;
}
return {
w: _w,
x: _x,
y: _y,
z: _z
};
}
function lookAt(eye, target, up) {
// This routine does not support objects with rotated and/or translated parent(s)
var m1 = lookAt2(target, eye, up);
return setFromRotationMatrix(m1);
}
function lookAt2(eye, target, up) {
var elements = new Float32Array([
1, 0, 0, 0,
0, 1, 0, 0,
0, 0, 1, 0,
0, 0, 0, 1
]);
var x = {
x: 0,
y: 0,
z: 0
};
var y = {
x: 0,
y: 0,
z: 0
};
var z = {
x: 0,
y: 0,
z: 0
};
var te = elements;
z = subVectors(eye, target);
z = normalize(z);
if (lengthSq(z) === 0) {
z.z = 1;
}
x = crossVectors(up, z);
x = normalize(x);
if (lengthSq(x) === 0) {
z.z += 0.0001;
x = crossVectors(up, z);
x = normalize(x);
}
y = crossVectors(z, x);
te[0] = x.x;
te[4] = y.x;
te[8] = z.x;
te[1] = x.y;
te[5] = y.y;
te[9] = z.y;
te[2] = x.z;
te[6] = y.z;
te[10] = z.z;
return te;
}
function lookatOld(vecstart, vecEnd, vecUp) {
var temp = new THREE.Matrix4();
temp.lookAt(vecEnd, vecstart, vecUp);
var m00 = temp.elements[0],
m10 = temp.elements[1],
m20 = temp.elements[2],
m01 = temp.elements[4],
m11 = temp.elements[5],
m21 = temp.elements[6],
m02 = temp.elements[8],
m12 = temp.elements[9],
m22 = temp.elements[10];
var t = m00 + m11 + m22,
s, x, y, z, w;
if (t > 0) {
s = Math.sqrt(t + 1) * 2;
w = 0.25 * s;
x = (m21 - m12) / s;
y = (m02 - m20) / s;
z = (m10 - m01) / s;
} else if ((m00 > m11) && (m00 > m22)) {
s = Math.sqrt(1.0 + m00 - m11 - m22) * 2;
x = s * 0.25;
y = (m10 + m01) / s;
z = (m02 + m20) / s;
w = (m21 - m12) / s;
} else if (m11 > m22) {
s = Math.sqrt(1.0 + m11 - m00 - m22) * 2;
y = s * 0.25;
x = (m10 + m01) / s;
z = (m21 + m12) / s;
w = (m02 - m20) / s;
} else {
s = Math.sqrt(1.0 + m22 - m00 - m11) * 2;
z = s * 0.25;
x = (m02 + m20) / s;
y = (m21 + m12) / s;
w = (m10 - m01) / s;
}
var rotation = new THREE.Quaternion(x, y, z, w);
rotation.normalize();
return rotation;
}
以下是Plunker中的相同代码:http://plnkr.co/edit/vgNko1fJu9eYYCnJbYVo?p=preview
答案 4 :(得分:0)
你的问题的确切/字面答案将是一个糟糕/不道德的答案。不要试图使用欧拉角。欧拉坐标系用于协调。做方向/旋转不是一个好的系统。虽然容易为人类阅读,但很容易Gimbal lock - 会产生不正确的结果。
有两种常见的方向系统:变换矩阵和四元数。 three.js lookAt()
使用四元数,Crag.Li的answer使用变换矩阵。
我觉得有必要强调这一点,因为我曾经低估了3D转型,并试图以“简单的方式”解决它,浪费了将近一个月做傻瓜的工作。 3D转换很难。没有快速,肮脏的方式,你只能以正确的方式做到这一点。拿一本书(3D数学入门书是一本很好的书),如果你真的想要这本书,可以花时间学习数学。