关于叉乘符号与向量的角度方向关系,请参考《算法导论》,我只给出结论:
p1 * p2 = x1y2 - x2 y1 = -p2 * p1
If p1 * p2 is positive, then p1 is clockwise from p2 with respect to the origin (0, 0); if this cross product is negative, then p1 is counterclockwise from p2.
另外考虑的是共线(collinear )的问题, arcsine很难处理这个问题, 不过arecosine却能够明确的区分0和pi,因此作为特殊情况提前得出结论。
ps.因为主要是openGL要用, 所以返回的是角度值
/************************************************************************/
/* author : Navy@hust
* file : angle.c
* date : 5/12/2007
* desc : vector rotate angle calculation
/************************************************************************/
/* author : Navy@hust
* file : angle.c
* date : 5/12/2007
* desc : vector rotate angle calculation
/************************************************************************/
#i nclude <stdio.h>
#i nclude <math.h>
double getRotateAngle(double x1, double y1, double x2, double y2);
int main(int argc, char **argv)
{
double x1, x2, y1, y2;
double dist, dot, degree, angle;
freopen("angle.in", "r", stdin);
while(scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2) == 4) {
printf("the rotate angle from p1 to p2 is %.3lf\n",
getRotateAngle(x1, y1, x2, y2));
}
}
/*
* 两个向量之间的旋转角
* 首先明确几个数学概念:
* 1. 极轴沿逆时针转动的方向是正方向
* 2. 两个向量之间的夹角theta, 是指(A^B)/(|A|*|B|) = cos(theta),0<=theta<=180 度, 而且没有方向之分
* 3. 两个向量的旋转角,是指从向量p1开始,逆时针旋转,转到向量p2时,所转过的角度, 范围是 0 ~ 360度
* 计算向量p1到p2的旋转角,算法如下:
* 首先通过点乘和arccosine的得到两个向量之间的夹角
* 然后判断通过差乘来判断两个向量之间的位置关系
* 如果p2在p1的顺时针方向, 返回arccose的角度值, 范围0 ~ 180.0(根据右手定理,可以构成正的面积)
* 否则返回 360.0 - arecose的值, 返回180到360(根据右手定理,面积为负)
*/
double getRotateAngle(double x1, double y1, double x2, double y2)
{
const double epsilon = 1.0e-6;
const double nyPI = acos(-1.0);
double dist, dot, degree, angle;
// normalize
dist = sqrt( x1 * x1 + y1 * y1 );
x1 /= dist;
y1 /= dist;
dist = sqrt( x2 * x2 + y2 * y2 );
x2 /= dist;
y2 /= dist;
{
double x1, x2, y1, y2;
double dist, dot, degree, angle;
freopen("angle.in", "r", stdin);
while(scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2) == 4) {
printf("the rotate angle from p1 to p2 is %.3lf\n",
getRotateAngle(x1, y1, x2, y2));
}
}
/*
* 两个向量之间的旋转角
* 首先明确几个数学概念:
* 1. 极轴沿逆时针转动的方向是正方向
* 2. 两个向量之间的夹角theta, 是指(A^B)/(|A|*|B|) = cos(theta),0<=theta<=180 度, 而且没有方向之分
* 3. 两个向量的旋转角,是指从向量p1开始,逆时针旋转,转到向量p2时,所转过的角度, 范围是 0 ~ 360度
* 计算向量p1到p2的旋转角,算法如下:
* 首先通过点乘和arccosine的得到两个向量之间的夹角
* 然后判断通过差乘来判断两个向量之间的位置关系
* 如果p2在p1的顺时针方向, 返回arccose的角度值, 范围0 ~ 180.0(根据右手定理,可以构成正的面积)
* 否则返回 360.0 - arecose的值, 返回180到360(根据右手定理,面积为负)
*/
double getRotateAngle(double x1, double y1, double x2, double y2)
{
const double epsilon = 1.0e-6;
const double nyPI = acos(-1.0);
double dist, dot, degree, angle;
// normalize
dist = sqrt( x1 * x1 + y1 * y1 );
x1 /= dist;
y1 /= dist;
dist = sqrt( x2 * x2 + y2 * y2 );
x2 /= dist;
y2 /= dist;
// dot product
dot = x1 * x2 + y1 * y2;
dot = x1 * x2 + y1 * y2;
if ( fabs(dot-1.0) <= epsilon )
angle = 0.0;
else if ( fabs(dot+1.0) <= epsilon )
angle = nyPI;
else {
double cross;
angle = acos(dot);
//cross product
cross = x1 * y2 - x2 * y1;
// vector p2 is clockwise from vector p1
// with respect to the origin (0.0)
if (cross < 0 ) {
angle = 2 * nyPI - angle;
}
}
degree = angle * 180.0 / nyPI;
return degree;
}
angle = 0.0;
else if ( fabs(dot+1.0) <= epsilon )
angle = nyPI;
else {
double cross;
angle = acos(dot);
//cross product
cross = x1 * y2 - x2 * y1;
// vector p2 is clockwise from vector p1
// with respect to the origin (0.0)
if (cross < 0 ) {
angle = 2 * nyPI - angle;
}
}
degree = angle * 180.0 / nyPI;
return degree;
}