3d 求两个球面的交点
考虑以下问题-我有两个长度为L0和L1的链接。P0是第一个链接开始的点，P1是我希望第二个链接的末端在三维空间中的点。我应该写一个函数，它应该把这些3-D点（P0和P1）作为输入，并且应该找到把第二个链接的端点放在P1的链接的所有配置 我对如何进行的理解是——每个链接L0和L1将在其自身周围创建一个球体S0和S1。我应该找出这两个球体（将是一个圆）的交点，并打印出该圆圆周上的所有点 我在网上看到了gmatt的第一个回复，但由于图像没有显示，我无法正确理解。我还看到了一个计算交叉点的公式 我可以用mathworld上给出的方法求交点半径。我还可以找到圆的中心，然后用圆的参数方程找到点。我唯一的疑问是，对于上面提到的点P0和P1，这种方法是否有效3d 求两个球面的交点,3d,geometry,intersection,3d,Geometry,Intersection,考虑以下问题-我有两个长度为L0和L1的链接。P0是第一个链接开始的点，P1是我希望第二个链接的末端在三维空间中的点。我应该写一个函数，它应该把这些3-D点（P0和P1）作为输入，并且应该找到把第二个链接的端点放在P1的链接的所有配置 我对如何进行的理解是——每个链接L0和L1将在其自身周围创建一个球体S0和S1。我应该找出这两个球体（将是一个圆）的交点，并打印出该圆圆周上的所有点 我在网上看到了gmatt的第一个回复，但由于图像没有显示，我无法正确理解。我还看到了一个计算交叉点的公式 我可以用
请评论并让我知道您的想法。两个球体的方程式可以写成：
<X-P0,X-P0> - L0^2 = 0 (Eq0)
<X-P1,X-P1> - L1^2 = 0 (Eq1)
rdist(C,P0)^2+r^2=L0^2, or equivalently
dist(C,P1)^2+r^2=L1^2
如果球体没有交点，它可能没有解。我附上解的代码。P0被认为是一个肩部点，P1被认为是位于空间中的点（考虑到我上臂和前臂的长度，我应该抓住这个点）。请评论一下，让我知道你的想法
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
struct point {
double x, y, z;
};
/*
* Used to represent a vector in
* Xi + Yj + Zk format
*/
struct vector {
double i, j, k;
};
/*
* Used to represent a plane in
* Ax + By + Cz + D = 0 format
*/
struct plane {
double A, B, C, D;
};
/*
* Represents the final assembly of the configuration. When two spheres
* intersect they form a circle whose center is stored at "center" and radius is
* stored at "radius". The circle also has a support which is defined by a plane
* called as "radical plane" and its equation is stored at "p".
*/
struct configuration {
struct point center;
double radius;
struct plane p;
};
/*
* Conversion functions between vector and point
*/
struct vector get_vector_from_point(struct point p) {
static struct vector v;
v.i = p.x;
v.j = p.y;
v.k = p.z;
return v;
}
struct point get_point_from_vector(struct vector v) {
static struct point p;
p.x = v.i;
p.y = v.j;
p.z = v.k;
return p;
}
int check_if_same_points(struct point p1, struct point p2) {
return ((p1.x == p2.x) && (p1.y == p2.y) && (p1.z == p2.z));
}
/*
* Distance formula
*/
double distance(struct point p0, struct point p1) {
return sqrt(pow((fabs(p1.z - p0.z)), 2) + pow((fabs(p1.y - p0.y)), 2) + pow((fabs(p1.x - p0.x)), 2));
}
/*
* Customized utility functions used for vector mathematics
*/
double dot_product(struct vector p0, struct vector p1) {
return (p0.i * p1.i + p0.j * p1.j + p0.k * p1.k);
}
struct vector scale_vector(double scale, struct vector v) {
static struct vector scaled_vec;
scaled_vec.i = scale * v.i;
scaled_vec.j = scale * v.j;
scaled_vec.k = scale * v.k;
return scaled_vec;
}
struct vector add_vectors(struct vector v1, struct vector v2) {
static struct vector v;
v.i = v1.i + v2.i;
v.j = v1.j + v2.j;
v.k = v1.k + v2.k;
return v;
}
struct vector subtract_vectors(struct vector v1, struct vector v2) {
static struct vector v;
v.i = v1.i - v2.i;
v.j = v1.j - v2.j;
v.k = v1.k - v2.k;
return v;
}
/*
* Takes the given assembly of points and links. Returns object of configuration
* structure with necessary information. The center and radius from the returned
* structure can be used find possible locations of elbow.
* Client can use following parametric equation of circle in 3-D
* X(t) = C + r (cos(t) * U + sin(t) * V)
* where 0 <= t < 2 * pie, C is the center, r is the radius, U and V are unit
* normals to the plane such that if N is a unit length plane normal then
* {U,V,N} are mutually orthogonal.
*/
struct configuration return_config(struct point p0, double l0, struct point p1, double l1) {
struct vector p0_v = get_vector_from_point(p0);
struct vector p1_v = get_vector_from_point(p1);
double dot_prd_p0 = dot_product(p0_v, p0_v);
double dot_prd_p1 = dot_product(p1_v, p1_v);
struct vector sub_vec = subtract_vectors(p1_v, p0_v);
double D = ((l0 * l0) - (l1 * l1) + dot_prd_p1 - dot_prd_p0) / 2.0f;
static struct plane p;
p.A = sub_vec.i; p.B = sub_vec.j; p.C = sub_vec.k; p.D = D;
static struct configuration c;
/*
* Special case when object point and shoulder point are same.
*/
if(check_if_same_points(p0, p1)) {
printf("object and shoulder are at same location \n");
c.center.x = p0.x; c.center.y = p0.y; c.center.z = p0.z;
c.radius = l0;
c.p.A = c.p.B = c.p.C = c.p.D = 0.0f;
return c;
}
double a0 = (1.0f - (((l0 * l0) - (l1 * l1)) / (distance(p0, p1) * distance(p0, p1)))) / 2.0f;
struct vector lhs = scale_vector(a0,p0_v);
struct vector rhs = scale_vector(1.0f - a0, p1_v);
struct vector ans = add_vectors(lhs, rhs);
struct point center = get_point_from_vector(ans);
double radius = sqrt((l0 * l0) - (distance(center, p0) * distance(center, p0)));
c.center.x = center.x; c.center.y = center.y; c.center.z = center.z;
c.radius = radius;
c.p.A = p.A; c.p.B = p.B; c.p.C = p.C; c.p.D = D;
return c;
}
/*
* The logic is as follows - Point P0 generates a sphere of radius L0 around it,
* P1 generates another sphere of radius L1 around it. The intersection(if any)
* will be a circle with a plane. If I can return the center and radius of that
* circle and equation of the plane, then the client can find out any possible
* location of the elbow by varying the value of theta in the parametric
* equation of the circle. Thus if the spheres meet to generate a circle then
* there will be infinite elbow positions. If it does not generate a circle and
* meet "externally" then there will be only single elbow position. Otherwise
* there will be no solutions at all.
*/
int main() {
struct point p0, p1;
p0.x = 0, p0.y = 0, p0.z = 0;
p1.x = 50, p1.y = 50, p1.z = 0;
double l0 = 50, l1 = 50;
printf("Shoulder coordinates : (%lf, %lf, %lf) \n", p0.x, p0.y, p0.z);
printf("Object coordinates: (%lf, %lf, %lf) \n", p1.x, p1.y, p1.z);
printf("link0 = %lf, link1 = %lf \n", l0, l1);
if(distance(p0, p1) > (l0 + l1)) {
printf("The given combination of the points and links cannot make a valid configuration");
return -1;
}
struct configuration c = return_config(p0, l0, p1, l1);
printf("Center = (%lf, %lf, %lf), radius = %lf \n", c.center.x, c.center.y, c.center.z, c.radius);
printf("Equation of the radical plane = %lfA + (%lf)B + (%lf)C + %lf = 0 \n", c.p.A, c.p.B, c.p.C, c.p.D);
return 0;
}
#包括
#包括
#包括
结构点{
双x，y，z；
};
/*
*用于表示中的向量
*席+YJ+ZK格式
*/
结构向量{
双i，j，k；
};
/*
*用于表示平面中的平面
*Ax+By+Cz+D=0格式
*/
结构平面{
双A、B、C、D；
};
/*
*表示配置的最终组装。当两个球体
*相交它们形成一个圆心存储在“中心”，半径为
*存储在“半径”处。圆还有一个由平面定义的支撑
*称为“根平面”，其方程存储在“p”处。
*/
结构配置{
结构点中心；
双半径；
结构平面p；
};
/*
*向量与点之间的转换函数
*/
结构向量从点（结构点p）获取点向量{
静态结构向量v；
v、 i=p.x；
v、 j=p.y；
v、 k=p.z；
返回v；
}
结构点从向量中获取点（结构向量v）{
静态结构点p；
p、 x=v.i；
p、 y=v.j；
p、 z=v.k；
返回p；
}
int检查点是否相同（结构点p1、结构点p2）{
返回（（p1.x==p2.x）和&（p1.y==p2.y）和&（p1.z==p2.z））；
}
/*
*距离公式
*/
双距离（结构点p0、结构点p1）{
返回sqrt（pow（（fabs（p1.z-p0.z）），2）+pow（（fabs（p1.y-p0.y）），2）+pow（（fabs（p1.x-p0.x）），2））；
}
/*
*用于向量数学的自定义实用函数
*/
双点积（结构向量p0，结构向量p1）{
回报率（p0.i*p1.i+p0.j*p1.j+p0.k*p1.k）；
}
结构向量比例\向量（双比例，结构向量v）{
静态结构向量缩放向量；
标度向量i=标度*v.i；
标度向量j=标度*v.j；
缩放向量k=缩放*v.k；
返回缩放向量；
}
结构向量加法向量（结构向量v1，结构向量v2）{
静态结构向量v；
v、 i=v1.i+v2.i；
v、 j=v1.j+v2.j；
v、 k=v1.k+v2.k；
返回v；
}
结构向量减去_向量（结构向量v1，结构向量v2）{
静态结构向量v；
v、 i=v1.i-v2.i；
v、 j=v1.j-v2.j；
v、 k=v1.k-v2.k；
返回v；
}
/*
*获取点和链接的给定集合。返回配置的对象
*包含必要信息的结构。返回的曲面的中心和半径
*结构可用于查找弯头的可能位置。
*客户可以使用以下三维圆的参数方程
*X（t）=C+r（cos（t）*U+sin（t）*V）
*其中0（l0+l1））{
printf（“给定的点和链接组合不能构成有效配置”）；
返回-1；
}
结构配置c=返回配置（p0、l0、p1、l1）；
printf（“中心=（%lf，%lf，%lf），半径=%lf\n”，c.Center.x，c.Center.y，c.Center.z，c.radius）；
printf（“根平面方程=%lfA+（%lf）B+（%lf）C+%lf=0\n”，C.p.A，C.p.B，C.p.C，C.p.D）；
返回0；
}
点向量点矢量半径矢量dist(C,P0)^2+r^2=L0^2, or equivalently
dist(C,P1)^2+r^2=L1^2
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
struct point {
double x, y, z;
};
/*
* Used to represent a vector in
* Xi + Yj + Zk format
*/
struct vector {
double i, j, k;
};
/*
* Used to represent a plane in
* Ax + By + Cz + D = 0 format
*/
struct plane {
double A, B, C, D;
};
/*
* Represents the final assembly of the configuration. When two spheres
* intersect they form a circle whose center is stored at "center" and radius is
* stored at "radius". The circle also has a support which is defined by a plane
* called as "radical plane" and its equation is stored at "p".
*/
struct configuration {
struct point center;
double radius;
struct plane p;
};
/*
* Conversion functions between vector and point
*/
struct vector get_vector_from_point(struct point p) {
static struct vector v;
v.i = p.x;
v.j = p.y;
v.k = p.z;
return v;
}
struct point get_point_from_vector(struct vector v) {
static struct point p;
p.x = v.i;
p.y = v.j;
p.z = v.k;
return p;
}
int check_if_same_points(struct point p1, struct point p2) {
return ((p1.x == p2.x) && (p1.y == p2.y) && (p1.z == p2.z));
}
/*
* Distance formula
*/
double distance(struct point p0, struct point p1) {
return sqrt(pow((fabs(p1.z - p0.z)), 2) + pow((fabs(p1.y - p0.y)), 2) + pow((fabs(p1.x - p0.x)), 2));
}
/*
* Customized utility functions used for vector mathematics
*/
double dot_product(struct vector p0, struct vector p1) {
return (p0.i * p1.i + p0.j * p1.j + p0.k * p1.k);
}
struct vector scale_vector(double scale, struct vector v) {
static struct vector scaled_vec;
scaled_vec.i = scale * v.i;
scaled_vec.j = scale * v.j;
scaled_vec.k = scale * v.k;
return scaled_vec;
}
struct vector add_vectors(struct vector v1, struct vector v2) {
static struct vector v;
v.i = v1.i + v2.i;
v.j = v1.j + v2.j;
v.k = v1.k + v2.k;
return v;
}
struct vector subtract_vectors(struct vector v1, struct vector v2) {
static struct vector v;
v.i = v1.i - v2.i;
v.j = v1.j - v2.j;
v.k = v1.k - v2.k;
return v;
}
/*
* Takes the given assembly of points and links. Returns object of configuration
* structure with necessary information. The center and radius from the returned
* structure can be used find possible locations of elbow.
* Client can use following parametric equation of circle in 3-D
* X(t) = C + r (cos(t) * U + sin(t) * V)
* where 0 <= t < 2 * pie, C is the center, r is the radius, U and V are unit
* normals to the plane such that if N is a unit length plane normal then
* {U,V,N} are mutually orthogonal.
*/
struct configuration return_config(struct point p0, double l0, struct point p1, double l1) {
struct vector p0_v = get_vector_from_point(p0);
struct vector p1_v = get_vector_from_point(p1);
double dot_prd_p0 = dot_product(p0_v, p0_v);
double dot_prd_p1 = dot_product(p1_v, p1_v);
struct vector sub_vec = subtract_vectors(p1_v, p0_v);
double D = ((l0 * l0) - (l1 * l1) + dot_prd_p1 - dot_prd_p0) / 2.0f;
static struct plane p;
p.A = sub_vec.i; p.B = sub_vec.j; p.C = sub_vec.k; p.D = D;
static struct configuration c;
/*
* Special case when object point and shoulder point are same.
*/
if(check_if_same_points(p0, p1)) {
printf("object and shoulder are at same location \n");
c.center.x = p0.x; c.center.y = p0.y; c.center.z = p0.z;
c.radius = l0;
c.p.A = c.p.B = c.p.C = c.p.D = 0.0f;
return c;
}
double a0 = (1.0f - (((l0 * l0) - (l1 * l1)) / (distance(p0, p1) * distance(p0, p1)))) / 2.0f;
struct vector lhs = scale_vector(a0,p0_v);
struct vector rhs = scale_vector(1.0f - a0, p1_v);
struct vector ans = add_vectors(lhs, rhs);
struct point center = get_point_from_vector(ans);
double radius = sqrt((l0 * l0) - (distance(center, p0) * distance(center, p0)));
c.center.x = center.x; c.center.y = center.y; c.center.z = center.z;
c.radius = radius;
c.p.A = p.A; c.p.B = p.B; c.p.C = p.C; c.p.D = D;
return c;
}
/*
* The logic is as follows - Point P0 generates a sphere of radius L0 around it,
* P1 generates another sphere of radius L1 around it. The intersection(if any)
* will be a circle with a plane. If I can return the center and radius of that
* circle and equation of the plane, then the client can find out any possible
* location of the elbow by varying the value of theta in the parametric
* equation of the circle. Thus if the spheres meet to generate a circle then
* there will be infinite elbow positions. If it does not generate a circle and
* meet "externally" then there will be only single elbow position. Otherwise
* there will be no solutions at all.
*/
int main() {
struct point p0, p1;
p0.x = 0, p0.y = 0, p0.z = 0;
p1.x = 50, p1.y = 50, p1.z = 0;
double l0 = 50, l1 = 50;
printf("Shoulder coordinates : (%lf, %lf, %lf) \n", p0.x, p0.y, p0.z);
printf("Object coordinates: (%lf, %lf, %lf) \n", p1.x, p1.y, p1.z);
printf("link0 = %lf, link1 = %lf \n", l0, l1);
if(distance(p0, p1) > (l0 + l1)) {
printf("The given combination of the points and links cannot make a valid configuration");
return -1;
}
struct configuration c = return_config(p0, l0, p1, l1);
printf("Center = (%lf, %lf, %lf), radius = %lf \n", c.center.x, c.center.y, c.center.z, c.radius);
printf("Equation of the radical plane = %lfA + (%lf)B + (%lf)C + %lf = 0 \n", c.p.A, c.p.B, c.p.C, c.p.D);
return 0;
}