以 3D 形式表示机器人的肘部角度

Question

我得到了 3-D 中两点的坐标。肩点和物体点（我应该到达的位置）。我还得到了从肩膀到肘部的长度以及前臂的长度。我正在尝试解决未知位置（关节肘的位置）。我使用余弦法则来找出肘部角度。这是我的代码 -

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

struct point {
     double x, y, z;
};

struct angles {
     double clock_wise;
     double counter_clock_wise;
};

double max(double a, double b) {
     return (a > b) ? a : b;
}
/*
 * Check if the combination can make a triangle by considering the fact that sum
 * of any two sides of a triangle is greater than the remaining side. The
 * overlapping condition of links is handled separately in main().
 */
int valid_triangle(struct point p0, double l0, struct point p1, double l1) {
     double dist = sqrt(pow((fabs(p1.z - p0.z)), 2) + pow((fabs(p1.y - p0.y)), 2) + pow((fabs(p1.x - p0.x)), 2));
     if((max(dist, l0) == dist) && max(dist, l1) == dist) {
          return (dist < (l0 + l1));
     }
     else if((max(dist, l0) == l0) && (max(l0, l1) == l0)) {
          return (l0 < (dist + l1));
     }
     else {
          return (l1 < (dist + l0));
     }
}
/* 
 * Cosine rule is used to find the elbow angle. Positive value indicates a
 * counter clockwise angle while negative value indicates a clockwise angle.
 * Since this problem has at max 2 solutions for any given position of P0 and
 * P1, I am returning a structure of angles which can be used to consider angles
 * from both direction viz. clockwise-negative and counter-clockwise-positive
 */
void return_config(struct point p0, double l0, struct point p1, double l1, struct angles *a) {
     double dist = sqrt(pow((fabs(p1.z - p0.z)), 2) + pow((fabs(p1.y - p0.y)), 2) + pow((fabs(p1.x - p0.x)), 2));
     double degrees = (double) acos((l0 * l0 + l1 * l1 - dist * dist) / (2 * l0 * l1)) * (180.0f / 3.1415f);
     a->clock_wise = -degrees;
     a->counter_clock_wise = degrees;
}
int main() {

     struct point p0, p1;
     struct angles a;
     p0.x = 15, p0.y = 4, p0.z = 0;
     p1.x = 20, p1.y = 4, p1.z = 0;
     double l0 = 5, l1 = 8;

     if(valid_triangle(p0, l0, p1, l1)) {
          printf("Three lengths can make a valid configuration \n");
          return_config(p0, l0, p1, l1, &a);
          printf("Angle of the elbow point (clockwise) = %lf, (counter clockwise) = %lf \n", a.clock_wise, a.counter_clock_wise);
     }
     else {
          double dist = sqrt(pow((fabs(p1.z - p0.z)), 2) + pow((fabs(p1.y - p0.y)), 2) + pow((fabs(p1.x - p0.x)), 2));

          if((dist <= (l0 + l1)) && (dist > l0)) {
               a.clock_wise = -180.0f;
               a.counter_clock_wise = 180.0f;
               printf("Angle of the elbow point (clockwise) = %lf, (counter clockwise) = %lf \n", a.clock_wise, a.counter_clock_wise);
          }
          else if((dist <= fabs(l0 - l1)) && (dist < l0)){
               a.clock_wise = -0.0f;
               a.counter_clock_wise = 0.0f;
               printf("Angle of the elbow point (clockwise) = %lf, (counter clockwise) = %lf \n", a.clock_wise, a.counter_clock_wise);
          }
          else
               printf("Given combination cannot make a valid configuration\n");
     }
     return 0;
}

但是，这个解决方案仅在二维中才有意义。因为如果没有轴和旋转方向，顺时针和逆时针就没有意义。仅返回角度在技术上是正确的，但为该函数的客户端以有意义的方式使用结果留下了很多工作。如何进行更改以获得旋转轴和方向？另外，我想知道这个问题有多少种可能的解决方案。

请让我知道您的想法！非常感谢任何帮助...

Answer 1

基本问题是您正在解决（必然）平面上两个手臂部分之间的角度，但平面本身未指定。该解决方案将涉及限制平面本身或将肘部的位置限制到某种状态，例如“尽可能低的位置”（并且这两个可能具有相同的解决方案）。

因此，定义平面，并将三个端点定位在该平面上。