我已经创建了一个开放的平面区域，上面有一个薄圆柱体，如圆盘，它们在该区域周围反弹，并对放置在飞机上的一些较大的圆柱体进行碰撞检测。我试图让他们现在使用转向方法朝着飞机上的设定点前进。
通过计算距障碍物的距离然后计算行进方向与障碍物方向之间的角度，转向工作用于避开障碍物的工作，当冰球太靠近时，使用距离障碍物的距离的计算它转向左或右在计算的角度上。相反的技术无法用于转向点，我尝试使用acos和atan2来计算行进方向和目标方向之间的角度，并且从输出相信这个位是正确的但是当我尝试使用该计算来确定何时为了引导目标，我得到了意想不到的结果。有时随意转动？
#include "Puck.h"
#include <iostream>
#include <fstream>
using namespace std;
#include <math.h>
ofstream fout("danna.txt");
#ifndef M_PI
#define M_PI 3.1415
#endif
class TranslateCB : public osg::NodeCallback
{
public:
TranslateCB() : _dx( 0. ), _dy( 0. ), _dirx(1), _diry(0), _inc(0.1), _theta(0) {}
TranslateCB(Puck** pp, Obstacle** ob, int count, double r, double x, double y) : _dx( 0. ), _dy( 0. ),
_dirx(2.0*rand()/RAND_MAX-1), _diry(2.0*rand()/RAND_MAX-1), _inc(0.3), _theta(0)
{
obstacles = ob;
ob_count = count;
_radius = r;
_x = x;
_y = y;
puckH = pp;
}
virtual void operator()( osg::Node* node,osg::NodeVisitor* nv )
{
osg::MatrixTransform* mt =
dynamic_cast<osg::MatrixTransform*>( node );
osg::Matrix mR, mT;
mT.makeTranslate( _dx , _dy, 0. );
mt->setMatrix( mT );
double ob_dirx;
double ob_diry;
double ob_dist;
double centerX=0, centerY =0;
_theta = 0;
double min = 4;
// location that I am trying to get the pucks to head towards
centerX = 1;
centerY = 5;
double tDirx = (_x+_dx) - centerX;
double tDiry = (_y+_dy) - centerY;
double tDist = sqrt(tDirx*tDirx+tDiry*tDiry); //distance to target location
// normalizing my target direction
tDirx = tDirx/tDist;
tDiry = tDiry/tDist;
double hDist = sqrt(_dirx*_dirx + _diry*_diry); //distance to next heading
_dirx= _dirx/hDist;
_diry= _diry/hDist;
double cAngle = acos(_dirx*tDirx+_diry*tDiry); //using inverse of cos to calculate angle between directions
double tAngle = atan2(centerY - (_y+_dy),centerX - (_x+_dx)); // using inverse of tan to calculate angle between directions
double tMin = tDist*sin(cAngle);
//if statement used to define when to apply steering direction
if(tMin > 3)
{
if(tDist < 1){ _theta = 0; } //puck is inside target location, so keep travelling straight
if(cAngle > M_PI/2){ _theta = -0.1; } //turn left
else{ _theta = 0.1; } //turn right
}
else{ _theta = 0; }
////// The collision detection for the obstacles that works on the same princables that I am using above
for(int i = 0; i < ob_count; i++)
{
ob_dirx = (_x+_dx) - obstacles[i]->x;
ob_diry = (_y+_dy) - obstacles[i]->y;
ob_dist = sqrt(ob_dirx*ob_dirx+ob_diry*ob_diry);
if (ob_dist < 3) {
//normalise directions
double ob_norm = sqrt(ob_dirx*ob_dirx+ob_diry*ob_diry);
ob_dirx = (ob_dirx)/ob_norm;
ob_diry = (ob_diry)/ob_norm;
double norm = sqrt(_dirx*_dirx+_diry*_diry);
_dirx = (_dirx)/norm;
_diry = (_diry)/norm;
//calculate angle between direction travelling, and direction to obstacle
double angle = acos(_dirx*ob_dirx + _diry*ob_diry);
//calculate closest distance between puck and obstacle if continues on same path
double min_dist = ob_dist*sin(angle);
if(min_dist < _radius + obstacles[i]->radius && ob_dist < min+obstacles[i]->radius)
{
min = ob_dist;
if(ob_dist < _radius + obstacles[i]->radius){ _theta = 0; }
else if(angle <= M_PI/2){ _theta = -0.3; }
else{ _theta = 0.3; }
}
}
}
//change direction accordingly
_dirx = _dirx*cos(_theta) + _diry*sin(_theta);
_diry = _diry*cos(_theta) - _dirx*sin(_theta);
_dx += _inc*_dirx;
if((_x+_dx > 20 && _dirx > 0) || (_x+_dx < -20 && _dirx < 0))
{
_dirx = -_dirx;
_diry += (0.2*rand()/RAND_MAX-0.1); //add randomness to bounce
}
_dy += _inc*_diry;
if((_y+_dy > 20 && _diry > 0) || (_y+_dy < -20 && _diry < 0))
{
_diry = -_diry;
_dirx += (0.2*rand()/RAND_MAX-0.1); //add randomness to bounce
}
traverse( node, nv );
}
private:
double _dx,_dy;
double _dirx,_diry;
double _inc;
double _theta;
double _radius;
double _x,_y;
Obstacle** obstacles;
Puck** puckH;
int ob_count;
};
Puck::Puck()
{
}
void Puck::createBoids (Puck** pucks, Group *root, Obstacle** obstacles, int count, double xx, double yy)
{
// geometry
radius = 0.2;
x = xx;
y = yy;
ob_count = count;
Cylinder *shape=new Cylinder(Vec3(x,y,0),radius,0.1);
ShapeDrawable *draw=new ShapeDrawable(shape);
draw->setColor(Vec4(1,0,0,1));
Geode *geode=new Geode();
geode->addDrawable(draw);
// transformation
MatrixTransform *T=new MatrixTransform();
TranslateCB *tcb = new TranslateCB(pucks, obstacles,ob_count,radius,x,y);
T->setUpdateCallback(tcb);
T->addChild(geode);
root->addChild(T);
}
任何帮助都会很棒！
答案 0 :(得分：0)
这里的问题是，当冰球驶向障碍物时，总是会发生“避免障碍”的“工作”技术。这种特殊情况使得冰球的方向和相邻象限中障碍物的方向。
当试图使冰球朝向障碍物转向时，该技术会发生故障，因为冰球最有可能远离障碍物，不再具有目标和方向向量位于相邻象限的条件。
确定转向方向的正确方法是将目标矢量旋转一个角度，使方向矢量点在象限（0,1）中笔直向上。既然目标矢量相对于方向矢量（0,1），则查看目标矢量的x分量将确定转向方向。如果目标矢量的x分量为负，则冰球必须向左转以朝向目标（增加角度）。如果目标矢量的x分量为正，则冰球必须向右转以朝向目标（减小角度）。
考虑下面用python编写的代码片段来验证这一点，它仍然应该易于阅读，以便您掌握这个概念：
from math import *
dirX = 0.0
dirY = 0.0
targX = 1.0
targY = 0.0
def dir():
global dirX, dirY, targX, targY
# get magnitiude of direction
mag1 = sqrt(dirX*dirX + dirY*dirY)
if mag1 != 0:
# normalize direction vector
normX = dirX / mag1
normY = dirY / mag1
# get magnitude of target vector
mag2 = sqrt(targX*targX + targY*targY)
if mag2 != 0:
# normalize target vector
targX = targX / mag2
targY = targY / mag2
# find the angle need to rotate the dir vector to (0, 1)
rotateAngle = (pi/2.0) - atan2(normY, normX)
# rotate targ vector by that angle (we only care about the x component)
relTargX = cos(rotateAngle) * normX + sin(rotateAngle) * normY
# if the target vector's x is negative
if relTargX < 0:
# turn left
print "Left!"
# otherwise the target vector is 0 or positive
else:
# turn right
print "Right!"
def out():
global dirX, dirY, targX, targY
# function just prints values to the screen
print "dir(%f, %f) targ(%f, %f)" % (dirX, dirY, targX, targY)
# for values 0 to 360
for i in range(360):
# pretend this is the pucks direction
dirX = sin(radians(i))
dirY = cos(radians(i))
# print dir and target vectors to screen
out()
# print the direction to turn
dir()
我想我可以用C ++编写这个，但与运行python提示相比，这是一个很大的痛苦。它与我可以编写的任何伪代码一样可读，并且无论语言如何，概念都将起作用。