我找到了以下算法来生成多边形轮廓:
void CGlShape::GenerateLinePoly(std::vector<DOUBLEPOINT> &input, int width)
{
OutlineVec.clear();
if(input.size() < 2)
{
return;
}
if(connected)
{
input.push_back(input[0]);
input.push_back(input[1]);
}
float w = width / 2.0f;
//glBegin(GL_TRIANGLES);
for( size_t i = 0; i < input.size()-1; ++i )
{
POINTFLOAT cur;
cur.x = input[i].point[0];
cur.y = input[i].point[1];
POINTFLOAT nxt;
nxt.x = input[i+1].point[0];
nxt.y = input[i+1].point[1];
POINTFLOAT b;
b.x = nxt.x - cur.x;
b.y = nxt.y - cur.y;
b = normalize(b);
POINTFLOAT b_perp;
b_perp.x = -b.y;
b_perp.y = b.x;
POINTFLOAT p0;
POINTFLOAT p1;
POINTFLOAT p2;
POINTFLOAT p3;
p0.x = cur.x + b_perp.x * w;
p0.y = cur.y + b_perp.y * w;
p1.x = cur.x - b_perp.x * w;
p1.y = cur.y - b_perp.y * w;
p2.x = nxt.x + b_perp.x * w;
p2.y = nxt.y + b_perp.y * w;
p3.x = nxt.x - b_perp.x * w;
p3.y = nxt.y - b_perp.y * w;
OutlineVec.push_back(p0.x);
OutlineVec.push_back(p0.y);
OutlineVec.push_back(p1.x);
OutlineVec.push_back(p1.y);
OutlineVec.push_back(p2.x);
OutlineVec.push_back(p2.y);
OutlineVec.push_back(p2.x);
OutlineVec.push_back(p2.y);
OutlineVec.push_back(p1.x);
OutlineVec.push_back(p1.y);
OutlineVec.push_back(p3.x);
OutlineVec.push_back(p3.y);
// only do joins when we have a prv
if( i == 0 ) continue;
POINTFLOAT prv;
prv.x = input[i-1].point[0];
prv.y = input[i-1].point[1];
POINTFLOAT a;
a.x = prv.x - cur.x;
a.y = prv.y - cur.y;
a = normalize(a);
POINTFLOAT a_perp;
a_perp.x = a.y;
a_perp.y = -a.x;
float det = a.x * b.y - b.x * a.y;
if( det > 0 )
{
a_perp.x = -a_perp.x;
a_perp.y = -a_perp.y;
b_perp.x = -b_perp.x;
b_perp.y = -b_perp.y;
}
// TODO: do inner miter calculation
// flip around normals and calculate round join points
a_perp.x = -a_perp.x;
a_perp.y = -a_perp.y;
b_perp.x = -b_perp.x;
b_perp.y = -b_perp.y;
size_t num_pts = 16;
std::vector< POINTFLOAT> round( 1 + num_pts + 1 );
POINTFLOAT nc;
nc.x = cur.x + (a_perp.x * w);
nc.y = cur.y + (a_perp.y * w);
round.front() = nc;
nc.x = cur.x + (b_perp.x * w);
nc.y = cur.y + (b_perp.y * w);
round.back() = nc;
for( size_t j = 1; j < num_pts+1; ++j )
{
float t = (float)j/(float)(num_pts+1);
if( det > 0 )
{
POINTFLOAT nin;
nin = slerp2d( b_perp, a_perp, 1.0f-t );
nin.x *= w;
nin.y *= w;
nin.x += cur.x;
nin.y += cur.y;
round[j] = nin;
}
else
{
POINTFLOAT nin;
nin = slerp2d( a_perp, b_perp, t );
nin.x *= w;
nin.y *= w;
nin.x += cur.x;
nin.y += cur.y;
round[j] = nin;
}
}
for( size_t j = 0; j < round.size()-1; ++j )
{
OutlineVec.push_back(cur.x);
OutlineVec.push_back(cur.y);
if( det > 0 )
{
OutlineVec.push_back(round[j + 1].x);
OutlineVec.push_back(round[j + 1].y);
OutlineVec.push_back(round[j].x);
OutlineVec.push_back(round[j].y);
}
else
{
OutlineVec.push_back(round[j].x);
OutlineVec.push_back(round[j].y);
OutlineVec.push_back(round[j + 1].x);
OutlineVec.push_back(round[j + 1].y);
}
}
}
}
POINTFLOAT multiply(const POINTFLOAT &a, float b)
{
POINTFLOAT result;
result.x = a.x * b;
result.y = a.y * b;
return result;
}
POINTFLOAT normalize(const POINTFLOAT &a)
{
return multiply(a, 1.0f/sqrt(a.x*a.x+a.y*a.y));
}
POINTFLOAT slerp2d( const POINTFLOAT &v0,
const POINTFLOAT &v1, float t )
{
float dot = (v0.x * v1.x + v0.y * v1.y);
if( dot < -1.0f ) dot = -1.0f;
if( dot > 1.0f ) dot = 1.0f;
float theta_0 = acos( dot );
float theta = theta_0 * t;
POINTFLOAT v2;
v2.x = -v0.y;
v2.y = v0.x;
POINTFLOAT result;
result.x = v0.x * cos(theta) + v2.x * sin(theta);
result.y = v0.y * cos(theta) + v2.y * sin(theta);
return result;
}
有一条评论说应该进行内部斜接计算,但我不确定这是什么或应该如何进行。现在这个算法产生圆边但是我想修改它来制作斜边(正方形),这怎么可能呢?它是否涉及内部斜接计算?
由于
答案 0 :(得分:2)
尝试修改我的other answer:
// v0 and v1 are normalized
// t can vary between 0 and 1
// http://number-none.com/product/Understanding%20Slerp,%20Then%20Not%20Using%20It/
Vector2f slerp2d( const Vector2f& v0, const Vector2f& v1, float t )
{
float dot = v0.dot(v1);
if( dot < -1.0f ) dot = -1.0f;
if( dot > 1.0f ) dot = 1.0f;
float theta_0 = acos( dot );
float theta = theta_0 * t;
Vector2f v2( -v0.y(), v0.x() );
return ( v0*cos(theta) + v2*sin(theta) );
}
void glPolyline( const vector<Vector2f>& polyline, float width )
{
if( polyline.size() < 2 ) return;
float w = width / 2.0f;
glBegin(GL_TRIANGLES);
for( size_t i = 0; i < polyline.size()-1; ++i )
{
const Vector2f& cur = polyline[ i ];
const Vector2f& nxt = polyline[i+1];
Vector2f b = (nxt - cur).normalized();
Vector2f b_perp( -b.y(), b.x() );
Vector2f p0( cur + b_perp*w );
Vector2f p1( cur - b_perp*w );
Vector2f p2( nxt + b_perp*w );
Vector2f p3( nxt - b_perp*w );
// first triangle
glVertex2fv( p0.data() );
glVertex2fv( p1.data() );
glVertex2fv( p2.data() );
// second triangle
glVertex2fv( p2.data() );
glVertex2fv( p1.data() );
glVertex2fv( p3.data() );
// only do joins when we have a prv
if( i == 0 ) continue;
const Vector2f& prv = polyline[i-1];
Vector2f a = (prv - cur).normalized();
Vector2f a_perp( a.y(), -a.x() );
float det = a.x()*b.y() - b.x()*a.y();
if( det > 0 )
{
a_perp = -a_perp;
b_perp = -b_perp;
}
// TODO: do inner miter calculation
// flip around normals and calculate round join points
a_perp = -a_perp;
b_perp = -b_perp;
size_t num_pts = 1;
vector< Vector2f > round( 1 + num_pts + 1 );
for( size_t j = 0; j <= num_pts+1; ++j )
{
float t = (float)j/(float)(num_pts+1);
if( det > 0 )
round[j] = cur + (slerp2d( b_perp, a_perp, 1.0f-t ) * w);
else
round[j] = cur + (slerp2d( a_perp, b_perp, t ) * w);
}
///////////////////////
// new outer miter code
float theta = acos( a.dot(b) ) / 2.0;
float val = w / tan( theta );
float miter_length = sqrt( w*w + val*val );
Vector2f miter = ((a_perp+b_perp)*0.5).normalized() * miter_length;
round[1] = cur + miter;
// end new outer miter code
///////////////////////
for( size_t j = 0; j < round.size()-1; ++j )
{
glVertex2fv( cur.data() );
if( det > 0 )
{
glVertex2fv( round[j+1].data() );
glVertex2fv( round[j+0].data() );
}
else
{
glVertex2fv( round[j+0].data() );
glVertex2fv( round[j+1].data() );
}
}
}
glEnd();
}
答案 1 :(得分:0)
代码在多边形轮廓上放置圆角(mitres),但仅在外边缘。
如果您可以将方块的轮廓概念化为一个大的黑色方框,其中包含一个较小的白色方框,此代码将围绕黑匣子的边缘,但不是白色的边缘 (这只是一个思考练习,代码实际上并不像这样工作)
那就是所有slerp
的东西;它在垂直于第一条边和第二条边的方向之间进行插值,并在弧上添加点。
评论建议有人应该对内部角落做同样的事情。