Question

Polygon Outlines : Small features at concave vertexes

Asked by: FalconZero

I have a function  to calculate a polygon which is outset a fixed distance from an input polygon (Fig 1). My function operates on a vector representing the input polygon by taking three consecutive vertexes and calculating new vertexes based on them.

For colinear vertexes it does not output any vertex.

For convex vertexes it calculates an arc of vertexes around the input vertex (Shown below)

For concave vertexes it calculates the intersection point of two lines offset from the segments described by the two vertexes. In Fig 2, The line AB becomes DE, the line BC becomes FG, the intersect of DE with FG (marked H) is the new vertex.

My problem is that the convex vertex C produces the new vertexes around K (circled green). which are too close to AB. The correct output would eliminate the vertexes around K, eliminate vertex H and add a new vertex at M.

The Figures below give a clearer explanation of the problem.

My program already has OpenGL tessalator functions available for CSG, which might be useful, but a pure solution would be much better.

bool polygon_c::outline_vertex(polygon_c* newpoly, doublePair prev, doublePair cur, doublePair next, double distance){	
	//Get vectors representing the lines
	doublePair A = cur-prev;
	doublePair B = next-cur;
 
	//Calculate the offset vectors for the lines (rotate by 90degrees, divide by length to get unit perp vector, multiply by 'distance')
	doublePair offsetA=doublePair(-A.y,A.x).unit()*distance;
	doublePair offsetB=doublePair(-B.y,B.x).unit()*distance;
 
	//Determine if the angle made by A-B is convex, concave, straight or reverse
	double ang=A.angle(B);
 
	//Clamp really small angles to zero
	(abs(ang)<10e-5) && (ang=0);
 
	//Straight, convex or concave?
	if ((ang<0) || (prev==next)){			//Concave - return arc at 'distance' around cur starting from cur+offsetA, ending at cur+offsetB
		//TODO : Draw arc		
		double incAng=PI2/18;
		double angle;
		double endAng;
		
		//Add arc startpoints
		newpoly->verts.push_back(cur+offsetA);
 
		//For straight lines (IE prev==next), 'fix' the angle to account for lack of 'clockwiseness' in doubleback nodes
		if (prev==next){ang=-PI;}
 
		//Determine start and end angles
		angle=atan2(offsetA.y, offsetA.x)-incAng;
		endAng=(angle+ang)+incAng;
 
		double points=ceil((endAng-angle)/incAng);
		incAng=(endAng-angle)/points;
 
		//Add arc
		while (angle>endAng){
			newpoly->verts.push_back(cur+doublePair(cos(angle)*distance,sin(angle)*distance));
			angle-=incAng;
		}
 
		//Add the arc endpoint
		newpoly->verts.push_back(cur+offsetB);
 
	}else if (ang>0){	//Convex - return intersect(prev+offsetA->cur+offsetA, cur+offsetB->next+offsetB)
		doublePair intersect=doublePair::line_intersect(prev+offsetA,cur+offsetA,  cur+offsetB,next+offsetB);
		newpoly->verts.push_back(intersect);
	}else{				//Straigh line - return cur+offsetA		
		//Comment out to simplify the polygon
		//newpoly->verts.push_back(cur+offsetA);
	}
 
	return true;
}

                                  
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  • Poly2.png
    • 31 KB

    Fig 2 : Extended graphical depiction of problem

    Fig 2 : Extended graphical depiction of problem
  • Poly1.png
    • 8 KB

    Fig 1 : Graphical depiction of problem

    Fig 1 : Graphical depiction of problem

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Asked On
2009-09-29 at 09:44:16ID24770733
Tags

Polygon Outline

,

Concave Vertex

,

Geometry

Topics

3D Game Programming

,

Algorithms

,

C++ Programming Language

Participating Experts
4
Points
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Comments
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Answers

 

by: BitlabPosted on 2009-09-30 at 00:22:04ID: 25456226

Hello FalconZero.

There are questions and draft of required algorithm: http://landkey.net/d/A/Polygons/outset_poligon.png

Possibly this library or its ideas can be used: http://landkey.net/d/z/Essays/PolygonsIntersection/S/Poly.cs.htm

in section "Auxiliary mehods for polylines. "

Apparenlty, the problem you pictured is that your algorithm do not check that point K and its segment is already in internity of P' while BF and FG have been drawing.

Hope to be helpful, Konstantin.

 

by: thehagmanPosted on 2009-09-30 at 01:48:13ID: 25456602

BTW, how would you want to treat the case where a whole (i.e. non-contiguous polygon) might be created?

 

by: NovaDenizenPosted on 2009-10-01 at 06:41:57ID: 25468506

I'm trying to rephrase your goal.  Do you get the same result if you create a circle (diameter d) for each vertex and a rectangle (width d, length centered on the line) for each segment, then took the union of all of these shapes, then traced the "left" edge of it?

 

by: JoseParrotPosted on 2009-10-03 at 13:24:46ID: 25486890

As you have noticed, for convex polygons, the algorithm solves the problem properly. For some non convex polygons solves quite well also. So, let me suggest to first classify the polygon as convex and non convex. This can be done simply finding the convex hull of the polygon. If the number of points of both polygon and convex hull the same, the the polygon is a convex one, then a simple algortithm solves the problem.

The non convex section (or more than one section) is the one with the sequential points which aren't in the convex hull (red), as the black lines in Fig. 1.

For non convex polygons, one of the approaches is to check the relative position of the intersections, which are different for external angles <180 degrees, as the ones at vertices 1 and 4, at Fig. 2.

At Fig. 2a, the point c, in red, appears after the line (b1,b2), thus showing that (b2,b3) is the line of the "contour" polygon { b1, b2, ..., bn } and (b2,b3) is the next line. As matter of checking, point c will be in the line (b1,b2) but with a factor of less than zero or greater than one, relative to the segment (b1,b2).

At Fig. 2b, point c is between (b1,b2) thus showing that point c replaces both b2 and b3 in the "contour" polygon.

Depending on the next condiction, say, if the angle at point 4 had less than 180 degrees, then another similar situation would repeat and the new point c is the intersection of (b1,b2) with (b5,b6) as in Fig. 3.

Jose

 

by: FalconZeroPosted on 2009-10-05 at 04:40:29ID: 25494243

Thanks for the comments so far. I'm investigating these now and I'll let you know how I get on.

 

by: FalconZeroPosted on 2009-11-05 at 05:18:27ID: 31634972

This was the 'best' solution for my problem. I was able to modify my code rather than rewriting so have accepted this as the 'solution'. Thanks for your help.

20120131-EE-VQP-002

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