Polygon border - a different approach

As far as I can tell this is the same border algorithm you were doing before, only you draw it as a strip around the original polygon rather than overdrawing the original on top of an inflated polygon. So yes, this can be done.

Assuming the inflated polygon has the same number of vertices as the original (ie. sharp corners), this is very easy. Say you have a 4-sided polygon with vertices "abcd" in a counter-clockwise order. set up the inflated vertices "ABCD" using the method of choice. Then just draw a triangle strip between them, ordered like this: aAbBcCdDaA.

yeah the old version does seem slightly overly complex. to get the effect you describe is pretty easy, especially in 2d.

it's easiest describe in pseudo code, something like this (obviously the poly has > 2 verts :) )

//start a triangle strip off in whatever API you're using//assuming a handy v2 structure here...StartTriangleStrip();v2 cur, prev, next;prev = poly[poly.size()-1];cur = poly[0];next = poly[1];for (int i=0;i<poly.size();++i){  v2 cur_edge = (next-cur).Normal();  v2 prev_edge = (cur-prev).Normal();  v2 out = (cur_edge+prev_edge).Normal().Right();  AddAVertex(cur+out); //this does whatever  AddAVertex(cur-out);   prev = cur;  cur = next;  next= poly[(i+1)%poly.size()];}DrawTriangleStrip();

and that's it. you might find you need a special case for when the angle is too sharp (to make it look nice), maybe round off the corner or something.

edit: normalise cur_edge and prev_edge

[Edited by - honnyjopper on February 15, 2007 4:59:15 AM]

oh and obviously multiply the "out" vector by half the fatness of the border you want.

double p.s., i have assumed that the polygon is already sorted... if not this won't work obviously. sort your poly!

You could draw the original arrow scaled up a bit in all black then render the red arrow on top of it.

Quote:

Original post by Fingers_ As far as I can tell this is the same border algorithm you were doing before, only you draw it as a strip around the original polygon rather than overdrawing the original on top of an inflated polygon.

The existing method leaves the polygon untouched, creating a border on the exterior. Drawing a fat perimeter along the edges as suggested will obscure some of the polygon.

Quote:

Original post by Todd Smith You could draw the original arrow scaled up a bit in all black then render the red arrow on top of it.

This was an early suggestion that led to andy82's current implementation. The problem is in the definition of 'scaled up'. A uniform scale will give stupid results for non-uniformly-shaped polygons. This doesn't get much better when you normalise out the scaling factors, and still stands no chance on a convex polygon. In fact, in order to make the border entirely uniform, you need to scale the vertices individually, along their 'normalised & pre-normalised' normals. This is exactly the implementation we ended up at in the last thread.

Back onto the topic, I can't imagine there's any way to draw a thick border that has non-orthogonal edges without going all the way through the per-vertex analysis again. In particular, to find the terminal point on the exterior of each edge, you'd need to extrapolate along the normals as you already are. So unless I've made a grave error in judgement, I'm afraid you'll need to settle for rounded-edges or leave things as they are.

Quote:

Original post by andy82 Also, to the nice people who have followed my other threads on this polygon border topic: Would this thick line-approach solve the floating point accuracy problems which we have discussed in the other threads?

There's no way to 'solve' the floating inaccuracy issue - using a tolerance as I described is very standard, and works great - but if your method needs no equality tests then you'll have sidestepped the issue altogether. If you were to implement the 'rounded edge' solution then this would be the case.

Admiral