View Full Version : Detecting Edges

KurtCob

01-25-2001, 12:12 PM

Hi people,

I have a model that is scanned in slices, and in each slice I Have two edges, how can I detect the point where these edges are ?

Thank you

Best regards

Kurt

Ps.: I can send a picture.

chennes

01-25-2001, 01:50 PM

I'm a bit confused by what you mean by "edge". Are you talking about an area with a high color gradient?

Chris

KurtCob

01-25-2001, 02:40 PM

No, Itīs like a corner ... you know ? I have a cloud of points, and I want to know the corners.

Thank you

Best regards

Kurt

I'm still not at all clear what you're trying to do? Something like a convex hull of the point cloud?

Hello

Calculate the delta between two adjacent points and if it is big enough then it is a corner. If you have "noise" in the points try using the delta between a point and another point maybe 5 points away...

Pseudo code :

for(x=0;x<n-1;x++)

{

if (abs(p(x)-p(x+1))>delta)

return x;

}

KurtCob

01-26-2001, 06:01 AM

Hello guys,

Rob.: I could send you some picture, do you want ?

Osku.: Could you be more clear, I don't understand what you explain.

Thank you for the moment

Best regards

Kurt

chennes

01-26-2001, 07:03 AM

Wait... you have a 3D set of points in space that define some shape, and you want the outer perimeter of the shape? Like a 3D version of a convex hull algorithm?

If you get a minute, send me a picture while you're at it - I'm currently doing a lot of work with models that are input in planes as well.

Chris

Sure, you can send me a picture and I'll try to take a look at it.

rblanding@hotmail.com

As I understand you have a collection of points that is like a graph? Please send a picture and maybe I can help you.

Osku

chennes

01-27-2001, 12:13 PM

If you have the points ordered sequentially this is fairly easy... loop through all the points and get the slopes of each pair of adjacent points. If the slope differs by a large enough amount, then you have found your corner.

If they're not ordered sequentially, sort them first.

Chris

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