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Finding area center of rectangle

asked 2013-06-12 14:25:34 -0500

Papercut gravatar image

updated 2013-06-12 15:23:26 -0500

I am trying to find an area center of various types of rectagles.

(Center of gravity and midpoint of 4 vertices never work so please think in different way)

Please see this image:

image description I have to find the position of position of red dots

Point vertices[4];
Point areaCenter;

I have to find areaCenter so (nearly)

area(areaCenter, vertices[0], vertices[1]) = area(areaCenter, vertices[1], vertices[2]) =

area(areaCenter, vertices[2], vertices[3]) = area(areaCenter, vertices[3], vertices[0])

I tried many different ways to find the mid point but none of them covered every types of rectangles.

Can any1 give me some idea?

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answered 2013-06-12 14:56:09 -0500

unxnut gravatar image

updated 2013-06-12 17:09:36 -0500

You need to add the positions of non-zero pixels in horizontal (vertical) directions and divide them by the number of nonzero pixels to get the center of gravity.

xsum = 0;
ysum = 0;
num_nonzero_pxl = 0;
for each pixel (p(x,y))
    if pixel p(x,y) is non-zero {
        xsum += x;
        ysum += y;
        num_nonzero_pxl++;
    }
cgix=xsum/num_nonzero_pxl;
cgiy=ysum/num_nonzero_pxl;
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Thanks for the answer. But I don't quite understand what you are trying to say. Can you write some lines of pseudocode please?

Papercut gravatar imagePapercut ( 2013-06-12 15:15:30 -0500 )edit

that's not gonna work. Equally shaped rectangle may have different midpoint depending over its rotation.

Papercut gravatar imagePapercut ( 2013-06-12 15:55:02 -0500 )edit

It should. Remember, you are adding x and y coordinates (the locations) if the pixel is nonzero and not the actual value of pixel.

unxnut gravatar imageunxnut ( 2013-06-12 15:57:59 -0500 )edit

Shouldn't it be cgix=xsum/total_num_nonzero; cgiy=ysum/total_num_nonzero; ?

Papercut gravatar imagePapercut ( 2013-06-12 16:06:57 -0500 )edit

And I am pretty sure the midpoint will be placed outside of rectangle in the rectangle #6 in the image as well as it takes too long.

Papercut gravatar imagePapercut ( 2013-06-12 16:09:31 -0500 )edit

Yes, you are right about the denominator. It should be total_num_nonzero. My bad. Fixed the answer above.

unxnut gravatar imageunxnut ( 2013-06-12 17:05:22 -0500 )edit
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answered 2013-06-12 17:27:31 -0500

Hansg91 gravatar image

Perhaps this is what you are looking for :

http://en.wikipedia.org/wiki/Centroid#Centroid_of_polygon

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Thanks for the link. I have implemented the Centroid of polygon in your link and even posted my code in the page. However it also has chance to be placed outside of polygon.. :(

Papercut gravatar imagePapercut ( 2013-06-13 12:30:48 -0500 )edit
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Asked: 2013-06-12 14:25:34 -0500

Seen: 2,831 times

Last updated: Jun 12 '13