Expanding object: how do I calculate the speed of expansion from the movement of the edges?

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Julian
Julian 2013-3-5
Hello everyone,
I have a 320x320X360 cell in which a video of a (more or less) rectangular object is saved. The size of the object increses over the 360 frames and I would like to claculate the speed of expansion of the object and its change over time.
To do that, I would like to track the position of the edges at a certain row and column.
The images in the cell are binarized and I used the Canny method to extract the edges. Also, I know how big a pixel is (80x80 µm)
Until now I only have
for k = 1:framenumber
vertical(:,k) = bin_edge{1,k}(:,160);
horizontal (:,k) = bin_edge{1,k) (160,:);
end
which returns two matrices with zeros and ones, the ones indicating the found edges.
But how do I now calculate the speed from that? How can I measure the "distances", or the number of zeros between the ones (for they represent the edges)?
And what makes it even harder (at least for me, as a beginner) is that in some columns I have more than 2 ones. Sometimes, "edge" finds more edges within the objects, which I would like to exclude for the calculations.
I realy hope you can help me, since I tried for quite some time now and don't have any ideas anymore.
Many thanks from Germany Julian

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采纳的回答

Image Analyst
Image Analyst 2013-3-5
Why are you tracking edges? Why not find the area and log how that changes over the frames?
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Image Analyst
Image Analyst 2013-3-6
Something like
for col = 1 : columns
thisColumn = yourImage(:, col);
topRow = find(thisColumn, 1, 'first');
bottomRow = find(thisColumn, 1, 'last');
height(col) = bottomRow - topRow; % Add 1 if you're doing pixel based instead of pixel center - to - pixel center.
end
Julian
Julian 2013-3-7
Ah, the 'first', 'last' does the trick. Thank you a lot!
This one here is working just fine:
if sum (vertical(:,k)) < 2
else
topedge (1,k) = find (vertical(:,k), 1, 'first');
bottomedge (1,k) = find (vertical(:,k), 1, 'last');
end
if sum (horizontal(:,k)) < 2
else
leftedge (1,k) = find (horizontal(:,k), 1, 'first');
rightedge (1,k) = find (horizontal(:,k), 1, 'last');
end
height = bottomedge - topedge;
diameter = rightedge - leftedge;
Thanks again!
Julian

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更多回答（1 个）

Julian
Julian 2013-3-6
编辑：Julian 2013-3-6
Here is an example image
http://imgur.com/CGywVam
The edges deform quite a bit over time.
Here is a link to the "vertical" matrix in which column 160 is saved for all frames ( http://www.speedshare.org/download.php?id=EFCFBB7C1 )
Then, I used
[rowIdx,colIdx] = find (vertical(1:160,:));
[rowIdx2,colIdx2] = find (vertical(161:320,:));
to find the indices of the upper and lower boundaries of the object.
Link to rowIdx of the upper edge: http://www.speedshare.org/download.php?id=FE0FBF4F1
Now, as you can see in "vertical" quite often, there is more than one 1 in a column. How do I only "detect" the ones with the lowest row number (for the upper edges) and the highest row numer (for the lower edges)?
Am I on the right way, to solve my problem, or do you have a better idea?
Thank you Julian
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Julian
Julian 2013-3-6
Yes, you're right of course. The boards I usualy frequent (not matlab related stuff) don't distinguish between answers and comments, so in this case I just klicked the answer button withouth thinking about it. Won't happen again.

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