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

조회 수: 1 (최근 30일)
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

채택된 답변

Image Analyst
Image Analyst 2013년 3월 5일
Why are you tracking edges? Why not find the area and log how that changes over the frames?
  댓글 수: 5
Image Analyst
Image Analyst 2013년 3월 6일
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 2013년 3월 7일
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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추가 답변 (1개)

Julian
Julian 2013년 3월 6일
편집: Julian 2013년 3월 6일
Here is an example image
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
  댓글 수: 3
Julian
Julian 2013년 3월 6일
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.
Image Analyst
Image Analyst 2013년 3월 6일
Did you notice the comment I posted above, under my answer, that had some code in it?

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