One pixel line length
in [Matlab]
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From: Lars-Göran Nord on 30 Jul 2010 14:50 "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v3js$ija$1(a)fred.mathworks.com>... > "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i2v36p$lvu$1(a)fred.mathworks.com>... > > "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v2ls$i6v$1(a)fred.mathworks.com>... > > > "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i2v1ar$ljj$1(a)fred.mathworks.com>... > > > > "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v0ro$ln1$1(a)fred.mathworks.com>... > > > > > I have an binary image with several one pixel lines crossing each other (crack pattern) whose total length I'm trying to find. > > > > > I'm stuck! Any help would be greatly appreciated! > > > > > Lars > > > > > > > > Well, if they're one pixel wide couldn't you just sum them? > > > > If this isn't what you meant, could you post a picture somewhere so we can have a better idea of what you meant? > > > > > > Well I'm not sure it's that easy, the lines are not straight. I'm new to this so please tell me where to post my image. > > > Thanks, > > > Lars > > > > http://drop.io/ > > Is a good site for images. > > > > Do you have image processing toolbox? > > Yes I have the toolbox, and will drop the image in 30 min. > > thanks again. The image is now on drop, onepixelline(a)drop.io This picture was made in photoshop and then processed in matlab, bwmorph(bw2, 'thin', inf); Lars
From: Sean on 30 Jul 2010 15:30 "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v6t1$kq1$1(a)fred.mathworks.com>... > "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v3js$ija$1(a)fred.mathworks.com>... > > "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i2v36p$lvu$1(a)fred.mathworks.com>... > > > "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v2ls$i6v$1(a)fred.mathworks.com>... > > > > "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i2v1ar$ljj$1(a)fred.mathworks.com>... > > > > > "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote in message <i2v0ro$ln1$1(a)fred.mathworks.com>... > > > > > > I have an binary image with several one pixel lines crossing each other (crack pattern) whose total length I'm trying to find. > > > > > > I'm stuck! Any help would be greatly appreciated! > > > > > > Lars > > > > > > > > > > Well, if they're one pixel wide couldn't you just sum them? > > > > > If this isn't what you meant, could you post a picture somewhere so we can have a better idea of what you meant? > > > > > > > > Well I'm not sure it's that easy, the lines are not straight. I'm new to this so please tell me where to post my image. > > > > Thanks, > > > > Lars > > > > > > http://drop.io/ > > > Is a good site for images. > > > > > > Do you have image processing toolbox? > > > > Yes I have the toolbox, and will drop the image in 30 min. > > > > thanks again. > > The image is now on drop, onepixelline(a)drop.io > This picture was made in photoshop and then processed in matlab, bwmorph(bw2, 'thin', inf); > > Lars Well you have to decide what exactly you want to do with this. How do you want to measure pixel line length? Here's a few ideas to wrap your head around: Should each individual pixel count as one unit? I.e. from opposing face to opposing face? Or should it be 3*sqrt(2) i.e. corner to corner? If two pixels share a face should it be 2 units? I.e. opposite face to opposite face? Or should it be 1 unit center to center and the other units dependent on the next pixels off each end? What I'm hinting at is some sort of Marching Cubes-esque algorithm. If you just use as I mentioned above it will give you an over estimate but depending on your application that may not be a bad thing. What are these lines measuring and what are you doing with the results? -Sean
From: Sean on 30 Jul 2010 15:52 "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message = > If you just use sum() as I mentioned above it will give you an over estimate but depending on your application that may not be a bad thing. What are these lines measuring and what are you doing with the results? > > -Sean Typo, that sum() was kind of important to that sentence meaning anything.
From: ImageAnalyst on 30 Jul 2010 17:13 On Jul 30, 2:50 pm, "Lars-Göran Nord" <lars-goran.nordh(a)uddeholm.se> wrote: > The image is now on drop, onepixelline(a)drop.io > This picture was made in photoshop and then processed in matlab, bwmorph(bw2, 'thin', inf); > > Lars- ------------------------------------------------------------------------------------------------------------------------------ Correction: Your image is at http://drop.io/onepixelline You can use bwarea() if you want a different way of calculating area than sum(). sum() sums up all the "set" pixels. bwarea() applies weighting factors to adjust for the fact that pixels, in a triangle say, may be considered to have partial areas in each pixel depending on whether neighboring pixels are set. For example 0 1 1 1 Is the area of this 3? Or is it sqrt(2)? Or something different?
From: Lars-Göran Nord on 30 Jul 2010 17:21
"Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message <i2vah4$gtb$1(a)fred.mathworks.com>... > "Sean " <sean.dewolski(a)nospamplease.umit.maine.edu> wrote in message = > > If you just use > sum() > as I mentioned above it will give you an over estimate but depending on your application that may not be a bad thing. What are these lines measuring and what are you doing with the results? > > > > -Sean > Typo, that sum() was kind of important to that sentence meaning anything. Very good comments, first of all the lines are measuring crack lengt (thermal fatigue of steel) and when I have the length of the one pixel lines (cracks) I will calculate an average crack width by dividing the total crack area from the original image with the length of cracks. From the start I had an idea of finding pixels connected corner by corner and giving them the value sqrt(2) and all others connected face to face 1. I don't follow your 3*sqrt(2). |
