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Re: tube diameter

Posted by karo03 on Aug 08, 2007; 10:06pm
URL: http://imagej.273.s1.nabble.com/tube-diameter-tp3698688p3698691.html

Tscha, you start with the binary image of your tubes, create the  
distance transform (dt) and the skeleton (sk) from your binary image.  
Than it depends upon you what you want to do with the pixels of the  
skelton in the distance transform. Multiplication of the two images  
and division by 255 gives an image with the crete line pixels having  
the dt value. You can extract all pixels not equal zero with their  
coordinates ...
run("Distance Map");
run("Skeletonize");

Karsten
Am 08.08.2007 um 22:52 schrieb Paxton Provitera:

> Hi Karsten.
> This sounds like exactly what I need, but I am very 0confused as  
> how to do
> it.
> Could you  give me some greater detail in exactly what to do?
>
> Thanks.
>
> -Paxton
>
>
> On 8/8/07, Karsten Rodenacker <[hidden email]> wrote:
>>
>> Take all pixels in the skeleton of your tubes. They should reside on
>> the let's say crete line (Gipfellinie) of the distance transform.
>> Beware of border problems!
>> Karsten
>> Am 08.08.2007 um 21:46 schrieb Paxton Provitera:
>>
>>> Hi.
>>> I have images that I have thresholded and then run through
>>> Process>Binary>Distance Map.
>>>
>>> The images are of tubes in which I would like to extract the radius
>>> value
>>> (or the largest and center value given by the distance map).
>>>
>>> The images vary in size but are mostly 640x480.
>>> From the following example pixel values I want to extract only the
>>> value of
>>> 6:
>>>
>>>
>>> Sample pixel values from a distance map image:
>>>
>>> 0 0 0 0 0 0 1 2 3 4 5 6 6 5 4 3 2 1 0 0 0 0 0
>>> 0 0 0 0 0 1 2 3 4 5 6 6 5 4 3 2 1 0 0 0 0 0 0
>>> 0 0 0 0 0 0 1 2 3 4 5 6 6 5 4 3 2 1 0 0 0 0 0
>>> 0 0 0 0 0 0 0 1 2 3 4 5 6 6 5 4 3 2 1 0 0 0 0
>>> 0 0 0 0 0 0 0 0 1 2 3 4 5 6 6 5 4 3 2 1 0 0 0
>>>
>>> I've been stuck on some algorithm to extract the relevant  
>>> information.
>>> Thanks for any help on this.
>>>
>>> -Paxton
>>