# neutral axis of a non-symetrical shape

I remember that it is very easy to find the neutral axis of a symmetric beam cross-section, just locating the centroid and
intersecting a line of symmetry.
However, I want to find the neutral axis of an area which is non- symmetrical. Does anybody know off-hand what to do?
I am tryingto use the principle in image processing- if i see an object which can be rotated through any angle 0-360 degrees, i want to find the neutral axis of the object and rotate the image to attenuate inaccuracy in my recognition algorithm.
If anyone has a link that would be great, all i can find on google is symemetric examples- doh!
Thanks for any help, it is greatly appreciated. Adam
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On Sun, 6 Apr 2008 05:40:13 -0700 (PDT), Adam Chapman wrote:

http://physics.uwstout.edu/StatStr/statics/Beams/bdsn47.htm
the neutral axis (an axis running     through the centroid of beam cross section).

Why locate an "intersecting a line of symmetry"?
The moment of inertia with respect to a line parallel to the axis of bending will change as you rotate the shape. Does that matter in your application? Are you searching for an orientation to minimize (or maximize) that moment of inertia?
How does beam flexure theory relate to image processing?
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Im making a character recognition algorithm, and my camera could see alphabetic characters rotated about any angle. My inputs to the recognition algorithm are effectively the moments of inertia of the black pixels that make up the image of the character. The problem I have is that, like you said, the moments of inertia change with rotation of the shape. I think recognition accuracy could be improved by finding a neutral axis and rotating the black pixel are to line up that axis will improve accuracy.
Sorry the only examples of neutral axes that i have studied in mechanics about 3 years ago were all for symmetrical beam sections.
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On Sun, 6 Apr 2008 12:43:49 -0700 (PDT), Adam Chapman wrote:

Interesting stuff. Thanks for elaborating.
In doing more reading, the statement quoted in my previous post seems incomplete.
It appears that a particular kind of symmetry is needed for the neutral axis to coincide with the centroid of a homogeneous and isotropic shape: "the plane of loading contains an axis of symmetry of the cross section", according to Cernica's book, _Strength of Materials_.
A teardrop-shaped beam is symmetrical about only one plane. It seems that the neutral axis will be through the centroid of the teardrop only if the vectors for all bending forces are in the plane of symmetry.
Further, the neutral axis coincides with the centroid of straight beams -- not curved beams, as shown here: http://www.roymech.co.uk/Useful_Tables/Beams/Curved_beams.html
Good stuff on unsymmetrical beams here: http://courses.washington.edu/mengr354/jenkins/notes/chap4.pdf
From what I understand of your description, I still think that the centroid is the point that you want in dealing with 2-dimensional shapes.
These might be useful: http://en.wikipedia.org/wiki/Second_moment_of_area#Parallel_axis_theorem http://en.wikipedia.org/wiki/Second_moment_of_area#Axes_rotation
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