From what I've learned here (https://blogs.mathworks.com/steve/2018/04/17/feret-properties-wrapping-up/?s_tid=srchtitle) it seems that the minimum Feret diameter of an object (measured on a binary image of an object) is the same as the smaller side of a minimum-area bounding rectangle.
However, it is not said explicitly that this is true.
After searching through other resources on the internet, I still can't find a conformation that this is true. So, is it generally true or not?
No matter what the shape of the object is , bounding box (always a rectangle) else you can't crop or define the region of interest . Feret diameter can be either side by definition : https://en.wikipedia.org/wiki/Feret_diameter
it is based on projected plane that is length or distance
Interesting q! Assuming my interpretation of q is correct. Suggest to consider following :
1 An ellipse in x-y plane oriented at different angles theta.
2. Obtain Feret diameters (xf/ yf) a la wiki definition analytically.
3. Minimize rectangle xf*yf wrt theta. Hopefully this will be the
majorand minor axis of the ellipse which I suppose is a min
area rectangle.
Cheers
As a trivial application of above consider:
1. A line segment (degenerate ellipse) w at origin
oriented at angle theta to x axis .
2. Its area is 0.w =0
3. Feret xf= w*cos(theta) yf=w*sin(theta)
4 Area of rectangle A = xf* yf functin of theta.
5.for modulus consider A^2 = w^2*(sin(2theta))^2
6. diff and set to zero we get
sin(2theta) cos(2 theta ) =0
so theta =0 for min
7. Subst in 3 we get xf=w max Feret
and yf= 0 , min Feret. = min side of rectangle.
Cheers
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