Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to know how thickness is defined.

Let me start with the simplest shape due to its symmetry. This is a circle. Assume a circle with radius of 1. If we draw a circle with the same center and radius 2, we will have created a ring with thickness 1.

How can we extend this to different shapes? For example let's take a rectangle with length of sides a=1 and b=2. I can think of 2 ways to create a ring of thickness 1.

  • create a rectangle with length of sizes 2 and 3.
  • create a shape like previous rectangle, but the corners would be rounded. That'a because the thickness in the corners would be $\sqrt2$, not 1.

So my definition of thickness something like

  • The minimum distance between any point of the inner shape and any point of the outer shape should be 1.
  • there should be no point (A) in inner shape where there is no point (B) in outer circle such as AB=1.
  • there should be no point (A) in outer shape where there is no point (B) in inner circle such as AB=1.

Can this be correct? Assuming that above is correct:

If inner shape is circle $(x-a)^2+(y-b)^2=r^2$, then the outer shape should be $(x-a)^2+(y-b)^2=(r+1)^2$.

How about if the inner shape is an ellipsis? Or a square?

share|improve this question

1 Answer 1

You will always get the set of all points outsude of the inner shape exactly distance 1 from the inner shape. For a circle or ellipse, you get anityer circle or ellipse with the radius or semiaxes increased in length by 1.

For a square, you add a rectangle of thickness 1 to each edge and then add quarter circles at the corner to fill in the gaps.

By the way, have you heard of extremal length of an annulus?https://en.m.wikipedia.org/wiki/Extremal_length

It is really cool. It assigns a number to each ring that tells how 'thick' it is using complex analysis. It is hard to compute but useful.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.