Jump to content

How do you just inverse a shape?


SamBridge

Recommended Posts

So I know most shapes have an inverse, some of which do and don't have defanite edges, I know that you can program a transformation of a shape to its inverse, but I don't get what you do exactly especially for 3-D shapes. If I have a square, the "inverse" of a square seems be some kind of infinitely stretched thing with an open hole in the middle with arc lengths whos endpoints correspond to the vertices of a square. How would I see what it looks like for say...an inverse of a tube?

Link to comment
Share on other sites

Could you explain more - perhaps with a few links exactly what you mean by inverse shape. Do you mean some form of inverse (that in parallel to algebra) operates through some form of addition on the original to create the identity (and what is the identity (a single point, the origin)

 

[latex]Shape \oplus Shape^{-1} = Identity[/latex]

 

I am using the term identity like you would use it in matrices

 

[latex]\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}[/latex]

Link to comment
Share on other sites

So I know most shapes have an inverse, some of which do and don't have defanite edges, I know that you can program a transformation of a shape to its inverse, but I don't get what you do exactly especially for 3-D shapes. If I have a square, the "inverse" of a square seems be some kind of infinitely stretched thing with an open hole in the middle with arc lengths whos endpoints correspond to the vertices of a square. How would I see what it looks like for say...an inverse of a tube?

 

Perhaps you mean as in a "dual"? Try this...

 

Duality_(projective_geometry)

 

...

Three dimensions

In a polarity of real projective 3-space, PG(3,R), points correspond to planes, and lines correspond to lines. By restriction the duality of polyhedra in solid geometry is obtained, where points are dual to faces, and sides are dual to sides, so that the icosahedron is dual to the dodecahedron, and the cube is dual to the octahedron. ...

 

 

PS Also check out the Peaucellier–Lipkin linkage. >> Peaucellier-Lipkin_linkage

 

Inversive geometry

Thus, by the properties of inversive geometry, since the figure traced by point D is the inverse of the figure traced by point B, if B traces a circle passing through the center of inversion O, then D is constrained to trace a straight line. But if B traces a straight line not passing through O, then D must trace an arc of a circle passing through O. Q.E.D.

...

 

Edited by Acme
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.