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2D Geometry Problem

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Hi guys, take a look at this:

 

walljut_zps14b71260.jpg

 

That's the profile of a 3D wall. Basically I have the light grey part and want to make sections that jut out (the dark grey). The problem I'm having is getting a uniform thickness (the blue lines should all be the same length), I know I have to adjust the two gold vertices but I'm not sure how, I think I need to use the Pythagorean theorem but I don't know how to set it up. Also, in this case the angle is 45 degrees but I need a general solution for other angles.

 

It would be nice if i could just do it in 3ds Max but I can't see how, so I'll do the computations myself.

 

Thanks for helping.

 

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The problem I'm having is getting a uniform thickness

The issue will be, that you have to change the topology of the mesh. The light gray mesh right silhouette consists of 3 lines, a uniform thickness hull will need circles to be uniform. Eg take the corner of the light-gray section between the two lower blue lines and draw a circle around it to see the corner part  of a uniform thickness hull.

 

With this in mind you can construct a hullmesh by a selection of geometry objects. In 3d you need a box for a plane, a cylinder for a edge and a sphere for a vertex. The combination (union including the original mesh) of all this will define the hull mesh with a uniform thickness.

Edited by Ashaman73

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The problem I'm having is getting a uniform thickness ...

Of course, Ashaman73 is right, but if "uniform thickness" is not taken literally, ...

... (the blue lines should all be the same length)

... so that differences in thickness are allowed just in corners (like in the illustration), then ray intersections can be used to determine the vertex positions. I.e.make copies of 2 consecutive edges, compute rays from them, displace them orthogonally by the wall thickness, and compute their intersection point to get the "golden vertex".

 

E.g. with the consecutive vertices v1, v2, v3 the rays would look like

   r12( k ) := v1 + w * perp( v2 - v1 ) + k * ( v2 - v1 )

   r23( k ) := v2 + w * perp( v3 - v2 ) + k * ( v3 - v2 )

where perp(v) means the (wanted) perpendicular normalized vector of v, and w denotes the wall thickness.

 

Then

   r12( k' ) = r23( k' )

gives a condition for the intersection. The condition denotes a point if the rays are not parallel, and hence solving for k' and setting back in one of the ray equations gives the vertex of interest. 

Edited by haegarr

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