^{®}tesseract would look like...

I realised that we tend to look at a cube so that we can see three faces and that this changes the dead centre view for a 2Der...

When a 2Der looks at a square they would probably hold it so that they can see at least 2 faces (that we call edges).

So it occured to me to think about what is the optimal viewing angle for a 4Der as they would hold the tesseract so that they can see 4 faces (that we call cubes) at once.

I drew up the following as a result:

This shows 2Der's optimal view of a square and a 3Der's optimal view of a cube.

It also shows a 2Der's slice of the 3Der's optimal view straight on.

It also hopefully conveys properly a 3Der's slice of a 4Der's optimal view of a tesseract straight on?

Initialy for the later I drew only the first image but I realised that the 4Der would have 360° of optimal view compared to our single primary optimal view.

So I'm guessing that our 3D slice of the 4Der's optimal view from straight on would be as shown; or anywhere between those as well for a full 360°?

The grey areas are the edges of each shape as experienced by the dimensional being to whom the object belongs.

Are my depictions correct? Or have I assumed something incorrectly?