I'll have a limited space like $1 km^3$ that have (through handwaving) more then three dimensions. But anyone entering it would perceive it as a three dimensional space. A little bit like this video, but more or less a flat version.

This results in two people walking around the same tree in clockwise or anti clockwise direction would end up at different positions.

What I would like to know is how big is my surface area on which I can walk at maximum inside the multidimensional space.

My first guess was, if I have a $1 km^n$ hypercube and ignore height I would have a surface area of $1^{n-1} $. With an 4 dimensional example I would have a surface area of $1km^3$. Unfortunately this is a unit of volume. So my formula is missing something to get $km^2$. I would also assume that the result would be something greater than $1km^2$


Well, what you seem to be after is the number of 2D faces in an nD-cube? A 3D cube has six 2D faces, and a 4D cube has 24 2D faces. You can find a table and the mathematics here.

  • $\begingroup$ Some how I feel dump now. It seems to be so obvious. What I actually searching is the number of faces someone can walk on. For a 3-cube that would be one, for a tesseract it would be 6. It seems that the number of faces someone can walk on for an $n$-cube is the number of faces of the $n-1$-cube. $\endgroup$ – lokimidgard Jun 26 '16 at 12:56
  • $\begingroup$ Anyone can miss something. Don't worry about it. But you need to think a bit more carefully about gravity for the number of faces that can be walked on; I'm not sure your answer is right. $\endgroup$ – John Dallman Jun 26 '16 at 13:34
  • $\begingroup$ hm... I need to think about this. In my head I'm unwrapping the multidimensional space to a simple bigger 3d space and applying normal physics to it. But your right that this is most likely wrong. $\endgroup$ – lokimidgard Jun 26 '16 at 14:31

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