Has any mathematician constructed a physical, (global), ball model of $\Bbb R^3?$

Question

Q: Has any mathematician constructed a physical, (global), ball model of $\Bbb R^3?$

For example, to map $\Bbb R^3$ inside $S^2$ you can do, $(x,y,z)\mapsto(x,y,z)/\sqrt{1+r^2}$ so, $\left(\dfrac x{\sqrt{1+x^2+y^2+z^2}},\dfrac y{\sqrt{1+x^2+y^2+z^2}},\dfrac z{\sqrt{1+x^2+y^2+z^2}}\right).$

Answer 1

I've made a small model of $\left(\dfrac x{\sqrt{1+x^2+y^2+z^2}},\dfrac y{\sqrt{1+x^2+y^2+z^2}},\dfrac z{\sqrt{1+x^2+y^2+z^2}}\right)$ for $x,y,z \in\{-2,1,0,1,2\}$ (ie. $5^3 = 125$ points in total) and the grid lines passing trough them with red length-wise parallel lines, blue width-wise parallel lines and green height-wise parallel lines. You can download the model here and view it with View3dscene.

Geometrically, I don't know if there's much to say about the model. The sphere which is the boundary of the ball represents infinity (the metric distance from a point in the interior of the ball to a hypothetical point on the boundary of the ball is infinite), so the closer you get to the boundary, the "denser" the points will be.

Also, parallel co-planar lines will intersect at infinity, at two radially opposite points of the sphere.

As expected, it looks "most" Euclidean when looking from the center.