3

Suppose i have the following three-dimensional object which is desribed as follows in three-dimensional coordinates: The base area is rectangular and described by the coordiantes $(a,b,0),(a,-b,0),(-a,b,0),(-a,-b,0)$. For two points $(a,x,0)$ and $(-a,-x,0)$ we construct a parabola which has the maximum at $(0,0,h)$. It basically looks like this:

enter image description here

I am searching for the name of this object and a formula for it. My attempt was to define a function $f:[-b,b]\times[-a,a]\rightarrow\mathbb{R}_{\geq 0}$ by $$f(x,y)=\frac{-h}{a^2 + x^2}\left(\sqrt{x^2 + y^2}\right)^2 + h)$$ if $\sqrt{x^2 + y^2} > 0 \wedge |\sin^{-1}(\frac{x}{\sqrt{x^2 + y^2}})| < \sin^{-1}(\frac{b}{\sqrt{a^2 + b^2}})$ and $$f(x,y)=\frac{-h}{b^2 + y^2}\left(\sqrt{x^2 + y^2}\right)^2 + h)$$ if $\sqrt{x^2 + y^2} > 0 \wedge |\sin^{-1}(\frac{x}{\sqrt{x^2 + y^2}})| \geq \sin^{-1}(\frac{b}{\sqrt{a^2 + b^2}})$

But then i get this for $a=12$ and $b=6$:

enter image description here

If $a=b=6$, it looks fine:

enter image description here

peer
  • 866
  • Looks like a tent. – jonsno Jun 20 '17 at 09:50
  • 1
    @peer: It is easier to work with unit coordinates first, i.e. with $a = b = h = 1$. If you find the $f(x, y)$ for this, then the general one is $h f(x/a, y/b)$. In the above image you have left and right side walls where the surface normal only depends on the $z$ coordinate; if so, there is an easy form for $f(x, y)$, I believe. – Nominal Animal Jun 20 '17 at 10:05
  • @NominalAnimal But if my construction were correct, then all points from for example from $(10,6,0)$ to $(-10,-6,0)$ on the surface should form a parabola. But based on the above example, it seems that they don't form a parabola. Instead, on the short side, it looks correct. For example the points from $(-12,4,0)$ to $(12,-4,0)$ form a parabola. So basically, i am wondering about the curvature on the front side of the above example. – peer Jun 20 '17 at 10:20
  • Maybe this can help (as a starting point): https://en.wikipedia.org/wiki/Minimal_surface – Emilio Novati Jun 20 '17 at 10:23
  • @NominalAnimal Thanks for your hint, it seems to be correct now. I will try to figure out why my initial formula is wrong. – peer Jun 20 '17 at 10:30
  • 1
    You might try $f(x, y) = (1 - \max\left( \lvert x \rvert ,, \lvert y \rvert \right))^2$ in the unit case. Scaled to $a, b, h$, that becomes $$z(x,y) = h \left( 1 - \max\left( \left\lvert \frac{x}{a} \right\rvert ,, \left\lvert \frac{y}{b} \right\rvert \right) \right)^2$$I believe (but have not verified or tested). – Nominal Animal Jun 20 '17 at 10:56
  • @NominalAnimal Thanks, but i think it should be $f(x,y)=1-\max(|x|,|y|)^2$. – peer Jun 20 '17 at 21:59
  • @peer: Right, dome-like, not spire-like. That makes $$z(x,y) = h \left ( 1 - \max \left( \left\lvert \frac{x}{a} \right\rvert ,, \left\lvert \frac{y}{b} \right\rvert \right )^2 \right)$$which does seem correct in gnuplot. – Nominal Animal Jun 21 '17 at 06:21
  • I concur with @NominalAnimal's solution. The solid is the intersection of the "interiors" of two parabolic cylinders (truncated by the $xy$-plane); specifically, these cylinders: $$\frac{z}{h} = 1 -\left(\frac{x}{a}\right)^2\quad\text{and}\quad\frac{z}{h} = 1 - \left(\frac{y}{b}\right)^2$$ (NA's use of "max" correctly describes the top surface.) Note that the intersection of such a cylinder with any plane gives a parabola, so the intersection of the cylinders above gives the corner-to-corner parabolic seams of your solid (since the seams are uniquely determined by the apex and corners). – Blue Jun 21 '17 at 23:51

0 Answers0