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I searched some time over the web and this mathematics site without finding what I am searching for.

What I need is the 2D function of a simple balloon. Not an hot air balloon but a simple single-colored party balloon (without the knot at the bottom). Has anyone an idea for this?

Here an example for what I mean:

balloon

I only want the shape, not a complex function including the light effects, the knot or something else.

Of course, the upper part is just half of a circle, but I need some more exact function for the lower part.

Thanks for your help!

Eman Yalpsid
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Christian
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3 Answers3

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There's no physics here, but the polar graph $$ r = 1 + 0.375\operatorname{sech}(2.75(\theta + \tfrac{\pi}{2})) $$ is a good visual match for your image:

A profile of a fake balloon


Edit: I hit on the parameters after a fortuitously small number of tries. Here are images showing the profiles if the "width" or "height" (respectively) of the sech hump is varied:

Balloon: Varying the width of the hump Balloon: Varying the height of the hump

  • Nice. You might provide some small examples showing how tuning the two parameters modifies the shape. – Ethan Bolker Oct 06 '16 at 23:53
  • The first graph is exactly what I need, but when trying it out there is just kind of a circle as you can see here:

    http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=8d8e2c27bcaa121d6ee0de4b98774bb4&title=Polar%20Graphs&theme=blue&i0=1%2B0.375sech(2.75(%CE%B8%2BPi%2F2))&i1=0&i2=2Pi&podSelect=&showAssumptions=1&showWarnings=1

    What am I doing wrong - just a hint?

    – Christian Oct 07 '16 at 06:33
  • Oh, I have found my mistake: Use -PI..+PI for angle, not 0..2PI. Thanks a lot!! – Christian Oct 07 '16 at 06:57
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The physics dictating the shape at the bottom is subtle, but assume for simplicity that the shape is a bit more than a semicircle on top, and a parabola at the bottom.

Then if we take the center of the circle to be the origin, there are two parameters to describe the balloon: The angle past a semicircle that is still in the "upper" part, and the circle radius. I will take the radius to be one; I assume you can scale for a larger balloon.

The angle $\theta$ beyond a semicircle is a bit arbitrary but it looks from the picture that it it goes $30^\circ$ past the semicircle. Then The parabola goes thru the points $(-\frac{\sqrt{3}}{2}.-\frac12)$ and $(+\frac{\sqrt{3}}{2}.-\frac12)$, and has slope $\sqrt{3}$ at $x=\frac{\sqrt{3}}{2}$. This specifies the parabola unicles, so the shape of the balloon is $$ y = \left\{ \begin{array}{ccccl} +\sqrt{1-x^2} &,& -\sqrt{1-x^2} & \mbox{when }& -1 \leq x \leq -\frac{\sqrt{3}}{2} \\ +\sqrt{1-x^2} &,& \frac{\sqrt{3}}{2} x^2 - \frac{4+3\sqrt{3}}{8} & \mbox{when }& -\frac{\sqrt{3}}{2} < x < \frac{\sqrt{3}}{2} \\ +\sqrt{1-x^2} &,& -\sqrt{1-x^2} &\mbox{when }& x \leq +\frac{\sqrt{3}}{2} \leq 1 \end{array} \right. $$

Mark Fischler
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  • Nice and easy function. I realized it, but it is not exactly what I need. Sorry, but thanks for your entry! – Christian Oct 07 '16 at 06:02
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Try this. A 2D graph of a balloon: $$2x^2 + y^2 - 3\sin(y) - 7 = 0.$$ Also, if you want a line under the balloon, try $$\sqrt{2\cos(0.5y + 2.6)} + x -0.2 = 0.$$ Hint: type in GeoGebra.