# A Math function that draws water droplet shape?

I just need a quick reference. What is the function for this kind of shape?

Thanks.

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In reality water droplets are more spherical than that, especially at the top. –  Henry Jul 15 '11 at 7:18
@Henry, indeed, a falling liquid droplet will be close to spherical with a slightly flattened bottom part. The shape with the tail is something you will only see for a droplet running down a surface. In that case the tail forms due to a balance of capillary and viscous forces. –  Michiel May 26 '13 at 10:04

## 6 Answers

Luckily for you, though the admonition of "there's no real reason to expect that an arbitrary shape admits a nice description" is true in general, there is such a curve that might suit your needs: the piriform of Longchamps (piriform - "pear-formed"), with the parametric equation (changed a bit from the equations in the links; I'll leave the proof of equivalence to you):

\begin{align*}x&=a(1-\sin\,t)\cos\,t\\y&=b(\sin\,t-1)\end{align*}

Here for instance is the case $a=1,\quad b=\frac52$:

On the other hand, remember that in a genuine physical drop, surface tension compels the lower bulbous portion of the drop to be (more or less) spherical; this is of course not the case with the piriform.

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If you're creating a solid version, it is trivial to make a surface of revolution from the piriform. –  Ｊ. Ｍ. Jul 15 '11 at 3:45
Why did you choose Piriform approximation instead of Joukowski? Is there some physical or mathematical reason for the choice? –  hhh Mar 25 '13 at 23:44
@hhh, the reasons were mostly aesthetic in nature. If you could do better than either the airfoil or the piriform, you're most welcome to try your luck. –  Ｊ. Ｍ. Apr 2 '13 at 13:15

Or you might like this one, which is a Joukowski airfoil.

\eqalign{x &= q\sin \left( t \right) -{\frac {q\sin \left( t \right) }{ \left( 1-q+q\cos \left( t \right) \right) ^{2}+{q}^{2} \left( \sin \left( t \right) \right) ^{2}}}\cr y &= q\cos \left( t \right) +{\frac {1-q+q\cos \left( t \right) }{ \left( 1-q+q\cos \left( t \right) \right) ^{2}+{ q}^{2} \left( \sin \left( t \right) \right) ^{2}}}\cr}

where $1/2 < q < 1$. Here is the case $q = 2/3$:

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As a note: it looks a bit complicated from the Cartesian point of view, but the equation looks much better when one switches to complex form: $$z(t)=i\left(q\exp(-i t)+\frac1{1-q+q\exp(-i t)}\right)$$ –  Ｊ. Ｍ. Jul 16 '11 at 4:44
Could you mind confirming the graph represents the case $q=2/3$? I ask this because I got a different shape. For $q=1/3$ I got a similar graph but all of it below $x=0$. But I may be wrong. –  Américo Tavares Jul 17 '11 at 13:34

In general, there's no real reason to expect that an arbitrary shape admits a nice description. It is certainly true that many shapes do come as the zero-locus to some nice expression, but when given a shape "in the wild", the only real thing a person can do is approximate it. To that end, here are some suggestions that may help you work out an expression that suits your needs.

I'm assuming that your figure has rotational symmetry about the axis through the tip and the bottom of the raindrop. This means that it suffices to be able to describe the cross section (the outline of a raindrop, if you will); you can recover the rest by working in spherical coordinates. Also notice that the bottom of the raindrop looks fairly spherical. This means that your cross-section should approximate a circle in the bottom half. From this point, we can just think about starting with a circle, and tweaking it a bit so that it starts to look like the outline of a raindrop. We can accomplish this in polar coordinates, so that we just need to describe the radius of our figure for a given angle; let's call this function $f(\theta)$. For simplicity, let's measure angle clockwise from the vertical, so that we are starting near the point of the raindrop.

Let's describe some of the properties that $f$ should have. The first thing is that since the outline of our drop has an additional symmetry about the $y$-axis, we should have that $f(\theta) = f(2\pi - \theta)$. Since the bottom of the drop is roughly spherical, on the range $[\pi/2, 3\pi/2]$, $f$ should be approximately $1$. On the range $[0, \pi/2]$, $f$ should be decreasing from $~2$ to $1$.

How do we get the "point" in the raindrop? Well, "pointiness" is nothing more than a condition on the slope of $f$ near $\theta = 0$; $\frac{dy}{dx}$ should be approaching $-\infty$. And it's a nice calculus exercise to work out how to compute $\frac{dy}{dx}$ in terms of $f$. From this point, it's just a matter of picking a function that satisfies these properties, and tweaking such a function to suit your needs.

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Thanks! that's what I need! –  Yonk Shi Jul 15 '11 at 0:07

You may also try Maple to find a kind of water droplet as follows:

  [> with(plots):
[> implicitplot3d(x^2+y^2+z^4 = z^2, x = -1 .. 1, y = -1 .. 1, z = -1 .. 0, numpoints = 50000, lightmodel = light2, color = blue, axes = boxed);


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Blue!! This is wonderful +1 –  amWhy Sep 2 '13 at 13:21
@amWhy: I knew you'd like it. :-) –  B. S. Sep 3 '13 at 1:12

You can see more about this under Piriform parametrization here.

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This can be considered as a trapezium of very small base and a arc of circle below it.

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trapezia (or trapezoids) generally have straight sides, no? –  Willie Wong Oct 17 '12 at 6:36