Can the Fibonacci Spiral be expressed as a polar equation?

Question

Related to this question: can the Fibonacci Spiral be expressed as a polar equation?

I know the Golden Spiral can be and the Fibonacci differs in that it uses consecutive arcs of a circle for each square.

Answer 1

The Fibonacci spiral is a member of what I call pseudospirals because they are composed of circular arcs, rather than a continuous curve. About ten years ago I set out to find the equation for the Fibonacci spiral only to realize that I could do a lot more with it. An early version of my work can be found here.

The pseudospiral equation will give the spiral for any sequence $S={s_1,s_2,s_3,...s_n}$ of arbitrary length and arbitrary turn angle $\Delta\theta$. This equation is expressed in the complex plane as follows:

$$ z(\theta)=s_1-\sum_{n=1}^N(s_n-s_{n-1})(e^{i\theta_n}-e^{i\theta})u(\theta-\theta_n) $$

where $s_n$ are the elements of the sequence, $\theta_n=(n-1)\Delta\theta$, and $u$ is the Heaviside step function. These spirals unfold in the counterclockwise direction from a point on the $x$-axis equal to the first number in the sequence. The complete derivation is in the link provided above along with many examples.

If you want to compare this with the golden spiral, say $z=e^{(b+i)\theta}$, where the flair coefficient, $b=2\text{ln}\varphi/\pi$ and $\varphi$ is the golden ratio, you will have to scale and rotate this to be compatible with the pseudospiral.

Answer 2

Can it be written in polar form? For sure, it's a plane curve so there will be a parameterisation in polar coordinates. However, there are a couple of issues:

1. The scale will depend on how many squares you choose to draw (i.e. how many terms of the Fibonacci sequence you want to use for the curve).

2. The curve will be defined as a piecewise collection of quarter-circles.

3. The circles won't be centred at the origin, so the equations won't necessarily be particularly nice.

The general equation for a circle with centre $(r_0, \gamma)$ and radius $a$ is

$$r^2 - 2r r_0\cos(\theta - \gamma) + r_0^2 = a^2$$

or equivalently

$$r = r_0 \cos (\theta - \gamma) + \sqrt{a^2 - r_0^2\sin^2(\theta - \gamma)}$$

Noting that to achieve a quarter circle we actually need $\theta - \gamma$ to cover the appropriate quadrant.

So you will wind up with an equation that looks something like:

$$r = r_n \cos(\theta - \gamma_n) + \sqrt{F_n^2 - r_n^2 \sin^2(\theta - \gamma_n)}$$

where $F_n$ is the $n$th Fibonacci number, and $r_n$ and $\gamma_n$ are the co-ordinates of the appropriate corner of the $n$th square, and where $\theta \in \left[\gamma_n + n \frac{\pi}{2}, \gamma_n + (n + 1) \frac{\pi}{2}\right)$ is I think the correct way to define the intervals of the angles.