Golden Angle Golden Spiral

Question

Note. The answer must produce something self-similar that progresses in such a manner regardless of whether positive or negative values are graphed.

Note. The "distance between 'arcs'" I'm talking about is defined in my image (the red dotted line represents such a distance between 'arcs').

The parametric equations, $x,\ y=\sin(t)·φ^{t/π},\ \cos(t)·φ^{t/π}$, yield a spiral where the distance between the 'arcs' grows (/shrinks) gradually reaching powers of $φ$ at every multiple of $π$ (every $180$ degrees). (See My Graph.)

I want to find parametric equations for a spiral where this distance grows (/shrinks) by powers of φ in exactly the same way relative to multiples of the golden angle, that is, $\frac{2π}{φ}$ (Instead of doing so relative to π). ($φ$ is defined as $φ=\left(\frac{1+5^{1/2}}{2}\right)$ herein!)

Such a thing seems totally possible; however I have had little luck thus far. So, I'd be thankful for some help! Thanks.

Graph:

https://www.desmos.com/calculator/0iotz6gmdu

Answer 1

Expanding-upon comment, if we consider polar coordinates $(r,\theta)=(f(t),t)$, with initial condition that the curve passes through $(x,y)=(b,0)$, then the curve's defining relation is $$f(k\omega) - f(k\omega-2\pi) = \phi^{k+c} = \exp_\phi(k+c) \qquad\qquad f(0) = b\tag{1}$$ where $\omega := 2\pi/\phi$, $k$ is an integer, and $c$ is there just to make sure the multiples of $\omega$ match with the appropriate powers of $\phi$. The continuous, $k$-free version of $(1)$ is

$$f(t)-f(t-2\pi) = \exp_\phi\left(\frac{t}{\omega}+c\right) \qquad\qquad f(0)=b \tag{2}$$

Suppose the solution has the form $$f(t) = a\left(\phi^{t/\omega}-1\right) + b \tag{3}$$ Then $(2)$ requires $$\exp_\phi\left(\frac{t}{\omega}+c\right) = a\exp_\phi\left(\frac{t}{\omega}\right) - a\exp_\phi\left(\frac{t}{\omega}-\frac{2\pi}{\omega}\right) = a\exp_\phi\left(\frac{t}{\omega}\right) - a\exp_\phi\left(\frac{t}{\omega}-\phi\right) \tag{4}$$ Dividing the left-most and right-most expressions by $\phi^{t/\omega}$, $$\phi^c = a - a\phi^{-\phi} \qquad\to\qquad a = \frac{\phi^{\phi+c}}{\phi^\phi-1} \tag{5}$$ so that $(3)$ becomes

$$f(t) = \frac{\phi^{\phi+c}}{\phi^\phi-1}\left(\phi^{t/\omega}-1\right) + b \tag{$\star$}$$

Answer 2

You may parametrize the proposed spiral as

$$x=\sin(t)·\phi^{\frac{t\phi}{2\pi}},\>\>\>\>\>y =\cos(t)·\phi^{\frac{t\phi}{2\pi}}$$

which yields the distance between the arcs grows/shrinks gradually reaching powers of $\phi$ at every multiple of $\frac{2\pi}\phi$.

Answer 3

If I understand correctly you are seeking a spiral that increases by a factor of $\varphi$ for every angular change of $2\pi/\varphi$. This is the very definition of the logarithmic spiral, which is given in complex and Cartesian coordinates as follows

$$ z=e^{(b+i)\theta}\\ x=e^{b\theta}\cos\theta\\ y=e^{b\theta}\sin\theta $$

where $b$ is the flair coefficient, which is defined as the logarithm of the growth rate, say $p$, divided by the angular change, say $\Delta\theta$, thus

$$b=\ln(p)/\Delta\theta$$

For example, for the golden spiral we have $b=2\ln\varphi/\pi$, or a growth of $\varphi$ per $90^{\circ}$. In your case, the solution is

$$b=\frac{\varphi\ln\varphi}{2\pi}$$

The figure below shows a comparison of these two spirals. Clearly, the golden spiral grows at a much faster rate, namely, $4/\varphi$.