Spiral function with constantly increasing separation between its loops

Question

Spirals following the polar equation $r=a+bθ$ are known to have constant separation between each of its loops, I show this by transforming the polar equation into Cartesian equation by using the transformations $r=(x^2+y^2)^{0.5}$ and $ θ=tan^{-1}\left(\frac{y}{x}\right)$. For computing the separation between loops $n$ and $n+1$, I get the following formula

$x_{n+1}-x_{n}=b π [(n+1)-(n)]= b π$

As shown, the separation between the loops is a constant. When I use the equations $r=a+bθ^p$ or $r=a+bp^θ$, and then compute the separation between loops, for instance, $n$ and $n+1$, I get the following results;

$x_{n+1}-x_{n}=b 2π^p [(n+1)^p-(n)^p]$ for $r=a+bθ^p$

$x_{n+1}-x_{n}=b(p^{2π}-1)p^{n 2π }$ for $r=a+bp^θ$

As one would realize, the separation between loops of those two spirals increase, but not with a constant rate. I tried many more formulae, but I didn’t get any spiral with constant increase rate between its loops, i.e., stratifying the formula;

$x_{n+1}-x_{n}=An$

Wherein $A$ is a constant. Does anyone here know which equation would give me the spiral shape I’m looking for? Thank you so much guys.

Answer 1

You can use $r = a + b\left(\left(\frac\theta\pi\right)^2 - \frac\theta\pi\right).$

This way,

$$\begin{equation} \begin{split} x_{n+1}-x_{n} &= b\left[\left(\frac{(n+1)\pi}{\pi}\right)^2 - \frac{(n+1)π}{\pi}-\left(\frac{nπ}\pi\right)^2 - \frac{nπ}{\pi} \right] \\ &=b[n^2 + 2n + 1 - n - 1 - n^2 + n] \\ &= b[2n] \end{split} \end{equation}$$

I think this was what you were looking for.

Answer 2

Placing $ \theta $ in position of exponent makes it a log spiral, making geometric progression of radial ray lengths. Taking $$ r= r_0 + \dfrac{ p \theta }{2 \pi}$$ we have $p$ constant as radial increase or pitch for each revolution. It is in arithmetic progression.

Since $$ r =\sqrt {x^2+y^2}$$ $$ \sqrt{x^2+y^2} = \sqrt{x_0^2+y_0^2} + \dfrac{ p \theta }{2 \pi}$$ Setting $y=0$ we must have for x-axis start of Archimedean spiral $$ x= x_0 + \dfrac{ p \theta }{2 \pi}$$ and setting $x=0$ we must have for y-axis start of the spiral $$ y= y_0 + \dfrac{ p \theta }{2 \pi}$$ and the same holds good for any radial ray $ \theta = \alpha$.

So $p$ is radial pitch at any arbitrarily chosen $\alpha.$