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So I have two logarithmic spirals in parametric form $$ x(t) = ae^{bt}\cos t \\ y(t)=ae^{bt}\sin t $$ and $$ x'(t) = \alpha e^{\beta t}\cos t \\ y'(t)=\alpha e^{\beta t}\sin t $$ With $\beta$ and $b$ having opposite signs so the spirals grow in opposite directions. Setting the $x$'s and $y$'s equal and solving for $t$ I get $$ t = \frac{1}{b-\beta}\ln\frac{\alpha}{a} $$ Evaluating $(x(t),y(t))$ for that value does produce a single point of intersection for the two spirals, but I can't seem to find the general form that would give all the points of intersection.

I've fiddled around with it in polar form, both as $r(\theta)$ and as $\theta(r)$, and solved for the intersection points, but I couldn't figure out the general form in those cases either. (I tried integer coefficients, and multiples of $\pi$ and $2\pi$ to no avail.)

Eventually I realized I would prefer the parameterized approach because it'll be easiest to work in terms of $t$.

But I still feel lost in how to find the other points of intersection.

Yrast
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    When you equated the x's, what about the sine and cosine terms? You have to equate them as well to zero. – Aniruddha Deshmukh Jul 14 '17 at 06:06
  • Ah! I simply canceled them out! I'm not sure yet how I should handle them, but I can see how that could be where I dropped whatever "factor" would return the other values! – Yrast Jul 14 '17 at 06:11
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    Yes @Yrast. You will get other points of intersection in terms of $\pi$ or $\dfrac{\pi}{2}$. And also, you might get something as $\left( \left( 2n + 1 \right)\dfrac{\pi}{2}, n\pi \right)$. I am not sure about this. But it should come somewhat like this. – Aniruddha Deshmukh Jul 14 '17 at 06:39
  • For instance $ae^{bt}\sin t = \alpha e^{\beta t}\sin t$ not only when $ae^{bt}= \alpha e^{\beta t}$ but also when values of $t=\pi-t'+k\pi $ or $t=t'+2n\pi$ and $t=-t'+2h\pi$ for cosine – Raffaele Jul 14 '17 at 08:02

2 Answers2

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You are incorrect when you say With β and b having opposite signs so the spirals grow in opposite directions. This parameter is called the flair coefficient, its sign indicates whether the spiral is growing or shrinking, but both evolve in the anticlockwise direction. Thus it's quite possible that they do not intersect at all other than at their mutual starting point of $\theta=0$.

You should run some plots.

Cye Waldman
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In this reply I will give a solution for the intersection of two logarithmic spirals.

Consider two spirals in the complex plane with different angular regions, e.g.,

$$ z=e^{(b+i)u}\\ w=e^{(\beta+i)v} $$

and we seek all the points where $z=w$. We can expand these and separate the real and imaginary parts to obtain

$$ e^{bu}\cos u=e^{\beta v}\cos v\quad \quad \quad (1)\\ e^{bu}\sin u=e^{\beta v}\sin v\quad \quad \quad (2) $$

If we multiply (1) by $\sin u$ and (2) by $\cos u$ and subtract, we obtain

$$0=e^{\beta v}[\sin u\cos v-\cos u\sin v]=e^{\beta v}\sin(u-v)\quad \quad \quad (3)$$

Similarly, if we multiply (1) by $\cos u$ and (2) by $\sin u$ and add, we obtain

$$e^{bu}=e^{\beta v}[\cos u\cos v-\sin u\sin v]=e^{\beta v}\cos(u-v)\quad \quad \quad (4)$$

Equation (3) tells us that $(u-v)=n\pi$. Equation (4) tells us that $n$ must be even since $\cos n\pi$ alternates in sign, but (4) must be positive. So we conclude that

$$v=u+2n\pi$$

(assuming that $v$ the larger of the two).

Finally, going back to the beginning, we must also have $|z|=|w|$, and therefore

$$e^{bu}=e^{\beta v}=e^{\beta(u+2n\pi)}$$

so that

$$ u=\frac{2n\pi\beta}{b-\beta}\\ v=u+2n\pi $$

The figure below shows a sample calculation (cleverly chosen to give simple results). Here

$$ b=\frac{2\ln\varphi}{\pi},\quad \text{the golden spiral}\\ \beta=\frac{\ln\varphi}{2\pi}\\ $$

$$ u=\frac{2n\pi}{3}\quad \{120,240,360,480,600,720^\circ...\}\\ v=\frac{8n\pi}{3}\quad \{480,960,1440,1920,2400,2880^\circ...\} $$

enter image description here

Cye Waldman
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  • Thank you so much! This is precisely what I was trying to do. After your comment last night I played around a bit and realized I should switch to a complex valued function, but then today I was still thoroughly stumped. I don't think I could have figured most of that out on my own, maybe back while taking complex variables, but my math knowledge has atrophied pretty severely in the last ten years. I'm still not sure I follow every step completely but I think I will with a bit more effort. Thanks again! – Yrast Jul 15 '17 at 21:04
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    @Yrast Your thanks are appreciated. Notice that these spirals are both anticlockwise, if one of them was clockwise, say e^{(\beta-i)v}$ then Eqs. (3) and (4) would be slightly different and the results would be different too, of course. – Cye Waldman Jul 15 '17 at 22:20