Determine when does a spherical curve closes.

82 Views Asked by At

So I'm plotting curves in three dimensions on my computer. I'm currently using equations in spherical coordinates to define them. These are some examples, using ISO convention ( r , θ , φ ):

$\big(a.sin(b.\theta), \theta, c\big)$

$\big(a.sin(b.\theta), \theta, \theta\big)$

$\big(a.abs(sin(b.\theta)), \theta, c.sin(d.\theta)\big)$

How could I find the closing angle for these variations?

Maybe a good starting point should be looking at 2D rhodoneas. According to wolfram, a two-dimensional rhodonea curve ($r = a.cos(\frac{r}{s}.\theta)$) closes at a polar angle of $\theta=\pi .s.p$, where $p=1$ if $r.s$ is odd and $p=2$ if $r.s$ is even. But I can't understand why that happens.

Ref: https://mathworld.wolfram.com/Rose.html

I'm probably missing something or maybe I got something wrong, so if you have any thoughts on this I would love to know and discuss it further.

1

There are 1 best solutions below

4
On BEST ANSWER

If i understand correctly you are asking for the angle at which the curves are 'back at their starting point'. Let's start with the rhodoneas. You might know that the $\sin$ and $\cos$ functions are periodic with a period of $2\pi$, meaning that they are back at their starting point at an angle of $2\pi$ radians. A curve of the form $r=a\cos(\frac{n}{s} \theta)$ is back at its starting point when $\frac{n}{s} \theta$ is a multiple of $2\pi$ for the first time. If $n$ is even (a multiple of 2) this happens at $\theta=s \cdot\pi$, because $s \cdot \pi \cdot \frac{n}{s} = n\pi$ and $n$ is a multiple of two.

If $n$ is not even this happens at $\theta=2s \cdot\pi$.

Back to your curves:

$(a \sin(b\theta), \theta, c)$ closes when $b\theta$ and $\theta$ are both a multiple of $2\pi$ for the first time. If $b$ is an integer this happens at $\theta=2\pi$, if $b$ is a rational number given by $b=\frac{p}{q}$ with $p,q$ integers, this happens at $\theta=q\cdot 2\pi$.

If $b$ is irrational, the curve never closes because an integer multiple of $2\pi$ cant also be an integer multiple of $2\pi \cdot b$.

Similar logic applies to the other two curves.

EDIT:

I forgot to mention that $\theta$ needs to be an integer multiple of $2\pi$, otherwise the curve won't end at a point equivalent to $(0,0)$.