# Lemniscate Polygon

Bob Sneidar bobsneidar at iotecdigital.com
Wed Nov 3 11:25:20 EDT 2021

```Ok, you googled that, didn't you?? ;-)

Bob S

> On Nov 3, 2021, at 24:29 , Mark Waddingham via use-livecode <use-livecode at lists.runrev.com> wrote:
>
> Hi Roger,
>
> On 2021-11-02 22:27, Roger Guay via use-livecode wrote:
>> Dear List,
>> Bernd has produced an absolutely beautiful animation using a
>> Lemniskate polygon that was previously provided by Hermann Hoch. Can
>> anyone provide some help on how to create this polygon mathematically?
>> Since the equation for a Lemniskate involves the SqRt of negative
>> numbers, which is not allowed in LC, I am stumped.
>> You can find Bernd’s animation here:
>> https://forums.livecode.com/viewtopic.php?f=10&t=36412
>> <https://forums.livecode.com/viewtopic.php?f=10&t=36412>
>
> In general lemniscates are defined as the roots of a specific kind of quartic (power four) polynomials of the pattern:
>
>    (x^2 + y^2)^2 - cx^2 - dy^2 = 0
>
> So the algorithms for solving them you are probably finding are more general 'quartic polynomial' solvers - just like solving quadratic equations, the full set of solutions can only be computed if you flip into the complex plane (i.e. where sqrt(-1) exists) rather than the real plane.
>
> However, there is at least one type of Lemniscate for which there is a nice parametric form - Bernoulli's lemniscate, which is a slightly simpler equation:
>
>    (x^2 + y^2)^2 - 2a^2(x^2 - y^2) = 0
>
> According to https://mathworld.wolfram.com/Lemniscate.html, this can be parameterized as:
>
>    x = (a * cos(t)) / (1 + sin(t)^2)
>
>    y = (a * sin(t) * cos(t)) / (1 + sin(t)^2)
>
> Its not clear what the range of t is from the article, but I suspect it will be -pi <= t <= pi (or any 2*pi length range).
>
> So a simple repeat loop where N is the number of steps you want to take, and A is the 'scale' of the lemniscate should give you the points you want:
>
>    repeat with t = -pi to pi step (2*pi / N)
>       put A * cos(t) / (1 + sin(t)^2) into X
>       put A * sin(t) * cos(t) / (1 + sin(t)^2) into Y
>       put X, Y & return after POINTS
>    end repeat
>
> Warmest Regards,
>
> Mark.
>
> --
> Mark Waddingham ~ mark at livecode.com ~ http://www.livecode.com/
> LiveCode: Everyone can create apps

```