Is the maximal temperature of the curlicue fractal acheived by $e\times\gamma$? The Curlicue Fractal is defined as follows:
Choose an irrational number $s$ and a horizontal unit segment with angle $\phi_0 = 0$. Define $\theta_{n+1} = \theta_{n} + 2 \pi s \pmod{2 \pi}$, with $\theta_0=0$. To the previous segment, add a new unit segment with angle $\phi_{n+1} = \theta_{n} + \phi_{n} \pmod{2 \pi}$. The resulting series of line segments is the curlicue fractal. The "temperature" of these fractals measures the boundedness of these curves.
$s = e\times\gamma$ creates a very high temperature for this fractal, far greater than all the millions of other irrational numbers I've tried.  Does any other irrational number beat it?


The value $s = e\times\gamma$ reaches a maximum temperature of 2433.73 at 460024 steps.  For awhile, this has been the highest achieved temperature. 
Three identical maximal squares can be placed in an equilateral triangle. The Calabi triangle is the only other triangle with that property. 

$x^3+x^2-7x+1=0$ with $x\approx 2.10278...$ has a temperature of 4408.7 after 2005399 steps, a new record.  The ratio of sides in a Calabi triangle is $(x+1)/2$.
Here's the Calabi Curlique Curve after 2005399 steps: 

Can anyone find an irrational value with a temperature higher than 4408.7?
And Somsky has the answer, go for extreme continued fractions. Best I've gotten so far is 12/(Root[-2 + 4 #1 - 6 #1^2 + #1^3 &, 1]^24 - E^(Pi Sqrt[163]), about
0.4999999999999999780983596122, continued fraction 0, 2, 11414670114816032, 39, 1, 2, 12, 1, 12178299, 1, 1, ... I estimate it will take 20 quintillion steps to break out.  Here's the first million steps:

Here's 2 million steps of Somsky's curve:

Wild.  
