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?

 A: 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.  
A: Refined Solution -- A Red Hot Curlicue
A more elegant variation on my previous solution is 
$s=55\sqrt{2}+92\sqrt{5}\>$ 
which reaches a temperature of approx $T=226\,000$ near $l=1\,060\,000$.
Note that numerically $s=(567/2)-\!1.39499...\!\times\!10^{-7}$.

And yes, that is an actual plot -- I didn't just make a huge Point[].
At $l=2\,000\,000$ I get the following. (Which for some odd reason differs from what Ed Pegg gets for what should be the same thing. I'm wondering if there are round-off issues somewhere.)


Previous Solution
Sort of a cheat, perhaps, but, assuming I've made no mistakes, $s=1/2+\pi/10^6$ will give a temperature of approx $T=10\,000$ at $l=47\,000$

Note that the slight deviation from $1/2$ will eventually come further into play, dropping the temperature. Eg, at $l=100\,000$ we have (not to the same scale)

The temperature for the curve for the first $100\,000$ steps is

Similar curves w/ increasing powers dividing the $\pi$ should allow arbitrarily high temperatures to be reached.
