TITLE: Diameter of random segment intersection graph?
QUESTION [9 upvotes]: I have an even number of points $n$ randomly distributed (uniformly) in a disk.
Then the points are randomly connected to form $n/2$ segments, a perfect
matching.
Finally, I form the intersection graph $G$ whose nodes are the segments,
and each of whose edges represents a proper intersection between a pair of segments.
An example is shown below:

 


My question is:

Q. What is the diameter of $G$ as $n \to \infty$?

I have some evidence that the diameter
approaches a constant $> 1$,
perhaps $4$ or $5$, 
but to be honest,
the evidence is not robust.

Added.
This may not be helpful, but for $n=1188$, $n_{\mathop{segs}}=n/2=594$, one
simulation resulted in $G$ having $37,117$ edges, 
mean vertex degree $125$, one connected component, and diameter $4$:

REPLY [2 votes]: Here is a way of thinking.  It may turn out to be misleading, but it may also suggest a
way to think properly about this problem.
Lets take a random instance of a matching and reconstruct it edge by edge. We will
start by maximizing the diameter of the intersection graph. To me this means making
a large cycle of crossing sticks, so that the intersection graph is a path or a cycle.
Suppose we are lucky, and manage to use 2/3 (say) of the sticks to create a subarrangement
of (induced) maximal possible diameter.
Now we place the rest of the sticks, which can only decrease the induced diameter.
With high probability, the diameter is reduced by a minimum ratio with each
new stick or perhaps a small number of new sticks.  In any case, I see O(log G)
many sticks needed to bring the diameter back down.  For large G, this still
leaves many sticks left.
It still leaves the possibility that the diameter grows like log log G, but the idea
of finding the subconfiguration inducing the largest diameter and then reducing
it appeals to me as an inroad to the problem.