[Schubert](https://en.wikipedia.org/wiki/Hermann_Schubert) 
showed that a plane algebraic curve of degree $d$ has
at most
$$
\tfrac{1}{2} d (d-2) (d-3) (d+2) 
=
\tfrac{1}{2}d^4-d^3-\tfrac{9}{2} d^2 +9 d
$$
bitangents (a.k.a., double tangents).
And this number of real bitangents is achievable for a quartic,
such as Trott's quartic: 
<hr />
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
[![Trott28bits][1]][1]
<br />
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
<sup>
(Image from [Mathworld](http://mathworld.wolfram.com/Bitangent.html).)
</sup>
<hr />
(I learned this from Jan-MagnusØkland [@MSE](https://math.stackexchange.com/q/2494999/237).)

I have a curve $C$ that is a subcurve of a real algebraic plane curve of degree $d$, with these restrictions:

 - $C$ is connected (unlike Trott's quartic).
 - $C$ is simple, i.e., non-self-intersecting, i.e., it is embedded.
 - $C$ has no cusp singularities.
 - $C$ is an open curve, i.e., it is not closed to a cycle.

My question is: 

> ***Q***. Under these restrictions, is there a smaller upperbound
than provided by Schubert's formula? In particular, are there curves $C$ meeting
my conditions still with $\Omega( d^4 )$ bitangents? Maybe only $O(d^3)$?





  [1]: https://i.sstatic.net/GQ4PH.jpg