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:

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          [![Trott28bits][1]][1]
          ^{(Image from [Mathworld](http://mathworld.wolfram.com/Bitangent.html).)}

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(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 instead $O(d^3)$ holds?