Let $F,E,B$ be manifolds (we may assume them to be compact, without boundary if necessary) and $$F\to E\to B$$ be a fibre bundle. Suppose $$w(F), w(B),$$ the Stiefel-Whitney class of $F$ and $B$, are known. I notice that in particular, if the bundle is trivial, then $E=B\times F$ and $w(E)=w(B)w(F)$.

Question: In general, are there any formulas to compute the Stiefel-Whitney class of the total space $E$ $$ w(E) $$ in terms of $w(B)$ and $w(F)$? Or even in terms of the cohomology ring $$ H^*(B;\mathbb{Z}_2), H^*(F;\mathbb{Z}_2) $$ And other factors?

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    $\begingroup$ Certainly not just in terms of $w(B)$ and $w(F)$; the Klein bottle is a circle bundle over the circle. $\endgroup$ – Mike Miller Sep 25 '15 at 3:24

Just to get started (a bit long for a comment):

For a fiber bundle $F\rightarrow E\overset{\pi}{\rightarrow}B$ one has an exact sequence of bundles over $E$

$$ 0\rightarrow T_vE\rightarrow TE\rightarrow \pi^{*}TB\rightarrow 0. $$

Here $T_vE:=\ker T\pi$ is the vertical part of the tangent space to $E$, i.e. these are the tangent vectors to the fibers. The choice of a complementary bundle amounts to the choice of a connection, but can be identified with $\pi^*TB$. So one sees that

$$ w(TE)=w(T_vE)\pi^*(w(TB)). $$

I think one needs more information if one wants to continue the computation. For example one can go on if $E$ is the projectivization of a vector bundle over $B$.

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