Connections on a manifold and principal connections on the frame bundle Suppose $M$ is a manifold, and $E$ a vector bundle over $M$ equipped with a connection $\nabla $. If $F$ is the frame bundle of $E$, is there an explicit construnction of a connection on $F$ associated with $\nabla$ such that in this way connections on $E$ and $F$ are $1$-$1$ correspondent?

Edit for the bounty:
I really need an answer to this question, and as it was already posted I think that putting a bounty on it is the most sensible way to go.
To rephrase the question in my own terms: Let $M$ be a smooth $n$-manifold. We can associate the following principal $GL(n)$-bundle to it:
$$F = \{(m,\theta)|m\in M, \theta:\mathbb{R}^n\to T_mM\mathrm{\ lin.\ isom.}\}$$
with right action given by $(m,\theta)g = (m,\theta g)$. Its tangent space is defined (as for any other manifold) as a quotient of the space of paths on $F$. In order to get a more concrete representation, we need a way to differentiate "paths of frames," but as such paths can be seen as tuples of paths of vectors on $M$, it is enough to specify a connection $\nabla$ on $M$ to obtain the identification
$$T_{(m,\theta)}F \cong \{(\hat{m},\hat{\theta})|\hat{m}\in T_mM,\hat{\theta}:\mathbb{R}^n\to T_mM\}$$
where we identify the equivalence class of paths $[\gamma(t),\theta(t)]$ with $(\dot{\gamma}(0),(\nabla_{\dot{\gamma}}\theta)(0))$. This gives us a map
$$\{\mathrm{connections\ on\ }M\}\longrightarrow\{\mathrm{principal\ connections\ on\ F}\}$$
mapping $\nabla$ to $A([\gamma,\theta]) = \theta^{-1}\nabla_{\dot{\gamma}}\theta\in\mathfrak{gl}(n)$.
I believe there should be a way to invert this map (maybe only on a subset of the principal connections, though) but I cannot see how. Does anyone have an idea or a solution?
Remark 1: My question is in fact a special case of the original question on vector bundles, namely if we take $E=TM$.
Remark 2: I took a look at Taubes' book, as suggested in the answers, but I didn't find what I need (or maybe I found it, but wasn't smart enough to realize it).
 A: Ok, I had an inspiration, and found the following answer.

Identify $TM$ as the associated bundle $E = F\times_{GL(n)}\mathbb{R}^n$, where the action of $GL(n)$ on $\mathbb{R}^n$ is given simply by left multiplication. The identification between the two is given by
$$[(m,\theta),v]\in E\longmapsto\theta v\in T_mM.$$
Then we have a bijective correspondence between vector fields on $M$ and sections of $E$ where to $X:M\to TM$ we associate
$$\overline{X}(m) = [m,\theta,\theta^{-1}X(m)]$$
for any choice $\theta$ of frame at $m$. Now a connection $A\in\Omega^1(F;\mathfrak{gl}(n))$ gives us a way to differentiate $\overline{X}$, namely $d_A\overline{X}\in\Omega^1(M;E)$ ($1$-forms on $M$ with values in $E$).
Claim: The assignment
$$d_A\overline{X}\longmapsto\nabla X$$
is the required bijection, where from $\nabla$ we can recover $A$ by noticing that $d_A = d + \rho(A)$.
Proof: We have the further identification of $T^*M$ with $E^*=F\times_{GL(n)}(\mathbb{R}^n)^*$, where the action of $GL(n)$ on $(\mathbb{R}^n)^*$ is given by $g\cdot v^* = v^*\circ g$. This gives a correspondence between $\Omega^1(M)$ with sections of $E^*$ by
$$\overline{\beta}(m) = [m,\theta,\beta(m)\circ\theta]$$
in a way similar to the above. We have a natural isomorphism
$$\Omega^1(M;E)\cong\Omega^0(M;E^*),$$
thus the only thing we are left to show is that
$$d_A\overline{X} = \overline{\nabla X}.$$
This goes:
$$\begin{align}\overline{\nabla X}(\hat{m}) = & \langle[m,\theta,\nabla_{\theta-}X],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&\langle[m,\theta,d_A\overline{X}(\theta-)],[m,\theta,\theta^{-1}\hat{m}]\rangle\\=&d_A\overline{X}(\hat{m})\end{align}$$
for $\hat{m}\in T_mM$, where $\langle\cdot,\cdot\rangle$ is the natural pairing.
