Can you determine the length of a curve by the lengths of its projections onto planes? If $\,\Gamma \subset \mathbb R^n$ is $1$–rectifiable, then its Hausdorff measure is equal to its integral geometric measure. That is, 
$$\displaystyle\mathcal H^1\left(\Gamma\right) = \int_{G\left({1,\mathbb R}^n\right)} \int_K \operatorname{Card}\left(\left\lbrace {y \in \Pi_K}^{-1}\left(\left\lbrace x\right\rbrace \right)\right\rbrace \right)\, {d\hspace{0.125ex}\mathcal H}^1\left(x\right)\, d \hspace{0.125ex}\Theta_{{1,\mathbb R}^n}\left(K\right),$$
where $\operatorname{Card}(S)$ means the number of points in ${S,\Pi}_K$ denotes orthogonal projection onto ${K,\mathcal H}^1$ denotes the one–dimensional Hausdorff measure, $G\left({1,\mathbb R}^n\right)$ denotes the Grassmanian of unoriented lines through the origin in $\mathbb R^n$, and $\Theta_{{1,\mathbb R}^n}$ is the unique (up to suitable constant) finite Borel measure on $G\left({1,\mathbb R}^n\right)$ which is invariant under the action of the orthogonal group.
I would like to know if the following is true:
$$\displaystyle\mathcal H^1\left(\Gamma\right) = \int_{G\left({2,\mathbb R}^n\right)} \int_V \operatorname{Card}\left(\left\lbrace {y \in \Pi_V}^{-1}\left(\left\lbrace x\right\rbrace \right)\right\rbrace \right)\, {d\hspace{0.125ex}\mathcal H}^1\left(x\right)\, d \hspace{0.125ex}\Theta_{{2,\mathbb R}^n}\left(K\right),$$
The more general question where the numbers $1$ and $2$ are replaced by $j$ and $k$ with $j<k<n$ is also of interest.
 A: The formula should work with general $1\leq j\leq k\leq n$, modulo a dimensional constant, using the Area Formula for rectifiable sets and Fubini. 
I rewrite here the Area Formula as can be found in Theorem 2.91 in Ambrosio, Fusco, Pallara - Functions of bounded variations and free discontinuity problems; I believe a similar theorem can be found in Federer's book.
Area Formula for rectifiable sets Let $f:\mathbb R^n\to\mathbb R^k$ be a Lipshcitz function, and $\Gamma\subset\mathbb R^n$ a countably $\mathcal H^j$-rectifiable set. Then the multiplicity function $\mathcal H^0(\Gamma\cap f^{-1}(\{x\}))$ is $\mathcal H^j$-measurable in $\mathbb R^k$ and
$$\int\limits_{\mathbb R^k} \mathrm{Card}\left(\Gamma\cap f^{-1}(\{x\})\right)d\mathcal H^j(x)=\int\limits_\Gamma (J_jd_y^\Gamma f)\, d\mathcal H^j(y)
$$
where $J_jd_y^\Gamma f$ is the $j$-dimensional Jacobian of the tangential differential $d_y^\Gamma f$ at the point $y$.
In the case when $f$ is differentiable and $\Gamma$ is a regular manifold of dimension $j$, $d_y^\Gamma f$ is just the restriction of the differential $df$ to $Tan^j(\Gamma,y)$, the $j$-dimensional tangent space at $y$. Moreover $J_j d_y^\Gamma f$ is intended as the Jacobian of this map as a linear map from the tangent space $Tan^j(\Gamma,y)$ to its image in $\mathbb R^k$. Clearly for rectifiable sets one has to prove first that all these things are well defined (for example that a Lipschitz function is tangentially differentiable at $\mathcal{H}^j$-almost every point of an $\mathcal H^j$-rectifiable set...) but I refer to AFP for the details.

Applying this to our case we obtain
$$\int\limits_V \mathrm{Card}\left(\Pi_V^{-1}(\{x\})\right) d\mathcal H^j(x)=\int\limits_\Gamma (J_j d_y^\Gamma \Pi_V)\, d\mathcal H^j(y).
$$
If we now integrate over $G(k,\mathbb R^n)$ and change the order of integration we obtain
\begin{align}
\int\limits_{G(k,\mathbb R^n)}d\Theta_{k,\mathbb R^n}(V)\int\limits_V \mathrm{Card}\left(\Pi_V^{-1}(\{x\})\right) d\mathcal H^j(x) &= \int\limits_{G(k,\mathbb R^n)}d\Theta_{k,\mathbb R^n}(V)\int\limits_\Gamma (J_j d_y^\Gamma \Pi_V)\,d\mathcal H^j(y)\\
&=\int\limits_\Gamma d\mathcal H^j(y)\int\limits_{G(k,\mathbb R^n)}d\Theta_{k,\mathbb R^n}(V)(J_j d_y^\Gamma \Pi_V)\\
& =\mathcal{H}^j(\Gamma) \,I(j,k,n)
\end{align}
where
$$I(j,k,n)=\int\limits_{G(k,\mathbb R^n)}d\Theta_{k,\mathbb R^n}(V)(J_j d_y^\Gamma \Pi_V)
$$
is a dimensional constant independent of $y$, since it is the average of the "$j$-dimensional stretch factors" $J_j\Pi_V(y)$ over all $k$-dimensional subspaces $V$, so whichever approximate tangent plane $\Gamma$ has at $y$ this factors integrate to the same constant (I hope this last part was clear enough).
