no reason a priori to assume that this should be the case. Why should choosing a homotopy class for a swaddling map have anything to do with choosing a framing for a knot? In both of these approaches, we reach the image on $S^{2}$ via the Gauss map of the tangent bundle to an embedding into $S^{3}$ of $S^{1}$, but in the first approach we come to this image by first embedding $S^{1}$ into $D^{2}$ and then getting to $S^{2}$ via the swaddling map $\beta$, while in the second approach we first map to $SO(3)$ by choosing a framing (we shall call this map $\phi$), and then map down from there onto $S^{2}$. The situation is schematically depicted in the commutative diagram below:

$$\xymatrix{D^2 \ar[drr]_(.6)\beta
&& SO(3) \ar[d]^{e_{1}} \\
S^{1} \ar@{^{(}->}[u] \ar[rr]_{\dot{\gamma}} \ar[urr]^(.6)\phi &&
S^{2}}$$

In this notation, $\tau_{\beta}(\gamma)=\frac{1}{2\pi}\int_{D^{2}}\beta^{*}\omega$, while $\tau_{\phi}(\gamma)=\frac{1}{2\pi}\int_{S^{1}}\phi^{*}\tau$ for $\tau$ a pre-image of omega via the map $e_{1}$. Equality of these terms would follow from the existence of a map $\sigma$ such that the following diagram commutes:

$$\xymatrix{D^2 \ar[drr]_(.6)\beta \ar@{-->}[rr]^\sigma
&& SO(3) \ar[d]^{e_{1}} \\
S^{1} \ar@{^{(}->}[u] \ar[rr]_{\dot{\gamma}}  \ar[urr]^(.6)\phi &&
S^{2}}$$ For in that case $$\tau_{\beta}(\gamma)=\frac{1}{2\pi}\int_{D^{2}}\beta^{*}\omega=\frac{1}{2\pi}\int_{D^{2}}\sigma^{*}d\tau=
\frac{1}{2\pi}\int_{S^{1}}\phi^{*}\tau=\tau_{\phi}(\gamma)$$ The second equality stems from the fact that the diagram is 