= Solution
Use the equal-time points to estimate <velocity> and <acceleration> with centered <finite differences>:
$$
V_k\simeq{P_{k+1}-P_{k-1}\over2\Delta t},\qquad a_k\simeq{P_{k+1}-2P_k+P_{k-1}\over(\Delta t)^2}.
$$
At the endpoints use one-sided formulas of the desired order. A differentiated path and its time parametrization can instead supply these quantities. Let $g_k$ denote the gravitational <acceleration>, directed downwards. In the accelerating camera, the effective downward vector is $g_k-a_k$, so the locally felt upward vector is $U_k=a_k-g_k$. For a stationary camera this gives $-g_k$, fixing the sign.
The <apparent-gravity camera frame> is obtained by <orthogonal projection> and a <cross product>:
$$
X_k={V_k\over\|V_k\|},\qquad W_k=U_k-(U_k\cdot X_k)X_k,\qquad Z_k={W_k\over\|W_k\|},\qquad \boxed{Y_k=Z_k\times X_k}.
$$
When $W_k\ne0$, $X_k,Z_k$ are perpendicular <unit vectors>, and $Y_k$ completes a right-handed <orthonormal basis>: $X_k\times Y_k=Z_k$. Also $U_k=(U_k\cdot X_k)X_k+\|W_k\|Z_k$, so the felt-up vector lies in the $XZ$ plane, with its transverse component pointing towards positive $Z$.
The stated exclusion of $a_k=g_k$ ensures $U_k\ne0$, but it does not ensure $W_k\ne0$. \b[If felt-up is parallel to travel, the roll angle is undetermined rather than the construction being impossible.] Every transverse <orthonormal basis> then has the required plane property. For continuity, project the previous $Z$ into $X_k^\perp$ and normalize it, using $Y_k=Z_k\times X_k$. If this projection is also too small, choose a coordinate <unit vector> least aligned with $X_k$ and project that vector instead. Switch to this transported choice near the parallel case to avoid numerical amplification. A stopped vehicle likewise needs a chosen limiting travel <tangent vector>, since its zero <velocity> itself has no direction. None of this requires a <Frenet frame>, which can fail where <curvature> vanishes.
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