Day Length on Double Planets Assume two similarly sized bodies tide-locked to one another orbiting a barycenter between the two. That barycenter orbits a star. Since the two are tide-locked, their sidereal rotational period and sidereal orbital period (about one another) are the same (right?). There's no axial tilts or eccentricities to screw with things either for the sake of simplicity.

What would I base the length of a conventional day on for a society living on these worlds?

My gut's telling me to calculate the sidereal period based of the barycentral semi-major axes (or would I use the separation between the two worlds still?) for each and then compute the synodic periods and just use those, but my brain's too stupid to tell whether that's going to actually be anything close to a "solar day". I don't need perfect - just close enough.
Yes, if the planets are tide locked to one another, their orbital period and sidereal day are the same. That's assuming their orbital plane about each other isn't much inclined to their orbital plane about the star.

From the Wikipedia article on orbital period:

$T_d=2\pi*\sqrt{a_p^3/(G*(M_1+M_2))}$

Where G is gravitational constant, $M_1$ is mass of 1st body and $M_2$ mass of 2nd body. $a_p$ is semi major axis of elliptical orbit of the planets about each other.

For the planet's year you'd use a very similar equation.

$T_y=2\pi*\sqrt{a_s^3/(G*(M_0+M_1+M_2))}$

Where $a_s$ is the radius of their orbit about the sun and $M_0$ is the mass of the sun. If you like, you can leave off $M_1$ and $M_2$ as their masses are probably neglible in comparison to the sun's mass.

I am guessing by "conventional day" you want the time an inhabitant sitting on his front porch would measure between one high noon and the next.

Solar day = sidereal day * (1 + (sidereal day/year))

For example our solar day is about 1/365 longer than a sidereal day.