Here I present my method of determining an estimate of the lower bounding mass for the primary component of a binary star using mass ratio ($q$), velocity semi-amplitude ($K$), orbital eccentricity ($e$), and orbital period ($P$).
Newton's formulation of Kepler's $3^{rd}$ law is
$$ P^2 = (2\pi)^2 \frac{a^3}{G(M_1+M_2)}$$let $M_2 = qM_1$ where $M_1$ is the primary mass and $M_2$ is the secondary.
$$ P^2 = (2\pi)^2 \frac{a^3}{G(M_1+qM_1)}$$factor out the common $M_1$ term
$$ = (2\pi)^2 \frac{a^3}{GM_1(1+q)}$$To find $a$ in terms of quantities measurable for an SB2, we use the relationship to orbital speed, v_o
$$ v_o = \frac{2\pi a}{T}\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 -\frac{5}{256}e^6 -\frac{175}{16384}e^8 - \dots \right] $$Truncated to the $3^{rd}$ term.
$$v_o = \frac{2\pi a}{P}\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]$$let $v_o = K$ (This is the iffy part. It's accurate for circular systems, but is decreasingly accurate for more eccentric systems)
$$K = \frac{2\pi a}{P}\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]$$and solve for a
$$ a = \frac{KP}{2 \pi \left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]} $$Substituting this back into the equation for P gives
$$ P^2 = (2\pi)^2 \frac{(KP)^3}{(2\pi)^3 GM_1(1+q)\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]^3}$$Some terms cancel giving
$$ 1 = \frac{K^3P}{2\pi GM_1(1+q)\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]^3}$$Finally
$$ M_1 = \frac{K^3P}{2\pi G(1+q)\left[1-\frac{1}{4}e^2-\frac{3}{64}e^4 \right]^3}$$This is a lower bound estimate becuase I've left off the $\sin{i}$ term (the assumption is edge on orientation, i.e. orbital speed is no faster than observed).