I think this would be considered a maths question rather than a physics one.
For context, I am creating a calendar for a fantasy game that has $2$ moons. In my current case moon #1 orbital time is $28=(2*2*7)$ days, and moon #2 orbital time $77=(7*11)$ days. Both cycles begin at day $0$, so both moons are full every $2*2*7*11 = 308$ days. A year is currently $336$ days.
for ease of calculation, let's assume they are rotating around the same axis as the planet is spinning. and their rotational direction being the same. That way we can use a circle instead of a sphere.
How would one calculate a moon's position in the sky at a given day/time. (given in days from the start of time).
e.x. DAY=$123452.34$
I am thinking something along the lines of
moon position = ((The amount the planet has spun around itself since beginning of time in radians) - (The amount the moon has spun around the planet in radians)) modulo $2\pi$
which would make it:
moon #1 position = $(2\pi * (floor(DAY)-DAY) - 2\pi*$ (DAY modulo moon1.orbitTime) / moon1.orbitTime) modulo $2\pi$
Is it just this simple? I am assuming the difference between standing on the planets surface and not at its core is negligible. Considering that the radius of the earth is $7,000$ km and the real moon is $400,000$ km away would that be a reasonable assumption?
Also, does the fact that a year is divisible by moon #1's orbital time affect how frequent lunar eclipses are?
Here are my notes. I've managed to find $\theta_2$ as a function of $t$. But I'd like to find $\theta_1$ as a function of $r_1,r_2,t$
This is a creative question and deserves an answer!
If you ignore the size of the planet, then your basic calculation is sound; the difference between the angle covered by the turning planet and the angle covered by the moving moon, modulo one revolution, would be the relative direction of the moon seen from the planet.
I'm going to change your notation somewhat. Names and labels need to be carefully chosen to save on effort in a big problem.
Let's call their star "Lux" because it is shorter than "their sun" or "their star". Likewise we can call the planet "Dirt".
Let's say $r_d$ is the radius of Dirt, and $\theta_d$ is the angle Dirt has turned, modulo $1$ revolution.
Let's call moon #$1$ "Nero". Let $r_n$ be the distance to Nero and $\theta_n$ be the angle Nero has revolved since "the start of time" at $t=0$.
Let's call moon #$2$ "Faro" and give it $r_f$ and $\theta_f$ similarly.
We need a starting position for everything, so let's say that Nero eclipses Faro which eclipses Lux at high noon at time $t=0$ in City $A$. Every angle is $0$ then.
We have to distinguish two kinds of "day". A sidereal day is the time for Dirt to rotate until the same stars are overhead as before. This is not the same as a "Lux day", the time from local noon to local noon, because Dirt moves around Lux. Dirt turns a rotation, a sidereal day, but then has to turn a little farther to catch up with Lux, which has (from Dirt's perspective) moved onward. The net effect is that Dirt has to make one extra rotation per year, so 337 rotations to experience 336 Lux days.
In a similar fashion, a "sidereal Nonth" is the time it takes for Nero to completely orbit Dirt, as seen by the distant stars. I assume $28$ Lux days is the time it takes to go from Full Nero to the next Full Nero, which is yet another complication. Nero, Dirt, and Lux are all in relative motion.
This is already getting long, so I'll have to finish another time!