Moon Phase Calculator

Seven of the table's ten rows put a formula in the formula column. The Synodic month row puts a number: S = 29.5306 d, asserted. Derive it instead — from the Sidereal month row directly below it, plus one given the tool never mentions.

1. A month is a race between two angular speeds, so start by converting one. The sidereal month is the Moon's full lap against the fixed stars, and that is a clean 360° in the printed 27.3217 d.

   $${\omega }_{\mathrm{M}}=\frac{360^{\circ }}{T_{\mathrm{sid}}}=\frac{360^{\circ }}{27.3217\mathrm{d}}=13.17634^{\circ }{\mathrm{d}}^{-1}$$

2. Phase, though, is not measured against the stars — it is measured against the Sun, and the Sun's direction is not fixed, because Earth is orbiting it. The Sun therefore drifts one full turn against those same stars per sidereal year, 365.2564 d. This is the one number that does not come from the page; you have to bring it.

   $${\omega }_{\odot }=\frac{360^{\circ }}{T_{\mathrm{E}}}=\frac{360^{\circ }}{365.2564\mathrm{d}}=0.985609^{\circ }{\mathrm{d}}^{-1}$$

3. Now run one sidereal month and see what is missing. The Moon is back beside its starting star, but the Sun has slid on by θ₀, and the Moon needs a further t₁ to close that gap and be lit the same way again.

   $${\theta }_0={\omega }_{\odot }{T}_{\mathrm{sid}}=0.985609\times 27.3217=26.9285^{\circ },t_1=\frac{{\theta }_0}{{\omega }_{\mathrm{M}}}=\frac{26.9285}{13.17634}=2.0437\mathrm{d}$$

4. The Sun does not wait during the catch-up. In those 2.0437 d it moves again, by the fraction r of the gap it had already opened, and again during the fix for that — a geometric series, which is why the true delay is t₁ divided by 1 − r rather than t₁. Because r is only 0.075, the whole infinite chase costs just 8% more than the first estimate.

   $$r=\frac{{\omega }_{\odot }}{{\omega }_{\mathrm{M}}}=\frac{0.985609}{13.17634}=0.074801,\Delta t=t_1\sum _{k=0}^{\infty }{r}^k=\frac{t_1}{1-r}=2.0437\times 1.080849=2.2089\mathrm{d}$$

5. Add the catch-up to the sidereal month — or skip the series entirely, since phase advances at the difference of the two rates and 360° over that difference gives the same answer. Both routes land on 29.5306 d, which is precisely what the Synodic month row asserts, and 29.531 d is what its value column prints.

   $$S=T_{\mathrm{sid}}+\Delta t=27.3217+2.2089=29.5306\mathrm{d}⟺S=\frac{360^{\circ }}{{\omega }_{\mathrm{M}}-{\omega }_{\odot }}=\frac{360}{12.190731}=29.5306\mathrm{d}$$

6. Then check the arithmetic a second way, using nothing from outside the table at all: subtract the two printed values.

   $$S-T_{\mathrm{sid}}=29.531-27.322=2.209\mathrm{d}\approx \Delta t=2.2089\mathrm{d}$$