Separate lunar illumination from the visible disk

Separate the sunlit hemisphere from the visible disk

Imagine a ball illuminated by a distant lamp. Mark the half facing the lamp, then walk around the ball without moving the lamp. The illuminated hemisphere stays on the lamp-facing side, but the part visible to you changes. This construction separates illumination from viewpoint. A paper circle divided in half is useful for the illumination map; it cannot by itself reproduce the curved boundary on a projected spherical disk.

The Moon reflects sunlight. Ordinary phases arise because observers on Earth see changing portions of its illuminated hemisphere, not because Earth’s shadow sweeps across it every month. NASA explains this distinction in its Moon phases guide. In this lab, illumination is deliberately unshadowed: half of the physical spherical surface is sunlit at every position. The displayed disk tells you how much of the apparent disk would be illuminated without an eclipse.

Identify the viewpoint before naming an angle

Earth is at the origin, and the Sun is far away in the positive x direction. The Moon’s position is represented by a unit vector: an arrow with length one that records direction. Its x component is +1 when the Moon lies directly toward the Sun, zero when the directions are perpendicular, and −1 when the Moon is directly opposite the Sun.

Elongation E is the angle between the Sun and Moon directions as seen from Earth. It runs from 0° to 180°. The phase angle α describes the Sun–Moon–observer geometry at the Moon. With the model’s distant, parallel sunlight, α = 180° − E. These are different angles at different vertices. Sketch those vertices before using either name.

At the default Moon angle 90° and node direction 0°, elongation is 90° and half the apparent disk is illuminated. The Moon is also above the reference orbital plane. That vertical position does not prevent a half-lit view; the relevant illumination angle is three-dimensional.

Calculate a projected fraction

The model’s apparent illuminated fraction is f = (1 − x)/2. Here f is a number between zero and one, while 100f is a percentage. Because x = cos E, the same relationship can be written f = (1 − cos E)/2. A calculator in degree mode helps connect the angle to the fraction.

For x = 0, f = 1/2. For x = +1, f = 0: the Moon’s illuminated side faces away from Earth. For x = −1, f = 1: its illuminated hemisphere faces Earth. None of these calculations says that the physical Moon produces its own light or that its whole surface is illuminated at full phase.

On paper, make a ledger with x values +1, +0.5, 0, −0.5 and −1. Calculate the five fractions before checking the model’s geometry. The results are 0%, 25%, 50%, 75% and 100%. These percentages describe projected apparent area, not a proportion of the entire spherical surface.

Inspect a controlled sequence

Keep Ascending-node direction at 0°. Compare Moon angles 0°, 90°, 180° and 270°. Record elongation, disk fraction and latitude separately. The 90° and 270° positions both show 50% illumination although they are different positions in the orbit. A single fraction therefore cannot tell you whether the visible illumination is increasing or decreasing.

The model is a constructed circular geometry, not a sequence of recorded telescope images. At exact alignment the phase disk still follows the unshadowed equation. It does not darken the Moon inside Earth’s shadow or draw the Moon covering the Sun. Those omissions must remain explicit when interpreting the endpoints.

Transfer without the slider

An unfamiliar position has x = −0.6. Calculate the unshadowed illuminated apparent fraction. Explain whether this establishes the waxing or waning branch, and whether it means 80% of the whole physical Moon receives sunlight.

Check the moon-illumination transfer

f = (1 − (−0.6))/2 = 0.8, or 80% of the apparent disk. The fraction alone does not identify waxing or waning; a direction of change or additional position information is needed. In the unshadowed spherical model, half of the physical surface receives sunlight, regardless of this projected fraction.