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SCIENCE · INVESTIGATE AND EXPLAIN

Evaluate eclipse claims against model evidence

Distinguish alignment diagnostics from finite-size eclipse calculations and dated observations.

Learning outcome: Write a bounded evidence claim and identify missing information needed for actual eclipse occurrence and local visibility.

Starting recap

The phase disk remains unshadowed. Centerline offset is a geometric diagnostic, not an eclipse forecast.

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Try

A near-full 99.80% disk can miss the centerline by about 34,472 km in this model.

Open the experiment

Predict

Does the fully illuminated disk at exact opposite-Sun alignment prove that an actual lunar eclipse is absent?

Experiment

Compare near-full and exactly aligned full cases, then repeat on the solar side. List which claims current outputs support and which need additional evidence.

Investigate Moon phases and orbital nodes

Worked teaching model · synthetic values · Circular orbit of radius 384400 km, inclination 5.145°, Earth at the origin and distant Sun along +x. Parallel sunlight illuminates half the sphere. Apparent disk fraction is unshadowed geometric area, not brightness. Changing node direction rotates the orbital plane. No dates, orbital timing, finite-disk overlap, shadow cones, eclipse attenuation or eclipse visibility are calculated.

Reset keeps your prediction, captures and practice history.

View down onto the ecliptic plane · equal x and y scalesOrbit radius 384400 km · body symbols exaggerated · height is hiddenSun direction (+x)Incoming sunlightAscending nodeEarthMoonCircle: ascending node · square: descending node · dashed red: node lineProjected alignment alone is insufficient: check latitude and axis offset below.
Apparent disk from Earth · unshadowed geometryHalf illuminated: 50.00% lit areaElongation at Earth: 90.000°Phase angle at Moon: 90.000°Local trend: IncreasingSunward direction drawn right; not a sky compass or a waxing/waning orientation.
Three-dimensional geometry and apparent phase
QuantitySelected valueMeaning
Normalized position (x, y, z)(0.00000, 0.99597, 0.08968)Earth-to-Moon vector divided by orbit radius
Latitude above ecliptic5.145°Signed angle above Earth's orbital plane
Height above ecliptic34471.67 kmNegative means below
Offset from Sun–Earth axis384400.00 kmTransverse distance, including hidden height
Angle from nearer axis direction90.000°No exact centerline alignment
Physical sphere illuminated50%Parallel sunlight; eclipse shadows omitted
Apparent disk illuminated50.00%Projected area, not reflected brightness

Try angle 90° with node 0°, then angle 270°: both disks are half illuminated, but height and the direction of phase change differ. Next set node 90° and angle 90°: the nearly full Moon misses the Sun–Earth axis by about 34472 km. An orbital node is a crossing of the ecliptic; it need not point toward the Sun.

Ordinary phases come from viewing different parts of the sunlit hemisphere, not from Earth's shadow. Even at exact centerline alignment this display omits eclipse shading. Real eclipse type, contacts, dates and local visibility require finite body sizes, distances and time-dependent geometry. No control step represents a fixed number of days. A stationary trend means zero instantaneous change at the selected position, not an unchanged phase throughout the orbit.

With normalized Sunward coordinate x, elongation E = arccos(x), phase angle = 180° − E, and the unshadowed illuminated disk fraction = (1 − x)/2. The top view hides z; the axis offset uses both transverse coordinates, 384400 √(y² + z²) km.

Half illuminated: 50.00% of the apparent disk illuminated (unshadowed). Sun–Moon elongation 90.000°.

No exact centerline alignment. Moon latitude 5.145°; distance from the Sun–Earth axis 384400.00 km. Increasing illuminated fraction as Moon angle increases with node direction fixed. Exact alignment is a geometric condition, not an eclipse forecast.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Decide what a geometric result can support

A nearly full Moon is not automatically a lunar eclipse, and a nearly new Moon is not automatically a solar eclipse. NASA distinguishes solar eclipses, in which the Moon obscures the Sun for suitable observers, from lunar eclipses, in which Earth’s shadow falls on the Moon. The inclined lunar orbit usually carries the Moon away from the needed alignment. NASA eclipse explanation.

The investigation here asks a narrower question: how closely does a constructed Moon position approach the Sun–Earth centerline, and on which side of Earth is it? That question has numerical answers without pretending to calculate shadows or observer visibility. Keep this boundary in your written conclusion rather than relying on an image to communicate it.

Build a claim-and-evidence ledger

Make three columns headed claim, supporting output and missing information. For the default u = 90°, Ω = 0° case, the 50% disk fraction supports a half-lit apparent view in the unshadowed model. The 90° elongation rules out exact centerline alignment. Neither output supplies a date, local sky position or measured observation.

Next compare u = 90°, Ω = 90° with u = 180°, Ω = 0°. Both look almost or completely full, but the first misses the centerline by about 34,472 km and the second has zero centerline offset. The contrast explains why phase appearance alone is insufficient evidence for alignment. It does not calculate whether a finite shadow reaches either position.

Repeat the comparison on the solar side using u = 270°, Ω = 90° and u = 0°, Ω = 0°. Write the same distinction in your own words. A consistent method should work on both sides instead of relying on the appearance of one selected example.

Do not let an omitted effect become a prediction

The phase disk is explicitly unshadowed, even at exact alignment. At an aligned opposite-Sun position it still shows the fully illuminated apparent disk predicted by f = (1 − x)/2. That display does not claim Earth’s shadow is absent in the physical situation. Shadow attenuation simply is not calculated.

Likewise, a zero fraction toward the Sun is an ordinary unshadowed viewing-geometry result. The model does not render the Moon’s disk crossing the Sun. It does not compare angular radii, identify total versus annular eclipses, construct a shadow cone, account for atmospheric effects or determine visibility from an observer’s location.

Exact centerline alignment is a useful diagnostic within this model. Real eclipses involve finite bodies, so some near-centerline configurations can also produce overlap. Do not turn the diagnostic into a universal claim that real eclipses require mathematically zero offset. Determining an event needs the finite-size geometry and accurate positions that this lab omits.

Design an independent test instead of reusing the equation

Suppose you wanted to evaluate a prediction about an actual eclipse. A stronger evidence plan would include dated positions from an authoritative ephemeris, distances and apparent sizes, an explicit shadow or overlap calculation, and observations tied to time and location. Those are proposed evidence sources, not data imported into this lab.

For a paper exercise, classify four statements: “the model reports zero offset”; “an actual eclipse occurred at this timestamp”; “a particular city saw totality”; and “the model’s unshadowed disk is full.” The first and fourth can be checked against current outputs. The second and third require additional evidence and calculations. A screenshot cannot silently fill those gaps.

The fixed 384,400 km distance and 5.145° inclination are representative values from the NASA Moon fact sheet. Real distance and orientation vary. The sliders are geometric angles, not a chronological orbit solution. Agreement between an equation and its own visualization checks implementation consistency, not independent observational validation.

Transfer by repairing an overclaim

A report says: “The Moon is 99.8% illuminated, so a lunar eclipse will be visible tonight from our school.” Identify two unsupported steps and rewrite the conclusion using only the model’s evidence.

Check the eclipse-evidence transfer

A near-full fraction does not establish suitable alignment; the lab includes a near-full case with a large centerline offset. It also provides no dated ephemeris, shadow calculation or observer-specific visibility. A supported conclusion is: this chosen geometry produces a nearly full unshadowed apparent disk; eclipse occurrence and local visibility remain unassessed.

Apply

A near-full fraction in this lab establishes:

Public practice checks a specific claim. It does not establish independent mastery.

Repair an eclipse overclaim and propose independent dated, finite-size and observer-specific evidence.

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

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