[{"data":1,"prerenderedAt":925},["ShallowReactive",2],{"layer:eclipses:investigate":3},{"layer":4,"contentHash":905,"dependencyHashes":906,"approval":919,"releaseId":924},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":43,"sourceIds":900,"reviewStatus":901,"authoring":902},1,"eclipses","en","investigate","Build it, test it, try to break it","A lamp-and-balls model, hands-on measurements, and predictions checked against real eclipses","Hands-on layer: build a scale model of the Earth-Moon-Sun system, test the new-moon\u002Ffull-moon rule and the shadow-width formula for yourself, find the tilt's hidden threshold, build a pinhole projector and check its numbers, and plan around three real upcoming eclipses.",[13,14,15,16,17],"Build a scale model of the Earth-Moon system with a lamp and two balls, and compute the model's Sun size and distance.","Test the claim that eclipses need new or full moon using the moon-phase lab, and explain why other phases fail.","Measure and predict shadow widths using width = blocker width × screen distance ÷ blocker distance.","Use the eclipse-lab and a tilt table to find the threshold tilt below which the eclipse count would not change.","Build a pinhole projector, measure its image size, and check the measurement against the 9.3 mm-per-metre rule.",35,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37,40],{"label":23,"value":24},"Depth","Investigate",{"label":26,"value":27},"Reading time","≈ 35 minutes, plus hands-on time",{"label":29,"value":30},"Prior knowledge","Umbra\u002Fpenumbra and node geometry (Understand)",{"label":32,"value":33},"Chapters","10",{"label":35,"value":36},"Labs","Shadow lab ×2, moon-phase, eclipse lab ×2, match",{"label":38,"value":39},"You will need","A lamp, two balls, card and a pin",{"label":41,"value":42},"Safety","All experiments use models or projection, never the real Sun",[44,50,75,81,87,105,118,131,136,141,154,159,188,193,198,203,206,245,261,274,297,302,305,319,330,341,367,372,377,390,422,434,447,452,457,469,490,503,514,519,522,553,565,570,574,579,583,586,596,609,614,619,622,646,659,662,675,680,696,702,726,755,877,889],{"id":45,"type":46,"markdown":47,"help":48},"i-intro","prose","Discover told you what an eclipse is. Understand gave you the geometry. Now it is your turn to **build it, test it and try to break it**.\n\nThis layer is full of things to actually do: a lamp-and-balls model you can build on a table, labs where you drag the Moon around and read off numbers, and predictions you should make *before* you check the answer, not after.\n\nThe plan: build a scale model, test whether eclipses really do need new or full moon, measure shadow cones for yourself, find out whether the tilt is really doing all the work, and check the arithmetic behind the safety rules by making your own pinhole projector.",{"simplerExplanation":49},"This layer is hands-on: build a model, run the labs, predict first, then check.",{"id":51,"type":52,"title":53,"items":54},"i-steps-fair-test","steps","Running a fair test in any of the labs below",[55,59,63,67,71],{"title":56,"tag":57,"text":58},"Predict first","before","Write down what you expect to happen, and why, before touching any slider.",{"title":60,"tag":61,"text":62},"Change one thing","one variable","Move only the slider named in the instructions. Leave everything else exactly where it was.",{"title":64,"tag":65,"text":66},"Read off the result","measure","Note the number the lab reports — a width, a percentage, a count — not just a general impression.",{"title":68,"tag":69,"text":70},"Reset, then change the other thing","one at a time","Put the first slider back before touching a second one, so you always know which change caused which result.",{"title":72,"tag":73,"text":74},"Compare with the worked example","check","If your reading and the worked example disagree by more than a rounding error, look for what you changed by accident.",{"id":76,"type":77,"variant":78,"title":79,"markdown":80},"i-how-to","callout","observation","How to use a prediction","Every **prediction** box below asks you to commit to an answer *before* you get the explanation. That is not decoration — guessing first and then finding out you were wrong is one of the most reliable ways to make an idea stick. If you skip straight to the explanation, do the experiment in the lab anyway and see if it matches what you would have guessed.",{"id":82,"type":83,"title":84,"eyebrow":85,"navLabel":86},"i-ch1","chapter","Three quick predictions","Chapter 01","1 Predict first",{"id":88,"type":89,"prompt":90,"options":91,"explanation":104},"i-predict-shadow-size","prediction","You shine a torch at a 4 cm ball held 20 cm from the torch, with a screen 60 cm from the torch. The shadow on the screen is 12 cm wide (check: 4 × 60 ÷ 20 = 12). If you slide the ball to 40 cm from the torch, keeping the screen at 60 cm, what happens to the shadow?",[92,95,98,101],{"id":93,"label":94},"a","It grows bigger than 12 cm",{"id":96,"label":97},"b","It shrinks to 6 cm",{"id":99,"label":100},"c","It stays exactly 12 cm",{"id":102,"label":103},"d","The shadow disappears","**It shrinks to 6 cm (b).** Shadow width = ball width × screen distance ÷ ball distance = 4 × 60 ÷ 40 = **6 cm**, exactly half of before because the ball's distance from the torch doubled.\n\nThe rule: move the blocking object **away from the light and towards the screen**, and its shadow shrinks. Move it towards the light, and the shadow grows. Test it in the shadow lab below before you build anything.",{"id":106,"type":89,"prompt":107,"options":108,"explanation":117},"i-predict-monthly","The Moon takes 29.53 days to go from new moon to new moon, and Earth's shadow is enormous. If the Moon's orbit had **no tilt at all**, how many solar eclipses would a year contain?",[109,111,113,115],{"id":93,"label":110},"0 — the Moon would always miss",{"id":96,"label":112},"About 4 to 7, same as now",{"id":99,"label":114},"About 12, one at every new moon",{"id":102,"label":116},"Exactly 1","**About 12 (c).** With zero tilt, the Moon would be exactly on the Sun–Earth line at every single new moon — twelve times a year — giving twelve solar eclipses (and twelve lunar eclipses at the full moons in between). You will test this directly with the tilt slider in Chapter 7.",{"id":119,"type":89,"prompt":120,"options":121,"explanation":130},"i-predict-annular","You point a pinhole card at the Sun and make an image on paper 1 metre away: it comes out about 9 mm across. If you move the paper to 3 metres away, roughly how big is the image, and is it brighter, dimmer or the same?",[122,124,126,128],{"id":93,"label":123},"About 27 mm, and dimmer",{"id":96,"label":125},"About 27 mm, and just as bright",{"id":99,"label":127},"Still about 9 mm, but dimmer",{"id":102,"label":129},"About 3 mm, and brighter","**About 27 mm, and dimmer (a).** The image grows in direct proportion to distance — 9 mm × 3 = 27 mm — because a pinhole image is just a small-angle projection of the Sun's own 0.53° width. But the same amount of light is now spread over roughly 9 times the area (three times the width, both ways), so it looks noticeably dimmer. You will measure this for real in Chapter 8.",{"id":132,"type":83,"title":133,"eyebrow":134,"navLabel":135},"i-ch2","Build it: a lamp and two balls","Chapter 02","2 Build the model",{"id":137,"type":46,"markdown":138,"help":139},"i-build-intro","You do not need a planetarium to make a real eclipse. You need a lamp, two balls of different sizes, and a metre or two of floor space.\n\n**What you need:** a bright torch or a bare bulb lamp (the Sun), a small ball about 2 cm across such as a large bead or a marble (the Moon), a bigger ball about 7 cm across such as an orange or a tennis ball (Earth), and a dark room.\n\n**What to do:**\n\n1. Set the lamp at one end of the room. This is the Sun. Do not move it again.\n2. Hold the big ball (Earth) about 2 metres from the lamp.\n3. Hold the small ball (Moon) between the lamp and Earth, about 6 cm from Earth's surface — close, the way the real Moon is close compared with the Sun.\n4. Look at Earth's surface, on the side facing the small ball. Can you see a tiny dark spot? That is your model **solar eclipse** — the Moon's shadow landing on Earth.\n5. Now move the small ball to the far side of Earth, in Earth's own shadow. Its surface should darken. That is your model **lunar eclipse**.",{"simplerExplanation":140},"Lamp = Sun, small ball = Moon, big ball = Earth. Put the Moon-ball between the lamp and Earth-ball for a solar eclipse; put it behind Earth, in Earth's shadow, for a lunar eclipse.",{"id":142,"type":143,"title":144,"problem":145,"steps":146,"help":152},"i-we-scale","worked_example","Getting the model's proportions right","If your Moon-ball is 2 cm across, how big should the Earth-ball be, and how far apart should they stand, to keep the same proportions as the real Earth–Moon system? (Moon = 3,475 km across, Earth = 12,742 km across, distance = 384,400 km.)",[147,148,149,150,151],"Find the scale factor from the Moon: 3,475 km must become 2 cm, so 1 cm in the model stands for 3,475 ÷ 2 = **1,737.5 km**.","Earth's model size = 12,742 ÷ 1,737.5 = **7.3 cm** across — about the size of an orange or a tennis ball.","Distance between them = 384,400 ÷ 1,737.5 = **221 cm**, a little over two metres.","Do the same for the Sun, just to see how it breaks the room: 1,392,700 ÷ 1,737.5 = **801 cm**, an 8-metre sphere, standing **861 metres** away.","**That is the real lesson of this model.** You can fit the Earth–Moon system on a table. You cannot fit the Sun in the same room, or even the same street, at the same scale.",{"simplerExplanation":153},"A 2 cm Moon-ball needs a 7.3 cm Earth-ball about 2.2 m away. The Sun, at the same scale, would be an 8-metre ball nearly a kilometre off — completely impossible indoors.",{"id":155,"type":77,"variant":156,"title":157,"markdown":158},"i-model-limit","model_limit","What the tabletop model leaves out","Your lamp-and-balls model gets the **shapes** right — two shadow cones, umbra and penumbra — but it cannot get the **sizes** right at the same time. To fit the Sun in a room, the model has to use a lamp far too small and far too close, which makes its rays fan out much more than the Sun's nearly parallel rays do.\n\nThat single mismatch is why the model's shadow looks fat and short, while a real shadow cone reaching from the Moon to Earth is long and needle-thin. Treat the model as a way to see *which way round* everything works, not as a scale drawing of the real distances — the worked example above shows just how different the true scale is.",{"id":160,"type":161,"component":162,"componentVersion":5,"config":163,"objective":182,"textAlternative":183,"help":184},"i-lab-build","interactive","shadow-lab",{"objects":164,"source":173,"maxDistanceCm":174,"challenges":175},[165,169],{"id":166,"label":167,"heightCm":168},"moon2cm","2 cm Moon-ball",2,{"id":170,"label":171,"heightCm":172},"earth7cm","7 cm Earth-ball",7,"point",250,[176,179],{"prompt":177,"targetRatio":178},"Place the Moon-ball so its umbra just reaches the Earth-ball's surface.",0.03,{"prompt":180,"targetRatio":181},"Now find a distance where the umbra falls short of Earth, leaving a bright ring.",0.08,"Recreate the lamp-and-balls model on screen and find the distance where the small ball's shadow just reaches the big one.","A lamp, a small ball and a large ball on a virtual table, matching the tabletop version: 2 cm Moon-ball, 7 cm Earth-ball, point source lamp.\n\nSlide the Moon-ball back and forth. Close to the lamp, its shadow cone is long and reaches well past the Earth-ball — like the real Moon's shadow arriving with room to spare at perigee. Slide it further from the lamp and the cone shortens until its tip lands exactly on the Earth-ball's surface, and a little further still, the tip falls short and a bright rim shows around the dark centre — your tabletop version of an annular eclipse.\n\nUse this to connect the model in your hands with the numbers in the worked example: the real Moon's shadow cone is about 374,000 km long and its distance is about 384,400 km, a much closer call than the model usually shows unless you place things carefully.",{"hints":185},[186,187],"Start with the Moon-ball close to the lamp and slide it slowly away from the lamp.","Watch the exact moment the dark spot's edges reach the far ball. A tiny move past that point changes everything.",{"id":189,"type":77,"variant":190,"title":191,"markdown":192},"i-tryit-penumbra","try_it","Test the penumbra with the same model","Swap your point-like torch for a lamp with the shade off, or several torches held close together, so the light source has real width — more like the Sun.\n\nHold the Moon-ball between the lamp and a wall. Instead of a crisp round shadow, you should now see a **dark centre with a soft grey edge**. Move your eye into that grey edge and look back towards the lamp past the Moon-ball's rim: you can see part of the lamp peeping past it. That is a penumbra, built with nothing but a lamp and a ball, and it is exactly the same reason a partial solar eclipse looks the way it does.",{"id":194,"type":77,"variant":195,"title":196,"markdown":197},"i-tryit-safety","careful","One safety note for the model itself","None of this needs the real Sun. Do the whole experiment with a torch or a lamp indoors. If you ever want to compare your model's shadow with a real one outdoors, do it with your own shadow or a tree's shadow in sunlight — **never by looking at the Sun itself**, not even for a second, model or no model.",{"id":199,"type":83,"title":200,"eyebrow":201,"navLabel":202},"i-ch3","Test it: does an eclipse really need new or full moon?","Chapter 03","3 Test the phase rule",{"id":204,"type":46,"markdown":205},"i-phase-test","Understand told you a solar eclipse needs new moon and a lunar eclipse needs full moon. Do not take that on trust — test it.\n\nThe Moon runs through eight named phases in one 29.53-day cycle: new, waxing crescent, first quarter, waxing gibbous, full, waning gibbous, last quarter, waning crescent, and back to new. At each phase, ask: **is the Moon between Earth and the Sun, behind Earth, or off to one side?**",{"id":207,"type":208,"caption":209,"columns":210,"rows":215},"i-table-phase-eclipse","table","Testing every phase: can an eclipse happen there?",[211,212,213,214],"Phase","Where the Moon is","Solar eclipse possible?","Lunar eclipse possible?",[216,221,226,229,232,236,239,242],[217,218,219,220],"New moon","Between Earth and Sun","Yes, if also near a node","No — wrong side of Earth",[222,223,224,225],"Waxing crescent","A quarter-turn from new","No — nowhere near the Sun–Earth line","No",[227,228,225,225],"First quarter","Side-on to the Sun",[230,231,225,225],"Waxing gibbous","Approaching full",[233,234,235,219],"Full moon","Behind Earth from the Sun","No — wrong side entirely",[237,238,225,225],"Waning gibbous","Past full",[240,241,225,225],"Last quarter","Side-on, other side",[243,244,225,225],"Waning crescent","Approaching new",{"id":246,"type":161,"component":247,"componentVersion":5,"config":248,"objective":255,"textAlternative":256,"help":257},"i-lab-moonphase","moon-phase",{"startDay":249,"views":250,"showNames":252,"showTithi":253,"quizRounds":254},0,[251],"both",true,false,6,"Step the Moon through a full month and check the claim: only new moon lines up for a solar eclipse and only full moon lines up for a lunar eclipse.","A dual view: the Moon orbiting Earth as seen from above (from-space), and what its lit shape looks like from Earth (from-earth), both driven by the same day-of-month slider from 0 to 29.5.\n\nStep through slowly and watch the from-space view. Only at day 0 (new moon) does the Moon sit on the Sun side of Earth, roughly in line with the Sun. Only at day ≈14.8 (full moon) does it sit on the far side, roughly in line with Earth's shadow. At every other day it is off to one side, and no shadow from either body can reach the other.\n\nThis confirms the rule from Understand: the phase condition is not a coincidence added on top of the geometry, it **is** the geometry, seen from a different angle.",{"hints":258},[259,260],"Pause exactly at day 0 and day 14.8 and check the alignment.","Try a day like 7.4 (first quarter) and see how far off the line the Moon has swung.",{"id":262,"type":89,"prompt":263,"options":264,"explanation":273},"i-predict-quarter","Could a \"solar eclipse\" ever happen at first quarter, when the Moon is exactly half-lit as seen from Earth?",[265,267,269,271],{"id":93,"label":266},"Yes, if the tilt happens to help",{"id":96,"label":268},"No — at first quarter the Moon is 90° around its orbit from the Sun–Earth line",{"id":99,"label":270},"Yes, but only in some years",{"id":102,"label":272},"Yes, it happens about once a decade","**No (b).** At first quarter the Moon is a quarter of the way round its orbit, roughly 90° from the Sun–Earth line as seen from Earth. No amount of tilt closes a 90° gap — the tilt only ever moves the Moon by up to 5.145° off the flat plane. The phase condition and the node condition are two completely separate requirements, and first quarter fails the phase condition by a huge margin, tilt or no tilt.",{"id":275,"type":276,"itemId":277,"prompt":278,"check":279,"hints":291,"feedback":294},"i-practice-phase-node","practice","eclipses.investigate-p-phase-node","It is new moon today, so the Moon is roughly between Earth and the Sun. Is a solar eclipse guaranteed?",{"kind":280,"options":281,"correct":290},"choice",[282,284,286,288],{"id":93,"label":283},"Yes — new moon is the only requirement",{"id":96,"label":285},"No — the Moon also has to be near one of its two orbital nodes",{"id":99,"label":287},"No — it also has to be a full moon at the same time",{"id":102,"label":289},"Yes, but only in December",[96],[292,293],"Two separate conditions must both be true: the right phase, and the right position in the tilted orbit.","Being new moon fixes only the phase. Check what the Moon's height above or below the ecliptic is doing at the same time.",{"correct":295,"incorrect":296},"Right: new moon fixes the phase, but the Moon must also be close to a node for its shadow to actually reach Earth.","New moon is necessary but not sufficient. The Moon also has to be near a node — within about 16.6° of it — or its shadow passes above or below Earth entirely.",{"id":298,"type":83,"title":299,"eyebrow":300,"navLabel":301},"i-ch4","Measure it: the shadow-width rule","Chapter 04","4 Measure shadows",{"id":303,"type":46,"markdown":304},"i-measure-intro","Now put numbers on what you have been watching. A shadow behind a small, close blocker with a point light source obeys one simple rule:\n\n**shadow width = blocker width × (screen distance ÷ blocker distance)**\n\nUse the lab to measure real values, then check them against the practice problems below.",{"id":306,"type":276,"itemId":307,"prompt":308,"check":309,"hints":313,"feedback":316},"i-practice-shadow1","eclipses.investigate-p-shadow1","A 4 cm ball sits 20 cm from a point lamp, with a screen 60 cm from the lamp. How wide is the shadow, in cm?",{"kind":310,"answer":311,"tolerance":249,"unit":312},"number",12,"cm",[314,315],"Use shadow width = ball width × screen distance ÷ ball distance.","4 × 60 ÷ 20 = ?",{"correct":317,"incorrect":318},"Right: 4 × 60 ÷ 20 = 12 cm.","Multiply the ball's width by the screen distance first, then divide by the ball's distance from the lamp: 4 × 60 ÷ 20 = 12.",{"id":320,"type":276,"itemId":321,"prompt":322,"check":323,"hints":325,"feedback":327},"i-practice-shadow2","eclipses.investigate-p-shadow2","Same lamp and screen (60 cm away), but now the ball is 30 cm from the lamp. How wide is the shadow, in cm?",{"kind":310,"answer":324,"tolerance":249,"unit":312},8,[326],"4 × 60 ÷ 30 = ?",{"correct":328,"incorrect":329},"Right: 4 × 60 ÷ 30 = 8 cm — smaller than before, because the ball moved further from the lamp.","4 × 60 ÷ 30 = 8 cm. Moving the ball away from the lamp always shrinks its shadow on a fixed screen.",{"id":331,"type":276,"itemId":332,"prompt":333,"check":334,"hints":336,"feedback":338},"i-practice-shadow3","eclipses.investigate-p-shadow3","A 5 cm ball sits 50 cm from the lamp, with the screen 100 cm away. How wide is the shadow, in cm?",{"kind":310,"answer":335,"tolerance":249,"unit":312},10,[337],"5 × 100 ÷ 50 = ?",{"correct":339,"incorrect":340},"Right: 5 × 100 ÷ 50 = 10 cm.","5 × 100 ÷ 50 = 10. Notice the ball is exactly halfway to the screen, so the shadow is exactly double the ball's width.",{"id":342,"type":161,"component":162,"componentVersion":5,"config":343,"objective":361,"textAlternative":362,"help":363},"i-lab-measure",{"objects":344,"source":173,"maxDistanceCm":353,"challenges":354},[345,349],{"id":346,"label":347,"heightCm":348},"b4","4 cm ball",4,{"id":350,"label":351,"heightCm":352},"b5","5 cm ball",5,200,[355,358],{"prompt":356,"targetRatio":357},"Make the shadow exactly 3 times the ball's own width.",3,{"prompt":359,"targetRatio":360},"Make the shadow exactly half the ball's own width (move the ball closer to the screen than to the lamp).",0.5,"Move a ball between a point lamp and a screen, read off the shadow width, and check it against the ball width × screen distance ÷ ball distance rule.","A lamp, a movable ball with a choice of two widths, and a screen with a ruler along it, reporting the shadow's exact width as you drag.\n\nTry the three practice problems above in the lab before or after solving them on paper: set the ball to 4 cm, put it 20 cm from the lamp with the screen at 60 cm, and check the readout says 12 cm. Then try 30 cm and 40 cm and watch the shadow shrink each time.\n\nA ratio readout also shows shadow width ÷ ball width, which is the same as screen distance ÷ ball distance — useful for the two challenges above.",{"hints":364},[365,366],"For 'triple the width', you need the ball at exactly one third of the way from lamp to screen.","For 'half the width', the ball needs to be closer to the screen than to the lamp — beyond the halfway point.",{"id":368,"type":83,"title":369,"eyebrow":370,"navLabel":371},"i-ch5","Investigate: total, annular, or nothing at all?","Chapter 05","5 Total or annular?",{"id":373,"type":46,"markdown":374,"help":375},"i-tan-intro","Set up an eclipse-lab experiment: fix the Sun and Earth, and change only the Moon's distance and its offset from the Sun–Earth line. Before every run, write down a prediction.",{"simplerExplanation":376},"Change one thing at a time — distance, then offset — and predict before you look.",{"id":378,"type":161,"component":379,"componentVersion":5,"config":380,"objective":384,"textAlternative":385,"help":386},"i-lab-solar-test","eclipse-lab",{"modes":381,"showShadowCones":252,"tiltDegrees":383},[382],"solar",5.1,"Change only the Moon's distance and offset, one at a time, and record whether you get total, annular, partial or nothing.","The same solar-eclipse model as before, but treated as an experiment: change the Moon's distance slider and note the result, then reset the distance and change the offset slider instead, and note that result too.\n\nSuggested runs: (1) offset = 0, distance = perigee — expect the umbra to reach the ground; (2) offset = 0, distance = apogee — expect the antumbra and a ring; (3) offset = 0, distance = mean — expect a very thin ring; (4) offset = small, distance = perigee — expect the black spot to slide off the globe while the grey penumbra still clips it, giving a partial eclipse for a wider area; (5) offset = large — expect nothing at all.\n\nKeep a simple table as you go: distance, offset, result. That table is the evidence behind every rule in Understand — you are re-deriving it instead of reading it.",{"hints":387},[388,389],"Change one slider at a time. Changing both together makes it hard to tell which one caused the result.","Write your prediction down before you drag the slider, then compare.",{"id":391,"type":208,"caption":392,"columns":393,"rows":398},"i-table-results-matrix","A results matrix worth filling in yourself as you run the five suggested trials",[394,395,396,397],"Moon distance","Offset from the line","Prediction","What the lab actually shows",[399,404,408,412,417],[400,401,402,403],"Perigee (363,300 km)","0 (centred)","Umbra reaches ground","Total — black disc, corona visible",[405,401,406,407],"Apogee (405,500 km)","Umbra falls short","Annular — bright ring, ≈18% of area left",[409,401,410,411],"Mean (384,400 km)","Only just falls short","A very thin ring — barely annular",[413,414,415,416],"Perigee","Small (a few thousand km)","Umbra clips the edge of Earth","Partial for a wide area; total only along a thin, curved track",[418,419,420,421],"Any distance","Large (tens of thousands of km)","Both cones miss Earth entirely","Nothing — an ordinary new moon",{"id":423,"type":276,"itemId":424,"prompt":425,"check":426,"hints":428,"feedback":431},"i-practice-annular","eclipses.investigate-p-annular-area","The Moon looks 30 arcminutes across and the Sun 32 arcminutes across in a particular annular eclipse. What percentage of the Sun's area is left shining? Give your answer to the nearest whole per cent.",{"kind":310,"answer":311,"tolerance":5,"unit":427},"%",[429,430],"Area scales as the square of the diameter. Fraction covered = (30 ÷ 32)².","30 ÷ 32 = 0.9375. Square it: 0.9375² = 0.879. That is the fraction covered.",{"correct":432,"incorrect":433},"Right: 0.9375² ≈ 0.879 covered, so about 1 − 0.879 = 0.121, which rounds to 12%.","First find the diameter ratio (30 ÷ 32 = 0.9375), then square it for the area covered (≈0.879), then subtract from 1 for the part left shining (≈0.12, or 12%).",{"id":435,"type":89,"prompt":436,"options":437,"explanation":446},"i-predict-hybrid","In a hybrid eclipse, the same event is total in the middle of its path and annular at both ends. What is different about the ends of the path compared with the middle?",[438,440,442,444],{"id":93,"label":439},"The Moon is moving faster there",{"id":96,"label":441},"Observers there are further from the Moon, because Earth's curve puts them slightly beyond the umbra's tip",{"id":99,"label":443},"The Sun is dimmer near sunrise and sunset",{"id":102,"label":445},"Nothing — hybrid eclipses are a myth","**(b).** Earth is a curved surface, so an observer near the edge of the illuminated disc is measurably further from the Moon than someone directly underneath — by up to Earth's own radius, 6,371 km. If the umbra's tip lands just at that borderline distance, it can reach the closer, central observers (total) but fall just short for those further round the curve (annular). It is a small, real effect, and it makes hybrid eclipses genuinely rare — a handful per century.",{"id":448,"type":83,"title":449,"eyebrow":450,"navLabel":451},"i-ch6","Investigate: how rare is it to stand in the path?","Chapter 06","6 How rare?",{"id":453,"type":46,"markdown":454,"help":455},"i-rarity-intro","You now know the umbra's footprint is a spot only about 160 km wide. Before you do the arithmetic, predict: out of every 1,000 people scattered randomly over Earth's surface, roughly how many would you expect to be standing in the path of totality during any one total solar eclipse?",{"simplerExplanation":456},"Compare the size of the dark spot with the size of the whole Earth.",{"id":458,"type":143,"title":459,"problem":460,"steps":461,"help":467},"i-we-path-share","What share of Earth does one totality path cover?","Model the path of totality as a strip 160 km wide and about 10,000 km long (a generous, typical length). Earth's surface area is 4 × π × 6,371² km². What percentage of Earth's surface does the path cover?",[462,463,464,465,466],"Path area ≈ 160 × 10,000 = **1,600,000 km²**.","Earth's surface area = 4 × π × 6,371² ≈ **509,900,000 km²**.","Share = 1,600,000 ÷ 509,900,000 × 100 ≈ **0.31%**.","So out of 1,000 people spread evenly over the globe, only about **3** would be standing in the path during any single total eclipse — and that is only if everyone stayed put and a total eclipse happened to be crossing Earth that day.","**Why 'once every 375 years for one spot' still fits:** total eclipses happen somewhere on Earth roughly every 18 months, so multiply the tiny per-eclipse share by the roughly 0.67 eclipses per year, and by very roughly how many independent chances a spot gets over centuries — the numbers work out to an average wait measured in centuries for any one fixed place, even though eclipses themselves are not rare at all.",{"simplerExplanation":468},"The path covers about 0.31% of Earth — roughly 3 people in 1,000 — which is why any one town waits centuries between visits, even though an eclipse is happening somewhere every year or two.",{"id":470,"type":471,"tone":472,"items":473},"i-spec-rarity","spec","neutral",[474,478,482,486],{"label":475,"big":476,"value":477},"Path width","≈160 km","Typical width of the umbra's footprint on the ground.",{"label":479,"big":480,"value":481},"Path share of Earth","≈0.31%","One path, modelled as 160 × 10,000 km, against Earth's full surface.",{"label":483,"big":484,"value":485},"Total eclipses somewhere","≈68 per century","About one total solar eclipse every 18 months on average, somewhere on Earth.",{"label":487,"big":488,"value":489},"Wait for one fixed spot","≈375 years","The average time between two total eclipses crossing the very same place.",{"id":491,"type":89,"prompt":492,"options":493,"explanation":502},"i-predict-rarity","Two friends argue. One says \"total eclipses must be incredibly rare — I might never see one.\" The other says \"they're not rare at all — one happens most years.\" Using the numbers above, whose claim is closer to being useful advice for planning a trip?",[494,496,498,500],{"id":93,"label":495},"The first friend — you should not expect to see one at all",{"id":96,"label":497},"The second friend — since one happens most years, you could travel to see many in a lifetime",{"id":99,"label":499},"Both are wrong — total eclipses do not really happen",{"id":102,"label":501},"Neither view helps; the two claims cancel out","**The second friend, for practical purposes (b).** Both statements are numerically true, but they answer different questions. \"Rare for one fixed spot\" (about once every 375 years) is true if you never travel. \"Not rare at all\" is true for the planet as a whole — roughly one total eclipse crosses somewhere on Earth every 18 months. Because you, unlike a fixed spot, **can travel**, the second view is the useful one for planning: a handful of trips over a lifetime can realistically put you in the path several times, which is exactly what eclipse-chasers do.",{"id":504,"type":143,"title":505,"problem":506,"steps":507,"help":512},"i-we-crossing-time","How long does the shadow take to cross a path?","A particular eclipse's path of totality runs about 4,800 km across the surface, and the shadow moves at roughly 2,000 km\u002Fh relative to the ground near the equator. Roughly how long does the whole event take to sweep from one end of the path to the other — and how does that compare with how long totality lasts at any one point?",[508,509,510,511],"Time = distance ÷ speed = 4,800 ÷ 2,000 = **2.4 hours** for the shadow to sweep the whole path end to end.","But that is not how long **totality** lasts for any one town — a town only sees darkness for the few minutes the narrow shadow takes to pass directly over it, not the hours it takes to cross the whole planet.","Using the path's own width (about 160 km) instead of its length: 160 ÷ 2,000 = **0.08 hours ≈ 4.8 minutes**, which matches the typical few minutes of totality quoted throughout this topic.","**The two numbers answer different questions:** 2.4 hours is how long the *event* takes to travel across the globe from start to finish; a few minutes is how long *any single spot* spends inside it.",{"simplerExplanation":513},"The shadow itself takes hours to cross the whole planet, but any one town is only inside it for a few minutes, because the dark spot is so narrow.",{"id":515,"type":83,"title":516,"eyebrow":517,"navLabel":518},"i-ch7","Investigate: how much does the tilt matter?","Chapter 07","7 Vary the tilt",{"id":520,"type":46,"markdown":521},"i-tilt-experiment","Understand explained why the real 5.145° tilt makes eclipses rare. Now run the experiment yourself and see how sensitive the answer is: does a *little* less tilt make a *little* more difference, or a *lot*?",{"id":523,"type":208,"caption":524,"columns":525,"rows":529},"i-table-tilt","The node window at different tilts, and how much of the whole orbit counts (both nodes)",[526,527,528],"Tilt","Window around each node","Share of the whole orbit",[530,534,538,541,545,549],[531,532,533],"0.0°","360° — the whole orbit","100% — every new moon and every full moon gives an eclipse",[535,536,537],"0.5°","360° — still the whole orbit","100% — a tilt below about 1.475° cannot keep the Moon out of range at all",[539,536,540],"1.0°","100% — the required separation (1.475°) is still bigger than the tilt itself",[542,543,544],"2.0°","≈95°","≈53% — now the tilt is bigger than the separation limit, and gaps start to open up",[546,547,548],"5.145°","≈33°","≈18.5% — the real value, split between the two nodes",[550,551,552],"10.0°","≈17°","≈9.4% — double the real tilt roughly halves the share again",{"id":554,"type":161,"component":379,"componentVersion":5,"config":555,"objective":559,"textAlternative":560,"help":561},"i-lab-tilt",{"modes":556,"showShadowCones":252,"tiltDegrees":558},[557],"why-not-monthly",5.145,"Drag the tilt slider through the values in the table and check the counter against your own predictions.","The same why-not-monthly lab as before, used here to test the table above directly: set the tilt to each value in turn (0°, 0.5°, 1°, 2°, 5.145°, 10°) and note how many eclipses per year the counter reports.\n\nThe pattern to look for: nothing changes at all between 0° and about 1°, because the tilt is still smaller than the 1.475° separation the Moon needs to clear. Only once the tilt passes that threshold does raising it start cutting the eclipse count down — and it keeps cutting hard as the tilt grows further.",{"hints":562},[563,564],"Try 0.5° and 1° first and see that the eclipse count barely moves from 24.","Then try 2°, 5.145° and 10° and see how much faster it falls once the tilt is 'big enough to matter'.",{"id":566,"type":77,"variant":567,"title":568,"markdown":569},"i-aha-threshold","aha","There is a hidden threshold, not a smooth slope","The table and lab both show something easy to miss: for any tilt **below** about 1.475°, absolutely nothing changes — the Moon can never get more than 1.475° off the ecliptic anyway if the tilt itself is smaller than that, so it is *always* within range and eclipses stay possible every month.\n\nOnly once the tilt **exceeds** the separation limit does raising it start to matter, and after that point every extra degree of tilt costs a bigger and bigger share of the orbit. The real Moon, at 5.145°, sits well past that threshold — more than three times past it — which is exactly why eclipses are rare rather than merely \"a bit less common than every month\".",{"id":571,"type":572,"prompt":573},"i-reflect-tilt","reflection","If the Moon's tilt were 1.2° instead of 5.145°, would eclipses be common (close to 24 a year), rare (4 to 7 a year), or somewhere in between? Use the table and the lab to support your answer, and explain what \"in between\" would actually mean here.",{"id":575,"type":83,"title":576,"eyebrow":577,"navLabel":578},"i-ch8","Test your eyes safely: build a pinhole projector","Chapter 08","8 Build a projector",{"id":580,"type":77,"variant":195,"title":581,"markdown":582},"i-safety-repeat","Before you do anything in this chapter","**Never look at the Sun directly, with or without any home-made equipment, even during a partial eclipse.** Every activity in this chapter works by projecting an **image** of the Sun onto paper. Your back stays to the Sun the entire time. If you ever want to test this on a non-eclipse day, that is fine — a plain pinhole makes a small round image of the ordinary Sun too, and it is exactly as safe.",{"id":584,"type":46,"markdown":585},"i-pinhole-build","**What you need:** two pieces of stiff card, a pin, and a sunny day (a total eclipse is not required — a pinhole always makes an image of the Sun, eclipsed or not).\n\n1. Make one small, clean pinhole in the centre of the first card.\n2. Stand with your back to the Sun. Hold the pinhole card up so sunlight passes through the hole.\n3. Hold the second card as a screen, some distance behind the first, and find the bright disc of light.\n4. Measure the disc's width with a ruler, and measure the distance between the two cards.\n5. Move the screen further away and measure again.",{"id":587,"type":143,"title":588,"problem":589,"steps":590,"help":594},"i-we-pinhole-check","Checking your own measurement against the formula","The Sun's angular width is 0.533°, so a pinhole makes an image about 9.3 mm across for every metre of distance to the screen. You measure your own projector at a screen distance of 2 metres and get an image 19 mm across. Is that close to the prediction?",[591,592,593],"Predicted size at 2 m: 9.31 × 2 = **18.6 mm**.","Your measurement, 19 mm, is only 0.4 mm off — well within the accuracy you can expect from a hand-held ruler and a slightly wobbly card.","**Conclusion:** the measurement supports the formula. If your own result were, say, 40 mm at 2 m, look for a mistake first — most likely the 'pinhole' was too big and ragged, which blurs and enlarges the image without following the clean angular-size rule.",{"simplerExplanation":595},"9.31 mm per metre, doubled for 2 metres, gives about 18.6 mm — close to a real 19 mm measurement, so the formula checks out.",{"id":597,"type":276,"itemId":598,"prompt":599,"check":600,"hints":603,"feedback":606},"i-practice-pinhole","eclipses.investigate-p-pinhole","Using 9.31 mm of image per metre of distance, how wide (in mm) would the Sun's image be on a screen 4 metres from the pinhole? Round to the nearest whole mm.",{"kind":310,"answer":601,"tolerance":5,"unit":602},37,"mm",[604,605],"Multiply 9.31 by the distance in metres.","9.31 × 4 = ?",{"correct":607,"incorrect":608},"Right: 9.31 × 4 ≈ 37 mm, a little under 4 cm.","9.31 mm\u002Fm × 4 m = 37.24 mm, which rounds to 37 mm.",{"id":610,"type":77,"variant":611,"title":612,"markdown":613},"i-misconception-bigger-hole","misconception","\"A bigger pinhole gives a bigger, better image\"","It gives a **bigger** image, but not a **better** one. A pinhole works because it only lets through a thin pencil of rays from each point of the Sun; make the hole bigger and you let through a wider spread of rays from each point, which blurs the crisp disc into a soft, washed-out patch and can hide a crescent shape completely during a partial eclipse.\n\nThe rule of thumb: keep the hole **small and clean** (a fine pin, not a nail), and get a bigger image by moving the screen **further away** instead, exactly as the formula says. You trade some brightness for size either way — a bigger hole trades away sharpness for nothing.",{"id":615,"type":83,"title":616,"eyebrow":617,"navLabel":618},"i-ch9","Test your predictions against real eclipses","Chapter 09","9 Real dates",{"id":620,"type":46,"markdown":621},"i-real-dates","The best test of anything you have learned is a real eclipse on a real date. Three are already on the calendar for India.",{"id":623,"type":208,"caption":624,"columns":625,"rows":630},"i-table-real","Three eclipses to plan around",[626,627,628,629],"Date","Kind","Where in India","What you should predict beforehand",[631,636,641],[632,633,634,635],"31 Dec 2028","Total lunar","Fully visible across the whole country","Roughly what time totality starts, and how dark (Danjon score) it might look",[637,638,639,640],"21 May 2031","Annular solar","Path crosses Kerala, north Sri Lanka, the Andaman & Nicobar Islands","What percentage of the Sun will be covered where you live, if you are outside the path",[642,643,644,645],"20 Mar 2034","Total solar","Path of totality crosses northern India, including Kashmir","Whether your town is inside the path, and if not, how much partial coverage to expect",{"id":647,"type":471,"tone":648,"items":649},"i-spec-countdown","amber",[650,653,656],{"label":632,"big":651,"value":652},"833 days","About 2.3 years from today (20 Sep 2026).",{"label":637,"big":654,"value":655},"1,704 days","About 4.7 years away.",{"label":642,"big":657,"value":658},"2,738 days","About 7.5 years away — you would be about seven years older.",{"id":660,"type":572,"prompt":661},"i-reflect-plan","Pick one of the three eclipses above. Write a short plan: what you would prepare in advance (equipment, viewing spot, who to invite), what you would predict will happen at your location, and how you would check afterwards whether your prediction was right.",{"id":663,"type":89,"prompt":664,"options":665,"explanation":674},"i-predict-2034","The path of totality for 20 March 2034 crosses northern India. If your town is 400 km south of that path, what would you expect to see?",[666,668,670,672],{"id":93,"label":667},"A total eclipse, just a little shorter",{"id":96,"label":669},"A partial eclipse — a bite out of the Sun, with filters needed throughout",{"id":99,"label":671},"No eclipse at all",{"id":102,"label":673},"An annular eclipse","**A partial eclipse (b).** The path of totality is only about 160 km wide; 400 km outside it puts you well beyond the umbra but still inside the much wider penumbra, which typically spans thousands of kilometres. You would see a large bite taken out of the Sun — how large depends on exactly how far south you are — but never full darkness, and filters or projection are needed for the entire event.",{"id":676,"type":83,"title":677,"eyebrow":678,"navLabel":679},"i-ch10","Wrap-up","Chapter 10","10 Wrap-up",{"id":681,"type":682,"items":683},"i-formulas-recap","formulas",[684,687,690,693],{"expression":685,"caption":686},"w = b × s ÷ d","Shadow width w from a point source: blocker width b, screen distance s, blocker distance d.",{"expression":688,"caption":689},"image (mm) ≈ 9.31 × distance (m)","Pinhole image size, from the Sun's 0.533° angular width.",{"expression":691,"caption":692},"share = path area ÷ Earth area","Fraction of Earth's surface covered by one totality path, about 0.31%.",{"expression":694,"caption":695},"time = path length ÷ shadow speed","How long the shadow takes to sweep a path: hours for the whole path, minutes for one spot.",{"id":697,"type":698,"conceptId":699,"relation":700,"explanation":701},"i-conn-gravity","connection","gravity","helps_understand","The Moon's elliptical orbit, which decides whether an eclipse is total or annular, is shaped by gravity.",{"id":703,"type":161,"component":704,"componentVersion":5,"config":705,"objective":724,"textAlternative":725},"i-lab-match-recap","match-pairs",{"prompt":706,"mode":707,"pairs":708},"Match each test you ran to what it showed.","connect",[709,712,715,718,721],{"a":710,"b":711},"Lamp-and-balls model","Earth and Moon fit on a table; the Sun does not",{"a":713,"b":714},"Moon-phase lab","Only new moon and full moon line the Moon up for either eclipse",{"a":716,"b":717},"Shadow-width measurements","Width = blocker width × screen distance ÷ blocker distance",{"a":719,"b":720},"Tilt slider","Below 1.475° nothing changes; above it, more tilt cuts eclipses hard",{"a":722,"b":723},"Pinhole projector","Image size grows about 9.3 mm per metre of distance, and dims as it grows","Connect five experiments from this layer to the finding each one produced.","A matching game pairing five hands-on tests with their results: the lamp-and-balls model showing the Sun cannot fit to scale in a room; the moon-phase lab confirming only new and full moon line up; shadow measurements confirming the width formula; the tilt slider revealing a hidden threshold near 1.475°; and the pinhole projector confirming the 9.3 mm-per-metre growth rate.",{"id":727,"type":728,"title":729,"terms":730},"i-glossary","glossary","Words from this layer's tests",[731,735,739,743,747,751],{"term":732,"meaning":733,"example":734},"Point source","A light source small enough that its rays fan out from one spot, giving shadows with sharp edges.","A bare, tiny bulb or a distant torch behaves like one.",{"term":736,"meaning":737,"example":738},"Extended source","A light source with real width, so different edges of it are blocked at different places, giving a soft penumbra.","The Sun is a large extended source, half a degree wide.",{"term":740,"meaning":741,"example":742},"Scale model","A model where every real distance and size is shrunk by the same factor.","A 2 cm Moon-ball needs a 7.3 cm Earth-ball 221 cm away to stay to scale.",{"term":744,"meaning":745,"example":746},"Projection","Forming an image of a light source without looking at it directly, by letting its light fall on a screen.","A pinhole, a colander, and gaps between leaves all work by projection.",{"term":748,"meaning":749,"example":750},"Path of totality","The narrow track on Earth's surface where a total solar eclipse's umbra actually lands.","About 160 km wide and roughly 0.31% of Earth's whole surface.",{"term":752,"meaning":753,"example":754},"Threshold","A value below which changing something makes no difference, and above which it starts to matter.","The 1.475° separation limit is a threshold for the Moon's tilt.",{"id":756,"type":757,"title":758,"questions":759},"i-quiz","quiz","Check what you found",[760,773,786,799,812,825,838,851,864],{"itemId":761,"prompt":762,"options":763,"correct":96,"why":772},"eclipses.investigate-q-shadow-rule","A 6 cm ball sits 30 cm from a point lamp; the screen is 90 cm from the lamp. How wide is the shadow?",[764,766,768,770],{"id":93,"label":765},"12 cm",{"id":96,"label":767},"18 cm",{"id":99,"label":769},"9 cm",{"id":102,"label":771},"3 cm","6 × 90 ÷ 30 = 18 cm.",{"itemId":774,"prompt":775,"options":776,"correct":96,"why":785},"eclipses.investigate-q-phase","In the phase test, which two phases allow any kind of eclipse at all?",[777,779,781,783],{"id":93,"label":778},"First and last quarter",{"id":96,"label":780},"New moon and full moon",{"id":99,"label":782},"All eight phases equally",{"id":102,"label":784},"Only full moon","Only new moon (Moon between Earth and Sun) and full moon (Moon behind Earth) put the three bodies anywhere near a line.",{"itemId":787,"prompt":788,"options":789,"correct":96,"why":798},"eclipses.investigate-q-model","Why can a lamp-and-balls model not show the Sun at true scale in an ordinary room?",[790,792,794,796],{"id":93,"label":791},"Lamps are not bright enough",{"id":96,"label":793},"At the scale where the Moon is a 2 cm ball, the Sun would need to be an 8-metre sphere 861 metres away",{"id":99,"label":795},"It is against the rules of physics",{"id":102,"label":797},"The Sun does not actually cast a shadow","The worked example computes it directly: 1,392,700 ÷ 1,737.5 ≈ 801 cm across, standing 861 m off.",{"itemId":800,"prompt":801,"options":802,"correct":96,"why":811},"eclipses.investigate-q-threshold","In the tilt experiment, why did the eclipse count stay at 24 a year for tilts of 0.5° and 1°?",[803,805,807,809],{"id":93,"label":804},"The lab was broken",{"id":96,"label":806},"Those tilts are still smaller than the 1.475° separation a solar eclipse can tolerate, so the Moon is always within range",{"id":99,"label":808},"The Moon's distance changed instead",{"id":102,"label":810},"It only updates every 5°","Below the 1.475° threshold, the Moon's maximum drift off the ecliptic is still small enough that every new moon qualifies.",{"itemId":813,"prompt":814,"options":815,"correct":93,"why":824},"eclipses.investigate-q-pinhole-image","Using 9.31 mm of image per metre, what image size would you predict at a screen distance of 0.5 m?",[816,818,820,822],{"id":93,"label":817},"About 4.7 mm",{"id":96,"label":819},"About 9.3 mm",{"id":99,"label":821},"About 18.6 mm",{"id":102,"label":823},"About 1 mm","9.31 × 0.5 ≈ 4.66 mm.",{"itemId":826,"prompt":827,"options":828,"correct":93,"why":837},"eclipses.investigate-q-annular-test","In the total-or-annular experiment, which single change turns a total eclipse into an annular one, all else being equal?",[829,831,833,835],{"id":93,"label":830},"Moving the Moon further from Earth",{"id":96,"label":832},"Moving Earth closer to the Sun",{"id":99,"label":834},"Changing the time of year only",{"id":102,"label":836},"Increasing the tilt","Moving the Moon towards apogee shrinks its apparent size until the umbra's tip falls short of Earth, leaving a ring.",{"itemId":839,"prompt":840,"options":841,"correct":96,"why":850},"eclipses.investigate-q-hybrid","What makes a hybrid eclipse total in the middle of its track but annular at the ends?",[842,844,846,848],{"id":93,"label":843},"The Moon speeds up",{"id":96,"label":845},"Earth's curved surface puts observers near the ends slightly further from the Moon",{"id":99,"label":847},"Clouds usually cover the ends",{"id":102,"label":849},"It never really happens","The extra distance near the ends, due to Earth's curvature, is just enough to let the umbra's tip fall short there while still reaching directly underneath.",{"itemId":852,"prompt":853,"options":854,"correct":96,"why":863},"eclipses.investigate-q-rarity","One totality path covers about 0.31% of Earth's surface. What does that best explain?",[855,857,859,861],{"id":93,"label":856},"Why total eclipses are impossible",{"id":96,"label":858},"Why any one fixed town waits centuries between visits, even though a total eclipse crosses somewhere every 18 months or so",{"id":99,"label":860},"Why the Moon is shrinking",{"id":102,"label":862},"Why lunar eclipses are rare too","A tiny share of Earth's surface, swept by a fast-moving path, means a fixed spot is rarely underneath it — but a traveller can catch several in a lifetime.",{"itemId":865,"prompt":866,"options":867,"correct":96,"why":876},"eclipses.investigate-q-2034","A town 400 km outside the 2034 path of totality — what should residents prepare for?",[868,870,872,874],{"id":93,"label":869},"A brief totality",{"id":96,"label":871},"A partial eclipse, filters needed the whole time",{"id":99,"label":873},"No eclipse",{"id":102,"label":875},"An annular ring","400 km is well outside the roughly 160 km-wide path but inside the much wider penumbra, giving a partial eclipse only.",{"id":878,"type":879,"title":880,"points":881},"i-cheat","summary","Cheat sheet",[882,883,884,885,886,887,888],"**Shadow-width rule (point source):** width = blocker width × screen distance ÷ blocker distance. Move the blocker away from the light and the shadow shrinks.","**Lamp-and-balls model:** a 2 cm Moon-ball needs a 7.3 cm Earth-ball about 2.2 m away to keep true proportions — but the Sun, at the same scale, would be an 8 m sphere 861 m off. The model shows the *idea*, not the true scale.","**Phase test:** only new moon (solar) and full moon (lunar) ever bring the three bodies near a line. Every other phase fails by tens of degrees, tilt or no tilt.","**Tilt has a threshold near 1.475°.** Below it, eclipses would happen every month regardless of the exact tilt. Above it, every extra degree of tilt cuts the eclipse-possible window hard. The real 5.145° is well past that threshold.","**Pinhole image size:** about 9.3 mm of image for every metre of screen distance, and dimmer the further you go. A bigger hole makes a bigger but blurrier, less crescent-shaped image — move the screen back instead.","**Three real eclipses to test yourself against:** 31 Dec 2028 (total lunar, all of India), 21 May 2031 (annular, Kerala\u002FSri Lanka\u002FAndaman & Nicobar), 20 Mar 2034 (total, path across northern India including Kashmir).","**SAFETY, always:** every activity here projects an image onto paper. Never look at the Sun directly, with or without home-made equipment.",{"id":890,"type":891,"sourceIds":892},"i-sources","sources",[893,894,895,896,897,898,899],"eclipses-nasa-eclipses","eclipses-nasa-safety","eclipses-timeanddate","eclipses-wiki-solar","eclipses-wiki-lunar","eclipses-britannica-kids","eclipses-ncert-curiosity",[893,894,895,896,897,898,899],"needs_review",{"generatedBy":903,"notes":904},"claude-code","Draft generated locally; pending owner review. Every number computed in scratchpad\u002Feclipses\u002Fnumbers.py and this generator's own header.","b1925e000f8929b78bce4f2413fde3c2dc9954f9b0d18799819203997fff8105",{"component:shadow-lab@1":907,"component:moon-phase@1":908,"logic:practice":909,"component:eclipse-lab@1":910,"component:match-pairs@1":911,"source:eclipses-britannica-kids":912,"source:eclipses-nasa-eclipses":913,"source:eclipses-nasa-safety":914,"source:eclipses-ncert-curiosity":915,"source:eclipses-timeanddate":916,"source:eclipses-wiki-lunar":917,"source:eclipses-wiki-solar":918},"7476ef546fdb1f398a07393483e549c923f9568dc1bb1c2561191bf47862c475","39afeb0bba7518b8118317655457b27214a4a2a315d236762bf1d6a6a40e18f3","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","284cb1682e3994af706a6cf907dd7747c9c141577245765bceff3a44fc81cf10","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b10d074ebe9d86c884d3f9aab5d21f1e9ba85495f2d6add6a9263b96a9eb0a6a","c0bbcfd69fe06099e49c296bba2b105973ec1a794b69292fb65b4d81e9a2d672","f585b6bff2c52e5ed1c593b6222bc1e2805b7024b64f26d49271a5f19aa6dff7","c8e27588447f45548af86f4ac4ceb632dd451d78252b13c94d725b85ce34a24d","1508f2bc1fe7211250c3bc94beb9fff17e79cf0168bd0bd30264ee1e68d0086a","a72372b1c84fe7003a52eaf5654f6b88dabc645b405b6df7578cea1bae5d55e2","e32e7e8bfd7f8db32db243544cfc60cb96c0c7f050e964450ae3fc99bb21d344",{"state":920,"reviewer":921,"selfReview":252,"reviewedAt":922,"method":923},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899599114]