[{"data":1,"prerenderedAt":839},["ShallowReactive",2],{"layer:tides:deepen":3},{"layer":4,"contentHash":819,"dependencyHashes":820,"approval":832,"releaseId":838},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":814,"reviewStatus":815,"authoring":816},1,"tides","en","deepen","Deepen: the mathematics and history behind a tide table","Newton, Laplace, harmonic waves, closed-pipe resonance, and the physics of a bore","Trace the two-hundred-year path from Newton's equilibrium theory to Laplace's ocean waves and Kelvin's tide-predicting machine, meet the harmonic constituents that a real tide is built from, derive why a bay resonates at a quarter wavelength, and quantify Earth's own solid and atmospheric tides.",[13,14,15,16,17],"Explain what Newton's equilibrium theory got right and wrong, and what Laplace's dynamic theory fixed.","Use the shallow-water wave speed rule to show why even a deep ocean counts as 'shallow' for a tide.","Calculate a tidal form factor F from constituent sizes and classify a port's tide from it.","Explain, using the closed-organ-pipe analogy, why a bay resonates at a quarter wavelength rather than a half.","Describe Earth tides and atmospheric tides and how they are detected.",45,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Deepen",{"label":26,"value":27},"Reading time","≈ 45 minutes",{"label":29,"value":30},"Prior knowledge","Investigate: wave speed, resonance, tide tables",{"label":32,"value":33},"Chapters","9",{"label":35,"value":36},"Maths used","Ratios, square roots, multi-step worked problems",{"label":38,"value":39},"History","Newton 1687 to Doodson 1921",[41,45,70,76,82,85,90,93,118,159,182,187,192,195,200,215,219,224,227,255,278,283,288,291,309,320,339,354,380,384,390,395,400,403,419,423,441,454,459,462,475,494,498,503,506,510,514,538,543,548,551,555,559,564,567,581,607,621,624,628,658,789,804],{"id":42,"type":43,"markdown":44},"d-intro","prose","Understand gave you the mechanism, and Investigate let you test it. This layer goes further still: into the history of how people worked all this out, the mathematics that actually predicts a real tide, and a few calculations that would satisfy a genuinely sceptical scientist.\n\nExpect it to feel more like proper physics than the earlier layers. That is the point: tides are one of the oldest problems in science, and the story of solving them properly took over two hundred years.",{"id":46,"type":47,"tone":48,"items":49},"d-spec-glance","spec","copper",[50,54,58,62,66],{"label":51,"big":52,"value":53},"Open-ocean wave speed","≈ 686 km\u002Fh","A tide travels as a shallow-water wave even in the deepest ocean.",{"label":55,"big":56,"value":57},"Fundy quarter wavelength","≈ 303 km","Matches the real Gulf of Maine–Bay of Fundy system's size.",{"label":59,"big":60,"value":61},"Fundy system's Q value","≈ 5","Modest by physics standards, huge for an ocean.",{"label":63,"big":64,"value":65},"Doodson's constituents","388","Distinct tidal frequencies found in 1921.",{"label":67,"big":68,"value":69},"Michelson's Earth-tide pipe","166 m","Detected a 20-micrometre level change matching an elastic solid Earth.",{"id":71,"type":72,"variant":73,"title":74,"markdown":75},"d-how-to","callout","observation","How to use this lesson","The chapters build on each other more tightly than in earlier layers. Chapter 2's harmonic constituents are used again in Chapter 3, and Chapter 4's resonance argument leans on the wave-speed idea from Chapter 5. Read in order.",{"id":77,"type":78,"title":79,"eyebrow":80,"navLabel":81},"d-ch1","chapter","Newton's tide: a good idea that isn't quite true","Chapter 01","1 Newton's tide",{"id":83,"type":43,"markdown":84},"d-newton","The first person to explain tides with gravity was Isaac Newton, in the *Principia* of **1687**. His **equilibrium theory** imagined the ocean as a layer of water thin enough, and responding fast enough, to sit permanently in balance with the Moon's and Sun's pull — always instantly bulging exactly where the two-bulge picture in Understand says it should.\n\nIt was a triumph: for the first time, tides were connected to the same gravity that holds the Moon in its orbit and drops an apple to the ground. It correctly predicts spring and neap tides, the rough size of the effect, and the existence of two bulges.\n\nIt also makes a testable, and wrong, prediction: that every ocean, everywhere, should have very nearly the same small tidal range (a few tens of centimetres), always at the moment the Moon crosses overhead. You already know that is false — Kochi's metre and the Bay of Fundy's sixteen metres are nothing alike, and high water at most ports arrives hours after the Moon passes overhead, not at that instant.",{"id":86,"type":72,"variant":87,"title":88,"markdown":89},"d-misc-equilibrium","misconception","“Newton's theory is wrong, so ignore it”","A model that makes a wrong quantitative prediction is not necessarily a useless model. Newton's equilibrium theory correctly identifies the **cause** (differential gravity), the **timing pattern** (two bulges, spring and neap), and gives the right **order of magnitude**. What it gets wrong is treating the ocean as if it could respond instantly and freely, ignoring the fact that water has to physically flow into a basin shaped by real coastlines. Scientists still teach equilibrium theory first, on purpose, because the cause has to be understood before the complications are worth adding.",{"id":91,"type":43,"markdown":92},"d-laplace","The fix came from Pierre-Simon Laplace, who published a **dynamic theory of the tides** in **1775**. Laplace's key move was to stop asking \"where would the water settle if it had time to reach equilibrium\" and start asking \"how does a real ocean, with real depth and real coastlines, actually respond when a tidal force pushes on it, moment by moment?\"\n\nThat turns the problem from simple geometry into the physics of **waves in water**, which is exactly the physics you already used for tidal bores. Laplace's equations describe how a tide sloshes, reflects and resonates inside an ocean basin — precisely the behaviour that explains why the Bay of Fundy is not like Kochi, and why high water is late almost everywhere.",{"id":94,"type":95,"caption":96,"columns":97,"rows":101},"d-table-newton-laplace","table","Two theories, compared",[98,99,100],"Question","Newton's equilibrium theory","Laplace's dynamic theory",[102,106,110,114],[103,104,105],"Ocean response","Instant — always in balance with the Moon and Sun","Takes time — the tide must physically flow in as a wave",[107,108,109],"Range everywhere","Predicts nearly the same small range everywhere","Correctly allows huge differences (Kochi 1 m, Fundy 16 m)",[111,112,113],"High water timing","Predicts it happens as the Moon crosses overhead","Correctly allows a delay set by each basin's own behaviour",[115,116,117],"What it explains well","The basic cause: gravity; spring and neap; the two bulges","Why real ports differ so much from that basic picture",{"id":119,"type":120,"component":121,"componentVersion":5,"config":122,"objective":157,"textAlternative":158},"d-lab-sort-theory","interactive","sort-game",{"prompt":123,"bins":124,"items":131,"seconds":156},"Equilibrium theory (Newton) or dynamic theory (Laplace)? Sort each claim.",[125,128],{"id":126,"label":127},"newton","Equilibrium theory",{"id":129,"label":130},"laplace","Dynamic theory",[132,136,140,144,148,152],{"id":133,"label":134,"bin":126,"why":135},"th1","The ocean always sits in instant balance with the Moon and Sun","This is the simplifying assumption at the heart of equilibrium theory.",{"id":137,"label":138,"bin":129,"why":139},"th2","Tides are waves that must physically travel through a basin of a given shape and depth","Treating the tide as a real wave, not an instant shape, is Laplace's key move.",{"id":141,"label":142,"bin":126,"why":143},"th3","High water should happen exactly when the Moon is overhead","This clean prediction follows from assuming instant equilibrium, and it is the part that turns out to be wrong.",{"id":145,"label":146,"bin":129,"why":147},"th4","Different basins can resonate differently, giving wildly different ranges","Resonance is a property of a wave sloshing in a real basin — pure dynamic-theory territory.",{"id":149,"label":150,"bin":126,"why":151},"th5","Every ocean should show nearly the same small tidal range","A clean but false prediction of treating the whole ocean as one instant equilibrium shape.",{"id":153,"label":154,"bin":129,"why":155},"th6","A port's high water can lag hours behind the Moon passing overhead","This lag is exactly what dynamic theory explains and equilibrium theory cannot.",0,"Sort six claims about the tide into Newton's equilibrium theory or Laplace's dynamic theory.","A sorting game with two bins. Equilibrium theory (Newton): the ocean sits in instant balance with the Moon and Sun; high water should happen exactly when the Moon is overhead; every ocean should show nearly the same small range. Dynamic theory (Laplace): tides are waves travelling through a basin of given shape and depth; different basins can resonate differently, giving wildly different ranges; a port's high water can lag hours behind the Moon passing overhead.",{"id":160,"type":161,"title":162,"items":163},"d-timeline-history","timeline","From Newton's geometry to a machine that could predict the tide",[164,167,170,174,178],{"time":165,"title":99,"text":166},"1687","The *Principia* connects tides to gravity for the first time, treating the ocean as if it could sit in instant balance with the Moon and Sun.",{"time":168,"title":100,"text":169},"1775","Laplace treats the tide as a wave problem: how a real ocean, with real depth and coastlines, actually responds and sloshes, rather than sitting in permanent equilibrium.",{"time":171,"title":172,"text":173},"1872","Kelvin's tide-predicting machine","William Thomson (later Lord Kelvin) designs the first tide-predicting machine, a mechanical analogue computer combining about 10 astronomical components to trace out a future tidal curve.",{"time":175,"title":176,"text":177},"1921","Doodson's harmonic constituents","Arthur Doodson distinguishes 388 separate tidal frequencies (the Doodson Numbers) hidden inside the Moon's and Sun's combined pull, most of them far too small to matter for any real port.",{"time":179,"title":180,"text":181},"today","Computers, not gears","The same harmonic method Kelvin's gears carried out mechanically now runs on computers at INCOIS and the Survey of India, still built from the same idea: add up many steady waves.",{"id":183,"type":72,"variant":184,"title":185,"markdown":186},"d-example-machine","example","Gears that could out-think a room full of clerks","Kelvin's later machines combined up to 24 astronomical components using stacked pulleys and wires, cranked by hand, and could trace a full year's tidal curve for a port in a few hours — work that would otherwise have taken a team of human 'computers' (that word originally meant a person who calculates) weeks of arithmetic. Only about thirty of these machines were ever built worldwide, between 1872 and the 1960s, because each one was a hand-built, expensive piece of precision engineering. Digital computers finally made them obsolete once they became fast enough to do the same sums directly.",{"id":188,"type":78,"title":189,"eyebrow":190,"navLabel":191},"d-ch2","Why an ocean sloshes: tides as very long waves","Chapter 02","2 Tides as waves",{"id":193,"type":43,"markdown":194},"d-ocean-shallow","Here is a fact that sounds backwards at first: for the purposes of a tide, **every ocean on Earth counts as shallow**.\n\nThe shallow-water wave-speed rule from Investigate, v = √(g × depth), applies to any wave whose length is much greater than the water's depth — and a tidal 'wave' is roughly half the width of an ocean basin long, thousands of kilometres, which utterly dwarfs even the deepest trench. So even the open Pacific, averaging about **3700 m** deep, counts as \"shallow\" to a tide, and the same simple formula applies.\n\nUsing that average depth: v = √(9.8 × 3700) ≈ **191 m\u002Fs**, which is about **686 km\u002Fh** — roughly the cruising speed of a jet airliner. A tidal bulge does not creep across the ocean; it races across it, and still takes many hours to cross a whole basin, because the basins are so enormous.",{"id":196,"type":72,"variant":197,"title":198,"markdown":199},"d-aha-shallow","aha","This is why the tide is Laplace's problem, not Newton's","If tides really were an instant equilibrium, ocean depth would not matter at all. The fact that ocean depth sets the **speed** at which a tide can actually travel is exactly why the real ocean cannot keep up with an idealised, instantly-adjusting bulge. Different basins, with different depths and different shapes, let the tide travel at different speeds and reflect off coastlines at different times — which is the whole reason every port needs its own tide table instead of one global formula.",{"id":201,"type":202,"title":203,"problem":204,"steps":205,"help":210},"d-we-crossing-time","worked_example","How long does a tide take to cross an ocean?","Using the open-ocean wave speed of about 191 m\u002Fs, roughly how long would a tidal wave take to cross 10,000 km of open Pacific Ocean — about the distance from the Philippines to Peru?",[206,207,208,209],"Convert the distance to metres: 10,000 km = 10,000,000 m.","Time = distance ÷ speed = 10,000,000 ÷ 191 ≈ **52460 seconds**.","Convert to hours: 52460 ÷ 3,600 ≈ **14.6 hours**, about half a day.","That is the same order of size as the tide's own period (12 h 25 min). Neither instant nor negligible: this is exactly why an ocean basin can end up in or out of step with the tide driving it, which is the seed of resonance in Chapter 4.",{"simplerExplanation":211,"hints":212},"Time = distance ÷ speed, same as any journey. A tide takes hours, not minutes, to cross a big ocean, and not far off the length of the tide's own cycle.",[213,214],"10,000 km is 10 million metres.","Divide seconds by 3,600 to get hours.",{"id":216,"type":72,"variant":184,"title":217,"markdown":218},"d-example-tsunami-speed","The same formula explains tsunami warning times","This is also, unexpectedly, the physics behind a **tsunami** warning network. A tsunami is not a tide — you met that misconception in Understand — but it is also a very long wave in shallow water, governed by the very same v = √(g × depth) rule used here for tides and, earlier, for river bores. A tsunami crossing the deep Indian Ocean at roughly the speed you just calculated can still take a few hours to reach a distant coast, and that travel time is exactly the warning window that INCOIS's tsunami centre, set up after the 2004 Indian Ocean tsunami, uses to alert coastal areas before the wave arrives. Same formula, completely different cause.",{"id":220,"type":78,"title":221,"eyebrow":222,"navLabel":223},"d-ch3","Four waves hiding inside every tide","Chapter 03","3 Four constituents",{"id":225,"type":43,"markdown":226},"d-constituents","You met the idea of harmonic analysis in Investigate: a real tide is the sum of many simple, steady waves. Here are the four biggest, with the periods oceanographers actually use.\n\n- **M2**, the principal lunar semidiurnal wave: period **12 h 25 min** — this is simply the half-lunar-day you calculated in Understand, given a name.\n- **S2**, the principal solar semidiurnal wave: period exactly **12.0 hours**, because it is tied to the ordinary solar day, not the Moon.\n- **K1**, the lunisolar diurnal wave: period **23.9345 hours** — precisely one **sidereal day** (Earth's spin relative to the stars), reflecting the combined pull of the Moon and Sun once each rotation.\n- **O1**, the principal lunar diurnal wave: period **25.8193 hours**.\n\nNotice something important: **O1's period is not the same as the lunar (tidal) day of 24 h 50 min** from Understand, even though both come from the Moon. They arise from different parts of the Moon's motion (M2 and O1 both depend on the Moon, but at different harmonics of its orbit), and only when you add several constituents together do you get the familiar 24 h 50 min pattern of two highs a day.",{"id":228,"type":95,"caption":229,"columns":230,"rows":235},"d-table-constituents","The four main tidal constituents",[231,232,233,234],"Constituent","Full name","Period","Driven mainly by",[236,241,246,251],[237,238,239,240],"M2","Principal lunar semidiurnal","12 h 25 min","The Moon",[242,243,244,245],"S2","Principal solar semidiurnal","12.0 h","The Sun",[247,248,249,250],"K1","Lunisolar diurnal","23.93 h (one sidereal day)","Moon and Sun together",[252,253,254,240],"O1","Principal lunar diurnal","25.82 h",{"id":256,"type":120,"component":257,"componentVersion":5,"config":258,"objective":276,"textAlternative":277},"d-lab-match-constituents","match-pairs",{"prompt":259,"mode":260,"pairs":261},"Match each constituent to its period.","connect",[262,264,266,268,270,273],{"a":237,"b":263},"12 h 25 min — half a lunar day, the biggest wave almost everywhere",{"a":242,"b":265},"Exactly 12 hours — tied to the ordinary solar day",{"a":247,"b":267},"23.93 hours — one sidereal day, Moon and Sun together",{"a":252,"b":269},"25.82 hours — the main lunar diurnal wave",{"a":271,"b":272},"Form factor F","(K1 + O1) ÷ (M2 + S2): classifies the tide type",{"a":274,"b":275},"Spring–neap beat","M2 and S2 slowly drifting in and out of step","Match each of the four main tidal constituents, plus the form factor and the spring–neap beat, to its description.","A matching game with six pairs. M2 goes with its period of 12 h 25 min, half a lunar day. S2 goes with exactly 12 hours. K1 goes with 23.93 hours, one sidereal day. O1 goes with 25.82 hours, the main lunar diurnal wave. Form factor F goes with the ratio (K1+O1)÷(M2+S2) that classifies tide type. Spring–neap beat goes with M2 and S2 slowly drifting in and out of step.",{"id":279,"type":72,"variant":280,"title":281,"markdown":282},"d-nuance-beat","nuance","Where the spring–neap beat comes from, exactly","In Understand you found the spring–neap cycle two ways: from the calendar (a fortnight is half a synodic month) and from \"two clocks drifting apart\" (M2's 12 h 25 min against a 12-hour rhythm). Properly, that second method is comparing **M2 and S2**: two waves of almost the same period, 12 h 25 min and 12.0 h, whose gentle drift in and out of step **is** the spring–neap beat. When M2 and S2 line up (both waves cresting together), you get a spring tide; half a beat later, when one crests as the other troughs, you get a neap. The same trick — comparing two very similar periods — appears throughout physics, from beating musical notes to radio tuning.",{"id":284,"type":78,"title":285,"eyebrow":286,"navLabel":287},"d-ch4","The form factor: naming a tide with one number","Chapter 04","4 The form factor",{"id":289,"type":43,"markdown":290},"d-form-factor","Investigate asked you to classify tides as semidiurnal, diurnal or mixed by eye. Oceanographers have a precise number for it, the **tidal form factor**:\n\nF = (K1 + O1) ÷ (M2 + S2)\n\n— the combined size of the two big **daily** waves, divided by the combined size of the two big **twice-daily** waves. A port dominated by M2 and S2 gets a small F and looks semidiurnal; a port dominated by K1 and O1 gets a large F and looks diurnal; anything in between is mixed. The standard boundaries, used worldwide, are:",{"id":292,"type":95,"caption":293,"columns":294,"rows":296},"d-table-form-factor","The standard tidal-type boundaries",[271,295],"Tide type",[297,300,303,306],[298,299],"F \u003C 0.25","Semidiurnal",[301,302],"0.25 ≤ F \u003C 1.5","Mixed, mainly semidiurnal",[304,305],"1.5 ≤ F \u003C 3.0","Mixed, mainly diurnal",[307,308],"F ≥ 3.0","Diurnal",{"id":310,"type":202,"title":311,"problem":312,"steps":313,"help":316},"d-we-form-factor","Classifying three made-up ports with the form factor","These heights (in arbitrary matching units) are made up to practise the sum, not real measurements of any named port. Port A: M2=100, S2=30, K1=10, O1=8. Classify it.",[314,315],"F = (K1 + O1) ÷ (M2 + S2) = (10 + 8) ÷ (100 + 30) = 18 ÷ 130 = **0.14**.","0.14 is below 0.25, so Port A is **semidiurnal** — dominated overwhelmingly by its twice-daily waves.",{"simplerExplanation":317,"hints":318},"Add the two diurnal numbers, add the two semidiurnal numbers, divide.",[319],"K1 + O1 goes on top; M2 + S2 goes on the bottom.",{"id":321,"type":95,"caption":322,"columns":323,"rows":328},"d-table-two-more-ports","Two more made-up ports: work out F and the type yourself first",[324,325,326,327],"Port","M2, S2, K1, O1","F","Type",[329,334],[330,331,332,333],"B","40, 15, 45, 30","1.36","mixed, mainly semidiurnal",[335,336,337,338],"C","10, 5, 40, 25","4.33","diurnal",{"id":340,"type":341,"itemId":342,"prompt":343,"check":344,"hints":348,"feedback":351},"d-pr-form-factor","practice","tides.deepen-form-factor","A fourth made-up port has M2=60, S2=20, K1=20, O1=20. What is its form factor F, to two decimal places?",{"kind":345,"answer":346,"tolerance":347},"number",0.5,0.02,[349,350],"F = (K1 + O1) ÷ (M2 + S2).","K1 + O1 = 40. M2 + S2 = 80.",{"correct":352,"incorrect":353},"Right: F = (20 + 20) ÷ (60 + 20) = 40 ÷ 80 = **0.50**, which is mixed, mainly semidiurnal.","Add the top pair, add the bottom pair, then divide: 40 ÷ 80 = 0.50.",{"id":355,"type":120,"component":356,"componentVersion":5,"config":357,"objective":378,"textAlternative":379},"d-lab-data-constituents","data-lab",{"datasets":358,"valueRange":367,"step":5,"challenges":369},[359],{"label":360,"unit":361,"values":362},"Port A constituent sizes","units",[363,364,365,366],100,30,10,8,{"min":156,"max":368},110,[370,374],{"measure":371,"target":372,"prompt":373},"range",92,"What is the range between Port A's biggest and smallest constituent?",{"measure":375,"target":376,"prompt":377},"mean",37,"What is the mean size of Port A's four constituents?","Treat Port A's four constituent sizes as a tiny dataset and find their range and mean.","This lab plots Port A's four made-up constituent sizes — M2=100, S2=30, K1=10, O1=8 — as four dots. The range (biggest minus smallest) is 92, and the mean is 37.00. Neither number is the form factor F; they simply describe how spread out and how large the four constituents are as a small dataset, a reminder that the same four numbers can be analysed in more than one way depending on the question you ask.",{"id":381,"type":72,"variant":280,"title":382,"markdown":383},"d-nuance-declination","Why the diurnal wobble waxes and wanes through the month","K1 and O1, the two daily constituents, are not fixed in size: they grow when the Moon (and the Sun) sit well north or south of the equator, and shrink towards zero when the Moon crosses the equator. This is exactly the mechanism behind Understand's 'declination' nuance about India's east coast having unequal high tides — that inequality is the visible fingerprint of K1 and O1 rising and falling as the Moon swings above and below the equator over its 27.3-day orbit, layered on top of the steadier M2 and S2.",{"id":385,"type":386,"conceptId":387,"relation":388,"explanation":389},"d-conn-data","connection","data-handling","applied_in","The form factor is a ratio computed from measured constituent sizes — the same kind of data-handling skill (combine several measurements into one meaningful number) used for an average or a range.",{"id":391,"type":72,"variant":392,"title":393,"markdown":394},"d-model-limit-form","model_limit","F tells you the type, not the size","The form factor only compares **ratios** between constituents, so it says nothing about whether a port's tide is big or small overall — a busy semidiurnal port could have F near zero with a 1 m range or a 16 m range. Type and size are answered by different numbers: F for the shape of the daily pattern, and the constituents' actual sizes (plus the local funnelling, shallowing and resonance from Understand) for how big it actually gets.",{"id":396,"type":78,"title":397,"eyebrow":398,"navLabel":399},"d-ch5","Resonance: a bay is a closed organ pipe","Chapter 05","5 Resonance, properly",{"id":401,"type":43,"markdown":402},"d-organ-pipe","Investigate showed that the Bay of Fundy's real size lands close to a calculated \"quarter wavelength\". This chapter explains **why a quarter**, not a half or a whole wavelength.\n\nA bay like the Bay of Fundy is closed at its head (the land) and open at its mouth (the ocean). That is exactly the shape of a musical instrument you may already know: a **closed organ pipe**, or a bottle you blow across — closed at one end, open at the other. Such a pipe resonates most strongly when it is a **quarter** of a wavelength long, because the closed end must be a point where the wave's motion is zero (water cannot slosh through solid land) while the open end is free to move as much as possible — and a quarter of a full wave is the shortest length that fits a zero at one end and a maximum at the other.\n\nA bay open at both ends (imagine a strait connecting two seas) instead resonates like a pipe open at both ends, at a **half** wavelength. The Bay of Fundy, closed at its head, is the quarter-wavelength kind — which is exactly the formula Investigate used.",{"id":404,"type":405,"items":406},"d-formulas-resonance","formulas",[407,410,413,416],{"expression":408,"caption":409},"closed–open pipe (a bay): L = λ ÷ 4","Resonates at a quarter wavelength; the Bay of Fundy's case.",{"expression":411,"caption":412},"open–open pipe (a strait): L = λ ÷ 2","Resonates at a half wavelength; a channel joining two seas.",{"expression":414,"caption":415},"λ = v × T","Wavelength = wave speed × period, for any wave.",{"expression":417,"caption":418},"v = √(g × depth)","Shallow-water wave speed, from Investigate.",{"id":420,"type":72,"variant":184,"title":421,"markdown":422},"d-example-q","How efficient is the Fundy 'swing'? The Q-value","Physicists measure how sharply a resonator responds with a number called **Q** (for \"quality\"): a high Q means a system rings on for a long time once excited and responds very strongly right at its resonant period, while a low Q means it damps out quickly and is much less fussy about the exact push it gets. Published studies of the Gulf of Maine–Bay of Fundy system put its Q at roughly **5** — modest by the standards of a fine physics instrument (some lab resonators have a Q of a million or more), but easily enough, applied twice a day for thousands of years, to build the world's largest tide.",{"id":424,"type":120,"component":425,"componentVersion":5,"config":426,"objective":439,"textAlternative":440},"d-lab-resonance","tide-lab",{"modes":427,"places":430,"challenges":438},[428,429],"spring-neap","tide-clock",[431,435],{"id":432,"label":433,"rangeM":434},"fundy","Bay of Fundy (about 16 m)",16,{"id":436,"label":437,"rangeM":365},"khambhat","Gulf of Khambhat (about 10 m)",3,"See the spring-neap beat riding on top of a resonant coast's already-large range.","This lab reruns the tide clock and spring–neap modes for the Bay of Fundy and the Gulf of Khambhat, both closed-end, quarter-wavelength resonators. Even their neap tides are large by world standards, because resonance amplifies the whole M2 and S2 signal, not just its spring peaks.\n\nThree challenges ask you to compare a neap tide at a resonant coast with a spring tide at an ordinary coast like Kochi, and to explain which effect — resonance or the spring–neap beat — has the bigger influence on the numbers you see.",{"id":442,"type":202,"title":443,"problem":444,"steps":445,"help":449},"d-we-kutch","Would the Gulf of Kutch behave like a closed or an open pipe?","The Gulf of Kutch, like the Gulf of Khambhat, is closed at its landward end and open to the Arabian Sea at its mouth. Which resonance rule applies, and does that make it more or less able to resonate strongly than a strait open at both ends of the same length?",[446,447,448],"It is closed at one end (land) and open at the other (the sea): the **closed–open** rule, L = λ ÷ 4, the same family as the Bay of Fundy.","For a *given* basin length L, a closed–open pipe resonates at a **longer** wavelength (λ = 4L) than an open–open pipe of the same length (λ = 2L), because it only needs to fit a quarter-wave instead of a half-wave.","A longer resonant wavelength means it takes a **slower** wave, or a shorter basin, to hit exact resonance with a fixed tidal period — which is one more reason a closed gulf like Khambhat or Kutch, not just an open strait, is where India's biggest ranges are found.",{"simplerExplanation":450,"hints":451},"Closed at one end, open at the other, is the same family as the Bay of Fundy: quarter-wavelength resonance.",[452,453],"Land at one end blocks water from sloshing through: that is the 'closed' end.","The sea at the other end lets water move freely: the 'open' end.",{"id":455,"type":78,"title":456,"eyebrow":457,"navLabel":458},"d-ch6","Bores again: how far, and how fast, does the front pull ahead?","Chapter 06","6 Bores, quantified",{"id":460,"type":43,"markdown":461},"d-bore-quant","Investigate showed that water in 4 m of depth outruns water in 1 m of depth by about **11.3 km\u002Fh**. Here is the harder question: given a head start, how long does it actually take for the deep-water front to catch up completely and turn a gentle slope into a bore?",{"id":463,"type":202,"title":464,"problem":465,"steps":466,"help":471},"d-we-bore-close","How long until the front becomes a wall of water? (an illustration)","Suppose (again, an illustration, not a survey of a real river) the leading edge of a tide is 20 km ahead in water 4 m deep, while water only 1 m deep sits behind it. Using the speed difference from Investigate, roughly how long before the shallower water catches all the way up?",[467,468,469,470],"Speed in 4 m water: 6.27 m\u002Fs. Speed in 1 m water: 3.13 m\u002Fs.","The gap closes at the **difference** in speed: 6.27 − 3.13 = 3.14 m\u002Fs ≈ 11.3 km\u002Fh.","Time = distance ÷ relative speed = 20 km ÷ 11.3 km\u002Fh ≈ **6.4 hours**.","A few hours is entirely realistic for the time a tide spends travelling up a shallowing river mouth, which is exactly why real bores form only over a real distance of shallowing channel, not the instant the tide starts to turn.",{"simplerExplanation":472,"hints":473},"Use the same trick as any 'catching up' problem: distance ÷ the difference in speed.",[474],"Convert the speed difference to km\u002Fh first, to match the km distance.",{"id":476,"type":95,"caption":477,"columns":478,"rows":483},"d-table-bores","Comparing the Hooghly and the Qiantang",[479,480,481,482],"River","Typical bore height","Extreme bore height","Top speed",[484,489],[485,486,487,488],"Hooghly, India","often over 2.1 m","2.4 to 6.1 m (March\u002FSeptember)","not commonly quoted",[490,491,492,493],"Qiantang, China","several metres","about 9 m","about 40 km\u002Fh",{"id":495,"type":72,"variant":280,"title":496,"markdown":497},"d-nuance-froude","Why some big tides make no bore at all","A large tidal range and a shallowing river are necessary for a bore, but they are not quite enough on their own: the river also needs to be shallow and narrow enough, and its own downstream flow weak enough, for the incoming tide's front to actually catch up within the length of the river mouth. Rivers with the same tidal range but a deeper or more open mouth — or a much stronger river current pushing back — never develop a bore at all, even though their ordinary tide can be just as large. This is why bores are rare even on some very large-range coasts.",{"id":499,"type":78,"title":500,"eyebrow":501,"navLabel":502},"d-ch7","Earth tides: the ground itself has a tide too","Chapter 07","7 Earth tides",{"id":504,"type":43,"markdown":505},"d-earth-tides","It is not just the ocean that flexes under the Moon's pull. The **solid rock of Earth** rises and falls too, by tens of centimetres, twice a day, everywhere — including under dry land far from any coast. You cannot feel it, because everything around you (the ground, the building, your own body) rises and falls together, but precise instruments detect it clearly: sensitive gravimeters and GPS stations record the ground itself moving by roughly the same **order of magnitude** as the open-ocean tide, a scale of tens of centimetres.\n\nThis matters for real engineering, not just curiosity. Particle accelerators and gravitational wave detectors are sensitive enough that Earth tides have to be corrected for in their measurements, and precise satellite positioning (the kind that underlies modern surveying and map-making) must account for the ground itself shifting under the receiver.",{"id":507,"type":72,"variant":184,"title":508,"markdown":509},"d-example-michelson","The physicist who measured a tide with a pipe of water","In **1913–1914**, Albert Michelson — the same physicist famous for measuring the speed of light — and his colleague Henry Gale set out to catch the solid Earth in the act of flexing. They laid a **166 m** pipe half-filled with water in a trench, sealed the ends with glass, and watched the water level with a microscope fitted with a micrometer. Over each tidal cycle they measured level changes of only about **20 micrometres** — a fraction of the width of a human hair — and found that the timing and size matched almost exactly what you would expect if Earth behaved like a single, highly elastic solid ball. It was one of the first direct confirmations that the 'solid' ground really does flex with the tide, just as this chapter describes.",{"id":511,"type":72,"variant":184,"title":512,"markdown":513},"d-example-atmosphere","Even the air has a tide","The atmosphere has a tide too, driven mostly not by the Moon but by the **Sun heating it** every day — a thermal effect layered on top of the same gravitational tugging that affects the ocean and the solid Earth. Lord Kelvin himself noticed a mysterious, extremely regular twice-daily wobble in barometric pressure readings and correctly guessed it was a genuine atmospheric tide, long before satellites could confirm it directly.",{"id":515,"type":95,"caption":516,"columns":517,"rows":521},"d-table-tide-sizes","Three tides on Earth, same cause, very different sizes",[518,519,520],"Tide","Typical size","What responds",[522,526,530,534],[523,524,525],"Open-ocean tide","About 0.5 m","Sea water, far from any coast",[527,528,529],"Coastal tide","About 0.5–16 m","Sea water, shaped by funnelling, shallowing, resonance",[531,532,533],"Earth (solid) tide","Tens of centimetres","The rock of the planet itself, flexing elastically",[535,536,537],"Atmospheric tide","A small, regular wobble in air pressure","The whole atmosphere, mostly Sun-driven",{"id":539,"type":386,"conceptId":540,"relation":541,"explanation":542},"d-conn-gravity","gravity","helps_understand","Earth tides are the cleanest possible proof that 'solid' is a matter of degree: given a strong enough differential pull and enough time, even rock behaves a little like a fluid, flexing rhythmically rather than staying perfectly rigid.",{"id":544,"type":78,"title":545,"eyebrow":546,"navLabel":547},"d-ch8","Phase lag: why the response comes late","Chapter 08","8 Phase lag",{"id":549,"type":43,"markdown":550},"d-phase-lag","Understand's 'age of the tide' — the day or two's delay between a new or full moon and the biggest tides that follow — has a proper name in physics: **phase lag**, and it is a general property of *any* system that is pushed rhythmically rather than dragged instantly into place.\n\nPush a playground swing exactly in time with its own natural rhythm and it swings almost directly under your hand. Push it at a rate quite different from its natural rhythm and its highest point lags noticeably behind your push, arriving late. An ocean basin driven by the Moon's tidal force behaves the same way: it responds like a mass on a spring being pushed by an outside rhythm, not like a puppet moved exactly in step with the puppeteer's hand.\n\nThis is also why resonance and lag are two sides of the same coin. A basin driven very close to its own natural period (like the Bay of Fundy) both resonates strongly **and** shows a particular, fairly extreme phase relationship; a basin driven far from its natural period responds weakly and with a different lag. Every real port's 'age of the tide' and its typical range are both consequences of the same underlying physics: how well its own natural rhythm matches the rhythm the Moon and Sun are pushing it at.",{"id":552,"type":72,"variant":197,"title":553,"markdown":554},"d-aha-swing-lag","The swing analogy, made precise","If you push a swing at exactly its natural frequency, its peak lags your push by a quarter of a cycle — a fixed, predictable delay set entirely by the physics of a driven oscillator, not by anything mysterious. Push it much slower or much faster than its natural frequency and the lag shrinks towards nothing or grows towards half a cycle instead. Ocean basins driven by the tide show exactly this family of behaviour, which is why 'the age of the tide' is not a random quirk of geography but a calculable consequence of how close each basin sits to its own resonance.",{"id":556,"type":557,"prompt":558},"d-reflect-lag","reflection","A basin driven exactly at its resonant period gets both the biggest amplification and a very particular phase lag. Using the swing analogy, explain in three or four sentences why you might expect resonant coasts like the Bay of Fundy to have a fairly consistent 'age of the tide' from one spring cycle to the next, rather than a randomly varying one.",{"id":560,"type":78,"title":561,"eyebrow":562,"navLabel":563},"d-ch9","Put several ideas together","Chapter 09","9 Harder problems",{"id":565,"type":43,"markdown":566},"d-harder-intro","These problems each need more than one idea from this layer or an earlier one. Work through them properly before checking the steps.",{"id":568,"type":202,"title":569,"problem":570,"steps":571,"help":576},"d-we-multi","A multi-step port problem","A newly studied gulf is closed at its head, 220 km long, with an average depth of 45 m. Its form factor, from measured constituents, comes out at F = 0.18. (a) What tide type is it? (b) Estimate its resonant wavelength and quarter-wavelength length, and say whether 220 km is close to resonance.",[572,573,574,575],"(a) F = 0.18 is below 0.25, so it is **semidiurnal**.","(b) Wave speed: v = √(9.8 × 45) ≈ **21.0 m\u002Fs**.","Resonant condition for a closed–open basin: quarter-wavelength ≈ v × T ÷ 4, using the M2 period T ≈ 44,712 s: 21.0 × 44,712 ÷ 4 ≈ **235,000 m ≈ 235 km**.","The gulf's actual length, 220 km, is close to this 235 km resonant length — within about 6% — so this gulf should show **noticeably enhanced** range, a real-world sibling of the Gulf of Khambhat and the Bay of Fundy, even though it was entirely made up for this problem.",{"simplerExplanation":577,"hints":578},"Work out F to get the type; work out the resonant length the same way as the Bay of Fundy example, and compare it with the given length.",[579,580],"√(9.8 × 45) is a little over 21.","The M2 period in seconds is about 44,712.",{"id":582,"type":341,"itemId":583,"prompt":584,"check":585,"hints":601,"feedback":604},"d-pr-multi","tides.deepen-multi","A second gulf has the same 220 km length and the same closed shape, but is only 8 m deep on average. Compared with the 45 m gulf above, is it closer to or further from quarter-wavelength resonance at the M2 period?",{"kind":586,"options":587,"correct":600},"choice",[588,591,594,597],{"id":589,"label":590},"a","Closer — shallower water always resonates better",{"id":592,"label":593},"b","Further — its wave speed is much lower, so its resonant length is much shorter than 220 km",{"id":595,"label":596},"c","Exactly the same — depth does not affect resonance",{"id":598,"label":599},"d","There is no way to tell without measuring it directly",[592],[602,603],"Work out the new wave speed with v = √(g × depth).","A slower wave has a shorter resonant length for the same period.",{"correct":605,"incorrect":606},"Right: v = √(9.8 × 8) ≈ 8.9 m\u002Fs, giving a resonant length of about 8.9 × 44,712 ÷ 4 ≈ 99 km — far short of the gulf's actual 220 km, so it is well off resonance and should show a much smaller enhancement than the 45 m gulf.","√(9.8 × 8) ≈ 8.9 m\u002Fs is much slower than 21.0 m\u002Fs, so the resonant length shrinks to about 99 km, a poor match for a 220 km gulf.",{"id":608,"type":202,"title":609,"problem":610,"steps":611,"help":616},"d-we-third","A third problem: combining form factor and range","A port has a spring tidal range of 3.0 m and constituents M2=70, S2=25, K1=8, O1=5 (made up, for practice). (a) Find its form factor and type. (b) If its neap range follows the Moon-to-Sun ratio of about 2.18 to 1 from Understand, estimate its neap range.",[612,613,614,615],"(a) F = (8 + 5) ÷ (70 + 25) = 13 ÷ 95 ≈ **0.14** — semidiurnal, dominated by M2 and S2, matching the fact that its constituent list barely has any K1 or O1 at all.","(b) neap ÷ spring = (2.18 − 1) ÷ (2.18 + 1) ≈ 0.370, from the Understand worked example.","Neap range ≈ 3.0 × 0.370 ≈ **1.1 m**.","Notice how the two calculations use completely different inputs — harmonic constituent sizes for the type, and the Moon-versus-Sun tidal-force ratio for the spring–neap size — and yet both are just different views of the same underlying M2\u002FS2\u002FK1\u002FO1 waves added together in different combinations.",{"simplerExplanation":617,"hints":618},"Do the form-factor sum first, then reuse the spring-to-neap ratio method from Understand on the given spring range.",[619,620],"13 ÷ 95 is a little under 0.15.","The neap-to-spring ratio was worked out fully in Understand.",{"id":622,"type":557,"prompt":623},"d-reflect-model","Newton's equilibrium theory is simpler than Laplace's dynamic theory and is still taught first in every course, including this one. Argue, in four or five sentences, whether a simpler-but-less-accurate model like Newton's is good science or bad science, using tides as your example.",{"id":625,"type":72,"variant":184,"title":626,"markdown":627},"d-example-india-machine","India, tide science, and a very old connection","Coastal India has been part of world tide science for a long time: ports here needed reliable predictions for exactly the reasons Investigate described — deep-draught cargo ships, shallow harbour bars, and rivers with bores — from the earliest days of steamship trade. Today's computer-based harmonic predictions from INCOIS and the Survey of India descend directly from the same Doodson-style harmonic method that once turned inside Kelvin's brass and steel gears.",{"id":629,"type":630,"title":631,"terms":632},"d-glossary","glossary","Vocabulary for the mathematics of tides",[633,635,637,640,643,646,649,651,654],{"term":127,"meaning":634},"Newton's original tide theory, treating the ocean as if it always sat in instant balance with the Moon and Sun. Right about the cause, wrong about the details.",{"term":130,"meaning":636},"Laplace's theory treating tides as real waves sloshing through ocean basins of real shape and depth, which is why every port needs its own table.",{"term":638,"meaning":639},"Tidal constituent","One of the simple, steady waves (such as M2, S2, K1 or O1) that add up to make a real, measured tide.",{"term":641,"meaning":642},"Tidal form factor","F = (K1 + O1) ÷ (M2 + S2): a single number that classifies a tide as semidiurnal, mixed or diurnal.",{"term":644,"meaning":645},"Closed–open resonance","The quarter-wavelength resonance of a basin closed at one end (like a bay) and open at the other — the same physics as a closed organ pipe.",{"term":647,"meaning":648},"Earth tide","The twice-daily flexing of the solid ground itself, by tens of centimetres, detected by precise instruments rather than felt directly.",{"term":535,"meaning":650},"A twice-daily wobble in air pressure, driven mainly by the Sun heating the atmosphere rather than by gravity alone.",{"term":652,"meaning":653},"Phase lag","The delay between a rhythmic push and a system's peak response, seen in a pushed swing and in an ocean basin driven by the Moon and Sun alike.",{"term":655,"meaning":656,"example":657},"Q value","A number describing how sharply and strongly a resonator responds; the Bay of Fundy–Gulf of Maine system has a Q of roughly 5.","A high-Q system rings on for a long time once excited; a low-Q one damps out quickly.",{"id":659,"type":660,"title":661,"questions":662},"d-quiz","quiz","Check yourself: the mathematics and history of tides",[663,676,689,702,712,725,737,750,763,776],{"itemId":664,"prompt":665,"options":666,"correct":592,"why":675},"tides.deepen-q-newton","What did Newton's equilibrium theory get wrong?",[667,669,671,673],{"id":589,"label":668},"That gravity causes tides at all",{"id":592,"label":670},"That every ocean should have nearly the same small range, always under the Moon",{"id":595,"label":672},"That the Moon and Sun both contribute to tides",{"id":598,"label":674},"That there are two bulges","Newton correctly identified the cause and the basic pattern, but wrongly assumed the ocean could respond instantly, predicting a uniform small range everywhere — false, as Kochi and the Bay of Fundy show.",{"itemId":677,"prompt":678,"options":679,"correct":592,"why":688},"tides.deepen-q-laplace","What did Laplace add?",[680,682,684,686],{"id":589,"label":681},"The idea that gravity causes tides",{"id":592,"label":683},"Treating tides as waves moving through real ocean basins, not instant equilibrium",{"id":595,"label":685},"The invention of the tide table",{"id":598,"label":687},"The discovery of the Moon's phases","Laplace's dynamic theory treats the tide as a genuine wave problem, explaining why basin shape and depth matter so much.",{"itemId":690,"prompt":691,"options":692,"correct":592,"why":701},"tides.deepen-q-oceanspeed","Why does the open-ocean tide count as a 'shallow-water' wave even in the deep Pacific?",[693,695,697,699],{"id":589,"label":694},"The Pacific is not actually very deep",{"id":592,"label":696},"A tidal wave's length is thousands of kilometres, far bigger than any ocean depth",{"id":595,"label":698},"Shallow water waves are always fast",{"id":598,"label":700},"The formula only works in shallow water","'Shallow' compares depth with wavelength, not with an absolute number. A wave thousands of kilometres long makes any ocean depth 'shallow' by comparison.",{"itemId":703,"prompt":704,"options":705,"correct":592,"why":711},"tides.deepen-q-form","A port has M2=80, S2=20, K1=15, O1=10. What tide type is it?",[706,707,708,709],{"id":589,"label":308},{"id":592,"label":299},{"id":595,"label":305},{"id":598,"label":710},"No tide","F = (15+10) ÷ (80+20) = 25 ÷ 100 = 0.25, right at the boundary but classed semidiurnal for F ≤ 0.25 — dominated by its twice-daily waves.",{"itemId":713,"prompt":714,"options":715,"correct":592,"why":724},"tides.deepen-q-pipe","Why does a bay closed at one end resonate at a **quarter** wavelength, not a half?",[716,718,720,722],{"id":589,"label":717},"Because bays are always exactly a quarter of the tide's wavelength",{"id":592,"label":719},"The closed end needs zero motion and the open end needs maximum motion, which a quarter-wave shape provides",{"id":595,"label":721},"Because the Moon only pulls a quarter as hard on bays",{"id":598,"label":723},"There is no real reason; it is just an approximation","This is the same physics as a closed organ pipe: fitting zero motion at one end and maximum at the other takes a quarter of a full wave, not a half.",{"itemId":726,"prompt":727,"options":728,"correct":595,"why":736},"tides.deepen-q-doodson","About how many separate tidal frequencies did Doodson identify in 1921?",[729,731,733,734],{"id":589,"label":730},"4",{"id":592,"label":732},"10",{"id":595,"label":64},{"id":598,"label":735},"1,000,000","388 — though as this layer showed, a handful of the largest (M2, S2, K1, O1) usually do almost all the work for any given port.",{"itemId":738,"prompt":739,"options":740,"correct":595,"why":749},"tides.deepen-q-earthtide","How is the solid Earth's own tide normally detected?",[741,743,745,747],{"id":589,"label":742},"People can feel the ground move",{"id":592,"label":744},"It cannot be detected at all",{"id":595,"label":746},"By precise instruments such as gravimeters and GPS stations, since everything nearby moves together",{"id":598,"label":748},"Only by astronauts in orbit","Everything around you — ground, buildings, your body — rises and falls together by tens of centimetres, so only sensitive instruments comparing against a fixed reference can detect it.",{"itemId":751,"prompt":752,"options":753,"correct":592,"why":762},"tides.deepen-q-atmosphere","What mainly drives the atmospheric tide?",[754,756,758,760],{"id":589,"label":755},"The Moon's gravity alone",{"id":592,"label":757},"The Sun heating the atmosphere each day",{"id":595,"label":759},"Ocean currents",{"id":598,"label":761},"Volcanic eruptions","Unlike the ocean tide, the atmospheric tide is mostly a thermal effect: daily solar heating, with gravity as a smaller contributor.",{"itemId":764,"prompt":765,"options":766,"correct":592,"why":775},"tides.deepen-q-lag","What is 'phase lag', as used in this layer?",[767,769,771,773],{"id":589,"label":768},"A type of tidal bore",{"id":592,"label":770},"The delay between a driving push (like the Moon's pull) and a system's peak response, seen in any rhythmically driven system",{"id":595,"label":772},"A unit of tidal range",{"id":598,"label":774},"Another name for a neap tide","Phase lag is general driven-system physics, seen in a pushed swing just as much as in an ocean basin driven by the Moon and Sun.",{"itemId":777,"prompt":778,"options":779,"correct":592,"why":788},"tides.deepen-q-q","A resonator with a high Q value…",[780,782,784,786],{"id":589,"label":781},"Damps out almost instantly",{"id":592,"label":783},"Rings on strongly and responds sharply at its own resonant period",{"id":595,"label":785},"Cannot resonate at all",{"id":598,"label":787},"Only exists in a laboratory, never in nature","A high Q means a strong, long-lasting response concentrated near the resonant period; the Fundy system's more modest Q of about 5 is still enough to build the world's biggest tide over thousands of repetitions.",{"id":790,"type":791,"title":792,"points":793},"d-cheat-sheet","summary","Cheat sheet: the mathematics and history behind a tide table",[794,795,796,797,798,799,800,801,802,803],"**Newton (1687):** equilibrium theory — right about the cause, wrong to assume the ocean responds instantly.","**Laplace (1775):** dynamic theory — tides are waves sloshing through real ocean basins, which is why every port needs its own table.","The open ocean, averaging about 3700 m deep, still counts as 'shallow' to a tide wave thousands of kilometres long, giving a wave speed of about 686 km\u002Fh.","Four main constituents — **M2** (12 h 25 min), **S2** (12 h), **K1** (23.93 h) and **O1** (25.82 h) — add together to make a real tide.","The **form factor** F = (K1+O1) ÷ (M2+S2) turns 'semidiurnal, diurnal or mixed' into one calculable number.","A bay closed at one end resonates like a **closed organ pipe**, at a quarter wavelength — which is why the Bay of Fundy's real size matches the calculation so well.","A tidal bore forms when the time for deeper water to catch up with shallower water ahead of it fits within the length of a shallowing river mouth.","**Earth tides** flex solid rock by tens of centimetres twice a day; the **atmosphere** has its own, mostly Sun-driven tide too.","**Phase lag** — the delay between a push and a peak response — explains the 'age of the tide' from Understand as ordinary driven-oscillator physics, the same as a pushed swing.","**Kelvin (1872)** built the first tide-predicting machine; **Doodson (1921)** identified 388 separate tidal frequencies — the same harmonic method now runs on computers.",{"id":805,"type":806,"sourceIds":807},"d-sources","sources",[808,809,810,811,812,813],"tides-wikipedia-tide-machine","tides-wikipedia-tidal-resonance","tides-wikipedia-bay-of-fundy","tides-noaa-tides-tutorial","tides-incois-tide-forecasting","tides-wikipedia-earth-tide",[808,809,810,811,812,813],"needs_review",{"generatedBy":817,"notes":818},"claude-code","Draft generated locally; pending owner review.","fbcb610553e3398fced00bbf0b80135ccc4d01d426ea9173a8d7e9a530a61660",{"component:sort-game@1":821,"component:match-pairs@1":822,"logic:practice":823,"component:data-lab@1":824,"component:tide-lab@1":825,"source:tides-incois-tide-forecasting":826,"source:tides-noaa-tides-tutorial":827,"source:tides-wikipedia-bay-of-fundy":828,"source:tides-wikipedia-earth-tide":829,"source:tides-wikipedia-tidal-resonance":830,"source:tides-wikipedia-tide-machine":831},"b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","466896cc37735f48db03875fe9c9ce42fc8bcb7e5f937c9779d70513703b91bd","5fef8b331bba35d6df96a31b84dd1f98200bcd0242eeed843d22914391057b6c","c8bcfd0dc5b9671b4b89f5c1abc1f9dbf9e3f3caf1325102661e5d41a455803f","82683db912324c1f40c9e2f4d4cc0312c637b4af2490f13ca404549ec5e414a0","a11570fb567503cb357092d9ba5ad5cc3f29fa50d9094d46c5a4b2435e012a92","2d3845d00db006c355708d6f372329ed5223e57f6b03ea15cdd7c5c85b859bdc","7f84f7fc92ea0f8bd091c0a1e41d05af8277c84577f2ad7f3eeadb32d479dc99","87798cb144ca35c480ce860c5f7586591c363fb456f5eddd91f1de994cdacb4c",{"state":833,"reviewer":834,"selfReview":835,"reviewedAt":836,"method":837},"approved","The library owner",true,"2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899598010]