the wall  /  0018

Thirty-six centimetres, and sixteen metres

Newton's tide is a bulge of water that follows the Moon around the Earth, and you can work out how tall it is on the back of an envelope: 36 cm. The Bay of Fundy has 16 metres. Tahiti has about a foot, and takes it from the Sun rather than the Moon. Nowhere on Earth has 36 centimetres of tide, high tide almost nowhere happens when the Moon is overhead, and — the fact that ends the argument — no bulge could keep up with the Moon in the first place. This page derives the 36 cm anyway, because it is the correct answer to a different question; derives the tide's entire clock from five astronomical periods; then runs a rotating sea until the thing real oceans do appears on its own. The bill for all of it is paid in the length of the day.

The tide is a subtraction

The Sun pulls on the Earth times harder than the Moon does, and raises less than half the tide. Both of those are true because the tide is not made by gravity; it is made by the difference in gravity across the width of the Earth. The Earth is in free fall around the Moon, so what an ocean feels is only what is left after the fall is subtracted — the pull at your feet minus the pull at the centre.

pull  ∝  M / d2      difference across 2R  ≈  2 M R / d3

One power of distance is the whole story. It costs the Sun a factor of — how much further away it is than the Moon — and that is why an object 27 million times heavier loses. The arrows below are the real residual field, computed without that approximation: the pull of the Moon at each point, minus the pull at the centre. Note that most of them do not point up or down. The tide is raised mainly by the sideways component, which has a whole ocean to push on and nothing to lift.

distance
residual force equilibrium surface, exaggerated the undisturbed sphere

Let the water settle under that field and you get an ellipsoid whose height is arithmetic with no fudge in it: h = (M/M⊕)(R⁴/d³), which for the Moon is above mean and half that below, a range of . The Sun manages . They add at new and full moon and fight at the quarters:

Everything on this page after this point is about why that answer, which is correct, describes no coastline on Earth.

Five periods, and the whole clock falls out

Before the disagreement, the agreement — and it is total. The tide is a sum of pure sine waves, and their frequencies are not measured from the sea at all. Every one of them is a whole-number combination of six rates, and the six rates are just how fast the sky turns. Feed in five periods that have nothing to do with water — a day, a month, a year, the 8.85-year wobble of the Moon's perigee, the 18.6-year wander of its orbital plane — and the entire published catalogue of tidal constituents comes out to seven figures.

The tide-raising potential is a product of things that turn, and a product of cosines is a sum of cosines whose arguments are sums and differences. That is the entire reason. Doodson catalogued 388 of these in 1921 by hand. Below, the six rates are combined into the ones that matter, and checked against the speeds in the tide tables — which were fitted to water, by other people, and never told about this arithmetic.

Three of those deserve saying out loud. The lunar day is 24h 50m because the Moon has moved on by the time the Earth comes back around, so the semidiurnal beat is 12h 25m and the tide is late by roughly 50 minutes a day — the number in every fisherman's head. S2 is exactly 12 hours, by construction, because it is the Sun's, and our clocks are the Sun's. And K1's period is the sidereal day to nine figures: it is the one constituent tied to the Earth's rotation against the fixed stars rather than against anything in particular, which is why the Moon and Sun both contribute to it.

The spring–neap cycle is not a constituent at all. It is the beat between M2 and S2, at days, which is half a synodic month to five figures — and it has to be, because that is the same statement twice.

Now build a tide out of them

Sum the constituents and you get a curve that looks like a tide, because it is one. What changes from port to port is not the frequencies — those are the sky's — but the amplitudes and phases, which are the ocean's answer, and which have to be measured. Drag them. The single number that classifies a coastline is the form factor F = (K1+O1)/(M2+S2): below 0.25 two even tides a day, above 3 one a day, and in between the mess that makes tide tables necessary.

the sum M2 alone spring–neap envelope

The solar-dominated preset is not a curiosity. At Tahiti the solar tide beats the lunar one, so high water arrives at roughly noon and midnight every day of the year, and stays there — the one place where the folk rule is a clock rather than a moon. A tide gauge there disagrees with Newton's picture in the most direct way available: the Moon goes overhead and almost nothing happens.

The bulge that cannot keep up

Here is the fact that makes the 36 cm unrescuable, and it needs no ocean data. A long wave in water of depth h travels at √(gh). The point under the Moon travels at the speed of the Earth's surface. Set them against each other:

At the mean depth of the ocean a free wave manages . The sub-lunar point at the equator is doing . The bulge is being asked to run at times the fastest it can go, which means the ocean's response is not a bulge in equilibrium but a forced oscillation, permanently behind — and, in the deep water where the forcing does its work, nearly upside down with respect to it. To make a tide of a metre, let alone sixteen, the water needs the one trick a forced oscillator has: resonance, in a basin whose size happens to suit.

Which basins? The natural period of a gulf open at one end is the time a wave takes to travel its length four times — in and back, twice: T = 4L/√(gh). The ocean forces it near 12.42 hours. Anything whose quarter-wave period lands near there gets a tide out of all proportion to the 36 cm driving it, and anything that does not, does not.

That table is one formula and two measurements per row, and where it works it works well: the two bays with the largest tides on Earth are the two whose fundamental lands on top of the tide. But read the fifth column and not the fourth. A channel does not have a resonance, it has a ladder of them — the quarter wave fits an odd number of times, so the modes go 4L/(2n+1)√(gh) — and being far from the fundamental is not the same as being far from resonance.

The Mediterranean is the row that makes the point, and it caught this page out. Its fundamental is 34 hours, which looks safely detuned; its third mode is at 11.3 h, which is nearer the tide than the Adriatic's fundamental is. Solved as an open channel it would have a bigger tide than the Adriatic, and it does not. The reason it has centimetres is not detuning at all — it is that the Strait of Gibraltar is 14 km wide, so the basin is very nearly closed and there is almost nothing arriving to be amplified. A resonator with no driver is quiet whatever its period.

Sixteen metres, from first principles

So stop arguing and run one. Below is a rectangular sea on a rotating Earth, open at the left, closed at the other three sides, with the ocean's tide imposed at the mouth and nothing else put in by hand. It solves the linear shallow-water equations — the same three that give √(gh) — on a staggered grid, then harmonically analyses the last few cycles the way a tide gauge would.

length
width
depth
latitude
forcing
friction

Two things in that picture were not programmed in, and are the reason the page exists. The first is the amplification: the mouth is driven with a two-metre ocean tide and the head comes back with thirteen, from nothing but a length, a depth and a period that agree with each other. The real bay manages 16.3 m, and the missing part is named and not mysterious — this box has parallel walls, and Fundy narrows from about 90 km at its mouth to 30 km at the head. Green's law says a converging channel adds a factor of √(width ratio), which is another 1.7×. A rectangle was never going to get the whole way. The second only appears once rotation is on. Sweep the latitude to zero and the tide becomes a plain standing wave — high everywhere at once, a line of no tide across the basin. Turn rotation back on and that line collapses to a point, and the tide rotates around it. Those are amphidromic points. There are about a dozen in the real Atlantic, the tide goes round each one once per cycle, and the range at the point itself is zero.

Amplification against forcing period, from the exact solution

That curve is the non-rotating solution, and the grid above will not sit exactly on it whenever the latitude is off the equator: rotation both lowers the peak and shifts it to a shorter period, and the effect grows with the basin's width measured in Rossby radii. At 45° in a basin 80 km wide the peak drops by about a fifth and moves in by 4%. Set the latitude to zero and the two agree to a tenth of a per cent, which is the check that the grid is solving what it claims to be solving.

Who pays

A tide that lags is a tide that is pulled on, and the pull is off-centre. The Earth's bulges sit slightly ahead of the Moon, because the Earth turns faster than the Moon orbits and friction drags the water forward. So the bulge tows the Moon, and the Moon brakes the Earth. This is not a small effect measured by delicate instruments; it is the dominant term in the Earth–Moon system's future, and it is measured by bouncing lasers off reflectors the Apollo crews left behind.

Take the one measured number — the Moon receding — and the rest is conservation of angular momentum with no free parameters:

    The tidal braking is per century, and the day-length record says the observed figure is nearer 1.7 ms. The gap is not an error. Some of it is the Earth still rebounding from the ice that was on it 20,000 years ago, which pulls mass toward the axis and speeds the spin up — the planet is being braked and un-braked at once.

    And the bill has a witness that could have refused. If the day has been lengthening, then clocks running at a uniform rate get steadily ahead of the turning Earth, and the discrepancy grows as . Compute an ancient eclipse with uniform time and it lands in the wrong place — not vaguely, but by an amount you can convert into longitude:

    clock error from tidal braking eclipses people wrote down

    The Babylonians recorded a total eclipse at their own city on 15 April 136 BC. Computed with uniform time and no tidal braking, its path of totality falls of longitude to the west — over the Atlantic, and Babylon sees nothing at all. The clay tablet is the instrument. It says the ocean has been dragging its feet for 2,160 years, and it says by how much.

    What could have contradicted this page

    Every line below is an outside number — measured by somebody else, from something other than this arithmetic — checked against what the page computes. They are recomputed in your browser each time it loads, so a red mark is a real one.

    The eight-figure agreement between five astronomical periods and the tide tables is the honest headline here, and so is the size of the disagreement below it: the same theory that nails every frequency to a part in ten million gets the amplitude wrong by a factor of forty-five, in a bay that has been on charts since 1607. Both facts are about the same physics. The sky sets the clock; the shape of the sea sets everything else.