A triangle reads the geoid's curvature, and that one converges

The geoid’s curvature, measured against where its series stops, found that the surface every orthometric height is measured above has no Gaussian curvature that belongs to it. The geoid is only ever given as a spherical-harmonic series stopped at some degree, and each degree of it moves the curvature by the degree squared times its own height. Under Kaula’s rule the heights fall too slowly for that to settle: a model complete to degree 360 carries 0.36 per cent of the sphere’s own curvature at a point, EGM2008 at degree 2190 carries 2.19, and a model stopped at 36,000 would carry 36. The curvature at a place is a property of the model’s cutoff.

It ended by asking whether that matters to anything a survey actually does — whether any computation ever reads a curvature off the geoid rather than off the ellipsoid. It also said, in passing, what the answer would have to look like. The integral of the curvature over the whole geoid is exactly $4\pi$, whatever the cutoff, and an integral over a patch reduces to something on the patch’s boundary. A surveyor never measures a curvature at a point. Measuring curvature from inside is how a surveyor measures it at all: draw a triangle, add its angles, and the excess over 180° is the curvature integrated over the triangle.

So the question has a precise form. The triangle’s excess has a part from the geoid’s undulations. Is that part a quantity the Earth has, or another property of the model’s cutoff?

What a triangle’s excess is made of #

To first order in the geoid’s height $N$ above a sphere of radius $R$, Gaussian curvature changes by a term in the Laplacian of $N$, and the change in area element absorbs the rest. Integrated over a triangle $T$,

$$\iint_T K\,dA \;=\; \Omega_T \;-\; \frac{1}{R}\iint_T \nabla^2 N\,dA \;=\; \Omega_T \;+\; \frac{1}{R}\oint_{\partial T} \boldsymbol{\varepsilon}\cdot\boldsymbol{\nu}\,ds,$$

where $\Omega_T$ is the triangle’s excess on the sphere, $\boldsymbol\nu$ the outward normal of its edge, and $\boldsymbol\varepsilon = -\nabla N$ is the deflection of the vertical — the angle between the plumb line and the normal to the reference surface, which a deflection is the slope of a mass prices for a mountain. The second equality is the divergence theorem.

So the geoid’s share of a triangle’s excess is the flux of the deflection through its three sides, divided by the Earth’s radius. On the stated patch the three sides carry −1.321, −1.369 and −1.904 metres of it, and their sum over 6,371 kilometres is −0.149 seconds of arc of excess. That is a first derivative of the geoid’s height, averaged along three lines. The curvature at a point was a second derivative, taken at one place. Averaging along a line of length $L$ suppresses a wave much shorter than $L$ by about the square root of the number of its wavelengths the line holds, and the one derivative saved and the averaging together are the whole difference between a quantity that diverges and one that need not.

Whether it actually converges is a question about the spectrum, and it can be answered exactly.

A triangle keeps the long waves and filters the short ones #

Written in Fourier terms, a triangle sees each undulation of the geoid through its own transform — $F_T(\mathbf k) = \iint_T e^{i\mathbf k\cdot\mathbf x}\,dA$, which for a triangle has a closed form in its three vertices and here matches direct numerical integration to a part in a thousand. For a wave much longer than the triangle, $F_T$ is simply the triangle’s area: the whole triangle sits on one part of the wave and reads it as its value times the area, which is the curvature at a point times the area. For a wave much shorter, the interior cancels itself out and only the edges are left, and averaged over direction $|F_T|^2$ falls as the wavenumber to the minus third power — measured, −2.96.

The geoid’s contribution to the excess is then a sum over degrees of three factors: the degree’s own undulation in height, which under Kaula’s rule falls as $n^{-3/2}$ in amplitude; two powers of $n$ from the Laplacian, which make it a curvature; and the triangle’s window. Squared and summed,

$$\operatorname{Var} E \;=\; \frac{1}{R^2}\sum_n c_n\,k_n^4\,\big\langle |F_T(k_n)|^2\big\rangle,$$

with $c_n$ the degree variance of the height and $k_n = (n + \tfrac12)/R$. At low degrees the window is the area squared and the terms grow with $n$, as the curvature at a point does. At high degrees the window falls as $n^{-3}$ against a growth of $n$, and the terms fall as $n^{-2}$. A sum of $n^{-2}$ converges.

It converges where the curvature at a point does not #

Computed from degree 2 to degree 36,000, the geoid’s rms contribution to a fifty-kilometre triangle’s excess is 0.0353 seconds of arc with the series stopped at EGM2008’s degree 2190 and 0.0377 with it carried to 36,000, a change of 7 per cent. Over the same range the curvature the model contains at a point grows sixteenfold, from 2.19 to 36 per cent of the sphere’s own. As a fraction of the triangle’s own excess, the geoid’s contribution is 0.64 per cent at degree 2190 and 0.68 at 36,000.

Every triangle behaves the same way on its own scale. Its reading follows the curvature at a point while the model is coarser than the triangle — the triangle cannot tell the model’s waves from uniform curvature — then leaves that line and flattens once the model resolves waves shorter than the triangle’s side. A hundred-kilometre triangle settles at 0.34 per cent, a ten-kilometre one at 3.37. A one-kilometre triangle is still climbing at degree 36,000, whose half-wavelength of 560 metres is barely shorter than the triangle itself: at 2190 it reads 2.19 per cent, the same as the curvature at a point, and at 36,000 it reads 27.

That is the answer to the question the earlier essay could not settle. A triangle’s excess on the geoid is a property of the geoid, not of the model, as soon as the model resolves the triangle: a converged number, supplied by the waves about the triangle’s own size and a little shorter. The curvature at a point is the limit of that number as the triangle shrinks, and since the number grows without bound as the triangle shrinks, the limit does not exist. The triangle’s reading exists at every size; its limit does not.

The waves about a triangle’s own size supply its reading #

Split by degree, the fifty-kilometre triangle’s reading comes overwhelmingly from undulations about its own size. Degrees 300 to 999, half-wavelengths of 20 to 67 kilometres, supply 56 per cent of its variance; degrees 100 to 299, 15 per cent; 1000 to 2999, 18 per cent. Every degree below 100 — half-wavelengths longer than two hundred kilometres, the continental-scale undulations that make up almost all of the geoid’s height — supplies 2 per cent between them, because a triangle fifty kilometres across sees a wave of a thousand kilometres as a nearly uniform tilt, and a tilt has no curvature. Every degree above 3000 supplies 8.5 per cent between them, because the triangle’s edges average those waves away.

That is why the reading converges and why its value is modest. It is taken from a band of the spectrum a few octaves wide, centred on the triangle, and the geoid model stops at a degree is the reminder that EGM2008’s degree 2190 — a half-wavelength of nine kilometres — already covers the whole of that band for any triangle a first-order survey would build.

The resolution rule was the right shape and a little low #

The earlier essay estimated what a triangle would read with a rule of thumb: a triangle of side $s$ resolves the geoid down to a half-wavelength of about $s$, so it reads the curvature a model stopped at that resolution contains — 0.2 per cent at a hundred kilometres, 2 at ten, 20 at one. The exact computation confirms the rule’s slope, the inverse first power of the side, and puts it low by a factor between 1.3 and 1.7: 0.34 per cent at a hundred kilometres rather than 0.20, 3.37 at ten rather than 2.00.

The reason is visible in the spectrum. A triangle does not stop reading at a wave its own size; it reads the waves a little shorter through its edges, which is where the degrees from 1000 to 3000 get their 18 per cent for a fifty-kilometre triangle. The rule counted the triangle’s interior and forgot its boundary, and the boundary is exactly where the flux is measured.

Against everything else in a triangle’s excess #

A reading that exists is not necessarily a reading anyone could take. How big a triangle it takes priced the noise: the sum of three angles each good to one second of arc has a standard deviation of 1.73 seconds, and a fifty-kilometre triangle’s whole excess is 5.50. The geoid’s contribution is 0.038 seconds, forty-six times smaller than the noise of one-second angles. No triangle observed with a theodolite has ever seen it, and none will: the noise does not shrink with the side, while the geoid’s share of the excess shrinks as the side’s inverse and the excess grows as its square, so their product grows only as the side — 0.075 seconds at a hundred kilometres — and would reach the noise of one-second angles only for triangles thousands of kilometres across, larger than any triangle anybody has observed and larger than the flat approximation here allows. How many triangles it takes found that averaging many triangles buys accuracy only slowly; it would have to buy a factor of forty-six.

The comparison with the ellipsoid is the surprise. At 45° the ellipsoid’s Gaussian curvature differs from the mean sphere’s by about 0.2 per cent, which over a fifty-kilometre triangle is 0.012 seconds of arc of excess. The geoid’s undulations contribute 0.038 — three times as much. A triangle that could measure its excess to a hundredth of a second would be reading the geoid more than the ellipsoid, and the ellipsoid’s curvature, the thing the earlier essays on this ground have treated as the surface’s, would be the smaller correction.

A network of triangles reads its outline #

The flux form has a consequence for networks that is exact and worth stating, because the triangle-counting essays assumed the opposite without saying so. How many triangles it takes averaged the curvature read by many triangles and treated each triangle’s error as independent of its neighbours’. For the angle noise that is right. For the geoid’s contribution it is not.

Two triangles that share a side see the deflection’s flux through that side with opposite signs: what leaves one enters the other. Added together, the shared side cancels exactly, and the pair’s combined contribution is the flux through the outline of the quadrilateral they make. The same holds for any number of triangles tiling a region. However many triangles a network has, the geoid’s contribution to its total excess is the flux of the deflection through the network’s outer boundary, divided by the Earth’s radius, and nothing inside counts at all.

So the geoid’s share of a network’s total excess behaves like one large triangle’s, not like many small ones averaged. A network of triangles filling a triangle three hundred kilometres across reads the geoid exactly as that one large triangle would — about 0.12 per cent of its excess, rms — whether it is made of six triangles or six hundred. More triangles average the angle noise down; they do nothing whatever to the geoid’s part, which was never noise in the first place but a property of the region’s edge.

So a computation does read the geoid’s curvature #

The earlier essay’s question was whether any real computation reads a curvature off the geoid. A triangle observed with instruments levelled to the plumb line does: its angles are measured in the plane perpendicular to the local vertical, which is the geoid’s normal, so the excess it would show, measured perfectly and to first order, is the total curvature of the level surface over the triangle. That reading is a converged property of the Earth, fixed by the geoid’s undulations between about one and ten times the triangle’s size. It is also far below what any angle measurement can resolve, so every triangulation ever computed has absorbed it silently into its angle residuals.

What does not exist, for the reason the earlier essay gave, is the curvature at a point. The two statements are consistent and they are the integral and the integrand of one function. The geoid’s curvature is a perfectly good quantity over any region of stated size and not over a region of no size; a statement about “the geoid’s curvature” has to carry an area, as a statement about its height has to carry nothing and a statement about its slope has to carry a cutoff.

How the reading was checked #

The triangle’s transform must be the triangle’s. The closed form in the three vertices must match direct numerical integration over a ten-kilometre triangle at a stated wavenumber; it matches to 0.1 per cent.

A triangle much smaller than every wave kept must read the curvature at a point. A one-kilometre triangle with the series stopped at degree 36 must return the pointwise curvature of that model, since its window is then its area squared; it returns 0.0390 per cent against 0.0386.

The finding must be there. From degree 2190 to 36,000 the fifty-kilometre triangle’s reading must change by under fifteen per cent while the curvature at a point grows more than fivefold. It changes by 7 per cent against sixteenfold.

Where the reading stops #

Kaula’s rule is the spectrum. As in the earlier essay, the degree variances are Kaula’s $10^{-5}/n^2$ per coefficient, extrapolated beyond degree 2190 where no global data exist. The convergence depends on the spectrum falling faster than $n^{-1}$ in the combination above, which every proposed geoid spectrum does; the values would move with the spectrum.

A flat triangle on a curved sphere. The triangle’s window is computed in the plane, which is exact to well below everything here for triangles of a few hundred kilometres; at a continental size the curvature of the sphere itself would enter the window.

First order in the height. The geoid departs from the sphere by a hundred metres at most against a radius of six million, and the terms dropped are smaller by that ratio.

An rms, not a value. Every number is the root mean square over all placements of the triangle. A particular triangle over a trench or a mountain range reads more, one over an abyssal plain less.

Still open: whether a levelling loop reads the same thing #

A triangle of angles is one way to integrate the geoid’s curvature. There is another in every levelling network: carry a height round a closed loop and, because level surfaces are not parallel, the orthometric heights misclose by an amount that depends on gravity along the way. A levelled height is not a distance found the correction the converging level surfaces demand of a levelling line; the question it raises here is whether a levelling loop’s misclosure and a triangle’s excess, taken round the same region, are two readings of related integrals of the same field.

They are built from the same deflections — the triangle from their flux through the loop, the levelling from their circulation along it — and the flux and the circulation of a field are the two halves of its line integral round a closed curve. Whether the geoid’s contribution to a region’s excess and its contribution to a levelling loop’s misclosure are correlated, independent, or the same number seen twice, and whether a network that measured both would learn anything about the geoid neither measures alone, are questions a triangle on its own cannot ask.