A strip of triangles carries the sphere's error to the next base

The correction is the smaller of the two corrections measured what substituting a sphere for the ellipsoid does to one triangle of a triangulation, and set it beside Legendre’s correction for the spherical excess. On a thirty-kilometre triangle at 45° north, the two spheres right at first order — the conformal and the geocentric — left the computed side 228 and 156 millimetres wrong, about eight parts per million. That is below what a nineteenth-century baseline could detect and above what a modern re-observation would miss, and the essay ended on the possibility that some of the scale differences documented between old national triangulations and their modern replacements are exactly this: a wrong choice of auxiliary sphere, misread as a baseline standardisation error.

What would settle it, the essay said, is the pattern. A baseline error is one number. A sphere’s error depends on latitude — as the sine of twice the latitude for the good substitutions, as the square of its cosine for the bad ones — so a network spanning enough latitude should carry a signature a baseline cannot mimic.

That is right in outline and wrong in the part that matters. Every number in the earlier essays was one triangle, solved from a measured base. A triangulation is not one triangle. It is a strip, each computed side becoming the base of the next triangle, and a substituted sphere’s error behaves in a strip in a way no single triangle shows: it is made afresh in every triangle and the strip adds it up.

A strip hands each computed side on #

The strip is the simplest a first-order chain could be. Its vertices lie on two parallel edges, thirty kilometres of cross side apart, and every triangle is equilateral; each takes the cross side its predecessor computed as its base and computes the next cross side from the two measured angles at that base, reduced by Legendre’s theorem. Each triangle carries the strip half a side along its heading, fifteen kilometres. On each substituted sphere the three angles are what that sphere’s latitude rule makes of the true geodesic triangle, exactly as the earlier essays computed them, and everything is referred to the WGS84 ellipsoid computed directly.

The scale error of a computed side is a ratio, so down a strip the ratios multiply, and in parts per million they add. If every triangle hands on its side 7.9 parts per million too long, the twelfth side is 95 parts per million too long, the twentieth 158. A bad base does nothing of the kind. It makes the first side wrong by some factor and every side after it wrong by the same factor, because every later side is computed from the first by ratios the base does not enter.

The drift is the sphere’s own scale #

On a strip running due north, every substitution adds a nearly constant error per triangle, and the totals diverge in straight lines. The conformal and geocentric spheres add almost the same, 7.90 and 7.94 parts per million a triangle, and reach 158 after twenty. The parametric sphere adds the same amount with the opposite sign. The authalic adds −2.62. The geodetic-latitude sphere, which spoiled a single triangle by 1,435 parts per million, adds −23.7. And the rectifying sphere adds almost nothing, 0.01 a triangle.

Those are six unrelated-looking numbers, and there is one quantity that predicts all of them. Each substitution maps the ellipsoid onto a sphere by replacing geodetic latitude with an auxiliary one, and in doing so it stretches the meridian by a factor that depends on latitude: $m(\varphi) = R\,\dfrac{d\chi}{d\varphi}\big/M(\varphi)$, with $\chi$ the auxiliary latitude and $M$ the ellipsoid’s meridian radius of curvature.

What the strip has accumulated at its far end is the change in that meridian stretch between the far end and the base, $\ln m(\varphi_1) - \ln m(\varphi_0)$. For all six substitutions the two agree to within a per cent, and for the rectifying sphere — whose latitude is constructed to make meridian arcs exact, so its meridian stretch is one everywhere — the prediction is zero and the strip drifts by three tenths of a part per million in twenty triangles.

That turns a sequence of per-triangle errors into a statement about two points. A strip computed on a substituted sphere does not accumulate an error that depends on how many triangles it has, how they are shaped or how they were computed. It accumulates the sphere’s own scale, read at the base and at the far end, and nothing else. The per-triangle error is only that difference divided into steps.

The reason is the one a bearing on a sphere is decided by its latitude, not its radius found for a single direction. A sphere’s radius sets the size of everything on it and moves no angle, so it cannot enter a strip that is computed entirely from angles and one measured length; the only thing a substitution can get wrong in a strip is how its latitude rule spaces the parallels. The strip measures that spacing the way a surveyor would — by carrying a length from one latitude to another and comparing — and reports the ratio between the spacing at the two ends. What makes that spacing change with latitude at all is the ellipsoid’s own changing curvature, the 1.35 per cent from equator to pole that the curvature of the Earth is not one number measured, and each auxiliary latitude is a different attempt to absorb it.

The first-order spheres pay and repay #

The result has a surprise in it for the first-order spheres, which the earlier essays found wrong by hundreds or thousands of parts per million in a single triangle. The geodetic-latitude sphere spoiled one thirty-kilometre triangle turned 45° by 1,435 parts per million. On a strip running north it adds 23.7 a triangle. The large error has not been computed away. It has cancelled.

A first-order substitution stretches the meridian and the parallel by different amounts — at 45° the geodetic-latitude sphere’s two stretches differ by 3,364 parts per million — so a triangle whose base and computed side make different angles with the meridian has one of them stretched more than the other, and the ratio between them is wrong by a share of that difference. On a strip running due north the two cross sides make equal angles with the meridian, 60° either side, and the difference falls equally on both. Turn the strip 30° and they no longer do: the first triangle hands on a side 2,508 parts per million wrong. But the next triangle takes that side as its base and computes a side in the direction the first triangle’s base had, so it pays the same difference back. Every even side is back on the drift, −24.5 parts per million after two triangles, −148 after twelve.

So a first-order sphere’s single-triangle error, the thing that made it “announce itself” in the earlier measurements, is a property of which side of a triangle is measured and which computed. Along a strip it is a zigzag about the drift, and what a check base finds depends on which side of the zigzag it happens to be — an odd-numbered side on a strip turned 30° from the meridian would show two and a half thousand parts per million, and the next side along would show twenty-five.

Heading and latitude decide the drift #

Because the strip accumulates a change in scale with latitude, how fast it drifts depends on how fast it changes latitude. The conformal sphere’s drift per triangle follows the cosine of the heading almost exactly: 7.90 due north, 6.85 at 30°, 5.60 at 45°, 3.97 at 60°. The cosine is the share of each triangle’s half-side advance that goes north. Along a parallel every strip on every sphere drifts by nothing — two hundredths of a part per million per triangle at most — because a strip that never changes latitude never reads the sphere’s scale at two different places.

The geodetic-latitude sphere falls off faster than the cosine, from −23.7 due north to −12.5 at 30° and −8.0 at 45°. Its meridian and parallel stretches change with latitude at different rates, and a strip at an angle to the meridian hands its scale along sides that sample both; the conformal sphere has one stretch in every direction, which is what conformal means, and follows the cosine because it has nothing else to follow.

Latitude sets the rate as the earlier essay expected: both families follow, roughly, the sine of twice the latitude, largest at mid-latitudes and small at the equator and near the pole. That is the derivative of the stretch. The earlier essay’s signatures — $\sin 2\varphi$ for the good substitutions, $\cos^2\varphi$ for the bad ones — were single-triangle errors, where the bad spheres’ size came from the difference between their meridian and parallel stretches, and that difference cancels along a strip. What survives in a strip is the rate at which each sphere’s meridian stretch changes with latitude, and for good and bad spheres alike it peaks in the middle latitudes.

A check base sees it within three triangles #

A strip’s observed angles have their own errors, and they accumulate too, but differently. If each angle is observed with a standard error of one second of arc and each triangle’s three angles are closed to their proper sum by equal shares, each computed side carries a random scale error of 3.96 parts per million — $\sigma\sqrt{2}\cot 60°$, which the simulation reproduces — and after n triangles the accumulated random error has a standard deviation of $3.96\sqrt{n}$. The conformal sphere’s drift grows as $7.90\,n$.

A drift in proportion to the count against noise in proportion to its square root must win, and the question is only when. With one-second angles, three standard deviations are reached at 2.3 triangles. A check base at the side the third triangle computes — forty-five kilometres north of the first, well inside any first-order chain — would find the side computed on the conformal sphere 24 parts per million long, against a random error of 7. With three-second angles it takes twenty triangles; with five-second angles, fifty-six.

This is where the earlier essay’s arithmetic goes astray. Eight parts per million in one triangle is below what a nineteenth-century baseline could detect. Eight parts per million a triangle reaches a hundred and sixty in the three hundred kilometres between two bases, and a check base measured to a part in a million, as the best were, would have found it at once — and found it growing along every meridian strip and absent from every strip along a parallel, which is not how a baseline standardisation error behaves.

What the signature is, and where it would be read #

The answer to the earlier question is therefore sharper than it expected. A baseline error and a sphere error cannot be confused in a network that has more than one base, because they are different kinds of quantity: one is a constant factor, the other the difference of a known function of latitude between two points. A sphere error is zero between two bases at the same latitude, whatever lies between them, and between two bases at different latitudes it is a number any handbook of auxiliary latitudes lets anybody compute in advance.

It also means a strip is blind to anything the two ends share. An angle is a difference, and the difference doubles the error found a substituted sphere turning every direction at a point and the turn not cancelling in an angle; the strip adds a third stage to that sequence, where the angle errors of successive triangles do cancel, completely, into a function of the two ends. What a closed figure cannot see is the general form of that blindness, and on the conformal sphere, whose stretch is the same in every direction, it is exact: a closed loop of strips that returns to its own base, whatever latitudes it passes through, accumulates nothing at all.

That makes the archival test easy to state. Where an old triangulation was computed with a substituted sphere, the scale difference between it and a modern re-observation should be a smooth function of latitude that vanishes at each measured base and matches the sphere’s meridian stretch between them. Where it was computed on the ellipsoid, that function is absent, and whatever scale difference remains is the base’s. Four radii of the Earth is the reminder that the handbooks offered several spheres, and each of them predicts a different function.

It also changes what the documented differences of eight or so parts per million can be. A difference that size, spread uniformly across a network that spans several degrees of latitude and several bases, is not what any substitution here leaves: a sphere error between two bases a few degrees apart in latitude is hundreds of parts per million, or nothing, and is the same size in every strip only if every strip happens to span the same latitudes. The explanation the earlier essay offered for a uniform eight parts per million does not survive the strip.

How the drift was checked #

The drift must be the meridian scale. For all six substitutions, the scale error twenty triangles north from 45° must match the change in the sphere’s meridian stretch between the strip’s ends to within a per cent, or to within half a part per million where that change is nothing. The worst is the rectifying sphere’s 0.29 against zero.

A strip along a parallel must drift by nothing. Its ends share a latitude, so on every sphere it must accumulate less than half a part per million in twenty triangles. The largest is 0.37, on the conformal sphere.

The simulated noise must be the sine rule’s. Closed by equal shares, one second of arc in each angle leaves $\sigma\sqrt 2\cot 60° = 3.958$ parts per million in a computed side. Forty thousand simulated triangles give 3.961.

The single-triangle numbers must be the earlier essays’. The strip is built from the same triangles, computed the same way; the conformal sphere on the triangle turned 45° still leaves 7.6 parts per million and the geodetic-latitude sphere 1,435. The strip changes what is added up, not what one triangle does.

Where the strip stops #

One strip, straight, of equilateral triangles. A real first-order chain has triangles of varied shape, braced figures, diagonals and more than one route between bases. The prediction does not depend on any of that — the accumulated error is a function of the two ends — but the zigzag of the first-order spheres does, and a chain whose triangles vary in orientation would show a noisier version of it.

No adjustment. A real chain between two bases is adjusted to close on the second, and an adjustment spreads a misclosure through the chain as the two ways to spread a misclosure describes. It would remove the drift at the check base and, spread evenly, leave a scale that varies along the chain between the bases — close to the stretch function itself over a few degrees, so the signature would move from a misclosure at one base into the adjusted coordinates between them. How closely is the question below.

The sphere is substituted in every triangle. That is the model the earlier essays set up — each triangle’s angles computed on the substitution’s sphere — and it is the case in which the error accumulates. A computer who used the sphere only for the excess, and solved each triangle’s sides on the ellipsoid, would carry the sphere’s error only through the excess, which is the smaller of the two corrections.

Thirty-kilometre triangles from 45°. The drift per triangle scales with the side — 1.8 parts per million for ten-kilometre triangles and 10.5 for sixty — but per kilometre of latitude it is the same, because it is the sphere’s stretch that is being read.

Still open: whether an adjusted network still carries the stretch #

The unadjusted strip reads the sphere’s meridian stretch at its two ends, and the check is clean. Real networks were adjusted, and the adjustment is precisely the operation that takes a misclosure at a check base and distributes it. Distributed evenly per triangle, the drift of a meridian strip becomes a scale that varies linearly with position along it — very nearly the stretch function itself between the bases, because the stretch is nearly linear over a few degrees. Distributed by weight, it becomes something else.

The network’s answer is decided before it is measured found an adjustment’s treatment of an error fixed by its design rather than its data, which suggests the answer is computable in advance for any stated network. Whether a network adjusted by the methods of its day retains the sphere’s stretch as a recognisable function of latitude, whether the second-order part of the stretch — its curvature between two bases, a few parts per million over five degrees — survives an adjustment that spreads the linear part, and whether a re-observation could read that curvature through the noise of the old angles, are questions an unadjusted strip cannot ask.