How big a triangle it takes

Measuring curvature from inside is the rung this whole field rests on. A surface-dweller with no view from outside can read the Gaussian curvature by building a triangle out of geodesics and adding up its angles: Gauss–Bonnet makes the excess over 180° exactly the integral of K over the interior, so

$$\hat{K} = \frac{E}{A}$$

is an estimator of the curvature that needs nothing but angles and an area. It is the argument that makes Theorema Egregium a fact about the world rather than about coordinates, and every essay in this field cites it.

That rung measures exact angles. This one measures real ones.

The signal #

The excess of a triangle of area $A$ on a surface of curvature $K$ is $KA$, in radians. On WGS84 at 45° north, $K = 1/(MN) = 2.4582\times10^{-14}\ \mathrm{m^{-2}}$, so an equilateral triangle of side $s$ has

$$E = K\cdot\frac{\sqrt{3}}{4}s^2.$$

A fifty-kilometre triangle is about the largest anybody routinely observed, and its excess is five and a half seconds of arc — about a thirtieth of the apparent width of a full moon, distributed over three angles.

The noise #

Three measured angles with standard deviation σ each give a sum with standard deviation $\sigma\sqrt{3}$, and the excess is the sum minus a constant, so

$$\operatorname{sd}(\hat{K}) = \frac{\sigma\sqrt{3}}{A},\qquad \frac{\operatorname{sd}(\hat{K})}{K} = \frac{\sigma\sqrt{3}}{KA}.$$

The relative error falls as the inverse square of the side, because the excess grows as the area and the noise does not grow at all. That is the entire reason geodetic triangles are as large as topography and intervisibility permit, and it is also the reason the method is so hard: quadrupling the accuracy requires doubling every leg.

The competing error that turns out not to compete #

There is an obvious reason a triangle could be too large, and it is worth reporting that it is not the reason.

The Earth is not a sphere, so K is not constant: the curvature of the Earth is not one number measures it running from $2.4747\times10^{-14}\ \mathrm{m^{-2}}$ at the equator to $2.4417\times10^{-14}$ at the pole. A big triangle therefore returns the area mean of K over its interior rather than K at its centre, and the difference grows with the triangle — as its square, since K’s variation is smooth and the leading term is the second derivative times the spread.

So there ought to be a trade, and there is: fitted slopes of −2.000 for the noise and +1.989 for the variation, which is the arithmetic behaving exactly as the two formulas say.

What there is not is a crossing anywhere near a buildable triangle. At 50 km the variation contributes $8.7\times10^{-11}$ of relative error against the noise’s 0.32; at 800 km it is $2.2\times10^{-8}$ against $1.2\times10^{-3}$. The two arms cross somewhere past a hundred thousand kilometres, which is fifteen Earth radii.

The obstruction to measuring curvature from inside is the instrument, and it is not the shape of the Earth. That is worth stating as a result rather than leaving as an absence, because the intuition runs the other way: it feels as though the trouble with a large triangle must be that the surface stops being uniform, and the arithmetic says the surface is uniform to a part in ten million over distances at which the angles are useless.

What a survey would have needed #

A one-second theodolite is the instrument of the great nineteenth-century arcs, and reading K to one per cent with it needs a triangle 281 kilometres on a side — roughly London to Newcastle, with every vertex visible from the other two.

Every one of those surveys measured the figure of the Earth, several of them to remarkable accuracy, and not one of them did it this way. They measured the length of a meridian arc and the difference in astronomical latitude between its ends, which is the method the figure of the Earth was measured by — a ratio of a distance to an angle, where the angle is degrees rather than seconds and the arithmetic is correspondingly forgiving.

The required side has a closed form, and it is scale-free #

The table of sides is four numbers per row and it is one expression.

The relative error is σ√3 divided by KA, and an equilateral triangle’s area is (√3/4)s², so the two square roots cancel and

$$\varepsilon = \frac{4\sigma}{K s^2}, \qquad s = 2\sqrt{\frac{\sigma}{K\varepsilon}}.$$

With σ in radians. For a one-second instrument and a one-per-cent target that gives 280.9 kilometres against the table’s 281, and it reproduces every other entry to the digit.

Writing K as 1/R² turns it into something better:

$$s = 2R\sqrt{\frac{\sigma}{\varepsilon}}.$$

The required side is twice the body’s radius times the square root of the ratio of the instrument’s angular error to the target accuracy — and dividing through by R removes the body entirely:

$$\frac{s}{R} = 2\sqrt{\frac{\sigma}{\varepsilon}}.$$

So the triangle has to span the same fraction of any sphere, whatever its size. A one-second instrument reading K to one per cent needs a triangle 0.044 radians on a side — 2.5° of arc — on the Earth, on the Moon, on a marble and on a star. The difficulty is not that the Earth is large; it is the same difficulty on every body, and it is a statement about the ratio of two angles.

That reframes the historical table in a way the kilometres hide. Gauss’s 85-kilometre triangle is 0.0133 radians of the Earth, and the arithmetic says he needed 0.044 — a triangle three and a third times larger in angular terms, which is the same shortfall the table reports in per cent and is now a number about geometry rather than about Hanover.

It also gives the impossibility a clean form. Reading K to an accuracy equal to the instrument’s own angular error — ε = σ, which is what a naive reading of the theorem determines K from the angles would suggest is available — requires s = 2R, a triangle twice the radius on a side. That is not a large triangle; it is not a triangle. The estimator’s noise is a fixed angle divided by an area, and an area is bounded on a sphere, so there is a floor on ε that no instrument and no survey can pass: at s = πR, the largest a spherical triangle’s side can be, the best available accuracy is 4σ/π² — about four tenths of the instrument’s own error, on any body at all.

The two readings together are the rung’s real content. The method is scale-free, so nothing about the Earth is to blame, and it is bounded, so nothing about the instrument rescues it. A flatlander on any sphere, with any theodolite, can read the curvature of their world to about half their own angular precision and no better — and to do even that they must build a triangle spanning most of it.

Gauss’s own triangle #

The largest of these is Gauss’s, and it is the one with a story attached.

The Hanover triangulation of the 1820s included the triangle Hohehagen–Brocken–Inselsberg, with sides of roughly 70, 85 and 107 kilometres — the largest carefully observed triangle of its era. A durable piece of folklore says that Gauss measured it in order to test whether space is Euclidean, and that he found the angles summed to 180° within the error.

The arithmetic here says something more precise about what such a test could have meant. Its excess is about 15.9 seconds of arc, from the Earth’s own curvature, and Gauss knew that quantity to far better than his observations could check. With one-second angles the triangle constrains K to about eleven per cent, so any additional curvature of space would have had to be a tenth of the Earth’s surface curvature to be visible — which is an absurdly weak bound, and Gauss, who understood the estimator better than anybody, would have known it was.

The triangle was observed because Hanover needed a survey. The estimator says that as a test of anything cosmological it could not have worked, and being able to say so with a number rather than a suspicion is the point of the rung.

Why the arc method wins, in one comparison #

The two routes to the figure of the Earth can be put on the same footing, and the comparison is stark.

The excess route reads a curvature from an angle sum. The signal is $KA$, which at fifty kilometres is 5.5 seconds of arc, and the instrument’s noise on it is 1.7. Signal to noise: about three.

The arc route reads a radius from the ratio of a measured distance to a measured difference of astronomical latitudes. Over a degree of latitude the distance is 111 kilometres and the latitude difference is 3,600 seconds of arc. Measured with the same one-second instrument, the angular signal to noise is about 2,500 — three orders of magnitude better — and the distance is measured by a baseline and a triangulation chain rather than by the angles of one figure.

The difference is entirely in where the small quantity sits. The excess route puts the whole of the information into a second-order quantity that vanishes with the area; the arc route puts it into a first-order one that grows with the distance. A method whose signal is second order in the size of the survey is a method that cannot be rescued by a better instrument, only by a much larger survey, and the surveys were already as large as the ground allowed.

That is why the figure of the Earth was measured by arcs, and why the intrinsic route Gauss proved — the one that makes the curvature a fact about the surface rather than about its embedding — has never been the practical one. The theorem is about what is determined. The survey is about what is measurable, and a degree is not a unit of length is the arc method’s own arithmetic.

What was computed, and how #

Everything comes from ellipsoid.js’s own $M$ and $N$ on WGS84, so K is exact at every latitude and no sphere appears anywhere. The excess is $KA$ with $A$ the equilateral triangle’s planar area, which is accurate to a part in $10^{5}$ at these sizes and is stated rather than hidden.

The area mean of K over a triangle is a one-dimensional integral, because K on an ellipsoid depends on latitude alone: the triangle’s width at each latitude is its own triangular profile, and weighting by that width is what makes the result an area mean rather than a band mean.

Three assertions carry the rung and each could fail alone. The noise arm must have slope −2 to within a twentieth — it is a fixed angular error divided by an area, and any other slope means the estimator has been written wrongly. The variation arm must exist and grow, or the ellipsoid has been replaced by a sphere somewhere. And the variation must be more than a thousand times below the noise at every side up to a thousand kilometres, which is the finding stated as a check: a version of this file that had confused metres with radians would fail it.

The historical claim has its own assertion: at each survey’s own stated precision, the side needed for one per cent must exceed the side that survey used, for every one of them, with the closest short by a factor of more than two.

The estimator, stated as this collection states them #

The rung is another instance of a habit this collection has been building for a while, and it is worth naming.

K̂ = E/A is an estimator. It has a bias, which is the difference between the area mean of K and K at the centre, and it has a variance, which is $3\sigma^2/A^2$. Both are computed above and both are functions of one design parameter, the triangle’s size. Neither number appears anywhere in the classical statement of the theorem, because the theorem is about exact quantities and estimators are about noisy ones.

What makes the pairing informative here is that the two run in opposite directions in the size and are separated by seven orders of magnitude at every practical size. There is a trade in principle and none in practice, and a rung that had reported only the trade would have implied a design problem where there is none. The design problem is one-sided: build the largest triangle the ground permits, and it will still not be large enough.

Where the model stops #

One triangle, not a chain. A triangulation of $n$ triangles gives $n$ estimates, and averaging them reduces the noise as $\sqrt{n}$ — so a network of a hundred fifty-kilometre triangles reaches the accuracy of a single 158-kilometre one. That helps, and it is what a real survey has, and it does not change the conclusion: the Great Trigonometrical Survey’s thousands of triangles would still be reading a quantity nobody was trying to measure by a route nobody was using.

The angles are assumed independent and unbiased. Neither is quite true. Refraction bends every sight and its effect on a horizontal angle is systematic rather than random; the deflection of the vertical means a levelled instrument is not referred to the ellipsoid normal, which the plumb line is not the normal prices; and both are correlated between neighbouring stations. All of those make the real situation worse than the estimate here, not better.

And this is the excess route only. K is measurable from inside by other means — the circumference deficit of a small circle, the second derivative of a geodesic spread, the holonomy of a loop — and each has its own error budget. A direction carried round a loop is the holonomy version, and it has exactly the same problem, because the holonomy of a loop is the same enclosed curvature by a different name.

The generalisation #

The rule is one this collection keeps arriving at from different directions: a theorem that says a quantity is determined does not say it is measurable.

Gauss–Bonnet determines K from the angles. It says nothing about how much of K is in the angles, and the answer is $KA$ — a quantity proportional to the area, which is why the theorem is easy to state and the measurement is hard to make. Every intrinsic quantity has this shape: it is exactly recoverable from data on the surface, and the amount of it present in any finite patch goes to zero with the patch.

That is the same relation as the one between how small is flat enough and this rung, and the two are the same statement read in opposite directions. A patch is flat enough when the curvature in it is below the tolerance of the job; a patch is big enough to measure the curvature when the curvature in it is above the noise of the instrument. They are the same inequality with the sign turned round, and a surveyor working in the first regime and a geometer working in the second are asking one question.

It also explains why the two communities almost never argue. A surveyor and a geometer looking at the same fifty-kilometre triangle disagree about nothing: one of them is correct that the curvature in it is negligible, and the other is correct that it is present, and the quantity they are both describing is the same $KA$ read against two different thresholds.

Where this field goes next #

Eleven rungs have established that the curvature cannot be removed, that it can be measured from inside, that it varies, that it changes sign on real ground, and now how much of it a finite triangle actually contains. What none of them has touched is the third possibility: that the surface being measured is not a surface at all at the scale of the measurement. Topography has structure down to the roughness of a rock, and the Gaussian curvature of the actual ground is not the curvature of anything a survey could triangulate — which is a question about what a surface is, and it belongs to the field’s other anchor.