The radius of curvature is two numbers

There is no such thing as the radius of the Earth at a latitude. The Earth is a sphere, and when it is not, it has two. There are two, and the difference between them is larger than most people’s estimate of the whole uncertainty.

Two curvatures, because a surface has two #

Any smooth surface has two principal curvatures at a point, and their product is the curvature the impossibility theorem is about: the largest and the smallest normal curvature over all directions through it. On a sphere they are equal, which is why a sphere has one radius and why the habit of asking for “the radius” survives.

On an ellipsoid of revolution they are the curvature of the meridian section, whose radius is written $M$, and the curvature of the section perpendicular to it, whose radius is $N$:

$$M = \frac{a(1 - e^2)}{W^3}, \qquad N = \frac{a}{W}, \qquad W = \sqrt{1 - e^2\sin^2\varphi}.$$

Both are radii of curvature and neither is a distance to the centre. $N$ is the length of the normal from the surface to the axis of rotation — which is why it appears in the conversion between geodetic coordinates and Cartesian ones — and $M$ is not the length of anything; it is a curvature reciprocal.

Their product is the reciprocal of the Gaussian curvature, so the ellipsoid is not a surface of constant curvature and the two of them together are what makes it so.

The size of the disagreement #

At the equator $M = 6{,}335.44$ km and $N = 6{,}378.14$ km. The gap is 42.70 kilometres, or 0.67 per cent.

At 45° it is 21.46 kilometres. At the pole it is zero, because there the meridian and its perpendicular are indistinguishable and the surface is locally spherical.

Nothing about those numbers is small. Two thirds of one per cent is 6.7 metres on a kilometre, and a first-order survey works to about twenty parts per million — so the difference between the two radii is three hundred times the tolerance of the work that uses them.

Euler’s radius, and why the direction is the missing argument #

The right radius for a line is neither of them and depends on which way the line runs. Euler’s theorem gives the normal curvature in a direction $\alpha$ from north as an interpolation between the two principal ones:

$$\frac{1}{R(\alpha)} = \frac{\cos^2\alpha}{M} + \frac{\sin^2\alpha}{N}.$$

So $R$ runs continuously from $M$ due north to $N$ due east, passing through every value between. There is no direction in which the answer is 6,371 km unless the latitude happens to make it so.

The Gaussian mean $\sqrt{MN}$ is the value at 45° to second order and is the radius of the sphere with the same Gaussian curvature at that point. It is the right choice when the quantity being computed is an area or a curvature, and the wrong one when it is a distance in a stated direction.

What each substitution costs #

Five and a half metres on a kilometre is not a subtlety. It is the difference between a computation that meets a survey specification and one that misses it by two orders of magnitude, and it arises from taking a number out of a table because it is the number in the table.

Turn the line ninety degrees and the ranking changes completely. Due east at the equator the prime-vertical radius is exact by definition, the mean radius is out by 1.12 metres per kilometre, and the meridional radius — which was exact a paragraph ago — is out by 6.74.

The sign changes, and that matters #

An error that is always the same sign can be absorbed into a scale factor and forgotten. This one cannot: the mean radius is larger than Euler’s in some directions and smaller in others, so a survey computed with it accumulates an error whose sign depends on the bearing of each leg.

That is the worst arrangement available for a traverse. Legs running north and legs running east pick up errors of opposite sign, so a closed figure partly cancels them and reports a closure better than the underlying computation deserves — which is exactly the failure what a closed figure cannot see is about, arriving from a source that essay did not consider.

Which radius each formula wants #

The practical content is a short list, and every item on it is a place where somebody has to choose.

A north–south distance from a latitude difference needs $M$, integrated — which is the meridian arc, by series and by quadrature: $\mathrm{d}s = M\,\mathrm{d}\varphi$. This is where a degree of latitude gets its length, and it is why a degree of latitude is 110.57 km at the equator and 111.69 at the pole.

An east–west distance from a longitude difference needs $N\cos\varphi$, which is the radius of the parallel.

The conversion to and from Cartesian coordinates needs $N$, which is where a datum transformation does its arithmetic, because $N$ is the length of the normal to the axis and the Cartesian conversion is written along that normal.

An area needs neither, and an equal-area map of the ellipsoid needs the authalic latitude instead: it needs the authalic radius, or a closed form, or a quadrature over the ellipsoid’s own area element.

A local sphere for a survey needs Euler’s radius at the azimuth of the work, or the Gaussian mean when the work runs in every direction — and the choice between those two is a decision about the job rather than about the Earth.

Why the mean radius exists at all #

The number 6,371.0088 km is the arithmetic mean $(2a + b)/3$, and it has a defensible purpose: it is the radius of the sphere with the same mean of the three semi-axes, which is a reasonable single number for a body that is nearly spherical.

It also happens to sit within a thousandth of a per cent of the authalic radius — the radius of the sphere with the same surface area — so a calculation of a global area with it is right to five significant figures. That is where a single radius belongs.

Where it does not belong is in a distance. A distance is a line integral of a metric and the metric is anisotropic, and no scalar summarises an anisotropic quantity without a direction.

The other radius nobody names #

The Gaussian mean is the answer to a different question and is worth separating out, because it is the one a projection cares about.

A conformal map’s scale factor is fixed by the Gaussian curvature and by nothing else — that is the content of the Theorema Egregium as this collection uses it — so when a projection of a small region is approximated by a projection of a sphere, the sphere that should be used is the one with the matching curvature, not the matching radius in any direction. That sphere’s radius is $\sqrt{MN}$, and using it is what makes a local conformal approximation good to second order rather than first.

So there are three defensible radii and each answers a different question. $M$ and $N$ answer how long is this line, separately, according to which way it runs. $\sqrt{MN}$ answers what does the curvature do to a map here. The mean radius answers what is a single number for the size of the Earth, which is a question about tables rather than about geometry.

Where this is silently right, and where it is silently wrong #

Most software is silently right, because most software never takes a radius. A geodesic on the ellipsoid, a Krüger series, a Vincenty inverse and a proper conversion to Cartesian coordinates all carry $M$ and $N$ where each belongs, and none of them offers a single-radius option.

The places it goes wrong are the improvised ones, and they are common: a haversine distance on the “Earth’s radius”; a buffer computed by dividing a ground distance by a radius to get an angle; a rule-of-thumb conversion from degrees to metres. Each of those is a substitution of one number for a directional quantity, and each is out by up to two thirds of a per cent.

Two thirds of a per cent is 6.7 metres in a kilometre and 670 in a hundred. It is not a rounding error; it is the size of a plot boundary.

The refusal: a sphere has one radius #

The measure has to be able to return zero, and it does. On the sphere declared in this collection’s library, whose flattening is exactly zero, $M = N = a$ at every latitude and Euler’s radius is that same number in every direction. The direction-dependence collapses, and the substitution error of every candidate is identically zero.

That is what makes the ellipsoid figures a measurement of the flattening rather than of the arithmetic. The gap between $M$ and $N$ is $a e^2 \cos^2\varphi / W^3$ to first order, so it is the eccentricity, appearing as a distance.

An error budget with a direction in it #

A survey error budget is a list of variances added in quadrature, and every entry on it is a scalar. That is the format’s own limitation and it is where this error hides.

The radius substitution does not produce a variance. It produces a bias whose size and sign depend on the azimuth of the leg being computed, so a budget that lists it as “spherical approximation, 5 ppm” has already thrown away the only thing about it that matters. Two legs of a traverse computed with the same wrong radius have correlated errors whose correlation depends on the angle between them, which is exactly the structure a quadrature sum assumes away.

The correct treatment is the one geodesy adopted long ago and improvised geometry did not: do not substitute. Carry $M$ and $N$ separately and let each formula take the one it needs, and the term never enters the budget at all because it is never made.

Where the model stops #

Everything here is the ellipsoid of revolution. On a triaxial body the two principal curvatures are still two numbers and their directions are no longer the meridian and its perpendicular — the coordinate lines are not orthogonal — so Euler’s theorem still holds and the frame it is stated in is not the graticule’s.

Nor is anything here about the geoid, whose curvature varies with the mass under it and is not a property of any ellipsoid. A radius taken from a table is a statement about a fitted surface; the ground has its own and it is nobody’s closed form.

The two exact azimuths #

There is a small structural point in the last figure worth drawing out.

At an azimuth of exactly zero the meridional radius is exact, and at exactly ninety the prime-vertical radius is. At every azimuth between, none of the five candidates is exact, because Euler’s radius there is a harmonic interpolation between two numbers and is not equal to either or to any fixed combination of them.

So the familiar advice — use $M$ for north–south and $N$ for east–west — is not a rule of thumb that happens to work. It is the exact answer at two azimuths and nowhere else, and the reason it is safe in practice is that those two azimuths are the ones a graticule-aligned computation actually uses. A computation along a graticule line takes the exact radius; a computation along anything else takes an approximation whether or not anybody says so.

Who found it, and when #

Euler’s theorem on normal curvature is from 1760 and the two radii of the ellipsoid were standard in geodesy long before anybody had to write software with them. The confusion is entirely modern: it arrives with the availability of a single number in a reference table and with the writing of geometry by people whose training was not geodetic.

That is worth stating without condescension, because the table is not wrong. 6,371 km is the mean radius and is correctly labelled as such wherever it is printed. What is missing is the sentence that says what a mean radius is for, and the sentence is missing because in 1760 nobody needed it.

One number, five meanings #

Collected in one place, because the list is the practical content of the essay.

$M$, the meridional radius, is 6,335 to 6,400 km and is the one a change of latitude wants. $N$, the prime-vertical radius, is 6,378 to 6,400 km and is the one a change of longitude and every Cartesian conversion wants. $\sqrt{MN}$, the Gaussian mean, is 6,357 to 6,400 km and is the one a curvature or a local projection wants. The authalic radius, 6,371.007 km, is the one an area wants. The mean radius, 6,371.009 km, is the one a table wants.

Five numbers spanning 65 kilometres, all correct, all called the same thing in ordinary speech. Choosing between them is not a refinement to be applied when the accuracy demands it; it is the first step of the computation, and the step where a formula that takes a single radius has already decided.

How to find out which radius a formula meant #

The list is the practical content and it leaves a practical question: given a formula, a library or a piece of code that takes one radius, how does a reader establish which of the five it wants?

Ask what the formula does for a purely north–south displacement. If the answer should be the meridian arc, the radius in it is $M$ — that is the definition of the meridional radius, and no other member of the list gives the right answer there.

Ask what it does for a purely east–west one. A change of longitude at a fixed latitude runs along a parallel whose radius is $N\cos\varphi$, so a formula computing a departure from a longitude difference wants $N$. Every Cartesian conversion wants $N$ for the same reason.

Ask whether the quantity is a curvature or an area. A local projection’s scale factor, a geodesic’s behaviour, and anything derived from the Gaussian curvature want $\sqrt{MN}$; a total area wants the authalic radius; a total meridian length wants the rectifying one.

And if the formula takes a single constant radius with no latitude in it at all, it wants a sphere, and the question is only which sphere its author had in mind — which is usually the mean radius, because that is what a table prints.

It is a check worth running on inherited code as much as on a paper.

The size of getting it wrong is the reason to check. The five span 65 kilometres, which is about one per cent of the radius, so a formula using the wrong one is wrong by up to one per cent: ten metres in a kilometre, a kilometre in a hundred. That is far above the tolerance of anything anybody would write such a formula for, and it is far below the size that would make the result obviously absurd — which is the band in which errors survive.

That also makes it a reviewable property rather than a matter of trusting the source, which is worth something when the source is a code comment.

The one reassurance is that the choice is decidable from the formula alone, without knowing what its author intended. Each of the five is the unique radius making one particular quantity exact, so asking which quantity the formula is supposed to get right answers the question every time.

Where the ladder goes next #

Twelve rungs of this anchor have worked on the ellipsoid and this one is about the most elementary quantity any of them uses. What none of them has asked is which ellipsoid — the fit that produced $a$ and $f$ is a fit, its parameters are correlated, and the two are quoted to a precision their covariance does not support.