Every projection is wrong, here is how much

Gauss closed the hard part of this in 1827 and we have spent two centuries arguing about it anyway, mostly online, mostly badly. So: the mathematics first, then six real projections with real numbers, then the two arguments people reliably get wrong.

The theorem that closes the case

Gauss's theorema egregium says the Gaussian curvature of a surface is invariant under local isometry. Curvature is intrinsic: you can measure it from inside the surface, using only distances and angles along it, without stepping off into the surrounding space. Bend paper into a cylinder and its curvature does not change, because bending without stretching is an isometry.

Now compare the two surfaces we care about.

$K_{\text{sphere}}=\frac{1}{R^2}>0,K_{\text{plane}}=0$

An isometry between them would have to preserve $K$. There isn't one. Not for the whole sphere, not for any patch of it however small, because curvature is a pointwise quantity and the mismatch exists at every point. No projection is waiting to be discovered that fixes this. Two corollaries people get wrong in public: no projection is both conformal (angle preserving) and equal area, and none is equidistant everywhere.

Tissot's little circles

Nicolas Auguste Tissot presented the tool for measuring the damage in 1859 and 1871. Take a circle of infinitesimal radius on the globe and project it. Because the projection is smooth, its effect on an infinitesimal circle is a linear map, and a linear map sends a circle to an ellipse. That ellipse is the indicatrix and it contains everything about the distortion at that point.

The axes point along the directions of greatest and least scale. If the ellipse is a circle, the projection is conformal there. If it has the same area as the original circle, area is preserved there, no matter how squashed it looks. Write $h$ for the scale along the meridian, $k$ for the scale along the parallel, and ${\theta }^{\prime }$ for the angle between them as drawn. The areal scale factor is

$s=hk\sin {\theta }^{\prime }$

and the projection is conformal at that point exactly when $h=k$ and ${\theta }^{\prime }=\pi /2$. Equal area means $s=1$ everywhere. Gauss's theorem is the statement that both cannot hold at once, and the indicatrix is how you watch it happen: force the ellipse into a circle and its area drifts, hold the area and it stretches.

So "which projection is best" is not a question. The question is which of $h$, $k$ and ${\theta }^{\prime }$ your analysis is about to consume.

Six projections and what each of them breaks

Projection	Preserves exactly	Gives up	A number you can check
Mercator (EPSG:3395)	angle, at every point	area, spectacularly	areal scale ${\sec }^2\phi$: 4.00 at 60°, 14.93 at 75°
Cylindrical equal area, standard parallels 45° (Gall's orthographic, Peters's map)	area, everywhere, exactly	shape everywhere except 45°N and 45°S	indicatrix axis ratio 2.00 at the equator (tall), 4.27 at 70° (wide)
Transverse Mercator in a UTM zone	angle	area, and all candour about the zone edge	$k_0=0.9996$ on the central meridian, about 1.0010 at the zone edge on the equator; distortion under 1 part in 1,000 inside a zone
Azimuthal equidistant	distance and bearing from one chosen centre	everything not measured from that centre	transverse scale $\rho /\sin \rho$: 1.571 at 90° from centre, 5.236 at 150°, divergent at the antipode
Winkel tripel	nothing whatsoever	a little of everything, deliberately	Goldberg and Gott's summed squared error 4.563, best of the known projections they scored
Web Mercator (EPSG:3857)	angle, nearly	area, the poles, and your dignity	latitude cut at ±85.051129°; up to 0.7% scale error against true ellipsoidal Mercator

Two of those you can derive in three lines, and a distortion figure you derived yourself is worth ten you read.

Cylindrical equal area with standard parallel ${\phi }_0$ is $x=R\lambda \cos {\phi }_0$, $y=R\sin \phi /\cos {\phi }_0$. Differentiate: parallel scale $k=\cos {\phi }_0/\cos \phi$, meridian scale $h=\cos \phi /\cos {\phi }_0$. Their product is 1 at every latitude, which is the equal area condition. Their ratio is ${\cos }^2{\phi }_0/{\cos }^2\phi$, which is the shape damage. At ${\phi }_0=45°$ that ratio is $1/2$ at the equator, so the indicatrix there is exactly twice as tall as it is wide; at 70° it is 4.27 times wider than tall.

Azimuthal equidistant is quicker. A circle at angular distance $\rho$ from the centre has true circumference $2\pi R\sin \rho$, and the projection draws it at radius $R\rho$, so the scale along it is

$k\left(\rho \right)=\frac{\rho }{\sin \rho }$

which is 1.571 at $\rho =90°$ and 5.236 at $150°$, and diverges at the antipode, where one point smears around the whole bounding circle. The UN emblem is this projection, centred on the North Pole and sensibly cut off at 60 degrees south, before the arithmetic gets embarrassing.

Mercator's area inflation, worked

Mercator is conformal, which forces the two scale factors to be equal, so one number does both jobs:

$k\left(\phi \right)=\sec \phi ,s\left(\phi \right)=k^2={\sec }^2\phi$

$s\left(45°\right)=2.00,s\left(60°\right)=4.00,s\left(75°\right)=14.93$

That is the whole Greenland business in three numbers with no politics attached. Greenland covers 2,166,086 km² and sits almost entirely between about 60°N and 83°N. Africa covers roughly 30,043,862 km², a factor of about 14 more, straddling the equator where $s=1$. Draw both on Mercator and Greenland is multiplied by something between 4 and 60 while Africa is multiplied by roughly 1. They come out looking comparable. Nobody chose that; it falls out of ${\sec }^2\phi$, and it would happen on any conformal cylindrical projection, because conformality plus a cylinder forces it.

Areal scale factor on Mercator, every twenty degrees

Latitude	$\sec \phi$	${\sec }^2\phi$
0°	1.000	1.00
20°	1.064	1.13
40°	1.305	1.70
60°	2.000	4.00
80°	5.759	33.16
85°	11.474	131.65

What Mercator buys for that price usually gets left out: a line of constant compass bearing is a straight line on the chart. That is the entire reason the projection exists, and it was designed for going to sea rather than for looking at.

Web Mercator, the compromise nobody chose

Nobody convened to select the projection now carrying the overwhelming majority of the world's map views. It accreted. It first shipped as EPSG:900913, an unofficial code chosen because it reads as GOOGLE in leetspeak, defined by Christopher Schmidt and baked into OpenLayers 2. It was registered properly as EPSG:3785 in 2008 and superseded the same year by EPSG:3857, "WGS 84 / Pseudo-Mercator". The registry itself notes it is not a recognised geodetic system, which is an unusual thing for a registry to say about one of its own entries.

Its defining oddity is that it pushes coordinates on the WGS 84 ellipsoid through the spherical Mercator formulas. Those do not compose, so the result is not conformal, merely close enough that no human eye will complain. The latitude cut at ±85.051129° is not a geodetic judgement either: it is the latitude that makes the projected world exactly square, since the equatorial half width is $\pi R=\mathrm{20,037,508.34}$ m and the northing runs to the same figure, so tiles divide into powers of four forever. Feed 3857 coordinates to something expecting 3395, true ellipsoidal Mercator, and the registry warns of 0.7% scale error and northing differences up to 43 km on the map, around 21 km on the ground. That is not a rounding artefact. That is a town.

Same species of decision as web typography settling on a justification algorithm nobody would have specified from scratch, which Gerard takes apart in @bindery/justified-text-on-the-web-is-a-lie-we-agreed-to. A default that shipped early, worked adequately, and then became structural.

There is no correct projection and I am tired of pretending that is a subtle point. There is a quantity you intend to measure and a distortion you can afford, and those two choose the projection for you. If you are computing areas in EPSG:3857 you are not making a small error. At the latitude of Oslo you are out by a factor of about four, and it is entirely your own fault, because the number was in the registry the whole time.