The Peters Projection

There are many ways to transform a spherical surface into a flat one; they all mess with the geometry - unavoidably, because a sphere isn't flat - and various of them have particular virtues for particular uses. One of the simplest to describe is an axial projection, made famous since the 1970s as the Peters projection, though it has a longer history.

The projection in question respects the areas of things on the sphere: two portions of the sphere's surface, of equal area, will be projected onto portions of the map having equal areas; generally, the ratios between areas of portions of the sphere are equal to the ratios between the areas of the corresponding portions of the projected image of the sphere. Use of such a projection in map-making is, I am told, credited to Lambert's work in 1772, although he used the `natural' π:1 aspect ratio rather than one contrived to fit nicely with publishers' convenience. I'm told Peters' aspect ratio wasn't used until 1855, when some British cartographer introduced it.

It should be noted that modern cartographic orthodoxy doesn't have much time for single-piece rectangular maps of the whole globe - after all, they mess up the geometry and do so very drastically in places. We should really use globes as our maps or, failing that, each page covers only as much of the world as can be done without drastic geometric distortion. However, there are various ways that folk use maps, particularly for `sound-bite' quality communication, for which we aren't about to persuade them to use anything other than a simple flat rectangle.

The other widely-known flat rectangle map of the world is the Mercator projection, whose particular virtue is that any straight line on it is a line of constant compass bearing (albeit this isn't a real, magnetic, compass, it's an astronomical or celestial `compass' which knows where true North is). Since maritime navigators had a use for these, they used to be quite widely published; which is presumably how they came to be widely used by, for instance, schools as the ubiquitous `map of the world' - albeit always leaving off the neighbourhood of the poles, since the magnification involved grows hugely as one gets closer to either pole. Mercator greatly exaggerates areas near the poles relative to areas near the equator.

Arno Peters pointed out the value of area-faithful maps in policy-making decisions: it shows you accurately the amount of land area in different pieces of the world, even if it does mess up their geometry somewhat. You can read more about that argument, and see illustration maps, at the web-site. Peters was particularly attentive to the issue of reporting areas of countries, but I've seen equally strong argument for area-faithful maps from oceanographers. I'd display the map image from that in-line here, but I haven't yet asked the site's maintainers whether they're fussy about me doing so ... yes, lawyers have managed to use copyright law to make a minefield even of the web.

Description of the projection

Description of the projection: take your sphere; chose a line through its centre (e.g., for the Earth, the axis about which it spins); construct a cylinder centred on that axis; draw rays out of the axis, at right angles to it; wherever such a ray passes through an interesting feature on the surface of the sphere, mark the cylinder with a representation of that feature at the point where the ray meets the cylinder; thereby project the sphere onto the cylinder. Near where the axis cuts the sphere (the poles), the sphere's geometry is very badly messed up; but most of the surface area of the sphere is roughly half way between these two, where the geometry is treated relatively gently. Now, make a cut in the cylinder, parallel to the axis, and un-roll the projected sphere off the cylinder; the un-rolled image will lie flat (just like the label off a can of beans will lie flat if you manage to remove it).

I believe Archimedes knew about this projection (though he wasn't using it for map-making) and proved that, if the cylinder's radius is equal to that of the sphere (so that, formally, the cylinder is tangent to the sphere at its equator), the transformation preserves area - that is, the area of any feature on the sphere is equal to that of its depiction on the cylinder. I'm deeply impressed, if he did manage to prove this, because he'll have had to have done his proof without using infinitesimals (even if he used them in the heuristics that lead him to the proof), because ancient Greek mathematics/philosophy had a big hang-up about them. What Archimedes certainly did prove was that the whole sphere's surface area was equal to the whole piece of cylinder onto which this projection maps the sphere: when unrolled, this is a rectangle, one side of which is the sphere's circumference, the other is its diameter - a sphere's surface area is π times that of a square on its diameter.

Of course, if you do that to the planet Earth, you get a very big rectangle indeed, so you'll be wanting to scale it down to some more practical size for use as a map. Given that the projection does a certain amount of violence to the geometry, there's no real harm in scaling the map's two directions by different amounts. Doing so is equivalent to using a cylinder which has a different radius from that of the sphere, but then scaling down uniformly. Since scaling either direction down by a large factor preserves the ratios between the areas of features in the map, any map obtained by scaling down the two directions, even by different factors, will at least show features' areas faithfully, albeit with their aspect ratios (i.e. height divided by width) severely messed up in a way that depends on position.

If you use Archimedes' choice of cylinder and apply the same scaling to both directions, you'll get a map that's a little over three times as wide as it is tall - or the other way round, but from here onwards I'll assume you've used the spin axis, so the short direction is North-South, the long one is East-West; and that you're following the orthodox tendency to take the former as `up and down' with the latter as `left to right'. The distance, along a meridian, from one pole to the other is half the length of the equator; if this persuades you that you want the map to be only twice as wide as it is tall, you'll scale East-West down by slightly more than North-South. Alternatively, you might want the map to look æsthetically pleasing, and there's a certain orthodoxy that says that a rectangle looks best if its sides are in `the golden ratio' to one another; that's (1+sqrt(5))/2, about 1.6, so requires you to scale East-West down by even more, compared to North-South. Choice of scalings to apply to the two directions - or, equivalently, choice of cylinder, before scaling down uniformly - is equivalent to chosing the aspect ratio (width / height) of the final map (give or take some uniform scaling).

Now, with Archimedes' choice of cylinder, the equator's geometry is not distorted by the projection; scaling down uniformly will thus represent the equator and its immediate neighbourhood faithfully; everywhere else, the map will exaggerate East-West distances relative to North-South ones; mildly so near the equator but drastically so near the poles. Scaling East-West distances down more drastically than North-South ones (e.g. as indicated above to obtain aspect ratios of 2 and the golden ratio, rather than Archimedes' π) will have the effect of exaggerating North-South distances, relative to East-West ones, near the equator; near the poles, East-West will still be exaggerated relative to North-South; but at some latitude in between (both North and South) the geometry will be undistorted. I'm not sure what aspect ratio Peters chose, but uses a map with an aspect ratio (376 to 233) suspiciously close to the golden ratio. That puts the undistorted latitudes at 44 degrees; an overall aspect ratio of 2 would make 37 degrees the undistorted latitudes.

Mathematics of the projection

To do this, we need to relate spherical and cylindrical co-ordinates to one another, so that we can relate the sphere to the cylinder neatly.

Define cylindrical co-ordinates [z, R, φ] with: z varying parallel to your axis and zero in the plane of the sphere's equator; R being distance from the axis; and φ being longitude, i.e. pick some radius of the sphere, in the plane of the sphere's equator, project the point, whose co-ordinates you're measuring, parallel to the axis onto that plane, connect this projected point to the centre of your sphere and measure the connecting line's angle with your picked line. This allows z to take arbitrary positive or negative values; R must be positive, but is otherwise arbitrary; and φ may be any angle, so its range is of width one turn (a.k.a. 2.π radians or 360 degrees), and I'm not fussy whether that range is from zero to one turn, from minus half a turn to plus half a turn, or what.

Define spherical co-ordinates [r, θ, φ], using the same φ, with r as distance from the centre of the sphere and θ being the angle between the equatorial plane and the line from the sphere's centre to the point whose co-ordinates you're after, measured as positive for points with positive z-values, negative for those with negative z-values.

We can express the former in terms of the latter, allowing that φ is the same in both systems, by

or, going the other way,

(using sqrt as the square root function). This gives us

dθ / radian
= d(arcTan(z/R)) / radian
= d(z/R) / (1 +z.z/R/R)
= R.R.(dz/R -z.dR/R/R) / (R.R +z.z)
= ( -z.dR) / (r.r)

and r.dr = + R.dR turns this into

= ( -z.r.dr/R / r / r
= (dz -z.dr/r) / R

Our sphere is a surface of constant r; Archimedes' cylinder tangent to its equator is a surface of constant R. The surface element on the sphere is r.dθ.r.Cos(θ).dφ/radian/radian and, on the sphere (so dr = 0), dz is r.Cos(θ).dθ/radian, so this is justφ/radian. Projecting this surface element onto the cylinder preserves z and φ (but changes R, r and θ) to give us an area dz.R.dφ/radian. Since R on the cylinder and r on the sphere are equal, this is equal to theφ/radian we had before projection: hence the projection preserves area.

Now, in the process, we've streched the φ-wards dimension of the element from r.Cos(θ).dφ/radian to R.dφ/radian, i.e. by a factor of 1/Cos(θ) and shrunk its z-wards dimension from r.dθ/radian to = r.dθ.Cos(θ)/radian, i.e. by a factor Cos(θ). Thus we exaggerate φ-wards distances by a factor of the square of Cos(θ) relative to z-wards ones. That gives a map with an aspect ratio of π:1; to get one with an aspect ratio q:1 requires scaling φ-wards down by a factor q/π times the scaling applied to z-wards; this will give undistorted geometry at latitude arcCos(sqrt(q/π)), as long as q < π. The same effect could have been achieved by projecting onto a cylinder with circumference q times the diameter of our sphere, then scaling uniformly; this cylinder cuts the sphere at latitude arcCos(q/π). For the golden ratio, this is 59 degrees (North and South); for an aspect ratio 2 to 1, this is 50 degrees.

Not only is the projection area-preserving (a detail which I believe Archimedes knew, albeit he wasn't using it in map-making); it is also easy to visualise (as cylindrical projection of a sphere) and produces a `flat' image of the sphere (by unrolling the cylinder). What still fascinates me is how Archimedes (without using infinitesimals) proved that it was area-preserving. Its use in map-making is, I am told, credited to Lambert's work in 1772, though it's commonly known as the Peters projection (let me guess, someone called Peters introduced the anglophone world to it).

There's a picture illustrating this projection at but I haven't yet asked the site's maintainers whether they're fussy about me using it as the SRC of an IMG tag ... yes, lawyers have managed to use copyright law to make even the web into a minefield.

While there's no real substitute for an actual (roughly) spherical map (I have a beach-ball which has a map of the world printed on it, though the writing isn't quite lined up right with the map), another quite illuminating view of our home planet is due to Richard Buckminster Fuller.

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