How many sheets an atlas needs

Every argument about which projection a map series should use assumes the series exists. Before that decision there is an earlier one that nobody frames as a projection question at all: how many sheets.

It is a projection question, it has an answer in closed form, and the answer says that the sheet count is settled by the tolerance long before anybody chooses a projection — and then multiplied by the projection anyway, by a factor of up to fifty-six.

The first half is a theorem #

Chebyshev’s criterion says that among all conformal projections of a region, the one with least scale variation is the one whose scale factor is constant on the boundary. For a spherical cap that projection has a name — the stereographic projection centred on the cap — and its scale spread has a closed form:

$$\frac{k_{\max}}{k_{\min}} \;=\; \sec^2\!\left(\frac{\rho}{2}\right)$$

where $\rho$ is the cap’s angular radius. That is a bound on every conformal projection of the cap, not a property of one of them, which is what Chebyshev’s criterion is about and what makes this arithmetic a lower bound rather than an estimate.

Inverting it costs one line. A stated tolerance $\tau$ on the scale error gives

$$\rho \;=\; 2\arccos\frac{1}{\sqrt{1+\tau}}$$

and for small $\tau$ that is $\rho \approx 2\sqrt{\tau}$: the radius goes as the square root of the tolerance, so the cap’s area goes as the tolerance itself.

The second half is a covering problem #

A sphere of area $4\pi$ covered by caps of area $2\pi(1-\cos\rho)$ needs at least $4\pi / 2\pi(1-\cos\rho)$ of them, and that number is achievable by nothing: discs overlap, and a disc is not a hexagon.

The thinnest covering of the plane by equal discs is the hexagonal arrangement, at a density of $2\pi/\sqrt{27} = 1.2092$ — every point covered 1.21 times over on average. The sphere’s covering numbers are not exactly the plane’s at any finite count, so both figures are carried: the ideal, which is a bound, and the hexagonal, which is what an arrangement achieves.

Covering, not packing. The two are different problems with confusingly similar constants — the hexagonal packing density is 0.9069, which is how much of the plane discs fill without overlapping — and using the packing number here would be asking for sheets that leave gaps between them.

The exponent is the whole of the first result. The count goes as $\tau^{-1}$, fitted at −0.989 across three decades, because the cap’s area goes as $\tau$ and the count goes as one over the area. Ten times the precision is ten times the atlas, and no cleverness anywhere in cartography changes that.

Against a real series #

The International Map of the World, agreed in 1913 and never finished, laid the world out in sheets of 4° of latitude by 6° of longitude at 1:1,000,000. A 4° × 6° cell at mid-latitude fits inside a cap of about 2.9° radius, which corresponds to a tolerance of one part in 1,560.

The arithmetic above predicts 1,888 sheets at that tolerance. The published series ran to roughly 2,500, with the excess accounted for by the two things this calculation deliberately leaves out — sheets are rectangles rather than discs, and the scheme merges cells above 60° instead of shrinking them.

Landing within a third of a real answer from two closed forms and no data is the check worth having. It says the tolerance-to-count chain is measuring the thing the sheet layout was actually constrained by, and not some idealisation of it.

The projection multiplies it, and the multiplier is not small #

The essay was written expecting to show that the projection choice is a rounding error beside the tolerance. That is false, and the machinery refused it.

Fifty-six times the sheets at the same tolerance is not a detail. At one part in a thousand it is the difference between 159 sheets over the region and 8,838.

So the count has two factors and they multiply:

$$N \;\approx\; \frac{1.209 \times 4\pi}{2\pi\left(1-\cos\rho(\tau)\right)} \;\times\; \frac{\epsilon_{\text{chosen}}}{\epsilon_{\text{optimal}}}$$

The first is arithmetic on the tolerance and nobody can improve it. The second is entirely the cartographer’s, it is available for free, and it is the one every argument about projections is actually about.

Why the second factor is so large #

A conformal projection’s scale spread over a region is set by how far the region is from where the projection is exact. The stereographic projection centred on the cap is exact at its centre and the spread grows as $\rho^2$; Mercator over the same cap at 45° north is exact at the equator, forty-five degrees away, and the spread over the cap is dominated by the variation of $\sec\varphi$ across it rather than by the cap’s own size.

That is the same mechanism where the worst point is measures: on a conformal map the extremes are always on the boundary, so the spread is a statement about the boundary’s distance from the point of exactness.

The same arithmetic for a country #

A national series covers a region rather than a sphere, so the covering runs over an area rather than over $4\pi$, and the numbers become recognisable.

Britain is about 244,000 square kilometres. At one part in a million — a tolerance a survey control network works to — a sheet may be 12.7 kilometres in radius and the country needs 579 of them. At one part in a hundred thousand a sheet may be 40 kilometres and 58 will do. At one part in ten thousand, six.

The last row is the one that explains national grids. A country the size of Britain fits inside a single sheet at one part in a thousand, so a national mapping agency that wanted only that would need no zone system, no seams, no sheet index and no edge-matching — which is exactly what a national grid is, and is why designing a grid for one region finds the gain over an international zone smaller than the argument sounds. The gain in distortion is 1.51; the gain in administration is the whole of the difference between one sheet and sixty.

Conversely, the 579-sheet row is why survey control is not published as a map at all. At the tolerance a control network needs, no map covers a country, and the answer is a coordinate list plus a projection formula rather than a series of sheets.

The tolerance is not a technical parameter #

The uncomfortable part of this arithmetic is what it does to the word “accuracy”. A series specified at one part in ten thousand is not a hundred times better than one at one part in a hundred; it is a hundred times larger, and the reader of a single sheet cannot tell which they are holding.

The trade is stated cleanly by the two exponents. Sheet radius goes as $\tau^{1/2}$ and sheet count as $\tau^{-1}$, so halving the sheet radius quadruples the atlas. A national series at 1:25,000 with a 10-kilometre sheet is not a decision about paper size; it is a decision that the country will be covered by some thousands of sheets rather than some tens.

Which is why the sensible question, and the one this arithmetic makes askable, is the inverse: given how many sheets a programme can afford to print and maintain, what tolerance does that buy? That is a division rather than a search, and it is the number that should be in the specification.

The bound is for conformal projections only, and that is a restriction #

Chebyshev’s theorem is a statement about the conformal projections of a region. It says nothing about the others, and the others are not obviously worse: an equal-area projection of a small cap has an areal factor of exactly one everywhere, which is a better result than any conformal projection manages about area.

What it does not have is a single scale factor. A conformal projection’s error can be stated as one number per point, so a tolerance on it is a tolerance on everything a sheet is used for; an equal-area projection has two different scales at each point and a tolerance has to name which. That is the distinction the two ways a map is wrong makes, and it is why sheet specifications are written in terms of scale error and therefore in terms of conformal projections.

So the count above is the count for a series whose sheets are meant to be measured on with a ruler in any direction. A series meant for area — a cadastral or a land-cover product — is a different optimisation with a different bound, and this site does not have the equal-area analogue of Chebyshev’s theorem because there does not appear to be one.

What this shares with the grid essays #

Designing a grid for one region asks a version of this question with the answer fixed at one sheet: given a country, how much distortion does a single projection impose, and is a national grid worth having over an international zone. The answer there was a factor of 1.51, which is real and much smaller than the argument is usually made to sound.

This is the same trade with the sheet count as the free variable rather than the distortion. A zone system is a covering: UTM and the zone system is sixty caps of a particular shape, chosen so that the worst scale error inside one is 1,000 parts per million, and sixty is what that tolerance costs at that shape.

The two schemes differ in shape — zones are gores and sheets are discs — and the arithmetic differs only in the covering density that shape implies. Gores tile exactly, with density 1, and pay for it by being long: the worst point in a zone is at its corner, far from the central meridian, so the tolerance a gore achieves is worse than a disc of the same area would.

The flat-Earth version of the same question #

There is a coarser calculation a builder makes that has the same shape and a different exponent, and putting them side by side says what the sheet count is actually a statement about.

The flat-Earth radius and the sheet radius obey the same law because they are the same quantity: how far a locally correct description survives. The difference is what is being described. A plane is correct at a point and wrong quadratically; a conformal projection is correct at a point and wrong quadratically; and the sheet count is the flat-Earth question asked over a whole sphere with the answers counted rather than quoted.

That also gives the honest reading of “the best possible projection”: it does not remove the second-order term, it moves the point where the term vanishes to the middle of the sheet. Chebyshev’s projection is the one that spreads the residual most evenly, and evenness is what makes a single tolerance describable at all.

What is deliberately not computed here #

The sheets’ shape. A real series uses rectangles in some grid, because sheets have to be printed, stored, indexed and joined, and a disc does none of those things. The covering density used here is the one a disc arrangement achieves and a rectangular scheme’s is different; the calculation is a statement about the projection’s limit rather than about the cartographer’s layout.

The overlap. Real sheets overlap deliberately so that a feature near an edge appears whole somewhere, which adds to the count by a factor that is a policy rather than a geometry.

Anything about content. The number of sheets a survey can afford is a question about money and staff, and the only thing this arithmetic contributes to it is the exponent — which is exactly the part that is not negotiable.

Interruption. A world map may be cut into lobes instead of into sheets, which reduces the distortion by a measurable factor and pays in continuity rather than in count. Giving up continuity measures that trade, and the two are alternatives: a lobed map is a covering whose pieces are joined along some edges and not others, and the sheet count above is the limit of cutting everywhere.

What a screen map does with the same arithmetic #

A tiling scheme is a covering too, and it answers the question the other way round: it fixes the sheets first and lets the tolerance fall where it may.

A tile at zoom $z$ covers $1/4^z$ of a square world, so the count is a power of four and the sheet radius halves at every level. Substituting into the law above, the scale tolerance a tile achieves improves by a factor of four per level — which is why a screen map is a pyramid of tiles never states a tolerance: it has a different one at every level, and the reader chooses the level.

That is a genuinely different design and it is worth naming the trade. A paper series picks a tolerance and pays a sheet count. A pyramid picks a sheet count per level and inherits whatever tolerance the level gives, which is why the top of the pyramid is a map nobody could measure anything on and the bottom is one nobody could print.

The law is linear, and that is the cheap direction #

The bound has a shape worth reading on its own, because it says which of an atlas’s two demands is the expensive one.

The count runs as $1/\tau$: one thousand and one sheets at one part in a thousand, ten thousand and one at one part in ten thousand, one hundred and one at one part in a hundred. Halving the tolerance doubles the sheet count and does nothing worse than that.

Now set that beside the other demand a series can make. A sheet of fixed paper size at scale $s$ covers ground area proportional to $1/s^2$, so doubling the scale — 1:50,000 to 1:25,000 — quadruples the number of sheets. That is the familiar arithmetic of a national series, and it is the reason a large-scale series runs to thousands of sheets while a road atlas runs to dozens.

So the two demands have different exponents, and the accuracy demand has the smaller one. Asking for twice the fidelity costs half as much as asking for twice the detail, measured in sheets, and the gap widens at every further doubling: four steps of tolerance is a factor of 16, four steps of scale is a factor of 256.

That inverts the intuition the words carry. Twice as accurate sounds like the severe request and twice as large a scale sounds like a routine one, and in sheet count it is the other way round.

The reason is that the two are demands on different things. Scale is a demand on how much of the sheet the ground occupies, which is an area, so it scales as a square. Tolerance is a demand on the scale factor’s spread over a cap, and the spread goes as the square of the cap’s radius — so inverting it gives a radius proportional to $\sqrt{\tau}$, an area proportional to $\tau$, and a count proportional to $1/\tau$. The square that appears in one law is the same square that cancels in the other.

None of this makes 1,001 sheets a small number. It does mean that a series which has already accepted a sheet count for its scale has room to be much more accurate than it usually is, at a price it has already shown itself willing to pay for something else.

Where this ladder goes #

The choosing field has now measured what a projection minimises, what a compromise buys, what a region does to a ranking, what a line does to it, what an aspect is worth, what interruption costs, and — here — what a tolerance costs before any of those are decided.

What remains is the case where the region is not a cap and not a country but a set of regions with different requirements, which is the atlas problem proper: not how many sheets, but which sheets, at which scales, with which projections, chosen together. That is an optimisation with a discrete part, and this site does not have it.

What it does have now is the floor: no arrangement of any conformal projections covers the world at one part in a thousand in fewer than 1,001 sheets, and the number is a theorem rather than an estimate.