What a Flat Map Costs You

Evidence this block produces: a written brief for one real map you would actually make — region, purpose, and the one question a reader must be able to answer from it.

The decision you're actually making

Almost everyone approaches projections backwards: they ask “which projection is most accurate?” The question has no answer, and the reason it has no answer is Week 2's threshold. The question that does have an answer is: given what this map is for, which distortion can I afford?

Snyder, whose 1987 USGS manual is the closest thing the field has to a standard reference, opens with exactly this. He writes that since a flat map cannot be made without distortion, the cartographer must choose what to show accurately at the expense of everything else, or accept a compromise across several properties Sourced fact S2. And he is blunt that it cannot be said there is one best projection, and that it is even risky to claim you've found the best one for a given application unless the parameters have been artificially constricted Sourced fact S2.

Note what that second sentence does. It's not modesty. It's a warning that “best for this application” is a claim that smuggles in a definition of best, and that you should be suspicious of your own version of it. Hold onto that; it becomes Threshold 3.

The five characteristics

Snyder lists the properties a projection can be built to preserve Sourced fact S2:

Two traps live in that table, and both catch experienced people.

Conformal does not mean shapes are right. Snyder is careful: conformality applies on a point or infinitesimal basis, while equal-area applies on a finite, map-wide basis Sourced fact S2. Mercator is perfectly conformal and Greenland is still monstrous. Nothing is broken. Continent-scale shape is not a local property, so a local guarantee says nothing about it.

Equidistant does not mean distances are right. No projection preserves all distances — equidistant projections hold true scale along selected lines, usually meridians from a centre Sourced fact S2. “Azimuthal equidistant” means distances from the centre point are true. Distances between two arbitrary points on it are not.

Your brief

Pick a real map. Not a hypothetical — a map you'd plausibly make, because you'll be living with it for six weeks. Something with a region and a job to do: a county's school-district boundaries, migration paths for a family line across two centuries, service coverage for a set of sites, a walking route.

Write three lines. Region and its rough latitude span. Purpose in one sentence. The one question a reader must be able to answer correctly from the map — and be specific, because “understand the area” is not a question.

Mark this block done

Evidence this block produces: one number — the area inflation your default web map applies at your region's latitude — computed by hand and checked against the bench.

Worked example — fully scaffolded

This week you copy a worked example exactly, changing only the latitude. The scaffolding comes off in Week 3 and is gone by Week 5; if you look back at this block in six weeks and it feels patronising, the fading worked.

Take a region centred at 42°N (roughly Chicago, Rome, or northern Utah).

1. cos 42° = 0.7431
2. sec 42° = 1 / 0.7431 = 1.3456
3. So h = k = 1.3456. Lengths are 34.6% too long in every direction.
4. Area scale = 1.3456² = 1.811. Areas are 81% too large.
5. Angular deformation ω = 0, because h = k. Mercator is conformal; that's the whole trade.

Now do it for your own region's centre latitude. Then set the bench to Mercator and that latitude and confirm all four readouts match your arithmetic. If they don't, it's almost always a degrees-versus-radians error in the calculator, which is a mistake you want to make now rather than in Week 6.

Mark this block done

Evidence this block produces: Decision Memo v0 — complete, one page, and knowingly bad.

Five sections. Write all five even where you have nothing yet; the empty ones are the syllabus for the next five weeks.

1. Brief. Region, purpose, the one question a reader must answer. From the last block.
2. Choice. Which projection, which aspect, which standard parallel(s) or central meridian. You don't know how to choose yet — choose anyway, and write down why you chose it, including “it's what the tool defaults to.” That sentence is data about how most maps get made.
3. Distortion table. Three points: your region's centre and its two most extreme corners. Columns for h, k, area, and ω. Fill them in from the bench. This week you can only do it honestly for the seven projections in the bench — if your choice isn't one of them, use the closest cylindrical and write “approximation” in the margin.
4. Rejected alternative. Name one other projection you could have used, and one sentence on why not. It will be a weak sentence. Leave it weak.
5. The cost. One sentence: what this map will systematically mislead a reader about. Every map has one. If you can't write this line, you don't yet know what you chose.

Mark this block done

Evidence this block produces: a one-paragraph written statement, in your own words, of why a cylinder can be unrolled flat and a sphere cannot.

Two kinds of curvature

Roll a sheet of paper into a tube. Nothing on the paper changed: distances between printed points are the same, angles are the same, a drawn circle is still a circle. Now try to wrap that same sheet around a ball. It creases, or it tears. The difference is not about effort or dexterity — it's a mathematical fact about the two surfaces.

Gauss's Theorema Egregium of 1827 says that a surface's Gaussian curvature can be determined entirely from measurements made within the surface — angles, distances, and their rates of change — without any reference to the surrounding space it sits in Sourced fact S1. Gauss's own statement of the consequence: if a curved surface is developed upon any other surface whatever, the measure of curvature at each point remains unchanged Sourced fact S1.

The payoff for mapmakers is immediate. A plane has Gaussian curvature 0 everywhere. A sphere of radius R has curvature 1/R² everywhere, which is not 0. Since curvature is preserved by any distance-preserving map, and these two curvatures differ, there is no distance-preserving map from any patch of sphere to any patch of plane — not for the whole Earth, not for a continent, not for a city block Sourced fact S1.

A cylinder, by contrast, has Gaussian curvature 0 — intrinsically it is as flat as a plane, which is exactly why the paper rolls without complaint Sourced fact S1. This is why cylinders, cones and planes are called developable surfaces, and why they are the three shapes projections are built around Sourced fact S2.

The word “impossible” is doing real work

Snyder states the consequence flatly in his list of properties: no map can be both equal-area and conformal Sourced fact S2. Not “no known map,” not “no practical map.” None, ever, for any region however small. There is no research programme that will produce one, and a claim to have found one is a claim to have refuted Gauss.

Do the paper experiment if you haven't. Take an orange, or a tennis ball, and a sheet of paper, and spend a minute genuinely trying. The tactile version of this is not a teaching gimmick; the feeling of the paper refusing is closer to the theorem than the sentence is.

Mark this block done

Evidence this block produces: a written four-step proof, in your own hand, that conformal and equal-area are mutually exclusive.

You need one fact from Week 3, stated now and earned later: at each point, a projection turns an infinitesimally small circle on the globe into an infinitesimally small but perfect ellipse on the map, with semi-axes a and b Sourced fact S2. That ellipse is Tissot's indicatrix, and it is the complete local description of what the projection does.

The argument

1. Conformal means the indicatrix is a circle. If local angles are preserved, the scale factor at a point must be the same in every direction — stretch more one way than another and you have bent the angles between them. So a = b. Snyder states this directly: on a conformal projection the ellipse is a circle, an ellipse of zero eccentricity Sourced fact S2.
2. Equal-area means the indicatrix has unit area. The ellipse's area is proportional to ab. Preserve areas everywhere and ab = 1 at every point.
3. Both at once forces an isometry. If a = b and ab = 1, then a² = 1, so a = b = 1. Every infinitesimal circle maps to an identical circle: every length, in every direction, at every point, is preserved. That is precisely the definition of a distance-preserving map.
4. Gauss forbids it. A distance-preserving map preserves Gaussian curvature. Sphere: 1/R². Plane: 0. Contradiction Sourced fact S1. Therefore no projection is both. Inference — the four-step chain is my assembly; each link is sourced, the chain is not quoted from anywhere.

Write it out yourself, longhand, without looking. Not because writing is virtuous, but because this specific argument is short enough to feel understood while reading and is routinely not.

Why this changes the job

Before: choose the most accurate projection. After: choose which of area, angle, and distance to sacrifice, in which regions, by how much. The second question is answerable, has a professional literature, and has numbers in it. The first is a category error.

It also reframes every argument you'll ever see about a map being “wrong.” A map is not wrong for distorting. A map is wrong when its distortion undermines the question it exists to answer — and that's a judgement about fit between distortion and purpose, which is why your brief had to name a question. Inference

Mark this block done

Evidence this block produces: two answered retrieval items, and a corrected sentence 5 in your memo.

Then go back to your memo and rewrite section 5 — the cost. Last week it was probably a hedge. Now it should name a specific property you gave up and a specific reader who could be misled by it.

Mark this block done

Evidence this block produces: a filled-in table of h, k, ab and ω at 0°, 30°, 60° for two projections, computed by hand and reconciled with the bench.

The indicatrix

Tissot published his analysis of projection distortion in 1859 and 1881 Sourced fact S2. Inference The reason it survived 150 years is that it turns a global, qualitative complaint into a local, quantitative one.

The claim is exact. An infinitely small circle on the Earth projects as an infinitely small but perfect ellipse on any map projection Sourced fact S2. Not approximately an ellipse. Any smooth map, zoomed in far enough, is a linear transformation, and linear transformations take circles to ellipses. Everything local about a projection is contained in that ellipse's two semi-axes, a and b.

From them you read off everything:

The ω formula is Snyder's equation 4-1 Sourced fact S2. Sanity-check it before trusting it: set a = b and the numerator is 0, so ω = 0 — conformal, as it must be. That two-second check is the habit; do it to every formula you're handed.

A note on provenance, since this course keeps insisting on it: I have read Snyder's presentation of Tissot, not Tissot. The 1881 Mémoire is a source I am citing at one remove S4, which is exactly the amplification failure this course warns about — disclosed here rather than laundered into a confident citation.

h and k: the practical shortcut

Finding a and b in general is work. But for the normal aspect of cylindrical, conic, and polar azimuthal projections, meridians and parallels cross at right angles and line up with the ellipse's axes — so the two scale factors you can compute directly are a and b Sourced fact S2:

• h — scale factor along the meridian (north–south)
• k — scale factor along the parallel (east–west)

Snyder: when the meridians and parallels coincide with a and b, ω may be calculated from equation 4-1 substituting h and k, and the scale factor in any other direction falls between them Sourced fact S2. That last clause is a free bound: h and k bracket every direction. For a general projection the computation of ω is much more complicated Sourced fact S2 — which is precisely why this course sticks to seven projections in their normal or polar aspects.

For any cylindrical projection given by x as a function of longitude and y as a function of latitude, on a sphere of radius R, Snyder's equations 4-2 and 4-3 are Sourced fact S2:

Read them as ratios, which is all they are: h compares how far you move on the map when you move a degree north to how far that is on the globe. k does the same going east — and the cos φ is there because a degree of longitude on the globe shrinks as you go poleward. That factor is the entire source of cylindrical distortion.

Mark this block done

Evidence this block produces: three derivations of h and k — one copied, one completed, one unaided — each checked numerically against the bench.

The scaffolding fades inside this single block. Notice it happening.

1. Fully worked: plate carrée

The simplest projection there is: x = Rλ, y = Rφ. Longitude becomes horizontal position, latitude becomes vertical position, in radians.

So h = 1 always — scale along meridians is perfect, which is why it's also called equidistant cylindrical — and k = sec φ. Areal scale hk = sec φ, so it is not equal-area. And h ≠ k away from the equator, so it is not conformal either. It is the projection that gives up both, which is why Goldberg and Gott use it as the baseline everything else is normalised against Sourced fact S6.

At 60°: h = 1, k = 2, area = 2, and sin(ω/2) = (2−1)/(2+1) = 1/3, so ω = 2 · arcsin(0.3333) = 38.94°. Check it on the bench.

2. Completion problem: Lambert cylindrical equal-area

Now the same for x = Rλ, y = R sin φ. Two of four steps are done for you.

The third line is the point: if you've done it right, the φ cancels and you get exactly 1, at every latitude. That's what “equal-area” means, made mechanical. Verify at three latitudes on the bench before moving on.

3. Unaided: Gall–Peters

Gall–Peters is the same cylindrical equal-area idea with its standard parallels moved to 45°: x = Rλ cos 45°, y = R sin φ / cos 45°. No steps given. Find h, k, and hk, then predict — before computing — what happens at exactly 45°, and why anyone would want that.

Then check on the bench. If hk didn't come out to 1, the arithmetic slipped; the cos 45° appears once in each factor and cancels.

Mark this block done

Evidence this block produces: a three-line verification table confirming, or refuting, three published claims from a 2021 peer-reviewed paper — using only your own arithmetic.

Auditing a paper with a pocket calculator

Gott, Goldberg and Vanderbei describe the behaviour of three azimuthal projections at the rim of a double-sided disk map — the rim being the equator, 90° from the pole at the centre. They state, in prose, three specific things Sourced fact S6:

1. On the stereographic version, areas at the edge are four times enlarged relative to the centre.
2. On the Lambert equal-area version, local shapes at the edge are distorted 2:1.
3. On the azimuthal equidistant version, areas at the edge are π/2 larger than at the centre, and shapes are π/2 : 1 elongated.

You can now check all three yourself. Set the bench to each projection in turn, drag latitude to 0° (the rim), and read off the areal scale hk and the elongation k/h.

Fill in the table. All three claims hold exactly — 4.0000, 2.0000, and 1.5708 twice, which is π/2 to four decimals.

Sit with what you just did. You verified quantitative claims in a peer-reviewed astrophysics paper from first principles, in about fifteen minutes, with a slider and arithmetic. Inference That capability is the actual product of Threshold 2: not knowing what Tissot said, but being unable to be told a distortion claim you can't check.

Now fill in your memo's section 3 properly. Three points, four columns, real numbers. That table was empty in Week 1. It is the difference between an opinion and a memo.

Mark this block done

Evidence this block produces: a one-page decision table in your own words, mapping purposes to properties, with the failure mode of each written next to it.

The four families

Snyder's manual ends its general section with a selection list organised by region and purpose rather than by mathematics Sourced fact S3. Some entries worth internalising, because they encode decades of practice:

• Region with predominant east–west extent, away from the equator: Lambert conformal conic if you need angles, Albers equal-area conic if you need areas Sourced fact S3. This is most mid-latitude countries and US states.
• Predominant north–south extent: transverse Mercator, or transverse cylindrical equal-area Sourced fact S3. Chile, Norway, Vietnam.
• Roughly equal extent in all directions: an azimuthal — stereographic for conformal, Lambert azimuthal equal-area otherwise, in whatever aspect centres your region Sourced fact S3. Africa, Australia, an ocean basin.
• Straight rhumb lines: Mercator, and only Mercator Sourced fact S3.
• Straight great-circle routes: gnomonic, for less than a hemisphere Sourced fact S3.

Notice the structure of that list. The first question is never “which projection” — it's the shape of the region, then the property. Two questions, in that order, and the projection falls out. Inference

The compromise family is not a cop-out

It's tempting to read “preserves nothing exactly” as failure. But if your map has no single dominant quantitative job — a wall map, a reference map, an illustration — then optimising one property to exactness while letting another run to infinity is the worse choice. This is the reasoning behind the professional recommendation to abandon rectangular world maps in favour of minimum-error or compromise projections Sourced fact S12 Contested — and you'll meet the objection to that recommendation in Week 6.

Mark this block done

Evidence this block produces: eight briefs classified, each with a one-line justification, and at least one you got wrong and can say why.

Mark this block done

Evidence this block produces: two retrieval items from earlier weeks answered cold, and section 4 of your memo — the rejected alternative — rewritten with numbers in it.

Memo work. Section 4 was one weak sentence in Week 1. Rewrite it: name the alternative, give its hk and ω at your three points, and reject it on a number rather than a preference.

Mark this block done

Evidence this block produces: a sketch, with labels, of your region under tangent and secant placement, marking where scale is exactly true in each.

Tangent and secant

A cylinder wrapped around the globe touches it along one line — the equator, in the normal aspect — and scale is true only there. But the cylinder or cone can instead cut the globe at two parallels rather than touching at one, which conceptually gives two standard parallels Sourced fact S2.

The consequence is worth deriving rather than accepting. Tangent placement puts zero error at one line and lets error grow monotonically away from it — so across a region the error is 0 at one edge and maximal at the other. Secant placement puts the true-scale lines inside the region, so scale is slightly too small in the middle and slightly too large at both edges. The maximum absolute error is roughly halved. Inference — and one you can check on the bench in ninety seconds.

Do check it. Take a band from 30° to 60°. Compare the Lambert cylindrical equal-area (standard parallel 0°, tangent, far from the region) with Gall–Peters (standard parallels 45°, sitting right in the middle of it), reading k at both 30° and 60°:

Same family, same guarantee, identical equal-area property — and less than half the worst-case linear error, purely from moving a parameter. Snyder makes the same point generally: by choosing the locations of the true-scale lines properly, scale errors elsewhere may be minimised Sourced fact S2.

Aspect

The axis of the cylinder or cone need not match the Earth's axis, and the plane need not touch a pole. Tilting them produces the oblique, transverse and equatorial aspects, in which meridians and parallels are generally no longer straight lines or circular arcs; what were standard parallels become standard lines that don't follow parallels at all Sourced fact S2.

This is the single highest-leverage move in the course. A transverse Mercator is a Mercator rotated 90°, so its line of true scale runs north–south along a chosen central meridian instead of east–west along the equator — which is why it, not Mercator, is the standard for north–south countries and for national grid systems Sourced fact S3.

The Universal Transverse Mercator system takes this to its logical end: divide the world into zones narrow enough that distortion within any one zone stays below a preset level Sourced fact S2, and apply a slight scale reduction along the central meridian — a factor of 0.9996 — so each zone is secant rather than tangent, splitting the error between the middle and the edges Sourced fact S16.

That's the whole idea of this week, industrialised: choose the family for the property, then spend your remaining freedom on aspect and standard lines.

Mark this block done

Evidence this block produces: a chosen standard parallel for your own region, with the worst-case error before and after, in writing.

No worked example this week. You have the tools; use them.

1. Take your region's latitude span, φ_south to φ_north.
2. For a cylindrical equal-area with standard parallel φ_s, the parallel scale factor is k = cos φ_s / cos φ. Confirm this against the bench by checking that Gall–Peters, where φ_s = 45°, reports k = 1 at exactly 45°.
3. Compute the worst |k − 1| across your band for φ_s = 0.
4. Now find the φ_s that minimises the worst error across your band. You can do this by search — try five values and take the best — which is what minimax optimisation looks like before anyone dresses it up.
5. Write down both worst-case numbers and the ratio between them. That ratio is what the parameter was worth.

Mark this block done

Evidence this block produces: a four-candidate comparison table for your own brief, with a defended shortlist of two.

Build the table you'd hand to a colleague. Rows: four candidate projections, chosen using Week 4's region-shape-then-property procedure, each with a specified aspect and standard line(s). Columns: hk and ω at your region's centre and at its two worst corners, plus one column for “what this costs the reader.”

Then cut it to two, in writing, on the numbers. Keep the cut ones visible — a shortlist that shows its rejects is a defensible artifact; one that shows only the winner is an assertion.

If your region needs a conic and the bench doesn't have one: the bench is deliberately limited to projections with exact closed forms you can verify by hand. Use the closest cylindrical or azimuthal as a bound, note the substitution in the table, and say which direction the real numbers would move. Naming a gap beats quietly filling it with something that looks like a number.

Mark this block done

Evidence this block produces: a written statement of the error measure, region, and weighting under which your Week 5 shortlist winner is optimal — and one alternative weighting under which it isn't.

Optimality needs three inputs before it means anything

The classical result here is Chebyshev's: the best conformal map of a closed region of the sphere — in the sense of minimising the sum of the squares of the scale errors — is one that has constant scale along the region's boundary Sourced fact S15.

Read the qualifiers rather than the conclusion. Best is defined only after fixing: the class of maps considered (conformal ones), the error measure (sum of squared scale errors), and the region (a specific closed boundary). Change any one and the theorem's answer changes. This is not a weakness of Chebyshev's result — it's what a well-posed optimality claim looks like, and it's the template for reading every “best map” claim you'll ever see. Inference

A modern attempt at a single score

Goldberg and Gott proposed six error measures for whole-world maps: isotropy, area, flexion (the bending of great-circle routes on the map), skewness (lopsidedness), distances, and boundary cuts Sourced fact S6. The first two are Tissot's a and b in disguise — isotropy is the RMS of ln(a/b), area the RMS of ln(ab) relative to its mean Sourced fact S6 — so Threshold 2 is literally the first two columns of the modern state of the art. The other four require first derivatives of the metric and are genuinely new.

Each error is normalised against the equirectangular projection and the squares are summed, so the equirectangular scores exactly 6 and lower is better Sourced fact S6:

Values as published Sourced fact S6. Read the Mercator row across: perfect isotropy, worst area error in the table, worst total. Every column is a real property; the total is a choice.

Note the zeros. Mercator scores 0 on isotropy because it is conformal; Eckert IV scores 0 on area because it is equal-area. No row has both zeros, and after Week 2 you know no row ever can.

Mark this block done

Evidence this block produces: a written position on each of three contested cases, each naming what would change your mind.

Case 1: the score that changed its mind

The preprint of the Goldberg–Gott paper concluded, in its v1 abstract, that the Kavrayskiy VII projection beat the popular Winkel Tripel in every category except boundary cuts, where they tied, and had the lowest normalised errors of all projections studied Sourced fact S7. The published version and every later citation report the Winkel Tripel as the winner Sourced fact S5, S6.

Same authors, same six measures, same projections, different answer. The authors are open about the mechanism: the combination of the six error terms depends on the choice of normalising projection Sourced fact S6. They also say, of their own later projection, that they tried only one candidate specifically to avoid “teaching to the test” Sourced fact S6 — an admission that the metric is gameable by anyone willing to optimise against it.

This is Threshold 3 in a single example, and it is not a scandal. It is what happens when a multi-objective problem is compressed into one number: the compression is a value judgement wearing a decimal point. Inference

Case 2: Peters

The Gall–Peters projection — a cylindrical equal-area with standard parallels at 45°, which you derived in Week 3 — became the centre of a long public argument about the politics of map design. Peters argued that Mercator was culturally unfair to developing countries, shown smaller than they should be Sourced fact S6. Academic cartographers responded largely by assessing the map as cartographic design rather than debating each claim Sourced fact S11 Contested. In 1989 and 1990, seven North American geographic organisations adopted a resolution rejecting all rectangular world maps — a category containing both Mercator and Gall–Peters — though the North American Cartographic Information Society declined to endorse it Sourced fact S12. Denis Wood later called it an idiotic resolution Sourced fact S13 Contested.

You now have the equipment to hold both halves at once. Peters' quantitative complaint is correct: Mercator's area error is real, unbounded, and systematically favours high latitudes — you computed sec² φ in Week 1. The professional objection is also correct: fixing area by adopting a cylindrical equal-area buys hk = 1 at the price of ω reaching 73.7° at 60° latitude on the tangent version, and shapes shear badly. Both parties were right about their own error term and disagreed about the weighting, which is exactly the structure of Case 1. Inference

Provenance note: I have not read Robinson 1985, the 1989 resolution text, or Wood 2003 directly — they are cited here at one remove S11 S12 S13. Treat the characterisations as second-hand and the mathematics as first-hand.

Case 3: the projection you use every day is broken in a way nobody minds

Web Mercator, EPSG:3857, is what almost every web basemap renders in. It applies the spherical Mercator formulas to ellipsoidal coordinates, taking the sphere radius to be the ellipsoid's semi-major axis. The EPSG registry's own entry for it says plainly: not a recognised geodetic system; uses spherical development of ellipsoidal coordinates; relative to true ellipsoidal Mercator it gives errors of 0.7 percent in scale and northing differences up to 43 km on the map, 21 km on the ground Sourced fact S8.

The consequence that should make you sit up: unlike either the spherical or the ellipsoidal Mercator, this method is not conformal — the scale factor varies with azimuth, which creates angular distortion Sourced fact S9. The single property Mercator exists to provide is the one the web version quietly gave up, in exchange for tile arithmetic that was faster in 2005.

Vendor guidance on this is worth reading with the chip attached: Esri's explanation frames the issue as positional accuracy declining with latitude and reassures that it is unlikely to cause problems unless a coordinate system is mislabelled Sourced fact S10 Vendor framing. That's true. It is also written by a company whose products default to it.

And the practical verdict is still: usually fine. For rendering tiles it is fine. For measuring areas or distances across latitudes it is not, and no amount of it being the default makes it so. Threshold 3 doesn't say all choices are equal; it says the defence has to name the purpose. Inference

Mark this block done

Evidence this block produces: Decision Memo v1 — and the Week 1 version placed beside it.

Same five sections, no scaffolding, no template beyond the headings. What's changed is what each section now has to survive.

1. Brief. Unchanged, unless six weeks taught you the question was wrong. If it did, say so — that's a finding, not a failure.
2. Choice. Family, aspect, standard line(s), each justified separately. Region shape first, property second, placement third.
3. Distortion table. Centre and two worst corners; hk, ω, and the worst |k − 1| across the region. Every number computed and checkable.
4. Rejected alternative. One serious candidate, rejected on a number, with the number.
5. The cost — and the criterion. This is the new part and the whole of Threshold 3. State what the map systematically misleads about, and state the error measure and weighting under which your choice is optimal. Then name one different weighting a reasonable colleague might prefer and what it would select instead.

Then open the Week 1 memo next to it. The distance between them is the course.

Where to go next, if you want to

• The ellipsoid and datums. The largest non-goal here, and the source of more real-world map errors than projection choice ever is.
• Conics properly. Lambert conformal conic and Albers cover most national mapping and were deliberately left out because their scale factors don't reduce to one-line closed forms.
• Snyder's manual end to end. Forward and inverse equations for every projection you'll meet, with numerical examples S2.
• Finite distortion. Tissot is infinitesimal; flexion and skewness are the first derivatives of the metric and catch continental-scale errors Tissot cannot see S6.

Mark this block done