Map Projection Distortion Lab

Greenland covers 2.166 million km² and sits near 74.5° N. Work out what Mercator shows it as. Then compare all three projections at that latitude and find what they have in common.

1. Mercator is built to preserve angles, and that fixes everything else. A degree of longitude spans a ground distance that has shrunk by cos φ along the parallel, so to keep the local shape true the map must stretch east–west by 1/cos φ = sec φ.

   $$\sec \phi =\frac{1}{\cos 74.5^{\circ }}=\frac{1}{0.2672}=3.742$$

2. At 74.5°, cos φ = 0.26724, so sec φ = 3.742. Every east–west distance up there is drawn 3.742 times too long.

3. Angles are preserved only if north–south is stretched by the same factor, so the area is multiplied by sec φ twice: sec² φ = 14.002, which the panel rounds to 14.00.

   $$A_{\text{Mercator}}={\sec }^2\phi =3.742^2=14.00$$

4. Apply it. 2.166 × 14.002 = 30.33 million km², so Mercator draws a patch at Greenland’s mid-latitude at fourteen times its real size. The shape ratio stays 1, so nothing looks wrong at any one point, and the country still bends across its own 24 degrees of latitude because the scale is not the same at its two ends.

   $$2.166\times 14.002=30.33{\text{million km}}^2$$

5. Now the other two at the same latitude. Equal-area keeps the area at 2.166 by squashing north–south to 0.2672 while east–west still stretches 3.742, so its shape ratio is 3.742 ÷ 0.2672 = 14.00. Plate carrée leaves north–south alone: area 3.742, shape 3.742.

   $$\text{equal-area:}\frac{3.742}{0.2672}=14.00\text{(shape)},\text{area}=1$$

6. Multiply each pair. Mercator 14.00 × 1; equal-area 1 × 14.00; plate carrée 3.742 × 3.742. All three are 14.00, and 14.00 is sec² φ — the number we started with.

   $$\text{area}\times \text{shape}={\sec }^2\phi =14.00\text{for all three}$$

At 78.2° the panel puts Mercator’s east–west stretch at 4.89 and its area factor at 23.91. Both are local. Work out where that latitude is drawn on the sheet, then decide how the paper gets shared out.

1. Both printed numbers describe a patch. They say how much a small piece of ground at 78.2° is enlarged, and nothing about where on the sheet that piece is drawn.

   $$\sec 78.2^{\circ }=\frac{1}{0.2045}=4.89,{\sec }^{2}78.2^{\circ }=23.91$$

2. Position is the accumulation of all the stretching below it. A step of dφ north covers a fixed distance on the ground and is drawn sec φ times longer, so the sheet’s y coordinate grows at exactly the rate the panel prints.

   $$dy=\sec \phi d\phi ⟹y(\phi )={\int }_0^{\phi }\sec {\phi }^{\prime }d{\phi }^{\prime }$$

3. That integral has a closed form, and it is the one Mercator’s own construction conceals. At 78.2° it comes to 2.270.

   $$y=\ln \tan \left(45^{\circ }+\frac{\phi }{2}\right);y(78.2^{\circ })=\ln \tan 84.1^{\circ }=\ln 9.677=2.270$$

4. The sheet has to be cut, because y runs to infinity at the pole. Cut at 85°, as this map does, and the half-height is 3.131, so Svalbard’s latitude sits 72.5% of the way up.

   $$y(85^{\circ })=\ln \tan 87.5^{\circ }=\ln 22.90=3.131,\frac{2.270}{3.131}=72.5\%$$

5. Which latitude is halfway up? Invert the same formula at half the height and it is 66.4°, within a fifth of a degree of the Arctic Circle.

   $$y=\frac{1}{2}(3.131)=1.566⟹\phi =2\arctan e^{1.566}-90^{\circ }=66.4^{\circ }$$

6. Now the ground underneath. The area between two parallels goes as the difference of their sines, so the upper half of that sheet, everything north of 66.4°, holds 8.0% of what is drawn. The band from the equator to 30°, which takes 17.5% of the height, holds 50.2%.

   $$\frac{\sin 85^{\circ }-\sin 66.4^{\circ }}{\sin 85^{\circ }}=\frac{0.9962-0.9166}{0.9962}=8.0\%,\frac{\sin 30^{\circ }}{\sin 85^{\circ }}=50.2\%$$