# How do I calculate the area of a specific place on the map?

Technical question warning !

I'm trying to figure out the size of countries and continents but there is something I do not understand. The map I'm using was made with the equirectangular projection (its width is twice as large as its height). I converted the file with G projector into a mollweide projection (equal area) and now the image is an ellipse of 1660 by 830 pixels high. At the equator, each pixel represent 24 km. Each pixel theoretically represent an area of 24 by 24 km or 576 km2.

I used the magic wand in photoshop and it tells me that the map is 1,082,100 pixels (borders excluded). With each pixel representing 576 km2, the total is over 623,000,000 km2. The problem is that my map is supposed to be as big as Earth. Earth is 510,000,000 km2, so it's 22% larger for no reason.

• What I am doing wrong?
• Should I use the % of area covered and then convert it in square kilometers?

Here's the map, I'm not allowed to post it directly here since it's not CC.

• I don't know how many cartographers we have here. Since it's a photoshop question, really, you might want to see what Graphic Design has to offer..... – Shokhet Nov 9 '14 at 4:48
• @Shokhet : Maybe, but the problem is not really with photoshop but more with the difference between the two areas. Maybe the problem is that I misunderstood the proprieties of the projection, I don't know for sure. I think I already asked on the Cartographer's Guild but didn't got a response. – Vincent Nov 9 '14 at 5:03
• Aha. Sorry, I don't have enough experience with either to be able to help you there.....as an aside, I just found out that SE has a cartographer's site, you might consider checking that out. – Shokhet Nov 9 '14 at 5:05
• @Shokhet Thanks, I'll check it out. – Vincent Nov 9 '14 at 5:08
• @Oldcat, Yes, but projections will often preserve some property (at the expense of others). For example, Mercator terribly distorts area, but preserves bearings, which is why it's useful for navigation. Mollweide distorts bearings terribly, but it preserves relative area, which is the property that the OP desires. – Caleb Hines Nov 11 '14 at 2:04

Huh... I just ran some numbers, and this is very interesting! (Note that in the below formulae, I'm using tau ($τ = 2π$) as the circle constant for convenience).

Let's say your world is a sphere with radius $R$. Then the surface area of that sphere is given by:

$$A_S = 2τR^2$$

But now you have your ellipse. It's width is your planet's circumference: $w = τR$, and it's height is half that $h = τR/2$. Then the area of the ellipse in your map is given by:

$$A_E = πab = \fracτ2 (\frac w2) (\frac h2) = \frac τ2 (\frac{τR}2) (\frac{τR}4) = \frac{τ^3}{16} R^2$$

If we factor out the formula for the surface area of the sphere, we get:

$$A_E = \frac{τ^2}{32} (2τR^2) = 1.2337 A_S$$

which accounts for your 22% (with a bit of rounding error). But... I don't know why the formulas work like that!

The quick hack solution? Compute the area of your pixels as:

$$\text{area per pixel} = \frac{\text{desired total surface area}}{\text{number of pixels in ellipse}}$$.

EDIT: It bothered me that I couldn't figure out why these areas were different, and then I realized where the difference was coming from. If you physically stretched out each circle of fixed latitude into a straight line, their circumferences would be proportional to the cosine of the latitude, and the resulting map projection would be a sinusoidal projection. The extra increase in area is a result of stretching this sinusoidal shape into an ellipse. This is easily demonstrated by comparing the area of half the ellipse to that of half the sinusoid. If you have the ability to create a sinusoidal projection, you might consider using it, and seeing if the numbers make more sense.

Mollweide is an equal area projection so each pixel reperesents the same area as every other pixel (ignoring rounding errors). However the pixels do not represent perfect squares on the surface so your initial attempt to figure out the area of each pixel from the width of the map is where you made a mistake.

The correct way to get an 'area per pixel' value for a Mollweide projection, or any other equal area projection is to divide the surface area by the number of pixel. If we assume the planet is a sphere of radius $R_p$, and the image is $2h$ by $h$ and representing an ellipse with semimajor and semiminor axes $a=h$ and $b=\frac h2$ (Mollweide is always twice as wide as it is tall), and let $\tau=2\pi$ for convenience then

$$\frac{2\tau R_p^2}{\frac{\tau}{2} ab}=\frac{4 R_p^2}{h\cdot\frac{h}{2}}=\frac{16 R_p^2}{wh}=\frac{16 R_p^2}{2h\cdot h}=\frac{8 R_p^2}{h^2}$$

With $R_p=R_ⴲ=6371\space\text{km}$ and $h=830$ then

$$\frac{8\cdot 6371^2}{830^2}=471.4\space{\text{km}^2}/{\text{px}}$$

And checking with your pixel count gives $$471.4\text{ km}^2/\text{px} \cdot 1082100\text{ px}=510101940\text{ km}^2$$