Is there a site, app, or technique that you can use to measure the area of your world’s continents. It is rather easy to find the surface area of your world if the map is equirectangular or just a grid.

I have already found the surface area of the planet using a radius to surface area of a sphere equation. I got the radius from imputing the mass of the planet into Artifexian’s Worldsmith Spreadsheet. That is the easy part.

  • 1
    $\begingroup$ You will have to fudge the numbers rounding up, my friend. en.wikipedia.org/wiki/Coastline_paradox $\endgroup$ Commented May 27, 2023 at 20:31
  • 4
    $\begingroup$ One traditional (pre-computer) approach if you're actually measuring on a physical map would be a "planimeter". Either a dot planimeter, a polar planimeter, or something else similar. $\endgroup$
    – R.M.
    Commented May 28, 2023 at 2:16
  • 6
    $\begingroup$ @ChristopherHostage, the coastline paradox involves the perimeter of an irregular two-dimensional region. The OP is asking about area, which behaves much better, and can be meaningfully calculated even for regions which are much too irregular to have meaningful perimeters. $\endgroup$
    – Vectornaut
    Commented May 28, 2023 at 2:34
  • $\begingroup$ @R.M. Those are fascinating. I will definitely try them but I will probably want a more accurate form of measurement. $\endgroup$
    – Martamo
    Commented May 28, 2023 at 12:13

4 Answers 4

  1. Export your map to PNG or something using an equal-area (aka authalic) projection; for example, use NASA's G.Projector to convert from equirectangular to cylindric equal-area.

  2. Use your favorite raster image editor to fill the area of interest with a distinctive color.

  3. Use the pixel counting function of said raster image editor to count the pixels of the chosen color.


This is a cheater method.

  1. Tessellate. Cover the map with hexagons or triangles or squares of roughly equal size. Call this the "grid" and the shapes "cells".

  2. Ensure the size of the cells is small compared to the thing you are measuring. Like, 80%+ or more of the cells that touch your shape are fully within the shape, and 20% or less are partly within the shape.

  3. Count. Treat any cell that is partly in or out of the shape you want to measure as being a half cell.

So long as your tessellation is aligned random relative to the geography you are measuring, this will get really close. You can increase precision by subdividing (all of hexagons/triangles/squares can be subdivided into smaller regions of sufficiently similar shape).

The projection of your geography onto the regular grid can warp the size of the cells, which can cause errors or add to the complexity of the math.

You can look at the "edge" (the ones partly covered) to determine if it is likely to over or under estimate the shape. A smattering of dots in a void will be over estimated, and a smattering of voids in a region will under estimate the area of the region. You can bound the amount of error, however, buy the total area of the "edge" region.

I own some transparent grid and hex sheets. You can place this above your map and use that to tessellate.

You can even just sample on a regular pattern. Unless your pattern is somehow correlated with your shape and the pattern is fine enough, counting how many spots on a regular pattern are "inside" will get you a good measure of the area.

Pay attention to the size of your "dots" you are measuring intersection with; the larger they are, the more it will overestimate the "boundary"s contribution to the area of the shape.

A less mechanical and more mathematical method is to start off by drawing two polygons. One fully within the region you want to measure, and one fully without. The area between is your "complex coastline".

The trick is the area of what you want to measure is between those two region's areas. As you add more detail the gap shrinks. Or you can look at the gap, and fudge a percentage of it being within the region.

A very mathematical method is to use Green's Theorem. The area can be viewed as a double-integral; this can be converted into a path integral over the boundary of the region.

Using the above "inside" and "outside" curves or polygons, you just integrate over the paths and get the upper and lower bounds of area.

But this is probably not what you want to do.

An amusing way to do it is to use Monte Carlo methods. Pick a random spot and see if it is inside or outside.


After a few 1000s of samples you'll have a pretty good estimate of the area.

This technique is actually useful when describing the region you want to measure is difficult.

Take a picture. Flood fill the region you want to measure.

Count pixels.

Take a picture. Flood fill the region you want to measure as white, and everything else as black.

Blur the entire image uniformly. If the resulting pixel is 65 white, the shape was 65/255 or 24.5% of the original image.

I mean, how accurate do you need to get?

Look for big bulky "filled" parts of your shape. Toss some rectangles in and find out how big they are.

Then measure the fuzzy perimeter. How thick is it? (Ie, from the point where you already measured, to the definitely outside) How dense is it? Is it like half-ish inside and about 50 km thick and about 1000 km long? Well, that is about 25,000 km^2 give or take. Add that to your 100,000 km^2 interior and get a decent estimate.

Or even worse, just measure total width and height of your shape. That is your upper bound. Then guess what fraction of that area looks to be inside. Multiply. If your shape is complex? Do it in chunks that are easier to estimate.

Map of Canada

So lets say 3 pieces to estimate this. One is the 45 to 60 degree band, the other the NWT/Yukon 60 degree to 70 degree band, and the last if the northern islands.

The southern band is from 60 to 135 degrees, or 75 degrees, or about 20% of the Earth. The northern band is from 90 to 140, about 50 degrees or 14% of the Earth. The northern islands look to be about 1/3 of the area of the northern band total.

Both "solid" bands look to be about 80% land ish.

At 52 degrees latitude on the Earth, 75 degrees is about 5000 km.

At 65 degrees latitude on the Earth, 50 degrees is about 2500 km.

Each degree latitude is about 110 km, so 15 degrees is 1650 km, and 10 degrees is 1100 km.

So my estimated square km using this method is (5000 * 1650 + 2500 * 1100 * 4/3)*.8 = 9,533,333 km^2

The actual area of Canada is 9.985 million km^2.

Within 5%.

Considering how fudgey most of my guesses where, that is really close - closer than I expected, and way closer than you'll need for any practical world building reason.

Map of Canada Image source: Wikipedia


A more artisanal approach:

  1. Export your map to PNG or something using an equal-area (aka authalic) projection; for example, use NASA's G.Projector to convert from equirectangular to cylindric equal-area. (As in AlexP's answer)
  2. Print it
  3. Cut the printed area and weigh it
  4. Cut each continent and weigh it
  5. Since you know the total area of the world and the weight of its printed image, with the ratio of the weight of the continent to the weight of the printed world you can quickly find the surface of the continent.

Let's say the whole printe world weighs 50 grams and the continent 5 grams, or 1/10 of the whole world. Since you know that the whole planet has a surface of X square meter, the surface of the continent is X/10 square meter.

Benefit of this approach: you can quickly determine the effect of adding/removing surface by weighing the related print

  • 1
    $\begingroup$ You would need a very accurate scale. I’m not seeing how this will find the surface area number. I feel like this method will more find the ratio of land to water. $\endgroup$
    – Martamo
    Commented May 27, 2023 at 19:57
  • 2
    $\begingroup$ @Martamo --- Use chipboard sheets (very thick and heavy cardboard (avail at Amazon) or 1/4 inch plywood. Accurate scales are easy to come by, able to weigh down to the mg (avail at Amazon). Since the relative weights and the relative areas are proportions, they'll be the same. If I draw a squiggly continent on a 1kg piece of plywood, and it weighs 500g, how much will a piece of plywood weigh that is simply a 50% piece cut off with a table saw? Right, 500g. This way, when you're done, you can paint your bits of plywood and you'll have a world puzzle! $\endgroup$
    – elemtilas
    Commented May 27, 2023 at 21:59
  • $\begingroup$ So, if you have the world SA you can find the he weight of the whole world and have SA=AW(All weight). If one of the cutouts is 25% AW then that continent is 25% the SA of the world. $\endgroup$
    – Martamo
    Commented May 28, 2023 at 0:07
  • $\begingroup$ "You would need a very accurate scale" .... You would also need very accurate scissors and hands !!! (not to mention very accurate who-knows-what-tool if you were to use plywood!!!) $\endgroup$
    – Cal-linux
    Commented May 28, 2023 at 12:36
  • $\begingroup$ Fun thing about ISO paper is that 80GMS paper is 80 grammes per square meter. So an A0 sheet is 80g, A1=40g, A2=20g, A3=10g and an A4 sheet weighs 5g. $\endgroup$
    – KalleMP
    Commented May 29, 2023 at 12:23

This is the realm of Geographic Information Systems (GIS), and it is surprisingly complex because planets are 3d spheres, and area is a 2d measurement.

Measuring the area of a large block of land is challenging because there are different ways to do ir, all seemingly valid:

  • is it the planar area - what you see when the land is projected onto a 2d map? If so, how do you do the projection?
  • is it the spherical area allowing for the curvature of the earth? If so, what sphere/ellipsoid are you using? Planet's aren't quite spheres so even for earth there are dozens of 'standards'

The larger the area, the more divergent the results are.

Anyway, you probably don't care too much if you're off by a couple square kilometers, so I'd suggest loading it up in QGIS, mapping it into a polygon and then getting it to tell you the area. I don't know if it lets you use custom planets though.

An alternate is to turn your map into vector form (ie a polygon) and then use one of the methods for calculating area of an irregular polygon. Eg: https://www.mathsisfun.com/geometry/area-irregular-polygons.html


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .