How Mercator projection works

Question

I found this web page that explains how Mercator projection works : https://personal.math.ubc.ca/~israel/m103/mercator/mercator.html

In the section "How to do it", I didn't understand this paragraph about the very small square piece of land at latitude 'theta':

"It takes up a fraction hR/(2piRcos(theta)) of the parallel at latitude theta, so its width on the map is (hR)/(2piRcos(theta))2pi = hsec(theta). So its height on the map must also be hsec(theta)".

Can someone rephrase this in a clearer way and if possible with a diagram ?

Answer 1

But on the map the parallels are all the same length.

For the unit sphere, all parallels measure 2π in the map (the same length as the equator). But they measure 2π*cos(θ) in the "Earth" (in the reference surface, a sphere in this case). Divide one by the other to get the deformation coefficient in the direction of parallels: 1 / cos(θ).

So a parallel arc with h lenght in the reference surface measures h / cos(θ) in the map.

That's all. There was no need to prove it in other way...

For the unit sphere, h is the lenght of an arc. But for a sphere with radius R, h is an angle in radians. So in the map (in the equator and in all parallels), the length of the arc is h.R.
To get the fraction of the parallel in a map from a sphere with radius R, divide by how much the parallel measures in the sphere (2π.R.cos(θ)).
To get how much measures h in the map, multiply its fraction by how much measures the whole parallel: 2π.

Simplify all. In any case, h measures h / cos(θ) = h . sec(θ), in the map of a unit sphere.

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So its height on the map must also be h . sec(θ)

This is interesting. Parallels and meridians intersect orthogonally on the reference surface and also on the map. That is, they can be considered as the directions of the maximum and minimum deformation of Tissot's theorem.

If the deformation in both directions is the same, one being the maximum and the other the minimum, then the deformation in all directions must be the same, so the projection is conformal.

That's how Mercator projection works. Find the deformation in one of the principal directions, and analytically force the deformation in the other principal direction to be the same.

Answer 2

I find it simpler to work this through if we start with a sphere that has a circumference of 1 at the equator (and thus has a radius of 1/2π).

Then the circle at latitude θ has length cos(θ). The projected length on the map, however, is 1, just like the equator, so we must scale any length along this circle by 1/cos(θ) = sec(θ).

Now let our square piece of land at latitude theta have edges of length h. We scale this length by sec(θ) and arrive at h*sec(θ), as in the original text, but without all the extraneous parameters floating around, just to be canceled out.