[Differential Geometry] How do we derive the Mercator's Projection of a sphere?

We begin with the familiar spherical parametrisation of the unit sphere \(X: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3\), where \[X(\theta, \varphi) = (\sin{\theta}\cos{\varphi}, \sin{\theta}\sin{\varphi}, \cos{\theta}),\] is a diffeomorphic map from an open neighbourhood \[U = \{(\theta, \varphi) \in \mathbb{R}^2; 0<\theta<\pi, 0<\varphi<2\pi \},\] which is obtained by rotating a semicircle on the xz-plane around the z-axis.

Interestingly, by making the change of coordinate, \[u = \log \tan{\frac{1}{2}\theta}\] and \[v=\varphi,\]

we can show that a new parametrisation of the coordinate neighbourhood \(X(U)\) can be derived.

First, we will start by putting \( a= \tan{\frac{1}{2}\theta}\). Then, \[X(\theta, \varphi) = (\sin{\theta}\cos{\varphi}, \sin{\theta}\sin{\varphi}, \cos{\theta}) = \left(\frac{2a}{1+a^2}\cos{\varphi}, \frac{2a}{1+a^2}\sin{\varphi}, \frac{1-a^2}{1+a^2} \right), \] where we have used the half-angle formulae for our tangent function.

Now \[u= \tan{\frac{1}{2}\theta} \implies exp^u = \tan{\frac{1}{2}\theta} = a\] so

\[ =\left( \frac{2e^u}{1+e^{2u}} \cos{\varphi}, \frac{2e^u}{1+e^{2u}} \sin{\varphi}, \frac{1-e^{2u}}{1+e^{2u}}\right) \] \[ = \left( \frac{2}{e^{-u}+e^u} \cos{\varphi}, \frac{2}{e^{-u}+e^u} \sin{\varphi}, \frac{e^{-u}-e^u}{e^{-u}+e^u} \right) \] \[= ( sech u, \cos v, sech u sin v, tanh u). \]

Hence, \(Y(u,v) = \left( sech (u), \cos(v), sech{u} \sin(v), \tanh(u)\right) \) gives a diffeomorphic map that parametrises the unit sphere.

In fact, we can also show that the coefficients of the first fundamental form in the parametrisation \(Y\) are isothermal, hence locally conformal to an open neighbourhood of \(\mathbb{R}^2\).

Now that you know that \(y^{-1} : S^2 \rightarrow \mathbb{R}^3\) is a conformal (angle-preserving) mapping, what can you say about the meridians (lines of constant longitude) of the unit sphere, mapped under the Mercator's projection?

Note by Tasha Kim
2 weeks, 1 day ago

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Given a sphere with a map on it, put it inside of a cylinder of the same diameter (usually aligned with the axis of the Earth), draw a ray from the center of the sphere through a given point on the map, and continue until it intersects the cylinder. Then a point on the map on the sphere will have been projected into a corresponding point on the cylinder. When all done, unroll the cylinder flat.

This is the reason why you never see the north or south pole of the Earth on a typical Mercator projection.

Michael Mendrin - 2 weeks, 1 day ago

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Thanks for your comment, it's a very clever way of preserving angles on a planar map of the sphere!

Tasha Kim - 2 weeks ago

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I realize now that you were just starting on this note when I answered. Obviously you know quite a bit about this subject. And I had forgotten about the fact Mercator projections preserves local angles, i.e., only the scaling changes locally at different latitudes, i.e. maps of cities look approximately normal except for size. That is an interesting property.

Michael Mendrin - 2 weeks ago

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@Michael Mendrin Yes, I realised that too... It is indeed!

Tasha Kim - 1 week, 6 days ago

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