Lambert conformal projection derivation
Pub. 2024 Jun 8Below is a simple derivation of the lambert conical conformal transform for a sphere.
Most resources online are like this wolfram mathworld page, which sites a a book from 1987 contains just formulas, no derivation. That book in turn cites the original 1772 paper by Lambert. To nobody's surprise, I couldn't find this book, nor its 1972 translation online.
This 1918 (?) paper gives a good overview of the derivation. The first section gives a derivation of a (fairly poor) approximation, but then later gives the full thing. But it's over an ellipsoid, and it's maybe a bit overcomplicated.
Here's mine. I tried to keep it as simple as I could. It's only valid for a sphere, which is what the NDFD and HRRR datasets are in.
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We have here a sphere with radius $R$, a reference point $G$ with parallel $\phi_0$ and meridian $0$, and another point on the sphere with parallel $\phi$ and meridian $\lambda$.
The transform has a couple properties:
- "Wedges" of the sphere, like in the left image above, get transformed into sectors of a circle, like in the right image above.
- Meridians get mapped to straight lines radiating out from a center "origin" point (the point where the north pole gets mapped). Parallels get mapped to concentric circles (or rather portions of circles).
- The length of the reference parallel is the same as the length of the circle that the reference parallel gets mapped to.
- A wedge with angle $\lambda$ get transformed into a wedge with internal angle $\lambda \sin{\phi_0}$. You could conceivably choose something different, but this choice minimizes the distortion around the reference parallel.
- From these two, the distance from $G'$ to the origin is $R \cot{\phi_0}$.
To derive the transformation, we're going to be looking at what happens to $N$ and $E$ in the above diagram.
Say you're standing on the sphere at $(\phi, \lambda)$. Taking a small step of $\Delta$ meters northward corresponds to a change in parallel $d \phi = \frac{\Delta}{R}$. Taking the same small step $\Delta$ eastward cooresponds to a change in meridian of $d \lambda = \frac{\Delta}{R \cos{\phi}}$.
The goal here is to express the cooreponding change to $N$ and $E$ in the projected map, $dN$ and $dE$. Conformal maps preserve scale in all directions at every point, so $dN = dE$ in a lambert conformal map.
Going off of the diagram above, a step eastward on the globe would correspond to a change in the projected map $dE = (R \cot{\phi_0} - N) d \lambda \sin{\phi_0} = (R \cot{\phi_0} - N) \frac{\Delta}{R \cos{\phi}} \sin{\phi_0}$.
We don't know what $N$ is as a function of $\Delta$. However, we can say that whatever $N$ is, $dN = \frac{dN}{d \phi} d \phi = \frac{dN}{d \phi} \frac{\Delta}{R}$.
Putting it all together:
$$\begin{align}dN & = dE \\\frac{dN}{d \phi} \frac{\Delta}{R} & = (R \cot{\phi_0} - N) \frac{\Delta}{R \cos{\phi}} \sin{\phi_0} \\\frac{dN}{d \phi} & = (R \cot{\phi_0} - N) \sec{\phi} \sin{\phi_0}.\end{align}$$
This is where we need to bring in calculus. This is a fairly simple seperable differential equaion:
$$\frac{dN}{R \cot{\phi_0} - N} = \sec{\phi} \sin{\phi_0} d \phi,$$
which solves to
$$\begin{align}-\ln \left| R \cot{\phi_0} - N \right| & = \sin{\phi_0} \sinh^{-1}(\tan{\phi}) + C \\\text{OR} & = \sin{\phi_0} \ln \left| \sec{\phi} + \tan{\phi} \right| + C \\\text{OR} & = \sin{\phi_0} \ln \left| \tan \left( \frac{\phi}{2} + \frac{\pi}{4} \right) \right| + C \\\text{OR} & = ...\end{align}$$
There are famously several equivalent integrals of secant. Sticking with the top one:
$$\begin{align}-\ln \left| R \cot{\phi_0} - N \right| & = \sin{\phi_0} \sinh^{-1}(\tan{\phi}) + C \\\left| R \cot{\phi_0} - N \right| & = C e^{-\sin{\phi_0} \sinh^{-1}(\tan{\phi})} \\N & = R \cot{\phi_0} + C e^{-\sin{\phi_0} \sinh^{-1}(\tan{\phi})}.\end{align}$$
Combining this with $N(\phi_0) = 0$ gives:
$$N = R \cot{\phi_0} \left( 1 - e^{\sin{\phi_0}(\sinh^{-1}(\tan \phi_0) - \sinh^{-1}(\tan \phi))} \right).$$
Completing it with E:
$$E = (R \cot{\phi_0} - N) \lambda \sin{\phi_0}$$