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# The Mollweide Projection

The Mollweide projection is an attempt to moderate the behavior of the equal-area sinusoidal projection.

This projection is also equal-area. This is achieved by locating each parallel where the proportion of the area of the ellipse between that parallel and the equator is the same as the proportion of the area on the globe between that parallel and the equator.

This involves using iterative methods to determine the inverse of a function that cannot be inverted analytically, but the process converges rapidly.

Since a sinusoid takes up less space than an ellipse, there is a slight, but hardly noticeable, stretching of countries along the equator.

Since this projection is a pseudocylindrical projection like the Sinusoidal projection, it can be interrupted.

And, like any projection, it can be drawn in its oblique case:

Since it presents the world on an ellipse, doing so with the Mollweide is more popular than with the Sinusoidal projection.

The diagram below illustrates the mathematical theory behind the Mollweide projection.

The Cylindrical Equal-Area projection illustrates the fact that the proportion of the Earth's surface area between a given latitude and the Equator is proportional to one-half the sine of the latitude.

Simple geometry shows what the proportion of the area of a circle is between a diameter and a line parallel to the diameter. In the diagram, it is expressed in terms of the angle psi rather than the height of the second line, for simplicity. The green triangular areas each have an area of sin(psi)*cos(psi)/2, and the yellow portions of the circle each have an area of psi/2. The circle as a whole has an area of pi.

Since the Mollweide projection places the world in an ellipse, the same proportionality law applies to the area between its parallels and the Equator.

The expression for this area, though, cannot be inverted simply. However, an iterative process can quickly determine the proper value for psi (from which the height can be directly calculated, as it is the sine of psi). Since the expression can be easily differentiated, instead of using a binary search, one can use Newton-Raphson iteration, which converges very quickly.

That is, to solve the equation:

```(psi+sin(psi)cos(psi))/pi = sin(lat)/2
```

for psi, given lat, one can start with lat as our first guess for psi, and then improve each guess by noting that the difference between the true area and that given by any guess is:

```error = ((guess+sin(guess)cos(guess))/pi - sin(lat)/2)
```

and, since the rate of change in that area for a change in the guess is the following function:

```(1+(cos(guess))^2-(sin(guess))^2)/pi
```

then a good next guess is the old guess minus the error divided by the rate of change in the error over the rate of change in the guess.

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