The **Mollweide projection** is a type of map projection that represents the world in an elliptical shape. It is an **equal-area projection**, meaning that areas on the map are proportional to their corresponding areas on the globe. This makes it particularly useful for displaying global data, such as population density, climate zones, or land use, where the preservation of area is essential.

Mollweide’s projection is often used in world maps to balance distortion of shapes and sizes, providing an accurate visual comparison of different regions.

## Key Features

**Equal-Area**: The Mollweide projection preserves areas across the entire map, ensuring that the relative sizes of countries and continents are displayed accurately.**Elliptical Shape**: The projection forms an ellipse, with the equator being the longest horizontal line.**Curved Meridians**: Unlike cylindrical projections, the meridians (lines of longitude) curve and converge toward the poles, reflecting the Earth’s curvature.

## The Mathematics Behind Mollweide’s Projection

The Mollweide projection transforms geographic coordinates (latitude and longitude) into planar coordinates (x and y) using specific equations. These equations ensure that the projection remains equal-area while providing a reasonable visual representation of the globe.

### Key Equations

**Auxiliary Angle Equation**:

$$2\theta + \sin(2\theta) = \pi \sin(\phi)$$

Here, $$(\theta)$$ is an auxiliary angle, $$(\phi)$$ is the latitude, and $$(\pi)$$ is the mathematical constant. This equation must be solved numerically for each latitude.**X-Coordinate (Longitude)**:

$$x = \frac{2 \sqrt{2}}{\pi} R (\lambda – \lambda_0) \cos(\theta)$$

Where:

- (R) is the Earth’s radius (or scale factor),
- $$(\lambda)$$ is the longitude,
- $$(\lambda_0)$$ is the central meridian,
- $$(\cos(\theta))$$ is the cosine of the auxiliary angle (\theta).

**Y-Coordinate (Latitude)**:

$$y = \sqrt{2} R \sin(\theta)$$

Where:

- (R) is the Earth’s radius,
- $$(\sin(\theta))$$ is the sine of the auxiliary angle $$(\theta)$$

### Step-by-Step Example

Let’s assume we want to project a point at **latitude **

$$( \phi = 30^\circ )$$

and **longitude $$( \lambda = 60^\circ )$$** using the Mollweide projection. For simplicity, we use a map scale where ( R = 1 ) (for relative distance).

#### Step 1: Solve for the Auxiliary Angle $$(\theta)$$

For $$( \phi = 30^\circ )$$, solve the equation $$( 2\theta + \sin(2\theta) = \pi \sin(30^\circ) )$$

This gives $$( 2\theta + \sin(2\theta) = \pi/2 )$$ which we can solve numerically to find $$( \theta \approx 0.6435 )$$ radians.

#### Step 2: Calculate X and Y Coordinates

- Using $$( \theta = 0.6435 )$$ radians, $$( \lambda = 60^\circ )$$, and $$( \lambda_0 = 0^\circ )$$:

$$x = \frac{2 \sqrt{2}}{\pi} (60^\circ – 0^\circ) \cos(0.6435) \approx 1.029$$ - For the y-coordinate:

$$y = \sqrt{2} \sin(0.6435) \approx 0.905$$

So, the point at $$((30^\circ \text{N}, 60^\circ \text{E}) $$ in geographic coordinates becomes $$((x \approx 1.03, y \approx 0.91) $$ on the Mollweide projection.

## Example

**Question:** Draw a graticule for the Mollweide’s Projection showing the entire world on a scale of 1:100,000,000. The graticules should have 30° intervals for both latitudes and longitudes.

**Solution:**

To draw the Mollweide projection as shown in the image with a scale of 1:100,000,000, follow these step-by-step instructions:

#### Step 1: **Set Up the Parameters**

**Radius of the Earth (R)**: 6,371,000 meters or 6,371 km.**Map Radius (R_map)**: The map scale is 1:100,000,000. So, convert the Earth’s radius to the map scale:

$$R_{map} = \frac{6,371,000 \, \text{cm}}{100,000,000} = 6.371 \, \text{cm}$$- Central meridian $$( \lambda_0 = 0^\circ ) (Greenwich Meridian)$$

#### Step 2: **Draw the Elliptical Boundary**

- Draw a horizontal axis that is 4 times the radius $$(2R_{map})$$ (use 12.742 cm as the total width of the ellipse, i.e., the major axis).
- Draw the vertical axis with a height of $$(2R_{map} \sqrt{2})$$ (use 9.01 cm as the total height, i.e., the minor axis).
- Use a compass to draw the elliptical boundary using these measurements.

#### Step 3: **Mark the Parallels (Latitudes)**

- The parallels should be spaced at 30° intervals (from 90°N to 90°S).
- Use the following table to calculate the distance of each latitude from the centre (equator) on the map. The formula to use is $$(y = \sqrt{2} \cdot R_{map} \cdot \sin(\theta))$$ where $$( \theta )$$ is the auxiliary angle derived from $$(2\theta + \sin(2\theta) = \pi \sin(\phi))$$
- The approximate values for each latitude (in cm) from the centre are:

Latitude (°) | $$( \theta )$$ (radians) | Distance from Equator (y) (cm) |
---|---|---|

90°N | 1.571 | 9.01 |

60°N | 1.047 | 5.53 |

30°N | 0.523 | 2.87 |

0° (Equator) | 0 | 0 |

30°S | -0.523 | -2.87 |

60°S | -1.047 | -5.53 |

90°S | -1.571 | -9.01 |

Mark the distances for each parallel on the vertical axis.

#### Step 4: **Mark the Meridians (Longitudes)**

- The meridians should be spaced at 30° intervals (from 180°W to 180°E).
- Use the following formula to calculate the x-coordinates of each meridian:

$$x = \frac{2\sqrt{2}}{\pi} R_{map} (\lambda – \lambda_0) \cos(\theta)$$

where $$( \lambda )$$ is the longitude and $$( \theta )$$ is the auxiliary angle corresponding to the latitude.

For example, for $$( \lambda = 30° )$$, use the values of $$( \theta )$$ calculated earlier. The x-coordinate values for 30° intervals will be:

Longitude (°) | X-Coordinate (cm) |
---|---|

0° | 0 |

30°E/W | ±2.54 |

60°E/W | ±5.08 |

90°E/W | ±7.62 |

120°E/W | ±10.16 |

150°E/W | ±12.71 |

180°E/W | ±15.25 |

Mark the points for each meridian along the horizontal axis.

#### Step 5: **Draw the Graticules**

- Draw the parallels (latitude lines) as horizontal lines spaced according to the distances calculated in Step 3.
- Draw the meridians (longitude lines) as curved lines that pass through the x-coordinates calculated in Step 4, ensuring they meet at the poles.