The **Sanson-Flamsteed Projection**, also known as the **Sinusoidal Projection**, is one of the earliest **equal-area projections** used in cartography. It was independently developed by two individuals: **Nicolas Sanson** (a French cartographer) in the 17th century and **John Flamsteed** (the first Astronomer Royal of England) in the 18th century.

## Why is it known as a Sinusoidal Projection?

The **Sanson-Flamsteed Projection** is called **Sinusoidal** because the shapes of the meridians (except the central meridian) follow a curve that resembles a **sine wave**.

The **Sanson-Flamsteed Projection** is called **Sinusoidal** because the shapes of the meridians (except the central meridian) follow a curve that resembles a **sine wave**. This mathematical property ensures that the projection maintains equal-area representation.

### Mathematical Explanation:

In the **Sinusoidal Projection**, the horizontal distance (x) of a point from the central meridian is proportional to the **longitude** and the **cosine of the latitude**. This means that the shape of each meridian, except the central one, follows a path similar to a sine wave.

The formula for the (x)-coordinate in a sinusoidal projection is:

$$x = R \, \lambda \, \cos \phi$$

$$Where: (x) = horizontal coordinate; (R) = radius of the Earth; (\lambda) = longitude of the point from the central meridian; (\phi) = latitude of the point$$

The **central meridian** is a straight line, but the other meridians curve in a way that resembles the periodic wave pattern of a **sine function** (hence the name **sinusoidal**).

### Visual Explanation:

Just like a sine wave fluctuates symmetrically around a central axis, the meridians in a sinusoidal projection curve symmetrically to the left and right of the **central meridian**, depending on how far from it they are. The projection ensures that the area between the meridians and parallels is accurate, even though shapes may get distorted.

The **curved meridians** that mimic a sine-like path are a defining characteristic. The name reflects how mathematical **cosine functions** (derived from the sine function) govern the projection’s geometry to preserve areas accurately.

## Construction:

**Question:** Construct a Sinusoidal Projection at a scale of 1:250,000,000 at 30° intervals.

**Solution:**

To find the radius of the Earth on the map based on the scale **1:250,000,000**, we can use the following steps:

**Radius of the Earth**: The Earth’s actual radius is approximately**6,371 km**(or 6,371,000 meters).**Map Scale**: The scale of**1:250,000,000**means that 1 unit on the map represents 250,000,000 units on the Earth’s surface.**Converting to centimetres**:- The Earth’s radius in centimetres is 6,371,000,000 cm (since 1 km = 100,000 cm).
- Now, divide this by the scale factor 250,000,000 to get the radius on the map.

$$\text{Radius on map (in cm)} = \frac{6,371,000,000}{250,000,000}$$

Radius = **2.55 cm**

4. Distance between the meridian can be found out by finding the length of the arc subtended by 30° on a circle with the radius

$$\text{Arc Length} = 2\pi r \times \frac{30^\circ}{360^\circ}$$

$$\text{Arc Length} = 2\pi \times 2.55 \, \text{cm} \times \frac{30}{360}$$

$$\text{Arc Length} \approx 1.336 \, \text{cm}$$

**5. Length of equator:** $$\text{Circumference} = 2 \pi r$$

Length of Equator = **16.01 cm**

- Draw a circle with a radius of 2.55 cm from point
*O*- From point
*O*make diameter*AB* *CD*intersects*AB*at*O*at 90°- From point,
*O*draws*OW*and*OY*at 30° and 60° respectively

- From point
- Draw an arc
*pqrs*with the length of arc 1.336 cm from point*O*- Drop perpendicular
*rr’*and*qq’*from where the arc*pqrs*intersects*OW*and*OY*

- Drop perpendicular

**Draw the projection:**- Draw line
*WE*representing 0° of length 16.01 cm. - Divide WE into 6 equal parts at 60° intervals (360°/30°)
*NS*is the central meridian bisecting*WE*from the centre. Its length is half of the equator (16.01/2=8 cm).*NS*is also bisected by*WE*.- Divide
*NS*into 3 equal parts on either side of*WE* *Os, rr’ and qq’*is the distance between the meridians on latitudes*0°, 30° and 60°*respectively

- Draw line