Gall’s projection, also known as the **Gall Stereographic Projection**, is a cylindrical map projection introduced by James Gall in 1855. Unlike the Mercator projection, it is not conformal (orthomorphic) and reduces the exaggerated area distortion near the poles by shortening the distance between parallels. The cylinder cuts through the sphere at 45° N and 45° S, which are the standard parallels. These lines are true to scale, and the scale decreases towards the equator and increases towards the poles, balancing area and shape distortions.

A key characteristic of Gall’s projection is that it offers a reasonable balance between shape and area preservation, making it useful for general-purpose world maps. While distortions increase at higher latitudes, it is still considered effective for mid-latitude regions and global-scale thematic mapping. Its geometric construction involves projecting the globe onto a secant cylinder, and it results in equidistant vertical meridians and horizontally projected parallels.

Despite not being equal-area or conformal, Gall’s projection minimizes distortions enough to make it visually appealing and suitable for educational purposes. It maintains consistent scaling along the equator and is simple to use, making it popular in atlases and general maps. However, its distortions, particularly near the poles, limit its precision for applications requiring exact measurements, such as polar region mapping.

## Construction

**Question**: Construct a graticule on Gall’s projection for the world map on RF 1:250,000,000 scale at 15° 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. Length of equator:**

$$\text{Circumference} = 2 \pi r$$

Length of Equator = **16.01 cm**

**5. Distance between longitudes at 15° intervals:**

Since the longitudes are at 15**°** intervals we need to draw 24 longitudes (360**°**/15). Hence at the given scale 16.01/24=0.67.

Distance between longitudes = **0.67 cm**

- With radius OB make a circle ADBC
- AB and CD intersect at O at 90°. (AB is Equator, CD is Prime meridian)
- Make Od and Oj at 45° from point O. These parallels are true to scale. (from these two parallels scales decrease towards 0° and increase towards 90°)
- From OB draw all the other parallels at 15° intervals.
- Draw line adjm to intersect the circle and Od and Oj.
- From point A (periphery), draw lines that intersect all the 15° latitudes and fall on line am.
- Points abcdefghijklm at line am are the points of latitudes at 15° intervals.
- Draw dD’ (16.01 cm) from point g, this is the equator. All the latitudes at 15° intervals can be drawn similarly from points abcdefhijklm of the same length.
- Divide the line gG’ in 24 equal parts, each of length 0.67 cm. With the help of a protractor draw all the longitudes 15° intervals.