After reading this article you will learn about the construction and classification of scale of a map.

**Construction of ****Scales of a Map: **

In plotting a survey, often an object to be represented on paper is so large that it would be inconvenient to make a full sized drawing of it. The drawing or map is then made to a reduced size, the operation being called **“drawing to scale “.** The Scale of a drawing or a map is the fixed proportion which the length of a line on the drawing or map bears to the corresponding distance on the ground.

Thus if a line one cm long represents 10 m on the ground, the drawing is said to be drawn to a scale of 10 m to a cm. It is usually written as 1 cm = 10m. It is evident that every line on this drawing is 1/1000th part of the corresponding line on the ground.

This fraction which represents the proportion of the drawing distance to the corresponding ground distance is called the Representative Fraction (R.F.) It is, therefore, the ratio of one cm map distance to the corresponding distance on the round converted into centimetres. It may be noted that in finding R.F., the numerator must always be one and denominator will be reduced to the same unit as the numerator.

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**For example, if the scale is 10 m to a cm, the R.F. of the scale is:**

The scale should be shown near the title of the drawing or map either graphically or by numerical relation. Scales are generally 15 cm to 25 cm long and never more than 30 cm. The right end of the extreme left division of the scale is invariably marked with 0; the secondary subdivisions starting from that point are marked from right to the left and the primary divisions from left to the right.

**Classification of Scale of a Map:**

**Scales may be classified as: **

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A. Plain Scales.

B. Diagonal Scales.

C. Comparative or Corresponding Scales.

D. Vernier Scales.

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E. Scales of Chords.

**A. Plain Scale: **

It is used to read in two dimensions only such as units and tenths and hundredths, kilometres and hectometres, metres and decimetres etc.

The scale consists of two lines about 3 mm apart, the bottom one being thicker than the top one. The whole length is divided into suitable number of equal parts or units, the first of which is further sub-divided into smaller parts or sub-units of the main unit. The primary and the secondary divisions are drawn perpendicular to the two lines and are made about 1.5 mm projecting above the top line.

ADVERTISEMENTS:

For convenience in reading, it should not involve any arithmetic calculation in measuring distance on the map. The main division should, therefore, represent one, the or hundred units etc.

**Example 1:**

Construct a scale 1 cm = 3 m to read to a metre and show on it 37 metres.

**Solution:**

Draw a line 20 cm long which represents 20 x 3= 60 m length. Divide it into 6 equal parts each representing 10 m length. Subdivide the extreme left division into ten equal parts, each sub-division reading 1 m. Place 0 at the right end of the extreme left division and mark the figures, counting from 0, in both the directions as shown in fig. 3.37. To read 37 metres, places one leg of the dividers at 30 and the other at 7.

**Caution:**

Avoid the common error of laying down centimetres and dividing the left hand centimetre into 3 parts and numbering the others as 3, 6, 9, 12 etc. Such a scale is an inconvenient for taking direct distances and will involve unnecessary counting.

**B. Diagonal Scale:**

This is used when it is required to read in three different dimensions such as units, tenths and hundredths; metres, decimetres and centimetres etc. The deficiency of plain scale of reading only in two dimensions is overcome by this scale.

**The principle of construction of a diagonal scale is based upon the fact that similar triangles have their like sides proportional and is explained below: **

Refer to fig. 3.39, suppose a short line AB is required to be divided into 10 equal parts.

Draw a line BC of any convenient length perpendicular to AB and divide it into 10 equal parts. Join AC and draw lines 1-1′, 2-2′, 3-3′ etc. parallel to AB.

It is obvious that the triangles CBA, C-1-1’, C-2-2’ etc. are similar.

**Now consider the ∆s CBA and C-9-9’: **

**Example 2:**

Construct a diagonal scale, 1 cm = 4 km and show on it 47.6 km.

**Solution: **

Draw a line 15 cm long which represents 15 x 4 = 60 km length. Divide it into 6 parts, each part represents 10 km. Sub-divide the extreme left hand division into 10 equal parts each sub-division reading up to 1 km. At the extreme left, draw a line perpendicular to the scale line and divide it into 10 equal parts of suitable length. Through each of these points, draw lines parallel to the scale line.

Project the points of sub-divisions of the extreme left division on the top most parallel line and then join them together diagonally.

Complete the scale as shown in fig. 3.40 measure 47.6 km, place one leg of the dividers at the intersection of vertical 40 and horizontal 6 and the other leg at the intersection of diagonal 7 and horizontal 6.

**C. Comparative or Corresponding Scale:**

When the given scale of a plain reads in a certain measure and it is required to construct a new scale for the same plan to read in some other measure such that the R.F. of both the scales remains the same, then the new scale is called the comparative or corresponding scale. The comparative scale has an advantage of taking measurements directly from the plan in the desired units without any calculation work.

**Example 3: **

The scale of a plan is, 1 inch = 200 ft, and it reads to a foot. Draw a comparative scale to read to a metre.

**Solution: **

The R.f of the new scale should also be the same so that it may correspond to the previous one.

i.e. new scale is, 1cm = 24m

To construct the required scale, take a line 25 cm long to represent a length of 25 x 24 = 600m. Divide it into 6 equal parts, each part reading to 100 m. Sub-divide the extreme left, hand part into 10 sub parts each representing 10m.

Then at the extreme left erect a perpendicular to the scale line and divide it into 10 equal parts of convenient length. Through each of these points, draw lines parallel to the scale line and complete the scale as shown in fig. 3.42.

Comparative Diagonal Scale, 1 cm = 24 m, corresponding to 1 in. = 200ft.

**D. Vernier Scale:**

The vernier is a device used for determining the fractional parts of the smallest division of the main scale more accurately than it can be done by simply estimating by eye. It consists of a small scale called the vernier scale which moves with its graduated edge along the graduated edge of a long fixed scale called the main scale. The scale may be either straight or curved.

**E. The Scale of Chords:**

It is used to make angles and to measure angles of any magnitude with high degree accuracy. It is generally marked on a rectangular protractor or on an ordinary box wood scale, the method of construction being given below in fig. 3.43.

Draw a line MN of suitable length. At N, draw a perpendicular NT to MN. With N as centre and radius NM, draw an arc MP cutting NT at P. Then the arc MP or the chord MP subtends an angle of 90° at the centre N.

Divide the arc MP into 18 equal parts each part therefore subtends at N an angle of = 5°. Now with M as centre turn down the divisions to 18 the line MR and complete the scale as shown. Then MR is the required scale of chords to read up to 5°.

It may be noted that the distance from M to each division on the scale is the chord of the angle containing that number of degrees e.g. M-30 is equal to the chord of 30° and M-60 is equal to the chord of 60° and so on. The chord of 60° (i.e. the distance M- 60) is always equal to the radius MN.

**Example 4:**

Construct angles of 20° and 35° by using scale of chords.

**Solution: **

(i) Draw a line OX of any suitable length as shown in fig. 3.44 (i). With O as centre, and radius equal to MN from the scale of chords fig. 3.43, draw an arc AD cutting OX at A. With A as centre and radius equal to 0-20 draw an arc cutting the previous arc AD at B. Join O with B. Then angle AOB is the required angle of 20°.

(ii) Similarly obtain the point B by drawing an arc with A as centre and radius equal to 0.35 as shown in fig. 3.44. (ii) Angle AOB is then the required angle of 35°.

Note:

If the angle to be constructed is greater than 90°, then it can be constructed in two parts so -than the sum of the two angles is equal to the given angle e.g. an angle of 105° can be constructed in two parts viz. 60° and 45°.