Use of Scales in Engineering Drawing


It is not possible always to make drawings of an object to its actual size. If the actual linear dimensions of an object are shown in its drawing, the scale used is said to be a full size scale. Wherever possible, it is desirable to make drawings to full size.

Reducing and Enlarging Scales

Objects which are very big in size can not be represented in drawing to full size. In such cases the object is represented in reduced size by making use of reducing scales. Reducing scales are used to represent objects such as large machine parts, buildings, town plans etc. A reducing scale, say 1: 10 means that 10 units length on the object is represented by 1 unit length on the drawing.

Similarly, for drawing small objects such as watch parts, instrument components etc., use of full scale may not be useful to represent the object clearly. In those cases enlarging scales are used. An enlarging scale, say 10: 1 means one unit length on the object is represented by 10 units on the drawing.

The designation of a scale consists of the word. SCALE, followed by the indication of its ratio as follows. (Standard scales are shown in Fig. 1)

Scale 1: 1 for full size scale
Scale 1: x for reducing scales (x = 10, 20 …… etc.,)
Scale x: 1 for enlarging scales.

Note: For all drawings the scale has to be mentioned without fail.

Fig. 1 Scales

Representative Fraction

The ratio of the dimension of the object shown on the drawing to its actual size is called the Representative Fraction (RF).

    \[RF = \frac{\text {Drawing of an Object}}{\text {Its actual size}}(\text {in same units})\]

For example, if an actual length of 3 metres of an object is represented by a line of 15 mm length on the drawing

    \[ RF= \frac{15mm}{3m} = \frac{15mm}{(3\times1000)mm} = \frac{1}{200} or 1 : 200\]

If the desired scale is not available in the set of scales it may be constructed and then used.

Metric Measurements

10 millimetres (mm) = 1 centimetre( cm)
10 centimetres (cm) = 1 decimetre(dm)
10 decimetre (dm) = 1 metre(m)
10 metres (m) = 1 decametre (dam)
10 decametre (dam) = 1 hectometre (bm)
10 hectometres (bm) = 1 kilometre (km)
1 hectare = 10,000 m2

Types of Scales

The types of scales normally used are:

1. Plain scales.
2. Diagonal Scales.
3. Vernier Scales.

Plain Scales

A plain scale is simply a line which is divided into a suitable number of equal parts, the first of which is further sub-divided into small parts. It is used to represent either two units or a unit and its fraction such as km and hm, m and dm, cm and mm etc.

Problem 1 : On a survey map the distance between two places 1 km apart is 5 cm. Construct the scale to read 4.6 km.

Solution: (Fig 2)

    \[RF = \frac{5cm}{1\times1000\times100cm} = \frac{1}{20000}\]

    \[\text {If x is the drawing size required x} = 5(1000)(10) \times \frac{1}{20000}\]

    \[\text {Therefore, x = 25 cm}\]

Note: If 4.6 km itself were to be taken x = 23 cm. To get 1 km divisions this length has to be divided into 4.6 parts which is difficult. Therefore, the nearest round figure 5 km is considered. When this length is divided into 5 equal parts each part will be 1 km.

1. Draw a line of length 25 cm.
2. Divide this into 5 equal parts. Now each part is 1 km.
3. Divide the first part into 10 equal divisions. Each division is 0.1 km.
4. Mark on the scale the required distance 4.6 km.

Plain Scale
Fig. 2 Plain Scale

Problem 2 : Construct a scale of 1:50 to read metres and decimetres and long enough to measure 6 m. Mark on it a distance of 5.5 m.

Construction (Fig. 3)

1. Obtain the length of the scale as:

    \[RF \times 6m =\frac{1}{50} \times 6 \times 100 = 12 cm\]

2. Draw a rectangle strip of length 12 cm and width 0.5 cm.
3. Divide the length into 6 equal parts, by geometrical method each part representing 1m.
4. Mark O(zero) after the first division and continue 1,2,3 etc., to the right of the scale.
5. Divide the first division into 10 equal parts (secondary divisions), each representing 1 cm.
6. Mark the above division points from right to left.
7. Write the units at the bottom of the scale in their respective positions.
8. Indicate RF at the bottom of the figure.
9. Mark the distance 5.5 m as shown.

Fig. 3

Problem 3 : The distance between two towns is 250 km and is represented by a line of length 50 mm on a map. Construct a scale to read 600 km and indicate a distance of 530 km on it.

Solution: (Fig 4)

1. Determine the RF value as

    \[\frac{50mm}{250km} = \frac{50}{250\times1000\times1000} = \frac{1}{5\times10^6}\]

2. Obtain the length of the scale as:

    \[\frac{1}{5\times10^6}\times600km = 120mm\]

3. Draw a rectangular strip of length 120 mm and width 5 mm.
4. Divide the length into 6 equal parts, each part representing 10 km.
5. Repeat the steps 4 to 8 of construction in Fig 2. suitably.
6. Mark the distance 530 km as shown.

Fig. 4

Problem 4: Construct a plain scale of convenient length to measure a distance of 1 cm and mark on it a distance of 0.94 em.

Solution: (Fig 5)

This is a problem of enlarged scale.
1. Take the length of the scale as 10 cm
2. RF = 10/1, scale is 10:1
3. The construction is shown in Fig 5

Fig. 5

Diagonal Scales

Plain scales are used to read lengths in two units such as metres and decimetres, centimetres and millimetres etc., or to read to the accuracy correct to first decimal.

Diagonal scales are used to represent either three units of measurements such as metres, decimetres, centimetres or to read to the accuracy correct to two decimals.

Principle of Diagonal Scale (Fig 6)

1. Draw a line AB and errect a perperrdicular at B.
2. Mark 10 equi-distant points (1,2,3, etc) of any suitable length along this perpendicular and mark C.
3. Complete the rectangle ABCD
4. Draw the diagonal BD.
5. Draw horizontals through the division points to meet BD at 1′ , 2′ , 3′ etc.

Considering the similar triangles say BCD and B44′

    \[\frac{B4'}{CD} = \frac{B4}{BC}; = \frac{4}{10} \times BC \times\frac{1}{BC} = \frac{4}{10}; 44' = 0.4CD\]

Principle of Diagonal Scale
Fig. 6 Principle of Diagonal Scale

Thus, the lines 1-1′,2 – 2′, 3 – 3′ etc., measure O.lCD, 0.2CD, 0.3CD etc. respectively. Thus, CD is divided into 1110 the divisions by the diagonal BD, i.e., each horizontal line is a multiple of 1/10 CD.

This principle is used in the construction of diagonal scales.

Note: B C must be divided into the same number of parts as there are units of the third dimension in one unit of the secondary division.

Problem 5 : on a plan, a line of 22 em long represents a distance of 440 metres. Draw a diagonal scale for the plan to read upto a single metre. Measure and mark a distance of 187 m on the scale.

Solution: (Fig 7)

Diagonal Scale
Fig. 7 Diagonal Scale


    \[RF = \frac{22}{440\times l00} = \frac{1}{2000}\]

2. As 187 m are required consider 200 m.
Therefore drawing size =

    \[\text {RF x actual size} = \frac{1}{2000} \times 200 \times 100 = 10 cm\]

When a length of 1 0 cm representing 200 m is divided into 5 equal parts, each part represents
40 m as marked in the figure.
3. The first part is sub-divided into 4 divisions so that each division is 10 cm
4. On the diagonal portion 10 divisions are taken to get 1 m.
5. Mark on it 187 m as shown.

Problem 6 : An area of 144 sq cm on a map represents an area of 36 sq km on the field. Find the RF of the scale of the map and draw a diagonal scale to show Km, hectometres and decametres and to measure upto 10 km. Indicate on the scale a distance 7 km, 5 hectometres and 6 decemetres.

Solution: (Fig. 38)

1. 144 sq cm represents 36 sq km or 12 cm represent 6 km

    \[RF = \frac{12}{6\times 1000 \times 100} = \frac{1}{5000}\]

    \[\text{Drawing size x = RF x actual size =} \frac{10\times1000\times100}{50000} = 20cm\]

Fig. 8

2. Draw a length of 20 cm and divide it into 10 equal parts. Each part represents 1 km.
3. Divide the first part into 10 equal subdivisions. Each secondary division represents 1 hecometre
4. On the diagonal scale portion take 10 eqal divisions so that 1110 ofhectometre = 1 decametre is obtained.
5. Mark on it 7.56 km. as shown.

Problem 7 : Construct a diagonal scale 1/50, showing metres, decimetres and centimetres, to measure upto 5 metres. Mark a length 4. 75 m on it.

Solution: (Fig 9)

1. Obtain the length of the scale as

    \[\frac{1}{50}\times5\times100 = 10cm\]

2. Draw a line A B, 10 cm long and divide it into 5 equal parts, each representing 1 m.
3-. -Divide the first part into 10 equal parts, to represent decimetres.
4. Choosing any convenient length, draw 10 equi-distant parallel lines alJove AB and complete the rectangle ABC D.
5. Erect perpendiculars to the line A B, through 0, 1,2,3 etc., to meet the line C D.
6. Join D to 9, the first sub-division from A on the main scale AB, forming the fIrst diagonal.
7. Draw the remaining diagonals, parallel to the first. Thus, each decimetre is divided into II 10th division by diagonals.
8. Mark the length 4.75m as shown.

Fig. 9

Vernier Scales

The vernier scale is a short auxiliary scale constructed along the plain or main scale, which can read upto two decimal places.

The smallest division on the main scale and vernier scale are 1 msd or 1 vsd repectively. Generally (n+ 1) or (n-l) divisions on the main scale is divided into n equal parts on the vernier scale.

    \[\text {Thus, 1 vsd} = \frac{(n-1)}{n} \text {msd or} \left[1-\frac{1}{n}\right]\text{msd}\]

When 1 vsd < 1 it is called forward or direct vernier. The vernier divisions are numbered in the same direction as those on the main scale.

When 1 vsd> 1 or (1 + lin), It is called backward or retrograde vernier. The vernier divisions are numbered in the opposite direction compared to those on the main scale.

The least count (LC) is the smallest dimension correct to which a measurement can be made with a vernier.

For forward vernier, L C = (1 msd – 1 vsd)

For backward viermier, LC = (1 vsd – 1 msd)

Problem 8 : Construct a forward reading vernier scale to read distance correct to decametre on a map in which the actual distances are reduced in the ratio of 1 : 40,000. The scale should be long enough to measure upto 6 km. Mark on the scale a length of 3.34 km and 0.59 km.

Solution: (Fig. 10)


    \[\text{RF = 1/40000; length of drawing} = \frac{6\times1000\times100}{40000} = 15cm\]

2. 15 em is divided into 6 parts and each part is 1 km
3. This is further divided into 10 divitions and each division is equal to 0.1 km = 1 hectometre.
Ims d = 0.1 km = 1 hectometre
L.C expressed in terms of m s d = (111 0) m s d
L C is 1 decametre = 1 m s d – 1 v s d
1 v s d = 1 – 1110 = 9110 m s d = 0.09 km
4. 9 m sd are taken and divided into 10 divisions as shown. Thus 1 vsd = 9110 = 0.09 km
5. Mark on itbytaking6vsd=6x 0.9 = 0.54km, 28msd(27 + 1 on the LHS of 1) =2.8 kmand Tota12.8 + 0.54 = 3.34 km.
6. Mark on it 5 msd = 0.5 km and add to it one vsd = 0.09, total 0.59 km as marked.

Forward Reading Vernier Scale
Fig. 10 Forward Reading Vernier Scale

Problem 9 : construct a vernier scale to read metres, decimetres and centimetres and long enough to measure upto 4m. The RF of the scale in 1120. Mark on it a distance of 2.28 m.

Solution: (Fig 11)

Backward or Retrograde Vernier scale
1. The smallest measurement in the scale is cm.
Therefore LC = O.Olm
2. Length of the scale = RF x Max. Distance to be measured

    \[=\frac{1}{20}\times 4m=\frac{1}{20}\times 400=20cm\]

Backward or Retrograde Vernier Scale
Fig. 11 Backward or Retrograde Vernier Scale

3. Draw a line of 20 cm length. Complete the rectangle of20 em x 0.5 em and divide it into
4 equal parts each representing 1 metre. Sub divide all into 10 main scale divisions.

1 msd = 1m/10 = 1dm.

4. Take 10+ 1 = 11 divisions on the main scale and divide it into 10 equal parts on the vernier scale by geometrical construction.

Thus 1vsd= 11msd/10= 1.1dm= 11cm

5. Mark 0, 55, 110 towards the left from 0 (zero) on the vernier scale as shown.
6. Name the units of the divisions as shown.

7. 2.28m = (8 x vsd) + 14msd)
= (8 x 0.11m) + (14 x O.lm)
= 0.88 + 1.4 = 2.28m.

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