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Digvijaysinh Gohil

The document provides information about different types of scales used in engineering drawings, including: - Plain scales which represent two units or a unit and its subdivision, up to a single decimal place. Examples and problems of constructing plain scales are provided. - Diagonal scales which represent units up to two decimal places. The principles of constructing a diagonal scale with examples are explained. - Vernier scales which also represent dimensions up to two decimal places. - Comparative scales for comparing two different units. An example of constructing a comparative plain scale is given. - Scales of cords used for measuring and constructing angles. The document provides step-by-step instructions on how to construct each type

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Project FUNDA H e a r i t L e a r n i t Let’s make engineering more easy engineering108.com
SCALE engineering108.com
Scales 1. Basic Information 2. Types and important units 3. Plain Scales (3 Problems) 4. Diagonal Scales - information 5. Diagonal Scales (3 Problems) 6. Comparative Scales (3 Problems) 7. Vernier Scales - information 8. Vernier Scales (2 Problems) 9. Scales of Cords - construction 10. Scales of Cords (2 Problems) engineering108.com
FOR FULL SIZE SCALE R.F.=1 OR ( 1:1 ) MEANS DRAWING & OBJECT ARE OF SAME SIZE. Other RFs are described as 1:10, 1:100, 1:1000, 1:1,00,000 SCALES DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE ON STANDARD SIZE DRAWING SHEET. THIS REDUCTION CREATES A SCALE OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION.. SUCH A SCALE IS CALLED REDUCING SCALE AND THAT RATIO IS CALLED REPRESENTATIVE FACTOR. SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASED FOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE. HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY. REPRESENTATIVE FACTOR (R.F.) = = = = A USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC. B LENGTH OF SCALE = R.F. MAX. LENGTH TO BE MEASURED.X DIMENSION OF DRAWING DIMENSION OF OBJECT LENGTH OF DRAWING ACTUAL LENGTH AREA OF DRAWING ACTUAL AREA VOLUME AS PER DRWG. ACTUAL VOLUME V V 3 engineering108.com
1. PLAIN SCALES ( FOR DIMENSIONS UP TO SINGLE DECIMAL) 2. DIAGONAL SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 3. VERNIER SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 4. COMPARATIVE SCALES ( FOR COMPARING TWO DIFFERENT UNITS) 5. SCALE OF CORDS ( FOR MEASURING/CONSTRUCTING ANGLES) TYPES OF SCALES: = 10 HECTOMETRES = 10 DECAMETRES = 10 METRES = 10 DECIMETRES = 10 CENTIMETRES = 10 MILIMETRES 1 KILOMETRE 1 HECTOMETRE 1 DECAMETRE 1 METRE 1 DECIMETRE 1 CENTIMETRE BE FRIENDLY WITH THESE UNITS. engineering108.com
0 1 2 3 4 510 PLAIN SCALE:- This type of scale represents two units or a unit and it’s sub-division. METERS DECIMETERS R.F. = 1/100 4 M 6 DM PLANE SCALE SHOWING METERS AND DECIMETERS. PROBLEM NO.1:- Draw a scale 1 cm = 1m to read decimeters, to measure maximum distance of 6 m. Show on it a distance of 4 m and 6 dm. CONSTRUCTION:- a) Calculate R.F.= R.F.= 1cm/ 1m = 1/100 Length of scale = R.F. X max. distance = 1/100 X 600 cm = 6 cms b) Draw a line 6 cm long and divide it in 6 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 4 m 6 dm on it as shown. DIMENSION OF DRAWING DIMENSION OF OBJECT PLAIN SCALE engineering108.com
PROBLEM NO.2:- In a map a 36 km distance is shown by a line 45 cms long. Calculate the R.F. and construct a plain scale to read kilometers and hectometers, for max. 12 km. Show a distance of 8.3 km on it. CONSTRUCTION:- a) Calculate R.F. R.F.= 45 cm/ 36 km = 45/ 36 . 1000 . 100 = 1/ 80,000 Length of scale = R.F. max. distance = 1/ 80000 12 km = 15 cm b) Draw a line 15 cm long and divide it in 12 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 8.3 km on it as shown. KILOMETERS HECTOMETERS 8KM 3HM R.F. = 1/80,000 PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS 0 1 2 3 4 5 6 7 8 9 10 1110 5 PLAIN SCALE engineering108.com
PROBLEM NO.3:- The distance between two stations is 210 km. A passenger train covers this distance in 7 hours. Construct a plain scale to measure time up to a single minute. RF is 1/200,000 Indicate the distance traveled by train in 29 minutes. CONSTRUCTION:- a) 210 km in 7 hours. Means speed of the train is 30 km per hour ( 60 minutes) Length of scale = R.F. max. distance per hour = 1/ 2,00,000 30km = 15 cm b) 15 cm length will represent 30 km and 1 hour i.e. 60 minutes. Draw a line 15 cm long and divide it in 6 equal parts. Each part will represent 5 km and 10 minutes. c) Sub divide the first part in 10 equal parts,which will represent second unit or fraction of first unit. Each smaller part will represent distance traveled in one minute. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a proper look of scale. e) Show km on upper side and time in minutes on lower side of the scale as shown. After construction of scale mention it’s RF and name of scale as shown. f) Show the distance traveled in 29 minutes, which is 14.5 km, on it as shown. 0 10 20 30 40 5010 MINUTESMIN R.F. = 1/100 PLANE SCALE SHOWING METERS AND DECIMETERS. KMKM 0 5 10 15 20 255 2.5 DISTANCE TRAVELED IN 29 MINUTES. 14.5 KM PLAIN SCALE engineering108.com
We have seen that the plain scales give only two dimensions, such as a unit and it’s subunit or it’s fraction. 1 2 3 4 5 6 7 8 9 10 X Y Z The principle of construction of a diagonal scale is as follows. Let the XY in figure be a subunit. From Y draw a perpendicular YZ to a suitable height. Join XZ. Divide YZ in to 10 equal parts. Draw parallel lines to XY from all these divisions and number them as shown. From geometry we know that similar triangles have their like sides proportional. Consider two similar triangles XYZ and 7’ 7Z, we have 7Z / YZ = 7’7 / XY (each part being one unit) Means 7’ 7 = 7 / 10. x X Y = 0.7 XY :. Similarly 1’ – 1 = 0.1 XY 2’ – 2 = 0.2 XY Thus, it is very clear that, the sides of small triangles, which are parallel to divided lines, become progressively shorter in length by 0.1 XY. The solved examples ON NEXT PAGES will make the principles of diagonal scales clear. The diagonal scales give us three successive dimensions that is a unit, a subunit and a subdivision of a subunit. DIAGONAL SCALE engineering108.com
R.F. = 1 / 40,00,000 DIAGONAL SCALE SHOWING KILOMETERS. 0 100 200 300 400 500100 50 10 9 8 7 6 5 4 3 2 1 0 KM KM KM 569 km 459 km 336 km 222 km PROBLEM NO. 4 : The distance between Delhi and Agra is 200 km. In a railway map it is represented by a line 5 cm long. Find it’s R.F. Draw a diagonal scale to show single km. And maximum 600 km. Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km SOLUTION STEPS: RF = 5 cm / 200 km = 1 / 40, 00, 000 Length of scale = 1 / 40, 00, 000 X 600 X 105 = 15 cm Draw a line 15 cm long. It will represent 600 km.Divide it in six equal parts.( each will represent 100 km.) Divide first division in ten equal parts. Each will represent 10 km.Draw a line upward from left end and mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and complete diagonal scale. DIAGONAL SCALE engineering108.com
PROBLEM NO.5: A rectangular plot of land measuring 1.28 hectors is represented on a map by a similar rectangle of 8 sq. cm. Calculate RF of the scale. Draw a diagonal scale to read single meter. Show a distance of 438 m on it. Draw a line 15 cm long. It will represent 600 m.Divide it in six equal parts. ( each will represent 100 m.) Divide first division in ten equal parts.Each will represent 10 m. Draw a line upward from left end and mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and complete diagonal scale. DIAGONAL SCALE SOLUTION : 1 hector = 10, 000 sq. meters 1.28 hectors = 1.28 X 10, 000 sq. meters = 1.28 X 104 X 104 sq. cm 8 sq. cm area on map represents = 1.28 X 104 X 104 sq. cm on land 1 cm sq. on map represents = 1.28 X 10 4 X 104 / 8 sq cm on land 1 cm on map represent = 1.28 X 10 4 X 104 / 8 cm = 4, 000 cm 1 cm on drawing represent 4, 000 cm, Means RF = 1 / 4000 Assuming length of scale 15 cm, it will represent 600 m. 0 100 200 300 400 500100 50 10 9 8 7 6 5 4 3 2 1 0 M M M 438 meters R.F. = 1 / 4000 DIAGONAL SCALE SHOWING METERS.engineering108.com
10 9 8 7 6 5 4 3 2 1 0 CENTIMETRES MM CM R.F. = 1 / 2.5 DIAGONAL SCALE SHOWING CENTIMETERS. 0 5 10 155 4 3 2 1 PROBLEM NO.6:. Draw a diagonal scale of R.F. 1: 2.5, showing centimeters and millimeters and long enough to measure up to 20 centimeters. SOLUTION STEPS: R.F. = 1 / 2.5 Length of scale = 1 / 2.5 X 20 cm. = 8 cm. 1.Draw a line 8 cm long and divide it in to 4 equal parts. (Each part will represent a length of 5 cm.) 2.Divide the first part into 5 equal divisions. (Each will show 1 cm.) 3.At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. 4.Complete the scale as explained in previous problems. Show the distance 13.4 cm on it. 13 .4 CM DIAGONAL SCALE engineering108.com
COMPARATIVE SCALES: These are the Scales having same R.F. but graduated to read different units. These scales may be Plain scales or Diagonal scales and may be constructed separately or one above the other. EXAMPLE NO. 7 : A distance of 40 miles is represented by a line 8 cm long. Construct a plain scale to read 80 miles. Also construct a comparative scale to read kilometers upto 120 km ( 1 m = 1.609 km ) SOLUTION STEPS: Scale of Miles: 40 miles are represented = 8 cm : 80 miles = 16 cm R.F. = 8 / 40 X 1609 X 1000 X 100 = 1 / 8, 04, 500 CONSTRUCTION: Take a line 16 cm long and divide it into 8 parts. Each will represent 10 miles. Subdivide the first part and each sub-division will measure single mile. Scale of Km: Length of scale = 1 / 8,04,500 X 120 X 1000 X 100 = 14. 90 cm CONSTRUCTION: On the top line of the scale of miles cut off a distance of 14.90 cm and divide it into 12 equal parts. Each part will represent 10 km. Subdivide the first part into 10 equal parts. Each subdivision will show single km. 10 100 20 305 50 60 70 MILES40 10 0 10 20 30 40 50 60 70 80 90 100 110 KM 5 R.F. = 1 / 804500 COMPARATIVE SCALE SHOWING MILES AND KILOMETERS engineering108.com
COMPARATIVE SCALE: EXAMPLE NO. 8 : A motor car is running at a speed of 60 kph. On a scale of RF = 1 / 4,00,000 show the distance traveled by car in 47 minutes. SOLUTION STEPS: Scale of km. length of scale = RF X 60 km = 1 / 4,00,000 X 60 X 105 = 15 cm. CONSTRUCTION: Draw a line 15 cm long and divide it in 6 equal parts. ( each part will represent 10 km.) Subdivide 1st part in `0 equal subdivisions. ( each will represent 1 km.) Time Scale: Same 15 cm line will represent 60 minutes. Construct the scale similar to distance scale. It will show minimum 1 minute & max. 60min. 10 100 20 305 50 KM40 10 100 20 305 50 MINUTES40 MIN. KM 47 MINUTES 47 KM R.F. = 1 / 4,00,000 COMPARATIVE SCALE SHOWING MINUTES AND KILOMETERS engineering108.com
EXAMPLE NO. 9 : A car is traveling at a speed of 60 km per hour. A 4 cm long line represents the distance traveled by the car in two hours. Construct a suitable comparative scale up to 10 hours. The scale should be able to read the distance traveled in one minute. Show the time required to cover 476 km and also distance in 4 hours and 24 minutes. COMPARATIVE SCALE: SOLUTION: 4 cm line represents distance in two hours , means for 10 hours scale, 20 cm long line is required, as length of scale.This length of scale will also represent 600 kms. ( as it is a distance traveled in 10 hours) CONSTRUCTION: Distance Scale ( km) Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 60 km) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 km) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE km. Time scale: Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 1 hour) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 minutes) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE minute. 10 5 0 kM kM 060 60 120 180 240 300 360 420 480 540 060 1 2 3 4 5 6 7 8 9 HOURS MIN. 10 5 0 KILOMETERSDISTANCE SCALE TO MEASURE MIN 1 KM TIME SCALE TO MEASURE MIN 1 MINUTE. 4 hrs 24 min. ( 264 kms ) 476 kms ( 7 hrs 56 min.) engineering108.com
Figure to the right shows a part of a plain scale in which length A-O represents 10 cm. If we divide A-O into ten equal parts, each will be of 1 cm. Now it would not be easy to divide each of these parts into ten equal divisions to get measurements in millimeters. Now if we take a length BO equal to 10 + 1 = 11 such equal parts, thus representing 11 cm, and divide it into ten equal divisions, each of these divisions will represent 11 / 10 – 1.1 cm. The difference between one part of AO and one division of BO will be equal 1.1 – 1.0 = 0.1 cm or 1 mm. This difference is called Least Count of the scale. Minimum this distance can be measured by this scale. The upper scale BO is the vernier.The combination of plain scale and the vernier is vernier scale. Vernier Scales: These scales, like diagonal scales , are used to read to a very small unit with great accuracy. It consists of two parts – a primary scale and a vernier. The primary scale is a plain scale fully divided into minor divisions. As it would be difficult to sub-divide the minor divisions in ordinary way, it is done with the help of the vernier. The graduations on vernier are derived from those on the primary scale. 9.9 7.7 5.5 3.3 1.1 9 8 7 6 5 4 3 2 1 0A 0B engineering108.com
Example 10: Draw a vernier scale of RF = 1 / 25 to read centimeters upto 4 meters and on it, show lengths 2.39 m and 0.91 m .9 .8 .7 .6 .5 .4 .3 .2 .1 .99 .77 .55 .33 .11 01.1 0 1 2 31.0 SOLUTION: Length of scale = RF X max. Distance = 1 / 25 X 4 X 100 = 16 cm CONSTRUCTION: ( Main scale) Draw a line 16 cm long. Divide it in 4 equal parts. ( each will represent meter ) Sub-divide each part in 10 equal parts. ( each will represent decimeter ) Name those properly. CONSTRUCTION: ( vernier) Take 11 parts of Dm length and divide it in 10 equal parts. Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle Covering these parts of vernier. TO MEASURE GIVEN LENGTHS: (1) For 2.39 m : Subtract 0.99 from 2.39 i.e. 2.39 - .99 = 1.4 m The distance between 0.99 ( left of Zero) and 1.4 (right of Zero) is 2.39 m (2) For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91 – 0.11 =0.80 m The distance between 0.11 and 0.80 (both left side of Zero) is 0.91 m 1.4 2.39 m 0.91 m METERS METERS Vernier Scale engineering108.com
Example 11: A map of size 500cm X 50cm wide represents an area of 6250 sq.Kms. Construct a vernier scaleto measure kilometers, hectometers and decameters and long enough to measure upto 7 km. Indicate on it a) 5.33 km b) 59 decameters. Vernier Scale SOLUTION: RF = = = 2 / 105 Length of scale = RF X max. Distance = 2 / 105 X 7 kms = 14 cm AREA OF DRAWING ACTUAL AREAV 500 X 50 cm sq. 6250 km sq.V CONSTRUCTION: ( vernier) Take 11 parts of hectometer part length and divide it in 10 equal parts. Each will show 1.1 hm m or 11 dm and Covering in a rectangle complete scale. CONSTRUCTION: ( Main scale) Draw a line 14 cm long. Divide it in 7 equal parts. ( each will represent km ) Sub-divide each part in 10 equal parts. ( each will represent hectometer ) Name those properly. KILOMETERSHECTOMETERS 0 1 2 310 4 5 6 90 70 50 30 10 99 77 55 33 11 Decameters TO MEASURE GIVEN LENGTHS: a) For 5.33 km : Subtract 0.33 from 5.33 i.e. 5.33 - 0.33 = 5.00 The distance between 33 dm ( left of Zero) and 5.00 (right of Zero) is 5.33 k m (b) For 59 dm : Subtract 0.99 from 0.59 i.e. 0.59 – 0.99 = - 0.4 km ( - ve sign means left of Zero) The distance between 99 dm and - .4 km is 59 dm (both left side of Zero) 5.33 km59 dm engineering108.com
100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 SCALE OF CORDS OA CONSTRUCTION: 1. DRAW SECTOR OF A CIRCLE OF 900 WITH ‘OA’ RADIUS. ( ‘OA’ ANY CONVINIENT DISTANCE ) 2. DIVIDE THIS ANGLE IN NINE EQUAL PARTS OF 10 0 EACH. 3. NAME AS SHOWN FROM END ‘A’ UPWARDS. 4. FROM ‘A’ AS CENTER, WITH CORDS OF EACH ANGLE AS RADIUS DRAW ARCS DOWNWARDS UP TO ‘AO’ LINE OR IT’S EXTENSION AND FORM A SCALE WITH PROPER LABELING AS SHOWN. AS CORD LENGTHS ARE USED TO MEASURE & CONSTRUCT DIFERENT ANGLES IT IS CALLED SCALE OF CORDS. engineering108.com
100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 OA OA B O1 A1 B1 x z y PROBLEM 12: Construct any triangle and measure it’s angles by using scale of cords. CONSTRUCTION: First prepare Scale of Cords for the problem. Then construct a triangle of given sides. ( You are supposed to measure angles x, y and z) To measure angle at x: Take O-A distance in compass from cords scale and mark it on lower side of triangle as shown from corner x. Name O & A as shown. Then O as center, O-A radius draw an arc upto upper adjacent side.Name the point B. Take A-B cord in compass and place on scale of cords from Zero. It will give value of angle at x To measure angle at y: Repeat same process from O1. Draw arc with radius O1A1. Place Cord A1B1 on scale and get angle at y. To measure angle at z: Subtract the SUM of these two angles from 1800 to get angle at z. SCALE OF CORDS 300550 Angle at z = 180 – ( 55 + 30 ) = 950 engineering108.com
100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 OA PROBLEM 12: Construct 250 and 1150 angles with a horizontal line , by using scale of cords. CONSTRUCTION: First prepare Scale of Cords for the problem. Then Draw a horizontal line. Mark point O on it. To construct 250 angle at O. Take O-A distance in compass from cords scale and mark it on on the line drawn, from O Name O & A as shown. Then O as center, O-A radius draw an arc upward.. Take cord length of 250 angle from scale of cords in compass and from A cut the arc at point B.Join B with O. The angle AOB is thus 250 To construct 1150 angle at O. This scale can measure or construct angles upto 900 only directly. Hence Subtract 1150 from 1800.We get 750 angle , which can be constructed with this scale. Extend previous arc of OA radius and taking cord length of 750 in compass cut this arc at B1 with A as center. Join B1 with O. Now angle AOB1 is 750 and angle COB1 is 1150. B1 750 1150 B 250 A O O C A To construct 250 angle at O. To construct 1150 angle at O. SCALE OF CORDS engineering108.com
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4.scale engineering108.com

  • 1. Project FUNDA H e a r i t L e a r n i t Let’s make engineering more easy engineering108.com
  • 3. Scales 1. Basic Information 2. Types and important units 3. Plain Scales (3 Problems) 4. Diagonal Scales - information 5. Diagonal Scales (3 Problems) 6. Comparative Scales (3 Problems) 7. Vernier Scales - information 8. Vernier Scales (2 Problems) 9. Scales of Cords - construction 10. Scales of Cords (2 Problems) engineering108.com
  • 4. FOR FULL SIZE SCALE R.F.=1 OR ( 1:1 ) MEANS DRAWING & OBJECT ARE OF SAME SIZE. Other RFs are described as 1:10, 1:100, 1:1000, 1:1,00,000 SCALES DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE ON STANDARD SIZE DRAWING SHEET. THIS REDUCTION CREATES A SCALE OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION.. SUCH A SCALE IS CALLED REDUCING SCALE AND THAT RATIO IS CALLED REPRESENTATIVE FACTOR. SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASED FOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE. HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY. REPRESENTATIVE FACTOR (R.F.) = = = = A USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC. B LENGTH OF SCALE = R.F. MAX. LENGTH TO BE MEASURED.X DIMENSION OF DRAWING DIMENSION OF OBJECT LENGTH OF DRAWING ACTUAL LENGTH AREA OF DRAWING ACTUAL AREA VOLUME AS PER DRWG. ACTUAL VOLUME V V 3 engineering108.com
  • 5. 1. PLAIN SCALES ( FOR DIMENSIONS UP TO SINGLE DECIMAL) 2. DIAGONAL SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 3. VERNIER SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS) 4. COMPARATIVE SCALES ( FOR COMPARING TWO DIFFERENT UNITS) 5. SCALE OF CORDS ( FOR MEASURING/CONSTRUCTING ANGLES) TYPES OF SCALES: = 10 HECTOMETRES = 10 DECAMETRES = 10 METRES = 10 DECIMETRES = 10 CENTIMETRES = 10 MILIMETRES 1 KILOMETRE 1 HECTOMETRE 1 DECAMETRE 1 METRE 1 DECIMETRE 1 CENTIMETRE BE FRIENDLY WITH THESE UNITS. engineering108.com
  • 6. 0 1 2 3 4 510 PLAIN SCALE:- This type of scale represents two units or a unit and it’s sub-division. METERS DECIMETERS R.F. = 1/100 4 M 6 DM PLANE SCALE SHOWING METERS AND DECIMETERS. PROBLEM NO.1:- Draw a scale 1 cm = 1m to read decimeters, to measure maximum distance of 6 m. Show on it a distance of 4 m and 6 dm. CONSTRUCTION:- a) Calculate R.F.= R.F.= 1cm/ 1m = 1/100 Length of scale = R.F. X max. distance = 1/100 X 600 cm = 6 cms b) Draw a line 6 cm long and divide it in 6 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 4 m 6 dm on it as shown. DIMENSION OF DRAWING DIMENSION OF OBJECT PLAIN SCALE engineering108.com
  • 7. PROBLEM NO.2:- In a map a 36 km distance is shown by a line 45 cms long. Calculate the R.F. and construct a plain scale to read kilometers and hectometers, for max. 12 km. Show a distance of 8.3 km on it. CONSTRUCTION:- a) Calculate R.F. R.F.= 45 cm/ 36 km = 45/ 36 . 1000 . 100 = 1/ 80,000 Length of scale = R.F. max. distance = 1/ 80000 12 km = 15 cm b) Draw a line 15 cm long and divide it in 12 equal parts. Each part will represent larger division unit. c) Sub divide the first part which will represent second unit or fraction of first unit. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale. e) After construction of scale mention it’s RF and name of scale as shown. f) Show the distance 8.3 km on it as shown. KILOMETERS HECTOMETERS 8KM 3HM R.F. = 1/80,000 PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS 0 1 2 3 4 5 6 7 8 9 10 1110 5 PLAIN SCALE engineering108.com
  • 8. PROBLEM NO.3:- The distance between two stations is 210 km. A passenger train covers this distance in 7 hours. Construct a plain scale to measure time up to a single minute. RF is 1/200,000 Indicate the distance traveled by train in 29 minutes. CONSTRUCTION:- a) 210 km in 7 hours. Means speed of the train is 30 km per hour ( 60 minutes) Length of scale = R.F. max. distance per hour = 1/ 2,00,000 30km = 15 cm b) 15 cm length will represent 30 km and 1 hour i.e. 60 minutes. Draw a line 15 cm long and divide it in 6 equal parts. Each part will represent 5 km and 10 minutes. c) Sub divide the first part in 10 equal parts,which will represent second unit or fraction of first unit. Each smaller part will represent distance traveled in one minute. d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a proper look of scale. e) Show km on upper side and time in minutes on lower side of the scale as shown. After construction of scale mention it’s RF and name of scale as shown. f) Show the distance traveled in 29 minutes, which is 14.5 km, on it as shown. 0 10 20 30 40 5010 MINUTESMIN R.F. = 1/100 PLANE SCALE SHOWING METERS AND DECIMETERS. KMKM 0 5 10 15 20 255 2.5 DISTANCE TRAVELED IN 29 MINUTES. 14.5 KM PLAIN SCALE engineering108.com
  • 9. We have seen that the plain scales give only two dimensions, such as a unit and it’s subunit or it’s fraction. 1 2 3 4 5 6 7 8 9 10 X Y Z The principle of construction of a diagonal scale is as follows. Let the XY in figure be a subunit. From Y draw a perpendicular YZ to a suitable height. Join XZ. Divide YZ in to 10 equal parts. Draw parallel lines to XY from all these divisions and number them as shown. From geometry we know that similar triangles have their like sides proportional. Consider two similar triangles XYZ and 7’ 7Z, we have 7Z / YZ = 7’7 / XY (each part being one unit) Means 7’ 7 = 7 / 10. x X Y = 0.7 XY :. Similarly 1’ – 1 = 0.1 XY 2’ – 2 = 0.2 XY Thus, it is very clear that, the sides of small triangles, which are parallel to divided lines, become progressively shorter in length by 0.1 XY. The solved examples ON NEXT PAGES will make the principles of diagonal scales clear. The diagonal scales give us three successive dimensions that is a unit, a subunit and a subdivision of a subunit. DIAGONAL SCALE engineering108.com
  • 10. R.F. = 1 / 40,00,000 DIAGONAL SCALE SHOWING KILOMETERS. 0 100 200 300 400 500100 50 10 9 8 7 6 5 4 3 2 1 0 KM KM KM 569 km 459 km 336 km 222 km PROBLEM NO. 4 : The distance between Delhi and Agra is 200 km. In a railway map it is represented by a line 5 cm long. Find it’s R.F. Draw a diagonal scale to show single km. And maximum 600 km. Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km SOLUTION STEPS: RF = 5 cm / 200 km = 1 / 40, 00, 000 Length of scale = 1 / 40, 00, 000 X 600 X 105 = 15 cm Draw a line 15 cm long. It will represent 600 km.Divide it in six equal parts.( each will represent 100 km.) Divide first division in ten equal parts. Each will represent 10 km.Draw a line upward from left end and mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and complete diagonal scale. DIAGONAL SCALE engineering108.com
  • 11. PROBLEM NO.5: A rectangular plot of land measuring 1.28 hectors is represented on a map by a similar rectangle of 8 sq. cm. Calculate RF of the scale. Draw a diagonal scale to read single meter. Show a distance of 438 m on it. Draw a line 15 cm long. It will represent 600 m.Divide it in six equal parts. ( each will represent 100 m.) Divide first division in ten equal parts.Each will represent 10 m. Draw a line upward from left end and mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and complete diagonal scale. DIAGONAL SCALE SOLUTION : 1 hector = 10, 000 sq. meters 1.28 hectors = 1.28 X 10, 000 sq. meters = 1.28 X 104 X 104 sq. cm 8 sq. cm area on map represents = 1.28 X 104 X 104 sq. cm on land 1 cm sq. on map represents = 1.28 X 10 4 X 104 / 8 sq cm on land 1 cm on map represent = 1.28 X 10 4 X 104 / 8 cm = 4, 000 cm 1 cm on drawing represent 4, 000 cm, Means RF = 1 / 4000 Assuming length of scale 15 cm, it will represent 600 m. 0 100 200 300 400 500100 50 10 9 8 7 6 5 4 3 2 1 0 M M M 438 meters R.F. = 1 / 4000 DIAGONAL SCALE SHOWING METERS.engineering108.com
  • 12. 10 9 8 7 6 5 4 3 2 1 0 CENTIMETRES MM CM R.F. = 1 / 2.5 DIAGONAL SCALE SHOWING CENTIMETERS. 0 5 10 155 4 3 2 1 PROBLEM NO.6:. Draw a diagonal scale of R.F. 1: 2.5, showing centimeters and millimeters and long enough to measure up to 20 centimeters. SOLUTION STEPS: R.F. = 1 / 2.5 Length of scale = 1 / 2.5 X 20 cm. = 8 cm. 1.Draw a line 8 cm long and divide it in to 4 equal parts. (Each part will represent a length of 5 cm.) 2.Divide the first part into 5 equal divisions. (Each will show 1 cm.) 3.At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. 4.Complete the scale as explained in previous problems. Show the distance 13.4 cm on it. 13 .4 CM DIAGONAL SCALE engineering108.com
  • 13. COMPARATIVE SCALES: These are the Scales having same R.F. but graduated to read different units. These scales may be Plain scales or Diagonal scales and may be constructed separately or one above the other. EXAMPLE NO. 7 : A distance of 40 miles is represented by a line 8 cm long. Construct a plain scale to read 80 miles. Also construct a comparative scale to read kilometers upto 120 km ( 1 m = 1.609 km ) SOLUTION STEPS: Scale of Miles: 40 miles are represented = 8 cm : 80 miles = 16 cm R.F. = 8 / 40 X 1609 X 1000 X 100 = 1 / 8, 04, 500 CONSTRUCTION: Take a line 16 cm long and divide it into 8 parts. Each will represent 10 miles. Subdivide the first part and each sub-division will measure single mile. Scale of Km: Length of scale = 1 / 8,04,500 X 120 X 1000 X 100 = 14. 90 cm CONSTRUCTION: On the top line of the scale of miles cut off a distance of 14.90 cm and divide it into 12 equal parts. Each part will represent 10 km. Subdivide the first part into 10 equal parts. Each subdivision will show single km. 10 100 20 305 50 60 70 MILES40 10 0 10 20 30 40 50 60 70 80 90 100 110 KM 5 R.F. = 1 / 804500 COMPARATIVE SCALE SHOWING MILES AND KILOMETERS engineering108.com
  • 14. COMPARATIVE SCALE: EXAMPLE NO. 8 : A motor car is running at a speed of 60 kph. On a scale of RF = 1 / 4,00,000 show the distance traveled by car in 47 minutes. SOLUTION STEPS: Scale of km. length of scale = RF X 60 km = 1 / 4,00,000 X 60 X 105 = 15 cm. CONSTRUCTION: Draw a line 15 cm long and divide it in 6 equal parts. ( each part will represent 10 km.) Subdivide 1st part in `0 equal subdivisions. ( each will represent 1 km.) Time Scale: Same 15 cm line will represent 60 minutes. Construct the scale similar to distance scale. It will show minimum 1 minute & max. 60min. 10 100 20 305 50 KM40 10 100 20 305 50 MINUTES40 MIN. KM 47 MINUTES 47 KM R.F. = 1 / 4,00,000 COMPARATIVE SCALE SHOWING MINUTES AND KILOMETERS engineering108.com
  • 15. EXAMPLE NO. 9 : A car is traveling at a speed of 60 km per hour. A 4 cm long line represents the distance traveled by the car in two hours. Construct a suitable comparative scale up to 10 hours. The scale should be able to read the distance traveled in one minute. Show the time required to cover 476 km and also distance in 4 hours and 24 minutes. COMPARATIVE SCALE: SOLUTION: 4 cm line represents distance in two hours , means for 10 hours scale, 20 cm long line is required, as length of scale.This length of scale will also represent 600 kms. ( as it is a distance traveled in 10 hours) CONSTRUCTION: Distance Scale ( km) Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 60 km) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 km) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE km. Time scale: Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 1 hour) Sub-divide 1st part in SIX subdivisions.( Each will represent 10 minutes) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length. And complete the diagonal scale to read minimum ONE minute. 10 5 0 kM kM 060 60 120 180 240 300 360 420 480 540 060 1 2 3 4 5 6 7 8 9 HOURS MIN. 10 5 0 KILOMETERSDISTANCE SCALE TO MEASURE MIN 1 KM TIME SCALE TO MEASURE MIN 1 MINUTE. 4 hrs 24 min. ( 264 kms ) 476 kms ( 7 hrs 56 min.) engineering108.com
  • 16. Figure to the right shows a part of a plain scale in which length A-O represents 10 cm. If we divide A-O into ten equal parts, each will be of 1 cm. Now it would not be easy to divide each of these parts into ten equal divisions to get measurements in millimeters. Now if we take a length BO equal to 10 + 1 = 11 such equal parts, thus representing 11 cm, and divide it into ten equal divisions, each of these divisions will represent 11 / 10 – 1.1 cm. The difference between one part of AO and one division of BO will be equal 1.1 – 1.0 = 0.1 cm or 1 mm. This difference is called Least Count of the scale. Minimum this distance can be measured by this scale. The upper scale BO is the vernier.The combination of plain scale and the vernier is vernier scale. Vernier Scales: These scales, like diagonal scales , are used to read to a very small unit with great accuracy. It consists of two parts – a primary scale and a vernier. The primary scale is a plain scale fully divided into minor divisions. As it would be difficult to sub-divide the minor divisions in ordinary way, it is done with the help of the vernier. The graduations on vernier are derived from those on the primary scale. 9.9 7.7 5.5 3.3 1.1 9 8 7 6 5 4 3 2 1 0A 0B engineering108.com
  • 17. Example 10: Draw a vernier scale of RF = 1 / 25 to read centimeters upto 4 meters and on it, show lengths 2.39 m and 0.91 m .9 .8 .7 .6 .5 .4 .3 .2 .1 .99 .77 .55 .33 .11 01.1 0 1 2 31.0 SOLUTION: Length of scale = RF X max. Distance = 1 / 25 X 4 X 100 = 16 cm CONSTRUCTION: ( Main scale) Draw a line 16 cm long. Divide it in 4 equal parts. ( each will represent meter ) Sub-divide each part in 10 equal parts. ( each will represent decimeter ) Name those properly. CONSTRUCTION: ( vernier) Take 11 parts of Dm length and divide it in 10 equal parts. Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle Covering these parts of vernier. TO MEASURE GIVEN LENGTHS: (1) For 2.39 m : Subtract 0.99 from 2.39 i.e. 2.39 - .99 = 1.4 m The distance between 0.99 ( left of Zero) and 1.4 (right of Zero) is 2.39 m (2) For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91 – 0.11 =0.80 m The distance between 0.11 and 0.80 (both left side of Zero) is 0.91 m 1.4 2.39 m 0.91 m METERS METERS Vernier Scale engineering108.com
  • 18. Example 11: A map of size 500cm X 50cm wide represents an area of 6250 sq.Kms. Construct a vernier scaleto measure kilometers, hectometers and decameters and long enough to measure upto 7 km. Indicate on it a) 5.33 km b) 59 decameters. Vernier Scale SOLUTION: RF = = = 2 / 105 Length of scale = RF X max. Distance = 2 / 105 X 7 kms = 14 cm AREA OF DRAWING ACTUAL AREAV 500 X 50 cm sq. 6250 km sq.V CONSTRUCTION: ( vernier) Take 11 parts of hectometer part length and divide it in 10 equal parts. Each will show 1.1 hm m or 11 dm and Covering in a rectangle complete scale. CONSTRUCTION: ( Main scale) Draw a line 14 cm long. Divide it in 7 equal parts. ( each will represent km ) Sub-divide each part in 10 equal parts. ( each will represent hectometer ) Name those properly. KILOMETERSHECTOMETERS 0 1 2 310 4 5 6 90 70 50 30 10 99 77 55 33 11 Decameters TO MEASURE GIVEN LENGTHS: a) For 5.33 km : Subtract 0.33 from 5.33 i.e. 5.33 - 0.33 = 5.00 The distance between 33 dm ( left of Zero) and 5.00 (right of Zero) is 5.33 k m (b) For 59 dm : Subtract 0.99 from 0.59 i.e. 0.59 – 0.99 = - 0.4 km ( - ve sign means left of Zero) The distance between 99 dm and - .4 km is 59 dm (both left side of Zero) 5.33 km59 dm engineering108.com
  • 19. 100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 SCALE OF CORDS OA CONSTRUCTION: 1. DRAW SECTOR OF A CIRCLE OF 900 WITH ‘OA’ RADIUS. ( ‘OA’ ANY CONVINIENT DISTANCE ) 2. DIVIDE THIS ANGLE IN NINE EQUAL PARTS OF 10 0 EACH. 3. NAME AS SHOWN FROM END ‘A’ UPWARDS. 4. FROM ‘A’ AS CENTER, WITH CORDS OF EACH ANGLE AS RADIUS DRAW ARCS DOWNWARDS UP TO ‘AO’ LINE OR IT’S EXTENSION AND FORM A SCALE WITH PROPER LABELING AS SHOWN. AS CORD LENGTHS ARE USED TO MEASURE & CONSTRUCT DIFERENT ANGLES IT IS CALLED SCALE OF CORDS. engineering108.com
  • 20. 100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 OA OA B O1 A1 B1 x z y PROBLEM 12: Construct any triangle and measure it’s angles by using scale of cords. CONSTRUCTION: First prepare Scale of Cords for the problem. Then construct a triangle of given sides. ( You are supposed to measure angles x, y and z) To measure angle at x: Take O-A distance in compass from cords scale and mark it on lower side of triangle as shown from corner x. Name O & A as shown. Then O as center, O-A radius draw an arc upto upper adjacent side.Name the point B. Take A-B cord in compass and place on scale of cords from Zero. It will give value of angle at x To measure angle at y: Repeat same process from O1. Draw arc with radius O1A1. Place Cord A1B1 on scale and get angle at y. To measure angle at z: Subtract the SUM of these two angles from 1800 to get angle at z. SCALE OF CORDS 300550 Angle at z = 180 – ( 55 + 30 ) = 950 engineering108.com
  • 21. 100 200 300 400 500 600 700 800 900 00 0 10 20 4030 7050 60 9080 OA PROBLEM 12: Construct 250 and 1150 angles with a horizontal line , by using scale of cords. CONSTRUCTION: First prepare Scale of Cords for the problem. Then Draw a horizontal line. Mark point O on it. To construct 250 angle at O. Take O-A distance in compass from cords scale and mark it on on the line drawn, from O Name O & A as shown. Then O as center, O-A radius draw an arc upward.. Take cord length of 250 angle from scale of cords in compass and from A cut the arc at point B.Join B with O. The angle AOB is thus 250 To construct 1150 angle at O. This scale can measure or construct angles upto 900 only directly. Hence Subtract 1150 from 1800.We get 750 angle , which can be constructed with this scale. Extend previous arc of OA radius and taking cord length of 750 in compass cut this arc at B1 with A as center. Join B1 with O. Now angle AOB1 is 750 and angle COB1 is 1150. B1 750 1150 B 250 A O O C A To construct 250 angle at O. To construct 1150 angle at O. SCALE OF CORDS engineering108.com
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  • 23. Thank You For more tutorials on engineering subjects visit is at engineering108.com engineering108.com