Download Engineering Drawing Notes and more Exercises Engineering Drawing and Graphics in PDF only on Docsity!
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,
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.
REPRESENTATIVE FACTOR (R.F.) =
A
USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC.
B LENGTH OF SCALE = R.F.^ X MAX. LENGTH TO BE MEASURED.
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
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.
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.
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, 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,
PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS
PLAIN SCALE
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.
PLAIN SCALE
MIN 10 0 10 20 30 40 50 MINUTES
R.F. = 1/
PLANE SCALE SHOWING METERS AND DECIMETERS.
KM 5 2.5 0 5 10 15 20 25 KM
DISTANCE TRAVELED IN 29 MINUTES. 14.5 KM
R.F. = 1 / 40,00,
DIAGONAL SCALE SHOWING KILOMETERS.
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
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.
10 9 8 7 6 5 4 3 2 1 0 M
M
M
438 meters
R.F. = 1 / 4000
DIAGONAL SCALE SHOWING METERS.
CONIC SECTIONS
ELLIPSE, PARABOLA AND HYPERBOLA ARE CALLED CONIC SECTIONS
BECAUSE
THESE CURVES APPEAR ON THE SURFACE OF A CONE
WHEN IT IS CUT BY SOME TYPICAL CUTTING PLANES.
Section Plane
Through Generators
Ellipse
Section Plane Parallel
to end generator.
Parabola
Section Plane
Parallel to Axis.
Hyperbola
OBSERVE
ILLUSTRATIONS
GIVEN BELOW..
These are the loci of points moving in a plane such that the ratio of it’s distances
from a fixed point And a fixed line always remains constant.
The Ratio is called ECCENTRICITY. (E)
A) For Ellipse E<
B) For Parabola E=
C) For Hyperbola E>
SECOND DEFINATION OF AN ELLIPSE:-
It is a locus of a point moving in a plane
such that the SUM of it’s distances from TWO fixed points
always remains constant.
{And this sum equals to the length of major axis .}
These TWO fixed points are FOCUS 1 & FOCUS 2
Refer Problem nos. 6. 9 & 12
Refer Problem no.
Ellipse by Arcs of Circles Method.
COMMON DEFINATION OF ELLIPSE, PARABOLA & HYPERBOLA:
A 3 2 1 B
C
D
Problem 2
Draw ellipse by Rectangle method.
Take major axis 100 mm and minor axis 70
mm long.
Steps: 1 Draw a rectangle taking major and minor axes as sides.
- In this rectangle draw both axes as perpendicular bisectors of each other..
- For construction, select upper left part of rectangle. Divide vertical small side and horizontal long side into same number of equal parts.( here divided in four parts)
- Name those as shown..
- Now join all vertical points 1,2,3,4, to the upper end of minor axis. And all horizontal points i.e.1,2,3,4 to the lower end of minor axis.
- Then extend C-1 line upto D- and mark that point. Similarly extend C-2, C-3, C-4 lines up to D-2, D-3, & D-4 lines.
- Mark all these points properly and join all along with ends A and D in smooth possible curve. Do similar construction in right side part.along with lower half of the rectangle.Join all points in smooth curve. It is required ellipse.
ELLIPSE
BY RECTANGLE METHOD
C
D
A 1 2 3 4 3 2 1 B
Problem 3:-
Draw ellipse by Oblong method.
Draw a parallelogram of 100 mm and 70 mm
long sides with included angle of 750.Inscribe Ellipse in
STEPS ARE SIMILAR TO^ it. THE PREVIOUS CASE (RECTANGLE METHOD) ONLY IN PLACE OF RECTANGLE, HERE IS A PARALLELOGRAM.
ELLIPSE
BY OBLONG METHOD
A B
D C
ELLIPSE
BY RHOMBUS METHOD
PROBLEM 5.
DRAW RHOMBUS OF 100 MM & 70 MM LONG
DIAGONALS AND INSCRIBE AN ELLIPSE IN IT.
STEPS:
- Draw rhombus of given dimensions.
- Mark mid points of all sides & name Those A,B,C,& D
- Join these points to the ends of smaller diagonals.
- Mark points 1,2,3,4 as four centers.
- Taking 1 as center and 1-A radius draw an arc AB.
- Take 2 as center draw an arc CD.
- Similarly taking 3 & 4 as centers and 3-D radius draw arcs DA & BC.
ELLIPSE
DIRECTRIX-FOCUS METHOD
PROBLEM 6:- POINT F IS 50 MM FROM A LINE AB.A POINT P IS MOVING IN A PLANE
SUCH THAT THE RATIO OF IT’S DISTANCES FROM F AND LINE AB REMAINS CONSTANT AND EQUALS TO 2/3 DRAW LOCUS OF POINT P. { ECCENTRICITY = 2/3 }
F ( focus )
DIRECTRIX
V
ELLIPSE
(vertex)
A
B
STEPS:
1 .Draw a vertical line AB and point F 50 mm from it. 2 .Divide 50 mm distance in 5 parts. 3 .Name 2nd part from F as V. It is 20mm and 30mm from F and AB line resp. It is first point giving ratio of it’s distances from F and AB 2/3 i.e 20/ 4 Form more points giving same ratio such as 30/45, 40/60, 50/75 etc. 5.Taking 45,60 and 75mm distances from line AB, draw three vertical lines to the right side of it.
- Now with 30, 40 and 50mm distances in compass cut these lines above and below, with F as center.
- Join these points through V in smooth curve. This is required locus of P.It is an ELLIPSE.
30mm
45mm
C
A B
PARABOLA
METHOD OF TANGENTS
Problem no.8: Draw an isosceles triangle of 100 mm long base and 110 mm long altitude.Inscribe a parabola in it by method of tangents.
Solution Steps:
- Construct triangle as per the given dimensions.
- Divide it’s both sides in to same no.of equal parts.
- Name the parts in ascending and descending manner, as shown.
- Join 1-1, 2-2,3-3 and so on.
- Draw the curve as shown i.e.tangent to all these lines. The above all lines being tangents to the curve, it is called method of tangents.
A
B
V
PARABOLA
( VERTEX )
F
( focus )
1 2 3 4
PARABOLA
DIRECTRIX-FOCUS METHOD
SOLUTION STEPS:
1.Locate center of line, perpendicular to AB from point F. This will be initial point P and also the vertex. 2.Mark 5 mm distance to its right side, name those points 1,2,3,4 and from those draw lines parallel to AB. 3.Mark 5 mm distance to its left of P and name it 1. 4.Take O-1 distance as radius and F as center draw an arc cutting first parallel line to AB. Name upper point P1 and lower point P2. (FP1=O1)
5.Similarly repeat this process by taking
again 5mm to right and left and locate P3P4.
6.Join all these points in smooth curve.
It will be the locus of P equidistance from line AB and fixed point F.
PROBLEM 9: Point F is 50 mm from a vertical straight line AB. Draw locus of point P, moving in a plane such that it always remains equidistant from point F and line AB.
O
P 1
P 2
