Engineering DrawingI Year B.Tech

ByNN..SSRRIINNIIVVAASSAA RREEDDDDYY..,, M.Tech.

Sr. Assistant ProfessorDepartment of Mechanical EngineeringVardhaman College of Engineering

SCALESUnit-I

Representative Fraction

- The Ratio of length of the drawing to the actual length of the object represented is called RepresentativeFraction (R.F).

- R.F. is less than unity is drawn on reduced scale, if greater than unity enlarging scale, if equal to one fullscale.

Scales on Drawing Sheeti) Representative fraction of the scale.ii) The units in which it must present i.e. (mm, cm, dm, inch, feet).iii) The maximum length which it must show (original length).

Types of Scales: 1) Plane Scale- It consists of a line divided into suitable no. of equal parts or units. The 1st of which is subdivided into

smaller parts.- Plane scale represent either two unit (or) a unit and it's subdivision.Procedure- The zero should be placed at the end of 1st main division between the unit and it's subdivision.- From the zero mark unit's should be numbered to the right and subdivision to the left.- The names of the units and the subdivisions should be stated clearly below or at the respective ends.- The name of the scale i.e. 1:10 or it's representative fraction should be mentioned below the scale.Problem-1

Construct a scale of 1:4 to show cm and long enough to measure upto 5dm and show distance of 3.7 dm onscale.

Procedure1. Determine R.F of the scale, here it is 1/4.2. Determine length of the scale

= R.F × max.length= 1/4 × 5dm= 12.5 cm

3. Draw a line of 12.5 cm long and divided it into 5 equal division each representing 1 dm.4. Mark O at the end of 1st division and 1, 2, 3 & 4 at the end of the each of each subsequent division to it's

right.5. Divide the 1st division into 10 equal parts each representing 1 cm.6. Mark cm to the left of zero and decimeter to the right corner.7. To distinguish the division clearly show the scale as a rectangle of small width (about 3mm).8. Draw the division line showing decimeter throughout the width of the scale.9. Draw the lines for subdivision slightly shorter as shown in figure.10. Draw thick and dark horizontal lines in the middle of all alternative division and subdivision.

2. Diagonal Scale

- It is used when every minute distances such as 0.1 mm etc are to be accurately measured or whenmeasurements are required in 3 units.Ex: DM, CM, MM, Yard, Foot and Inches.

- Small divisions of short are obtained by the principle of diagonal division.

Principle of Diagonal Scale- To obtain division of a given short line AB in multiples of 1/10 of it's length.

Ex: 0.1 of AB, 0.2 of AB etc.Procedure1. At one end say B draw a line perpendicular to AB and along it a step of 10 equal divisions of any length

(but uniform) starting from B and ending at C.2. No.of divisions points 9, 8, 7, 6 upto C. Join A with C.3. Draw lines parallel to AB and cutting AC at 1', 2' etc. It is evident that Δle 1'1C, 2'2C, 3'3C soon. ABC are

similar so 1'1 = 0.1 AB, 2'2 = 0.2 AB etc.

Problem

Construct a diagonal scale of 3:200 i.e. showing meters, decimeter and centimeter and to measure a

distance upto 6m. Show a distance of 4m, 5dm, 6cm.Procedure1. Draw a line AB = 9cm and divide it into 6 equal parts each part will show a meter.2. Divide the 1st part i.e. left side of O in 10 equal divisions each showing dm or 0.1m.3. At a draw a perpendicular and set off along it 10 equal divisions of any length ending at D.4. Complete the rectangle ABCD, draw perpendicular at meter division 0, 1, 2, 3 and 4.5. Draw horizontal lines through the division points on AD.6. Join D with the end of 1st division along AO i.e. AO at point 9.7. To the remaining points drawn parallel to D9.

10 equal division AD ; Draw line D to AO (9th part)

21: 66

3

8. In ΔOFE, FE represents 1DM or 0.1M each H.L below FE progressively diminishes in length by 0.1 FE.Thus, next line below FE = 0.9 FE and represents 0.9 DM or 0.09M or 9CM.

Geometrical Construction1. Bisecting a line.2. To draw perpendiculars.3. To draw parallel lines.4. To divide a line.5. To bisect an angle.6. To trisect an angle.7. To find centre of an arc.8. To construct equilateral Δle.9. To construct squares.10. To construct regular polygon.11. Special method for drawing regular polygon.12. Regular polygon inscribed in circles.13. To draw regular figures using T-square and set square.14. To draw tangents.15. To find length of arcs.16. Circles and lines in contact.17. Inscribed circles.

1. Bisecting a line (line length = 10 cm)Procedure- Let AB be the given line with center A and radius greater than half of AB draw an arcs on both sides of AB.- With centre and the same radius draw arcs intersecting the previous arcs at C and D. Draw a line joining C

and D, cutting A and B at E. Then,

- Further CD bisects AB at right angles.

- AB is thick line and also CD is thick but lighter than AB.

1AE EB AB

2= =

Arcs must be lighter

2. To bisect a given Arc (R = 5cm)

- Let AB be the given arc with centre A and radius greater D half of AB draw arcs on either side of AB.- Centre B with same radius draw arcs intersecting previous arcs at C and D. Draw a line joining C and D

upto O and cutting AB at E then

3. To draw perpendicular lines (10 cm, P = 4 cm from B)- To draw a perpendicular to a given line from a point within it.Procedure - Case-I- Let the point is near middle of the line. Let AB be the given line and P the point in it.- With P as centre and any convenient radius draw an arc cutting AB at C and D.- With any radius r2 > r1 and centre C and D draw arcs intersecting each other at O. Draw a line joining P and

O, then PO is required perpendicular.

Case-II (AB = 10 cm, P is at 2 cm)- When the point is near or end of the line.- Let AB be the given line and P the point in it.Method-I1. With any point O as centre and radius equal to OP draw an arc greater than semicircle cutting AB at C.2. Draw a line joining C and O and produce it to cut the arc at Q. Draw a line joining P and Q. then PQ is

required perpendicular.

1AE EB AB

2= =

AB and PQ are thick lines

AB is thick lineOP is thick line

centre of arc and arcs thick

Method-II- With P as center and any convenient radius draw an arc cutting AB at C.

- With the same radius cut the arc in two equal divisions CD and DE.- With the same radius with centre D and E draw an arcs intersecting each other at Q. Draw a line joining P

and Q then PQ is required perpendicular.

Case-III- To draw a perpendicular line to a point outside it (nearer to midpoint)Procedure (10cm, above AB 4cm)- Let AB be the given line and P is the point.- With centre P and any convenient radius draw an arc cutting AB at C and D.- With any radius greater than 1/2 CD with centre C and D draw arcs intersecting each other at E. Draw a

line joining P and E and cutting AB at Q. PQ is required perpendicular.

Case-IVWhen the point nearer to end then centre of line and outside of AB.

- With centre A and radius equal to AB draw an arc EF cutting AB (or) AB produced at C.- With centre C and radius equal to CP draw an arc cutting EF at D. Draw a line joining P and D and

intersecting AB at Q then PQ is required perpendicular.

4. To draw parallel lines- To draw a line through a given point parallel to a given straight line AB = 10cm parallel point = 4cm above

AB.Procedure- Let AB be the given and P is the given point with centre P and any convenient radius draw an arc CD cutting

AB at E.- With centre E with same radius draw an arc cutting AB at F. With centre E and radius equal to FP draw an

arc to CD at Q.- Draw a straight line through P and Q then this is the required line.

Case-II- To draw a line parallel to at a distance from a given.ProcedureLet AB is the given line and r is given distance, mark P, Q, S on AB of convenient distance.

- With P, Q and S as centre and radius r draw arcs on same side of AB. Draw the line CD just touching the 3arcs CD is required line parallel to AB.

Rule-1- For the force components is any quadrant away from origin and making angle θ (making an angle with

either of the axis).- The force makes an angle with either x-axis or y-axis, corresponding axis component is always cosθ and

the opposite axis is sinθ. Apply the sin value corresponding quadrant.

Rule-2- Force component towards horizon making an angle θ.I-quadrant

- Change the position into opposite quadrant and the angle with the same axis, then apply rule-1.

II-Quadrant

III-Quadrant

IV-Quadrant

Centre of gravity

1. For a line or uniform rod - mid point.2. Circle - centre3. Semi circle - 4r/3π from the base.4. Quadrant - 4r/3π from the centre.5. 11gm, square, rectangle - intersection of diagonals.6. Δle - h/3 from base and intersection of medians.

5. To divide a line- Let AB be the given line to be divided into 7 equal parts (AB = 7cm)1. Draw a line AC of any length inclined at some convenient angle to AB but less than 90°.2. From A and along AC cut off with a divider 7 equal divisions of any convenient length. Draw a line joining

B and 7th division on AC.3. With the aid of two set square's draw lines through 1, 2, 3 etc parallel to B7, intersecting AB at points 1',

2', 3' etc. Thus dividing it into 7 equal parts.

6. To divide a given straight line into unequal parts- Let AB be the given to be divided into unequal parts say 1/6, 1/5, 1/4, 1/3, 1/2 (AB = 12cm)1. Draw a line AB of given length say 120mm.2. Draw perpendicular AD and BC at the ends AB and complete rectangle ABCD.3. Join diagonals AC and BD intersecting at E. From E onto AB draw perpendicular then AF = 1/2 AB.4. Join D and F, the line DF intersect the diagonal AC at G. Drop perpendicular from G to AB then AH = 1/3

AB.5. Join D and H intersecting diagonal AC at I, from I draw perpendicular on to AB then AJ = 1/4 AB, similarly

for other fraction.

7. To bisect an angle- Let ABC be the given angle with B as centre and any radius draw an arc cutting AB at D and BC at E.- With centres B and E with the same or convenient radius draw arcs intersecting each other at F.- Draw a line joining B and F which bisects the ∠ABC. ∠ABF = ∠FBC

8. To draw a line inclined to a given line at an angle equal to a given angle- Let PQ be the given line and ∠AOB the given angle, with O as centre and any radius draw an arc cutting

OA at C and OB at D.- With the same radius and center P draw an arc EF cutting PQ at F. With F as centre and radius equal to CD

draw an arc cutting arc EF at G.- From P draw a line passing through G, then PR is required line.

9. To trisect an angle- Let ABC be the given right angle with centre B and any radius draw an arc cutting AD at B and BC at E.- With the same radius with centre B and E draw arc's cutting arc DE at P and Q. Draw lines joining B with

P and Q. BP and PQ trisect the right angle ABC. Thus angle ∠ABP = ∠PBQ = ∠QBC = 1/3 ∠ABC

10. To find centre of a given arc- Let AB be the given arc. In AB draw two parts CD and EF of any length.- Draw perpendicular bisectors of CD and EF intersecting each other at O, then O is required centre.

11. Construct Equilateral ΔΔles

Given length AB = 6 cm- With center as A and B and radius equal to AB draw arc intersecting each other at C. Draw lines joining C

with A and B, then ABC is the required Δle.

12. Construct a squareLength of side = 5 cm

- Draw a line AB equal to given line, at A draw a line AE perpendicular to AB with center A and radius ABdraw an arc cutting AE at D.

- With centre B and D with same radius draw arcs intersecting at C. Draw lines joining C with B and D.

13. Construct a Pentagon- Draw a line equal to given length.- Outside A and B with angle 72° draw two lines AE and BF.- With centre A and B and radius equal to AB, cuts the line AE at P and BF at Q.- With the same radius P and Q as centre's draw two arc's intersecting at R, then ABQRP is required pentagon.

Data required for construction of scale1. R.F of the scale2. The maximum length of scale to be drawn on drawing sheet.3. Least count of scale i.e. minimum length which the scale should show and measure.4. The maximum length of the scale to be drawn on sheet is determined by

Length of scale(L) = R.F × maximum lengthmostly it varies from 10 cm to 30 cm

Units of MeasurementMetric system1 km = 10 hecta meter1 hecto = 10 deca meter1 deca = 10 meter1 metre = 10 deci meter1 deci = 10 cm1 cm = 10 mmBritish System1 league = 3 miles1 mile = 8 furlongs1 furlong = 10 chains1 chain = 22 yards1 yard = 3 feet1 foot = 12 inch1 inch = eight 8

Conversion between metric and British1 mile = 1.609 km1 inch = 25.4 mm1 area = 100 m2

1 hectare = 100 acre1 acre = 1000 m2

1 sq. mile = 640 acres1 acre = 10 sq. chain = 4840 sq. yards

Problems6. Construct a scale to measure upto 50m if 1 cm represents 4m. Find it's R.F. Mark and measure a distance

of 37m on it.Solution

1R.F.

400=

Now,

= 12.5 cm

Geometrical Construction Problems

1. To Bisect a lineQ. Bisect an 80mm long line.

2. Perpendiculars to a lineQ. To draw a perpendicular to the given line from a point within it.3. Draw a perpendicular line to a 100 mm long straight lineQ. AB at point P lieing on the line at a distance of 40 mm from end A.Construction

Draw a 100 mm long line AB. Mark a point on it at 40 mm from end A.

Method-ISet the compass with convenient radius and P as centre strike arcs to intersect AB at C and D reset thecompass to any radius larger than 1/2 CD and with C and D as centres respectively strike arcs on one sideof the line to intersect a point E.

- Draw a line from P through E which is perpendicular to AB.Method-II- Set the needle point of the compass at any point above the line (say O) and draw a circle of radius OP.- The circle strikes the line AB at points P and C. Draw a line joining C and O and produce it to meet the

circle at point D. Join P and D which is perpendicular to line AB.Method-IIISet the compass to any convenient radius with P as strike arc CD to intersect AB at C. With the same radius (PC)

strike two divisions CE and EF respectively on arc.- Reset the compass to any radius larger than 1/2 EF and with EF as centres respectively strike arcs to

intersect at point G. Join GP which is perpendicular to line AB.To draw perpendicular to a given line

Method-I Method-II Method-III

AB → thick

1L

40=

08

5× 0

Q. Draw a perpendicular to 100mm long line AB from point P lieing 80mm from end A and 60mm from endB.

- Draw a 100mm long line AB draw arcs of 60 and 80mm from center A and B respectively to intersect atpoint P.

Method-I

Method-II

Method-I- Set the needle point of compass at P and strike an arc to intersect AB at C and D. With C and D as centre's

and any radius larger than 1/2 CD strike arcs to intersect at E.- Draw a line from P through E, the line PE perpendicular to AB.Method-II- Strike arcs with ends A and B as center's and radii AP and BP respectively to intersect at a point Q on other

side of line AB. Draw a line from P through Q. The line PQ is perpendicular to AB.Parallel linesQ. Draw a line parallel to the given straight line AB through a given point that is 50 mm from it.Method-I- Mark any point D on AB, with O as the centre and DC as the radius draw an arc CE to meet AB and point

E.- With E as centre and with same radius DC strike an arc DF. Set the compass to a chord of CE and lay off

the DF from point D. Thus locating point F. Draw a line through F and C which is parallel to AB.

Method-II- Set the compass to a 50mm radius (Distance parallel line from AB) and from any point C and D on AB

strike two arcs draw a line EF tangent to (Barely touching) the arcs which is parallel to line AB.

To Divide a lineQ. Divide a 100mm long line into 7 equal partsTo divide a line in proportionQ. Divide an 80mm long straight line into parts that are proportional as 2:3:4.Constructioni) Draw an 80mm long line AB. Draw a line AC at any convenient acute angle with

iv) Draw parallel lines to this (9' B) from the second and fifth interval to meet AB.

Bisecting an angle- Draw an angle of 75° and bisect it using geometrical construction.Construction- Draw the given angle ∠AOB = 75° set the needle point of the compass at O. Draw an arc CD of any radius

to strike OA and OB at point C and D respectively and with C and D as centers draw arcs of equal radiusto intersect arcs at P. Draw a line from O through P. The line OP bisects the angle AOB.

Finding centre of an arcQ. Determine the centre of an arc shown in figure.

ConstructionDraw the given arc AB

2. Draw two chords CD and EF of arbitrary lengths in the arc AB. Draw perpendicular bisectors PQ and RSto the chords CD and EF respectively.

3. To intersect each other at O which is required centre of the arc.

Draw a circle passing through points A, B and CDraw the line Joining AC and BC. Draw perpendicular bisectors of AB and BC to intersect each other at 'O'which is centre of given Ole

- With O as centre and OA as radius draw a circle.

Diagonal Scale- A diagonal scale is used to represent three units i.e. main unit and further sub-division of sub unit.- For example a diagonal scale can represent meters, dm and cm, km, hectometer, decameter, yards, feet,

inch.- The sub division of the small unit of a plain scale is done by diagonal principle as the third unit of scale is

measured with the help of diagonal scale is being named as diagonal scale.Principle of Diagonal Scale- Let the line AB be small length and has to divided into 10 equal divisionsConstruction of Diagonal Scale- Determine the R.F, length of scale and L.C. Draw a rectangle of length Ls and width 4 to 5 cm. Divide Ls

such that each division may represent main division.- Now divide 1st division of main division into such a no.of equal parts that each division may represent 2nd

unit.- Divide the 2nd unit into no.of equal parts using diagonal lines to represent the third unit.- Mark zero at the end of the first main division. Label main division of scale with primary unit towards the

right side of zero mark and at the last division.- Label the sub division of scale with secondary unit towards left side of zero mark.- Lebel the vertical line of scale with its 3rd units.- Write the main unit, sub unit and 3rd unit of scale at it's respective ends. Mention the R.F. below the labels.Procedure1. Draw a line AB.2. Draw a line BC perpendicular to AB at any convenient length.3. Divide the BC into equal parts, by the no.of sub divisions required say 10.4. Join the line AC. Parallel to AB draw lines from each division which cuts the AC at 1', 2'....upto 9'.5. Considering the Δles 1' 1C, 2'2C..... upto ABC where

Regular Polygons- They are defined as closed curves consisting of set of line segments connected such that no two segments

cross the straight line segments that make up the polygon are called it's sides or edges and the points wherethe sides meet are called polygon vertices.

- The simplest polygons are triangle, quadrilaterals, pentagons, hexagons etc.- A polygon is convex in any two points inside the polygon can be connected by a line segment that does not

intersect any side. If a side is intersected the polygon is called concave.- A polygon with all sides equal is equilateral. A polygon with all interior angles equal is equiangular. Any

polygon that is both equilateral and equiangular is a regular polygon. Ex: Equilateral Δle, square.- Construction of regular polygon is required in making drawing of engineering parts very frequently i.e. nut

bolt, assembly, spanner, gear etc.Construction of Regular Polygons- Draw a regular pentagon and regular heptagon having 40 mm long sides using general method.Method-I: (Construction is based on construction of n-gon)Pentagon1. Draw a 40 mm long line OA.2. With centre as O and radius OA draw a semi circular arc AP.3. Divide the arc AP into 5 equal parts.4. Number the division as 1, 2, 3... etc starting from point P.5. Join O with 26. With centre (2) and radius OA draw an arc to meet line O3 produced at point B.

7. Parallely with centre B and same radius OA draw another arc to meet line O4 produced at point C.

Method-II1. Draw a line AB 40 mm long. Draw a line PQ as the perpendicular bisector to the line AB. On diameter AB

describe a semicircle to meet line PQ at point 4, with either B as centre draw an arc of radius AB to meetline PQ at point 6. Bisect line 4-6 and obtain P on PQ.

2. Mark point 7 and 8 on PQ such that length 4 – 5 = 5 – 6 = 6 – 7 = 7 – 8. Draw a circumcircle with centre5 for pentagon and 7 for heptagon passing through point A and B. Lay off the circle into the parts chordlength AB and obtain the required polygon.

02BCA - Thick

222' AB...

10= ×

1 2 11C BC 2C BC......ar 11' AB

10 10 10= × = × = ×

HeptagonMethod-I1. Same point as 1 in pentagon.2. Same point as 2 in pentagon.3. Divide the arc AP into 7 equal parts.4. Same point as in (4) in pentagon.5. Join O with 26. 6 and 7 points are same.7. For heptagon also draw arcs of radius OA from C and D respectively to meet lines O5 and O6 produced at

point C and D respectively. Join the points to obtain required heptagon.

Method-IITrianglesQ. Draw a Δle having sides 80 mm, 60 mm and 50 mm.Equilateral ΔΔle of a given sideQ. Draw a equilateral Δle of 60 mm long sides.a. When one of the sides is horizontal.b. When one of the sides is vertical.

Rectangle / Square of given sides1. It is recommended to use drafter to draw rectangle / square when lengths of sides are given.2. Draw a line AB equal to the equal length of the drafter along AB erect perpendicular from A and B each

equal to the given width. Connect the ends of perpendicular to obtain required rectangle.Q. Draw a square with 50 mm long sides.

Wheni) one of the sides is horizontal.ii) A side is inclined at 30° to the horizontal.

02BCDEA are thick

Square of a given diagonalQ. Draw a square having 70 mm diagonal.

Regular Pentagon (Special Methods)Method-I1. Draw a 40 mm long line AB.2. Set the compass to radius AB and draw circle 1 and circle 2 with centre's A and B respectively. Let circle

and circle 2 intersect each other at points P and Q. With P as the centre and same radius PB, draw and arcto intersect circle 1 and circle 2 at points R and S respectively.

3. Draw a perpendicular bisector of line AB to intersect the arc RS at point M. Join R and M and produce tomeet circle 1 at point E.

4. With centre B and E radius AB draw arc to intersect each other at point D. Join ABCDE to obtain requiredpentagon.

Method-II1. Draw a 40 mm long line AB locate P as mid point of AB erect the line BQ perpendicular and equal to AB

with centre P and radius equal to PQ draw an arc to intersect the line AB produced at a point M. The lineAM is the diagonal of pentagon with centre A and radius AB, draw an arc to intersect the arc drawn with Bas the centre and radius AM at point C.

2. With centre B and radius AB draw an arc to intersect the arc drawn with A as centre and radius AM at pointE, with C and E radius AB draw arcs to intersect each other at point D. Join ABCDE to obtain the requiredpentagon.

Method-III- Draw a 40 mm long line AB draw 40 mm long line AE and BC both inclined at 108° to the line AB, with

centre's C and E and radius AB draw the arcs to intersect each other at point D.

Method-IV- Draw a 40 mm long line AB locate P as mid point of AB. Erect PD perpendicular AB with a length equal

to 1.54 times side AB with centre B and D radius AB. Draw arcs to intersect each other at point C.With centre A and D and radius AB draw arcs to each at point E. Join ABCDE to obtain required pentagon.

Conic Sections- In engineering practice we come across no.of objects containing plane curves such as ellipse, hyperbola,

parabola etc.- The curve which is obtained by cutting a right circular cone with the help of a plane in different positions

relative to the axis is called a conic section.Cone- A cone is formed if a right angled triangle with an apex angle proportional is rotated about its altitude as

the axis shown in figure.

Note: Between the slants of cone is 2∝

Circle- When the cutting plane is perpendicular to the axis of the cone (θ = 90°) the curve of intersection obtained

is a circle.

Application- Circle find their application in a vast no.of objects such as diaphragm, disks, rings, plates etc.- A circle revolving around it's axes forms a surface called sphere.

Isoceles ΔΔle

- When the cutting planes passed through apex and cuts the base of cone, then the curve of intersection isisoceles Δle.

Ellipse- When the cutting plane is inclined to axis and cuts all the generators of the cone, the section is an ellipse.

The inclination of the cutting plane for an ellipse must be greater than half of apex angle.

Application- Elliptical curves find their use in concrete arches, stone bridges, dams, monuments (.................. structures),

manholes, glands, stuffing boxes etc.- A planet travels around the sun in an elliptical orbit with sun at one of it's focii.- It is mostly applicable in medical for reflection of light and sound waves for treating kidney stones etc.Parabola- When the cutting plane is inclined to axis and is parallel to the one of generator of the cone, then section is

a parabola. The inclination of cutting plane is equal to 1/2 apex angle.

Application- Parabolic trajectory of a ball. It is used for reflecting the rays parallel to axis to produce a straight beam of

light like in head lamps torches and opposite principle is used in giant mirrors, reflecting telescope andantenna's to collect light and radio waves from outer space and focus them at focal point.Ex: Solar cooker, dish antenna

Hyperbola- When the cutting plane cuts the both parts of the double cone the section is a hyperbola (or) the cutting

plane inclination of the plane is less than 1/2 of the apex angle (θ < ∝ )

θ = ∝

θ >

Rectangular Hyperbola- When the cutting plane is parallel to the axes of the cone the section is a rectangular hyperbola (θ = 0°).Application- The curve graphically represent Boyle's Law (PV = K) - A hyperbola revolving around it's axis forms a surface called a hyperboloid. They are useful in design of

water channel, cooling tower in nuclear plant etc.- The principle (Reflecting telescope), the basis of a hyperbolic radio navigation system known as long range

navigation.Construction of EllipseMethod-I1. Eccentricity (General Method)2. Intersecting arcs method (or) arcs of circle.3. Concentric circle method4. Ablong method

1. Rectangle Method2. Parallelogram Method

1. Excentricity Method- An ellipse is defined as locus of a point P moving in a plane in such a way that ratio of it's distance from a

fixed point f1 to the fixed straight line dd1 is constant and is always < 1.

AB → principle axisV1V2 → vertices

1 2 → minor axis3 4 → major axisF1F2 → foci

Special Curves- These curves are generated by the rolling contact of one curve or line over another curve or line. There are

infinite types of curves.- For engineering purpose we use cycloids, involutes and spirals.Cyclodial Curves- They are generated by a point lying on circumference of a circle when it rolls along a fixed straight or a

curved path without slipping. The circle which rolls is called rolling circle and the generating and the fixedstraight line (or) the circle on it rolls is called directing line (or) the directing circle.

- Cyclodial curves are commonly used in kinematic (for study of motions) and in mechanism that work withrolling contact.

Cycloid- A cycloid is a curve traced on a circumference of a circle which rolls without slipping along a fixed straight

line.Epicycloid- It is a curve traced by a point on circumference which rolls along other circle outside without slipping.Hypocycloid- It is the curve traced by a point on circumference of circle which rolls an another circle inside it without

slipping.

Evolute- It is a curve traced by an enlarged string or thread when it is unwound from a circle (or) polygon, the thread

being kept tight, then evolute of circle is used as T-profile of a large gear wheel and reduced gear.

1. Draw a cycloid for 1 complete revolution of a circle having a 50 mm diameter. Draw a tangent and normalto the curve at a distance of 35 mm above the line.

Constructiona. Draw the generating circle with 50 mm diameter.b. Draw the directing line PQ = πD = 157 mm tangential to the circle.c. Dividing the rolling circle into 12 equal parts and mark 1, 2, 3.... etc. Draw lines through points 1, 2, 3....

parallel to PQ.d. Divide PQ into 12 equal parts and mark 1', 2', 3'..... on it. Draw perpendicular lines through these lines to

meet central line CB at points C1, C2, C3......C12.

e. Assume that circle rolls to the right through 1/12th required the point 1 of circle will get in contact with 1'.With centre C will move to new centre C and point P will move to the locus line through point 1. Also, thedistance of point t from P to the radius of circle 25 mm. Therefore, draw an arc with C, cutting locus linethrough point 1 at P1. Similarly draw arcs with centre's C2, C3.....C12 with radius 25 mm to meet the locus

lines to points 2, 3, 4......at point P2P3.....P4 respectively.

f. Draw a smooth curve passing through all the points the curve obtained is required cycloid.

Tangent and normal to a cycloida. Mark a point M on the cycloid at 35 mm from the directing line. With M as the centre and radius 25 mm

cut central line at point X. Through X draw a line perpendicular to PQ through point N. Join M and N andextended to N1 the line NN1 is required normal.

b. Through point M draw a line perpendicular NN1 the line TT1 is the required tangent.Q2. Draw an epicycloid generated by a rolling circle of 60 mm diameter for one complete revolution. The radius

of directric circle is 100 mm. Draw a tangent and normal at a distance of 150 mm from cente of circle.Constructiona. For generating circle covers an arc length of 2πr (where the radius of generating circle r = 30 mm). In one

revolution along directing circle.b. The angle subtended by arc PQ at centre O is given by

Therefore draw an arc PQ with O as centre and radius 100 mm subtending an angle of 108°.c. Join OP1 extended to C such that CP = 2r = 60 mm. Draw the generating circle CP as diameter. Divide this

circle into 12 equal parts and mark the divisions as 1, 2, 3, 4.....12. Also divide subtended angle of 108° into12 equal parts. Mark the divisions as ⊥1, 21......121.

d. Draw the arcs with O as centre and radius = O⊥, O2, O3..... O12 to meet OQ produced.

e. Draw an arc with O as centre and radius = OC. The arc CD is known as centre arc. Extend O1'........ to meet

r 30360 360 108

R 100θ = = ° = × = °

cycloid curve is thicktan and normal is the PQ

arc CD at points C1, C2, C3........

f. Assume that the circle rolls towards right to 1/12th revolution, thus point 1 of the Ole will get in contact with1' and centre C will move new centre C1 and point will move to arc P1. Also distance of P will be equal to

radius of circle (30 mm).g. Therefore, draw an arc with centre C1 to cut arc point 1 at P1. Parallel all other points P2......P12. Draw a

smooth curve passing through all these points. The required curve is hypicycloid.

Tangent and normal to hyptcycloid1. Mark a point M an hypicycloid at 150 mm from point 'O'.2. With M as centre and radius 30 mm cut the centre arc CD at point x. Join DX to meet the arc at point N.

Join NM and produce it to N'. The line NN1 is required normal.3. And through point M draw a line TT1 perpendicular to NN1 the line TT1 is required tangent.

Note: For all curves N is on PQ

M is on curvex is on centroid arc (or) central line