Mel110 part3

ENGINEERING CURVESP i t d i t t f di l tPoint undergoing two types of displacements

INVOLUTE: Locus of a free end of a string when it is gwound round a (circular) pole

CYCLOID L f i t th i h fCYCLOID: Locus of a point on the periphery of a circle which rolls on a straight line path.

SPIRAL: Locus of a point which revolves around a fixed point and at the same time moves towards it. p

HELIX: Locus of a point which moves around the f f i ht i l li d / d t thsurface of a right circular cylinder / cone and at the

same time advances in axial direction at a speed bearing a constant ratio to the speed of rotationbearing a constant ratio to the speed of rotation.

INVOLUTE OF A CIRCLEINVOLUTE OF A CIRCLE

• Problem:Draw Involute of

P3 P1Draw Involute of a circle. String length is equal to the circumference of i l

34

5

P44 to p

circle.

P

1

26

7 8P

APP8

P5

P7P6 πD

1 2 3 4 5 6 7 8

INVOLUTE OF A CIRCLEString length MORE than πD

Problem: Draw Involute of a circle. String length is MORE than the circumference of circle.

P2Solution Steps:In this case string length is more than Π D.

P3 P1

But remember! Whatever may be the length of string, mark Π D distance horizontal i.e.along the string and divide it in 8 number of d d v de 8 u be oequal parts, and not any other distance. Rest all steps are same as previous INVOLUTE. Draw the curve completely.

2

34

5

P44 to p

1 2 3 4 5 6 7 8P

16

7 8P5p8

1 2 3 4 5 6 7 8P7

P6165 mm

(more than πD)πD

INVOLUTE OF A CIRCLE

String length LESS than πD

Problem: Draw Involute of a circle.String length is LESS than the circumference of circle.

P2

String length LESS than πD

Solution Steps:In this case string length is Less than Π D.

P3 P1

But remember! Whatever may be the length of string, mark Π D distance horizontal i.e.along the string and divide it in 8 number of

4 t

d d v de 8 u be oequal parts, and not any other distance. Rest all steps are same as previous INVOLUTE. Draw the curve completely.

2

34

5

6

P44 to p

1 2 3 4 5 6 7 8

P1

6

7 8P5

P 1 2 3 4 5 6 7 8P7P6 150 mm(Less than πD)

πD

INVOLUTE OF A PENTAGON

ll

5l

5Thread should be taut

INVOLUTE OF COMPOSIT SHAPED POLE

Problem : A pole is of a shape of half hexagon (side 30 mm) and COMPOSIT SHAPED POLEg ( )semicircle (diameter 60 mm). A string is to be wound having length equal to the pole perimeter Calculate perimeter length

P1

P2

P

g q p pdraw path of free end P of string when wound completely.

1 to

P

P2

P

12

34

5

P3 3 to P

61 2 3 4 5 6

AP

πD/2P πD/2P4

P5P6

PROBLEM : Rod AB 85 mm long rolls over a semicircular pole without slipping from it’s initially vertical position till it becomes up-side-down vertical. Draw locus of both ends A & B.

4

B

A4

3

4B1

A3

OBSERVE ILLUSTRATION CAREFULLY!

when one end is approaching, other end will move away from

πR 2

3ypoll.

2

πR 2

A2 B2

1

23

4

1

A 4AA1

B3

BB4

CYCLOID:

DEFINITIONS SUPERIOR TROCHOID:If the point in the definition of cycloid is outside the circleCYCLOID:

LOCUS OF A POINT ON THEPERIPHERY OF A CIRCLE WHICH ROLLS ON A STRAIGHT LINE PATH.

cycloid is outside the circle

INFERIOR TROCHOID.:If it is inside the circleROLLS ON A STRAIGHT LINE PATH. If it is inside the circle

EPI-CYCLOIDIf the circle is rolling on another circle from outside

HYPO-CYCLOID.If the circle is rolling from inside theIf the circle is rolling from inside the other circle,

CYCLOIDPROBLEM: Draw locus (one cycle) of a point (P) on the periphery of a circle (diameter=50 mm) which rolls on straight line path.

p

p4

p4

C1 C2 C3 C4 C5 C6 C7 C8p2

p3 p5

p62

3 5

6C

p1 p71 7

P p8

πD

Point C (zero radius) will not rotate and it will traverse on straight line.

SUPERIOR TROCHOID

PROBLEM: Draw locus of a point (P), 5 mm away from the periphery of a Circle (diameter=50 mm) which rolls on straight line path.

Using 2Hi

p44

Using H

p3 p53 5

C1 C2 C3 C4 C5 C6 C7 C8p2 p6

2 6C

p1

p7

p8

1 7

DP p8πDP

INFERIOR PROBLEM: Draw locus of a point , 5 mm inside the periphery of aCi l hi h ll i h li h T k i l di t 50 TROCHOIDCircle which rolls on straight line path. Take circle diameter as 50 mm

C1 C2 C3 C4 C5 C6 C7 C8

p2

p3

p4

p5

p62

34

5

6C

P

p1 p7

p8

1

2 6

7

πD

CYCLOID

SUPERIOR TROCHOID

INFERIOR TROCHOID

12

EPI CYCLOID :

PROBLEM: Draw locus of a point on the periphery of a circle (dia=50mm) which rolls on a curved path (radius 75 mm).

Solution Steps:

Distance by smaller circle = Distance on larger circle

Solution Steps:1. When smaller circle rolls on larger circle for one revolution it

covers Π D distance on arc and it will be decided by included arc angle θangle θ.

2. Calculate θ by formula θ = (r/R) x 360°.3. Construct a sector with angle θ and radius R.4 Divide this sector into 8 number of equal angular parts4. Divide this sector into 8 number of equal angular parts.

EPI CYCLOID

C4 5

Generating/Rolling Circle

C2

2

3 6

7EPI-CYCLOIDIf the circle is rolling on

Pr = CP 1

2If the circle is rolling on another circle from outside

rR 3600=

Directing Circle

O

R

PROBLEM : Draw locus of a point on the periphery of a circle which rolls from the inside of a curved path Take diameter of Rolling circle

HYPO CYCLOIDrolls from the inside of a curved path. Take diameter of Rolling circle 50 mm and radius of directing circle (curved path) 75 mm.

P 7

P1

P21

6

P3

P

25

4 P4

P5 P6 P7

P8

34

O

OC R ( R di f Di ti Ci l )

rR

3600=

OC = R ( Radius of Directing Circle)CP = r (Radius of Generating Circle)

CYCLOID

SUPERIOR TROCHOID

INFERIOR TROCHOIDTROCHOID

16

SPIRALProblem: Draw a spiral of one convolution. Take distance PO 40 mm.

IMPORTANT APPROACH FOR CONSTRUCTION!FIND TOTAL ANGULAR AND TOTAL LINEAR DISPLACEMENTAND DIVIDE BOTH IN TO SAME NUMBER OF EQUAL PARTS.

2

P13

P2

P1

P3

P4

3

P4 O7 6 5 4 3 2 1

P4

P6P5

P7

O

5 7

6

SPIRAL of two convolutionsProblem: Point P is 80 mm from point O. It starts moving

towards O and reaches it in two revolutions around. It Draw locus of point P (To draw a Spiral of TWO convolutions).

2,10P2

1,93,11 P1

P3

P

16 13 10 8 7 6 5 4 3 2 1 P4 12 8,16P4 P8

P9P10

P11

4,12 ,P12

P13 P14

P15

5,13 7,15

P5

P6

P7

6,14

Problem: A link 60 mm long, swings on a point Oob e : 60 o g, sw gs o a po Ofrom its vertical position of the rest to the left through60° and returns to its initial position at uniformpvelocity. During that period a point P moves atuniform speed along the center line of the link from Op gat reaches the end of link. Draw the locus of P.

O, P

N

M

P8

HELIX (UPON A CYLINDER)

PROBLEM: Draw a helix of one convolution, upon a cylinder. Given 80 mm

6

7

8

P6

P7

8pitch and 50 mm diameter of a cylinder.

4

5

6

P4

P5

6

Pitch: Axial advance during one

2

3

P2

P3

4complete revolution .

1 P1

P

2

6FT

57

P

1 3

4

2

3

P8 HELIX (UPON A CONE)

PROBLEM: Draw a helix of one convolution, upon a cone, diameter of base 70 mm, axis

P5

P6

P7

( )p , ,90 mm and 90 mm pitch.

P4

P2

P3

6

PP1

X Y

57

P5P6

PP 4

P1P3

P4P7

P8

1

2

3P2

I i f S lid /Interpenetration of Solids /Intersection of Surfaces /Intersection of Surfaces /

Lines & Curves of Intersection

More points common to both the solids

Basic required knowledge:

~ Projections of Solid

~ Section of Solid22

~ Section of Solid

Problem: A cone 40 mm diameter and 50 mm axis is

.Projections of Solid

Problem: A cone 40 mm diameter and 50 mm axis is resting on one generator on Hp which makes 300

inclination with Vp. Draw it’s projections.

More number of generators Better approximation.

o’a’1h’ b’1

g’1

h’1

F

gX Ya’ b’ d’ e’c’ g

’f’h’ o’

g1g

c’1d’1e’1

f’1 o1

o1

30

h

a e

f

a1

h1f1

e1

a1

be

f1

g1 h1

o1

o1

T

o

23bc

d d1

c1

b1

c1

b1

d1

e11st. Angle

For TVFor TVSection of Solid

SECTION SECTION PLANEPLANE

I t t ti f S lidLi & Cxx yy

Apparent ShapeApparent Shape

Interpenetration of Solids

Intersection of Surfaces

Lines & Curves of Intersection

SECTION LINESSECTION LINES

Apparent Shape Apparent Shape of sectionof section

(45(4500 to XY)to XY)

SECTIONAL T.V.SECTIONAL T.V.

How to make these shapes

CURVES OF INTERSECTIONS are shown by WHITE ARROWS.

Machine component having Intersection of a Cylindrical two intersecting cylindrical surfaces with the axis atacute angle to each other.

ymain and Branch Pipe.Industrial Dust collector.

Intersection of two cylinders.

WHEN TWO OBJECTS ARE TO BE JOINED TOGATHER, MAXIMUM

Pump lid having shape of a h l P i d

SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC

REQUIREMENT FOR STRONGEST & LEAK PROOF hexagonal Prism and

Hemi-sphere intersecting each other.

Feeding Hopper.STRONGEST & LEAK-PROOF

JOINT.

Minimum Surface Contact.( Point Contact) (Maximum Surface Contact)

Li f I i C f I iLines of Intersections. Curves of Intersections.

Square Pipes. Circular Pipes. Square Pipes. Circular Pipes.

MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT.

Two plane surfaces (e-g. faces of prisms and pyramids) intersect in a straight line

Q

intersect in a straight line.

The line of intersection between two curved surfaces (e-g. of cylinders and cones) or between a plane surface and aof cylinders and cones) or between a plane surface and a curved surface is a curve.

30More points common to both the solids

How to find Lines/Curves of Intersection

Generator line Method Cutting Plane Method

T

F

T

FF

75

31

75

Problem: Find intersection curve .

Draw convenient number of lines on the surface of one of the solids.

Transfer point of intersection to their corresponding positions in other When one solid completely penetrates another, there will be two curves of intersection. 32

p p g pviews.

Problem: A cylinder 50mm dia. & 70mm axis is completely penetrated by another of 40 mm dia. & 70 mm axis horizontally. Both axes intersect & bisect each other. Draw projections showing curves of intersections

4” 1”3” 2”1’ 2’4’ 3’

Draw projections showing curves of intersections.

a”b”h”

a’

b ’h’

g” c”

f” d”

c’g’

d’f’

X Y

e”f” dd f

a’

4

1 3

2

33

Problem: CYLINDER (50mm dia.and 70mm axis ) STANDING & SQ.PRISM (25 mmsides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. All faces of

i ll i li d t H D j ti h i f i t ti

4” 1”3” 2”1’ 2’4’ 3’

prism are equally inclined to Hp. Draw projections showing curves of intersections.

a”

d”d” b”

Projections of Projections of Find lines of

X Yc”

jSolid 1.

jSolid2.Find lines of intersectionMissing critical points at which curve changes direction

4direction.

1 3

234

Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25 mm sides and 70 mm axis) PENETRATING. Both axes intersects & bisect each other. All faces of prisms are equally inclined to Vp Draw projections showing curves of

1’ 2’4’ 3’

faces of prisms are equally inclined to Vp. Draw projections showing curves of intersections.

a”a’ a’

1’ 2’4’ 3’ 4” 1”3” 2”

d” b”d’b’

d’b’

X Y

c”c’ c’

4Preference of object line over dash line

1 3dash line…

2 35

Problem. SQ.PRISM (30 mm base sides and 70mm axis ) STANDING & SQ.PRISM (25 mm sides and 70 mm axis) PENETRATING. Both axes Intersect & bisect each other. Two faces of penetrating prism are 300 inclined to Hp. Draw projections showing curves of

1’ 2’4’ 3’ 4” 1”3” 2”

p g p p p j gintersections in I angle projection system.

f”a’

f’

e’

c”c’

d’

b’

X Y4

300

d

1 3

2od"

36Other possible arrangements ???? b"

Problem: A vertical cylinder 50mm dia. and 70mm axis is completely penetrated by a triangular prism of 45 mm sides and 70 mm axis, horizontally. One flat face of prism is parallel to Vp and Contains axis of cylinder Draw projections showingprism is parallel to Vp and Contains axis of cylinder. Draw projections showing curves of intersections in I angle projection system.

4” 1”3” 2”1’ 2’4’ 3’

aaa

bc

d

bbcd

X Y

ed

f

e e

f f

4

1 3

2 37

Problem: A vertical cone, base diameter 75 mm and axis 100 mm long, is completely penetrated by a cylinder of 45 mm diameter and axis 100 mm long. The axis of the cylinder is parallel to Hp and Vp and intersects axis of the cone at a point 28 mm above the base. Draw

o”o’projections showing curves of intersection in FV & TV. in I angle projection system

111

2

3

46

7

8

33 7,

8,22

4

g

a’ b’h’ c’g’ d’f’ e’ g” f”h” a”e” b”d” c”

45

6

X Y5 5

64 428

h

g

f

a e

b

c

d 38

Problem: Vertical cylinder (80 mm diameter & 100 mm height) is completely penetrated by a horizontal cone (80 mm diameter and 120 mm height). Both axes intersect & bisect each other. Draw projections showing curve of intersections in I angle projection systemDraw projections showing curve of intersections in I angle projection system.

7’7

6’ 8’

1’ 5’

2’ 4’

X Y3’

1

2 8 Intersection of a curve with

3 7curve with another !!!!Generator lines..

4 6

5 39

Problem: CONE STANDING & SQ.PRISM PENETRATING(BOTH AXES VERTICAL)

2’

1’

5’

3’

X Ya’ b’h’ c’g’ d’f’ e’

4’6’

Y

h

g

f8

a b h c g d f e

a e

f

1 610

9 7

Problem: A cone70 mm base diameter and 90 mm axis is completely penetrated by a square prism from top

b d

23

4

5with it’s axis // to cone’s axis and 5 mm away from it.A vertical plane containing both axes is parallel to Vp.Take all faces of sq.prism equally inclined to Vp.Base Side of prism is 30 mm and axis is 100 mm long

c5 mm OFF-SET

Base Side of prism is 30 mm and axis is 100 mm long.Draw projections showing curves of intersections.

40

Intersection of two cylindersIntersection of two cylinders oblique to each other –Use PAVoblique to each other Use PAV

III angle projection

41

Intersection of two cylindersIntersection of two cylinders oblique to each other –Use AVoblique to each other Use AV

42

Intersection of Cone and ObliqueIntersection of Cone and Oblique cylinder using PAVcylinder using PAV

43

Intersection of irregular Prism & goffset Cylinder

III angle projection

44


InvisibleInvisibleVisible

45


InvisibleInvisibleVisible

46


47

DEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDSDEVELOPMENT OF SURFACES OF SOLIDS

Solids are bounded by geometric surfaces:Solids are bounded by geometric surfaces:

LATERLAL SURFACELATERLAL SURFACE IS SURFACE EXCLUDING SOLID’S TOP & BASE.IS SURFACE EXCLUDING SOLID’S TOP & BASE.

Development ~ obtaining the area of the surfaces of a solid.

Development of the solid when folded or rolled gives the solid

- Plane Prism, PyramidSingle curved Cone Cylinder

Development of the solid, when folded or rolled, gives the solid.

- Single curved Cone, Cylinder- Double curved sphere 48

Examples

Prism – Made up of same number of rectangles as sides of the base

One side: Height of the prismg p

Other side: Side of the base

Cylinder – Rectangle

One side: Height of the cylinder

Other side: Circumference of the baseOther side: Circumference of the base

Pyramid – Number of triangles in contact

h

The base may be included

if present

πdT. L.

49

Methods to Develop Surfacesp

1. Parallel-line development: Used for prisms (full or truncated), c linders (f ll or tr ncated) Parallel lines are dra n along thecylinders (full or truncated). Parallel lines are drawn along the surface and transferred to the development

Cylinder: A Rectangle

H

Cylinder: A Rectangle

πDD

H= Height D= base diameterg

Prisms: No.of Rectangles

H

50SS H= Height S = Edge of base

Ex:Ex:

DOTTED LINES are never DOTTED LINES are never shown on developmentshown on development

51

shown on developmentshown on development

4, d

2, b

Complete de elopment of c be c t b c tting plane (inclined to HP at

52

Complete development of cube cut by cutting plane (inclined to HP at 30 degrees and perpendicular to VP)

4, d

2, b

53

Cylinder cut byh

ij

k

φ50 Cylinder cut by three planes

a g

hlthree planes

G

T

F

bc

d ef

Cc’d'

e' f'g'

D

EF

G

45oH

IJ

K IA B Ca'

b'c

15o

100

AI

3 4 5 6 7 8 9 1041 51 61 71

54πx50

112

3 4 10 11 1211

2131 41

Problem: Development of a solid is a parabolaProblem: Development of a solid is a parabolawith a 180 mm base and a 90 mm height. Draw theprojections of solidprojections of solid.

180=π D

D=57.3 mm

55


1. Parallel-line development

2. Radial-line development: Used for pyramids, cones etc. in which the true length of the slant edge or generator is used as radius

Cone: (Sector of circle) Pyramids: (No.of triangles)

θ

R=Base circle radius. L= Slant edge

θ = RL 3600

L=Slant height.L Slant edge.S = Edge of base

56

Parallel vs Radial line method

Parallel line method

Radial line method 57

DEVELOPMENT OF DEVELOPMENT OF

FRUSTUMSFRUSTUMS

Base side

Top side

DEVELOPMENT OF FRUSTUM OF CONE

DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID

Top side

θ

θ = RL 3600

R= Base circle radius of coneL= Slant height of cone

L= Slant edge of pyramidL1 = Slant edge of cut part.

L1 = Slant height of cut part.

1. 1. Development Development is is a shape showing AREA, means it’s a 2a shape showing AREA, means it’s a 2--D plain drawing.D plain drawing.22 AllAll dimensions of it must be TRUE dimensionsdimensions of it must be TRUE dimensions

ImportantImportant pointspoints..

2. 2. All All dimensions of it must be TRUE dimensions.dimensions of it must be TRUE dimensions.3. As it is representing shape of an un3. As it is representing shape of an un--folded sheet, no edges can remain hiddenfolded sheet, no edges can remain hidden

and hence DOTTED LINES are never shown on development.and hence DOTTED LINES are never shown on development.

Development of Sphere using Frustum of Cones: Zone MethodFrustum of Cones: Zone Method

Zone 1: Cone

Zone 2: Frustum of cone

Zone 3: Frustum of cone

Z 4 F t fZone 4: Frustum of cone

θ = RL 3600

θ

59

Development by Radial MethodPyramids (full or Truncated) & Cones (full or Truncated)Cones (full or Truncated).

If the slant height of a cone is equal to its diameter of base then its development isIf the slant height of a cone is equal to its diameter of base then its development isa semicircle of radius equal to the slant height.

60

Ex:

61

Development of Oblique Objects

• Right regular objects – Axis of object perpendicular to base.

• Axis of any regular object (prismAxis of any regular object (prism, pyramid, cylinder, cone, etc.) inclined at angle other than right angle Obliqueangle other than right angle – Oblique OBJECT. Use ARC method.

62

Oblique prismOb que p s

c

de

fg j

a b

cf

h i

g

Parallel

j

a bf

a' b’f’ c'

f

h’ i'g' j'

h ig

gg

63

Draw the development of an oblique circular cylinder with base diameter 30 mm and axis inclined at 75o with the base. Height of the cylinder is 50 mm

• Divide the surface of the cylinder into equal parts as shown, with the generator lines parallel to the end generators

• Draw projection lines from top edge of cylinder

φ30

AGga

T

• Draw projection lines from top edge of cylinder such that they are perpendicular to end generator

• Mark distances AB, BC etc. from one projector

A

EB

CDF

ga

F Aline to the next to complete the profile

• Do the similar process for the bottom edgeA’G’B

C

A

75o

G50

A1

75

g’ a' aA1

64

A1

g a

Oblique Cone

65

Intersection of Plane & Pyramid. 14yDevelopment of resulting lateral

0

CD Develop

1-D-A-2-12-A-B-3-21

B Dg

truncated Pyramid23

AB 3-B-C-4-3

1-D-C-4-12

1

B

YX

B’

0

A

B

D’

4

231’o

3

266


1. Parallel-line development:2. Radial-line development:

3. Triangulation development: Complex shapes are divided into a number of triangles and transferred into the development

Tetrahedron: Four Equilateral TrianglesEXAMPLES:EXAMPLES:--Boiler Shells & chimneys, Boiler Shells & chimneys, Pressure Vessels Shovels TraysPressure Vessels Shovels Trays

All sides equal in length

Pressure Vessels, Shovels, Trays, Pressure Vessels, Shovels, Trays, Boxes & Cartons, Feeding Boxes & Cartons, Feeding Hoppers, Large Pipe sections, Hoppers, Large Pipe sections, B d & P t f t tiB d & P t f t tiequal in length Body & Parts of automotives, Body & Parts of automotives, Ships, Aero planes.Ships, Aero planes.

67

Transition Connect two hollow objects having different base.

PiecesThree surfaces

Triangulation Method: Dividing a surface into a Dividing a surface into a number of triangles and transfer them to the developmentdevelopment.

68

Ex: In air conditioning system, a square duct of 50mm by 50mm is connected to another square duct of 25mm by 25 mm by using a connector (transition piece) of height 25mm Draw development of lateral surface of the connector (Neglectheight 25mm. Draw development of lateral surface of the connector (Neglect thickness of connector). Pyramids: (No.of triangles)

O’O

L= Slant edge.S = Edge of base

a’ A

g

b’ B

a

b

O

69c

D l t f T iti Pi fDevelopment of Transition Piece for Difference Shapes and Sizes p

• Connect a Square pipe with circular pipe.• Ex: Imagine a transition piece (height = 25)

to connect a chimney of square crossto connect a chimney of square cross section 50mm * 50 mm to circular pipe of 30mm diameter Draw the projections and30mm diameter. Draw the projections and develop lateral surface of the transition piece.

70

1’2’, 8’

1/8 of i fc

a’, b’

circumferencebc

2

34 1

1

2

5

8

I angle p ojectionad

67

8

18

A B

71

I angle projectionad 18

Development of Sphere/HemisphereSphere/Hemisphere using Lune Method

i25% circ-umference

ence

rcu

mfe

re50

% c

ir

72

Summarizing DEVELOPMENT OF SPHERE

• All three views -> CIRCLE.All three views CIRCLE.• Approximate development by dividing

h i t i fsphere into a series of zones.

Cone FrustumCone, Frustum of cone, Symmetry

73

Draw the development of a hemispherical bowl of radius 3 cm by any method12 such developments.

42 6 8 10 4 8 106

o a cb d e1 3 5

42

7 9

O E

1 3 5 7 9

42

da cbo e

42

8 106

0

This “Lune method” is also

T

1 3 5 7 9

da cbo e

1 3 5 7 9

42

8 1 064 8 106

method” is also known as “Polycylindricmethod” or

2 OE Rπ=

o'F

9

da cbO E

1 3 5 7 9

42 6da cbo e

42

8 106

method or “Gore method”.

a'b'

c' 1 3 5 7 9

da cbo e

1 3 5 7 9

42

8 1064 8 106

For a reasonable accuracy circle

c

d' e'

7 9

da cbo e

1 3 5 7 9

42 6da cbo e

42

8 106

needs to be divided in 16 or more parts.

o

1 3 5 7 9

da cbo e

1 3 5 7

42

8 106

8 106

74

Draw the development of a hemispherical bowl of radius 3 cm by any methodo4

o4'

o a b c d eo360

'0 ×=oaoaθ

To3

o360'1

1 ×=

ocbo

obθ

o '

o'Fa'

b'c'

o2

o

o

360

360'2

2

×

×=

odco

oc

θ

θ

o'o1'

o2'

o3

o 'c

d' e'

o360

360'

4

33

×=

×=

oedo

θ

θ o

a'b'

'

o'

o1'

'44 eo c'

d'

e'

PROBLEM: A room is of size L=6.5m, D=5m, H=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch is placed in one g g pof the corners of the room, 1.5m above the flooring. Draw the projections and determine real distance between the bulb & switch.

Ceiling

TV

Bulb

TV

Switch

DD

76

PROBLEM:- A picture frame 2 m wide and 1 m tall is resting on horizontal wall railing makes 350 inclination with wall. It is attached to a hook in the wall by two strings The hook is 1 5 m above wall railing Determine length of eachwall by two strings. The hook is 1.5 m above wall railing. Determine length of each chain and true angle between them

350

Wall railingWall railing

77

PROBLEM: Line AB is 75 mm long and it is 300 & 400 Inclined to HP & VP respectively. End A is 12mm above Hp and 10 mm in front of VP. Draw

b’ b’1

projections. Line is in 1st quadrant.

TLFV

a’θ

aX Y

Ø 1LFV

TLTV

b b1

PROBLEM: Line AB 75mm long makes 450 inclination with VP while it’s FV makes 550. End A is 10 mm above HP and 15 mm in front of VP. If line is in 1st

b’1b’

quadrant draw it’s projections and find it’s inclination with HP.

LOCUS OF b1’

0

Ya’

550

X Y

a 1LFV

b1bLOCUS OF b

PROBLEM: FV of line AB is 500 inclined to XY and measures 55 mm long while it’s TV is 600 inclined to XY line. If end A is 10 mm above HP and 15 mm in front of VP draw it’s projections find TL inclinations of line with HP & VP

b’ b’1

front of VP, draw it’s projections,find TL, inclinations of line with HP & VP.

X a’Y

500

θ

X Y

aΦ

600Φ

bb1

PROBLEM :- Line AB is 75 mm long . It’s FV and TV measure 50 mm & 60 mm long respectively. End A is 10 mm above Hp and 15 mm in front of Vp. Draw

b’1b’

g p y p pprojections of line AB if end B is in first quadrant. Find angle with HP and VP.

X Ya’

1’LTVθ

X Y

a 1LFVΦ

b1b

PROBLEM.Length (L) depth (D) andLength (L), depth (D), and height (H) of a room are 6.5m, 5m and 3.5m especti el An elect ic b lbrespectively. An electric bulb

hangs 1m below the center of ceiling. A switch is placed

Ceiling

TVin one of the corners of the room, 1.5m above the flooring.

Bulb

gDraw the projections and determine real distance between the bulb and

Switch

Dbetween the bulb and switch.

D

PROBLEM: L=6.5m, D=5m, H=3 5m An electric bulbH=3.5m. An electric bulb hangs 1m below the center of ceiling. A switch isof ceiling. A switch is placed in one of the corners of the room, 1.5m above

6.5m

the flooring. Draw the projections and determine real distance between the

a’

b’ b’1

3.5m

1m

real distance between the bulb and switch.

ax y

1.5

b5m b5m

Answer: a’ b’1

PROBLEM: A PICTURE FRAME 2 mPICTURE FRAME, 2 m WIDE & 1 m TALL, RESTING ON HORIZONTAL WALLHORIZONTAL WALL RAILING MAKES 350

INCLINATION WITH WALL. IT IS ATTAACHED TO A

350A

B

HOOK IN THE WALL BY TWO STRINGS.THE HOOK IS 1.5 m

DTHE HOOK IS 1.5 m ABOVE WALL RAILING. DETERMINE LENGTH OF EACH CHAIN AND TRUE

Wall railingC

ANGLE BETWEEN THEM

PROBLEM- A picture frame 2 m wide & 1 m tall is resting onwide & 1 m tall is resting on horizontal wall railing makes 350

inclination with wall. It is attached t h k i th ll b t t i

h’to a hook in the wall by two strings.The hook is 1.5 m above the wall railing. a’b’

(chains)

DETERMINE LENGTH OF EACH CHAIN AND TRUE ANGLE BETWEEN

a b1.5 m

1m

THEM c’d’

(wall railing)

X Yad

h

a1

(frame)

h

(chains)

Answers:bcb1

Answers:Length of each chain= hb1True angle between chains =

PROBLEM:- Two mangos on a tree A & B are 1.5 m and 3.0 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If thethose are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite sides of it. If the distance measured between them along the ground and parallel to wall is 2.6 m, Then find real distance between them by drawing their projections.

TV

B

A0.3M THICK

86

PROBLEM:- Two mangos on a tree A & B are 1 5 m and 3 00

b’tree A & B are 1.5 m and 3.00 m above ground and those are 1.2 m & 1.5 m from a 0.3 m thick wall but on opposite

3.0a’

ppsides of it. If the distance measured between them along the ground and parallel to wall is 2 6 m Then find

1.5to wall is 2.6 m, Then find real distance between them by drawing their projections.

b2.6

B

1.5

1 2

a

1.2

PROBLEM:-Flower A is 1.5 m & 1 m from walls Q (parallel to reference line) & P (perpendicular to reference line) respectively. Flower is 1.5 m above the ground. Orange B is 3.5m & 5.5m from walls Q & P respectively. Drawing projection, find distance between them If orange is 3.5 m above ground.

b’ b’b 1

3.5 ma’

x y1.5 m

Ground

a1.5 m

Wall Q B

1 m3.5 m

b5.5 m

Wall P

88

PROBLEM :- A top view of object (three rods OA, OB and OC whose ends p j ( ,A,B & C are on ground and end O is 100mm above ground) contains three lines oa, ob & oc having length equal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line isrespectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod.

Tv

O

A

C

Fv

B

PROBLEM :- A top view of object (three rods OA, OB and OC whose ends A,B & C are on ground and end O is 100mm above ground) contains three lines oa, ob & oc having length

l t 25 45 d 65 ti l Th th li ll i li d d thequal to 25mm, 45mm and 65mm respectively. These three lines are equally inclined and the shortest line is vertical. Draw their projections and find length of each rod.

o’

TL1TL2

a’b’ c’ c1’b1’ a1’x y

a

o Answers:

b

c

Answers:TL1 TL2 & TL3

PROBLEM:- A pipeline from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same l l f i li f 0 f h d i li flevel of A. Pipe line from B runs 150 Due East of South and meets pipeline from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground.

1

5

Bearing of a LINE:

A

1 Bearing of a LINE: Horizontal angle between line &

idi ( th thAB12 M

Emeridian (north south line)----Measured in DEGREES (0 to 90◦).

C

Measured in Top View.S 45° E.S 15° E.

PROBLEM:- A pipe line from point A has a downward gradient 1:5 and it runs due South - East. Another Point B is 12 m from A and due East of A and in same level of A. Pipe line from B runs 150 Due East of South and meets pipe line from A at point C. Draw projections and find length of pipe line from B and it’s inclination with ground.

12m

15a’ b’

FV

x yN

c’2c’ c’1

a b

y

450

N

EASTW

150

TV

c= Inclination of pipe line BC

SOUTH

= Inclination of pipe line BC

PROBLEM: A person observes two objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450 respectively. Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find distance of objects from observer and from tower also.

OO

300

450

A

W

S

B

PROBLEM: A person observes two objects, A & B, on thetwo objects, A & B, on the ground, from a tower, 15 M high, at the angles of depression 300 & 450 Object A is due North-West

o’

300 450

45 . Object A is due North-West direction of observer and object B is due West direction. Draw projections of situation and find

a’1’

30

15M

projections of situation and find distance of objects from observer and from tower also. Na

b’a’

W Eob

Answers:

S

Answers:Distances of objects from observeo’a’1 & o’b’From tower SFrom toweroa & ob

PROBLEM:- A tank of 4 m height is to be strengthened by four rods from each corner by fixing their other ends to the flooring, at a point 1.2 m and 0.7 m from two adjacent walls respectively, as shown. Determine graphically length and angle of each rod with flooring.

TV

A

4 M

B

PROBLEM:- A tank of 4 m height is to be strengthened by four rods from h b fi i th i th d t th fl i t i t 1 2 deach corner by fixing their other ends to the flooring, at a point 1.2 m and

0.7 m from two adjacent walls respectively, as shown. Determine graphically length and angle of each rod with flooring.

a’

True Length

FV

True Length

Answers:Length of each rod= a’b’1Angle with Hp.

a

b’b’1

=X Y

a

b

TVTV

SHORTEST DISTANCE BETWEEN POINT and LINE

Measure distance between Point P and Line AB for given TV & FV arrangement. aa

p

b

yT

Fx

p

p’

ba

p

A BDa’

b’

ba A B

q

D

pD is ⊥ to ab

97

bq ab

a2, b2

Point view of linea2, b2

Shortest distance a2 p2

a1

b1

p2Point

x1

p1

a

1

y1b

yp

TF

x

’

p’

a’

b’

Shortest Distance between 2 skew Lines (AB & CD)Shortest Distance between 2 skew Lines (AB & CD)

Skew (oblique) lines Lines that are not parallel & do not intersect

c

•Find T.L. of one of the lines and project its point view Distance measured along

Line ⊥ to both

b

c p j pusing auxiliary plane method

Line ⊥ to both.

a

b

d •Project the other line also in each view

TF

a

b

•The perpendicular distance between the point view of

c

pone line and the other lineis the required shortest di t b t th t

99d distance between the two

lines

Primary auxiliary viewcTL

d d, cSecondary auxiliary view

c

Required distance

b

c a

a

b

d TF

d

a

b

ab B AP

c

qdP is ⊥ to ab

d In mines, this method might be used to locate a connecting tunnel.

True Angle between 2 Skew Lines (AB & CD)

Draw P.A. V. such that one line (AB) shows its True Length

Measure angle in view that Shows both lines in true length

Draw S.A.V. view with reference line perpendicular to the True Length of the line (AB) to get the point view of the linea

b

dT

F

Draw a tertiary auxiliary view with reference line parallel to the other line in order to get its True Length

x y

d’

cd

Since the secondary auxiliary view had the point view of the first line, the tertiary auxiliary view will have the True L th f th fi t li l

a’d

Length of the first line also.c’ b’

101

TRUE LENGTH OF

TERTIARY AUXILIARY VIEWANGLE BETWEEN TWO LINES

a3c3

TRUE LENGTH OF BOTH LINES

True Angleb h

ParallelLINES

b32

d3PRIMARY

AUXILIARY VIEWbetween the two lines

a2 ,b2

c2

Point view of one line

SECONDARY AUXILIARY VIEWTrue lengthb1

c1

d2

one line

Parallel

a1

b

d1

T

Parallela

cd Angle between two

nonintersecting linesT

Fx y

a’d’

nonintersecting linesis measurable in a view that shows both lines in true shapea’

c’ b’

lines in true shape.

Angle between 2 planesAngle between 2 planesLine of intersection of the 2 planes (here it is True Length)

e fe f

ac

d

x TF

y

a’ b’

ab • Obtain an auxiliary view such that the

reference line is perpendicular to the True Length of the line of intersection of the

la

planes

• In this case, the intersection line is parallel to both principle planes and hence is in True L th i b th f t d t i

e’

d’ c’

f’

Length in both front and top views

• Both planes will be seen as edge views in the auxiliary view.

e f’• The angle between the edge views is the angle between the planes

Line of intersection of the 2 planes (here it is True Length)x1

e f f1, e1

PRIMARY AUXILIARY VIEW

e f ,

ac

d

b1, a1c1, d1

x •Obtain an auxiliary view such that the reference line is perpendicular to the True Length of the line of intersection of the

l

TF

y

y1a’ b’

ab

b1, a1

planes

•In this case, the intersection line is parallel to both principle planes and hence is in True L th i b th f t d t i

1a

Length in both front and top views

•Both planes will be seen as edge views in the auxiliary view.

e’

d’ c’

f’•The angle between the edge views is the angle between the planes

e f’

Piercing point of a line with a plane

Edge view of the plane

True length of T A1

p1

gprincipal line

In a view showing the

Linep

T

gplane as an edge, the piercing point appears where the line

Planep’

Fwhere the line intersects the edge view.

Principal linePart of the line hidden by the plane h ld b h

Draw auxiliary view to get EDGE VIEW.

should be shown dotted

Find the shortest distance of point P from the body diagonal AB of the cube of side 50

1

mm as shown

a1b1

a1‘, b1’

d’g’ b’ e’

p1

p ’

10

1 , 1

p’

d g

10

50

b’, e’p1’Required distance

Draw an auxiliary view to get the 50Draw an auxiliary view to get the true length of the line

Draw an auxiliary view to get the point view of the diagonal c’, d’F, A

c,bf,d p

point view of the diagonal

Project the point P in these views to get the required distance

d,ea,g

Mel110 part3

Entertainment & Humor

Transcript of Mel110 part3