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MEL 110

Development of surfaces

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

One side: Height of the prism

Other side: Side of the base

Cylinder – Rectangle

One side: Circumference of the base

Other side: Height of the cylinder

Pyramid – Number of triangles in contact

The base may be included

if present

Development is a graphical method of obtaining the area of the surfaces of a solid. When a solid is opened out and its complete surface is laid on a plane, the surface of the solid is said to be developed. The figure thus obtained is called a development of the surfaces of the solid or simply development. Development of the solid, when folded or rolled, gives the solid.

h

dd

Examples

T. L.

3

Methods used to develop surfaces

1. Parallel-line development: Used for prisms, cylinders etc. in which parallel lines are drawn along the surface and transferred to the 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.

3. Triangulation development: Complex shapes are divided into a number of triangles and transferred into the development (usually used for transition pieces).

4. Approximate method: Surface is divided into parts and developed. Used for surfaces such as spheres, paraboloids, ellipsoids etc.

Note:- The surface is preferably cut at the location where the edge will be smallest such that welding or other joining procedures will be minimal.

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Parallel line development: This method is employed to develop the surfaces of prisms and cylinders. Two parallel lines (called stretch-out lines) are drawn from the two ends of the solids and the lateral faces are located between these lines.

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D

H

D

SS

H

L

= RL

+ 3600

R=Base circle radius.L=Slant height.

L= Slant edge.S = Edge of base

L

S

S

H= Height S = Edge of base

H= Height D= base diameter

Development of lateral surfaces of different solids.(Lateral surface is the surface excluding top & base)

Prisms: No.of Rectangles

Cylinder: A RectangleCone: (Sector of circle) Pyramids: (No.of triangles)

Tetrahedron: Four Equilateral Triangles

All sides equal in length

Cube: Six Squares.

Parallel-line development

Radial-line development

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L L

= RL

+ 3600

R= Base circle radius of coneL= Slant height of coneL1 = Slant height of cut part.

Base side

Top side

L1 L

1

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

DEVELOPMENT OF FRUSTUM OF CONE

DEVELOPMENT OF FRUSTUM OF SQUARE PYRAMID

FRUSTUMSFRUSTUMS

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Project, horizontally, the points of intersection of the cutting plane with the edges.

Mark distances 3M, 3N

2, b

4, d

1

2

3

4

A

B

C

1

D

Cube cut by section plane

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2

Draw the development of the lower portion of the cone surface cut by a plane. Cone base diameter is 40 mm and height is 50 mm. The cutting plane intersects the cone axis at an angle of 45o and 20 mm below the vertex

a

bc

d efa

b c d e f g g

4

23

1

o

431

a

bc

de

f

go

T

F

l

Length of arc = circumference of base of cone

2

b

2

True lengths b2, 2o obtained by auxiliary view method

o

• Divide the cone in the top view and project the corresponding generator lines in the front view

• Develop the complete surface of the cone by drawing an arc with radius = length of side generator of cone and length of arc = circumference of cone base

• Draw the corresponding generator lines

• Obtain true lengths of o1, o2 etc. by auxiliary view, rotation method OR by projecting onto one of the side generators (which are in true length)

• Mark the distances (true lengths) o1, o2…etc. in the development and join them to get the development of the lower portion of the cone

a

o

l = R

l

+ 3600

Radius of cone = R

2’3’

4’

True length of (o2, o3) = (o2’, o3’) etc.

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If R = 2r then θ = 180°, i.e., if the slant height of a cone is equal to its diameter of base then its development is a semicircle of radius equal to the slant height.

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Develop the surface of the symmetrical half of an oblique pyramid with a horizontal regular hexagonal base (side 20 mm and vertex 30 mm above one corner of the base)

o

a b c d

a

b c

o, d

c b

cb

True lengths

o

c

b

a

d

T

F

Obtain true lengths of the edges ob and oc by rotation or auxiliary view method

Edge oa is seen in true length in the Front View

ab = bc = cd = side of hexagonal base = 20 mm

od and dc can be constructed as they are perpendicular to each other

The lengths of bc, and ob are known and therefore these distances can be marked with the compass

After drawing triangles odc and ocb, triangle oba can be completed

a

d

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a b c

Develop the surface of the cylinder which is cut as shown

ab,l c,k d,j

e,i f,hg

d

ef

g

• Divide the base of the cylinder in the top and front views into the a certain number of equal parts (12 here)

• Develop the surface of the cylinder (rectangle with length x diameter and height = height of cylinder) and divide it into the same number of equal parts

• The projector lines from the top view intersect the cut portion of the cylinder at a, b, c…..f.

• Project these points onto the developed surface

15o

45o

x50

100

T

F

h i h h

a

a g

bc

d ef

hi

jk

l

hi

j k l

50

30o

a’, e’ b’, f’

d’, h’ c’, g’

a, i, d b, k, c

e, j, h f, l, g

i’, j’

k’, l’

b

i’, j’ b’, f’

d’, h’ k’, l’

j f

id

h

k

l

Oblique square prism

i’ b’

d’ k’

f’

l’

j’

h’

i’

d’

13

Oblique prism

a b

c

de

f

a bf

h i

g

h ig

a

h

b

i

f

g

f

g

Parallel to each other