ta 101 t hink and a nalyze anupam saxena associate professor mechanical engineering compliant and...

TA 101

Think and Analyze

Anupam SaxenaAssociate Professor

Mechanical EngineeringCompliant and Robotic Systems Lab

Indian Institute of Technology Kanpur

Organization of Lectures and Laboratory AssignmentsTopic Week (No. of Lectures) Lab

Intro and Basic Constructions Week 1 (2)

Orthographic Projections Week 2 (2) Lab 1

Orthographic Projections Week 3 (2) Lab 2

Isometric Projections Week 4 (2) Lab 3

Missing Views Week 5 (2) Lab 4

Sectional and Assembly Week 6 (2) Lab 5

Oblique Projections Week 7 (2) Lab 6

Perspective Projections Week 8 (2) Lab 7

Lines and Planes Week 9 (2) Lab 8

Lines and Planes Week 10 (2) Lab 9

Auxiliary Projections Week 11 (2) Lab 10

Intersection of lines/planes/solids

Week 12 (2) Lab 11

Intersection and Development Week 13 (2) Lab 12

TOTAL 26 12

Basic Construction and Conics

ANUPAM SAXENATA101 LECTURE II

POLYGONS: HEXAGONS

d

Vertex-to-vertex distance Distance between flats

d

60

Circumscribing circle Inscribing circle

ANUPAM SAXENATA101 LECTURE II

POLYGONS: PENTAGON

d

A B C

D

EF

rR

ANUPAM SAXENATA101 LECTURE IIPOLYGONS: OCTAGON

ANUPAM SAXENATA101 LECTURE II

POLYGONS: SEPTAGON OR ANY REGULAR POLYGON

Given a side

23 4

5

67

A B

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: CONCENTRIC CIRCLES

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: CONCENTRIC CIRCLES

(x, y)

q

P (bcosq, bsinq)

P

Q Q (acosq, asinq)

x = acosq y = bsinq

Exact Ellipse

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: CONCENTRIC CIRCLES

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: CONCENTRIC CIRCLES

P

Tangent-Normal at P

M

F G

|FM| + |MG| = 2a

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: CONCENTRIC CIRCLES

P

Tangent-Normal at P

M

F G

n

TA: Why is n the normal?

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: INTERSECTING ARCS

2a

2b

|FT| + |TG| = 2a

F G

M

QP

QP

T

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: INTERSECTING ARCS

2a

2b

F G

M

QP

QP

T

TA: What is the maximum radius of the arc possible

from F?

?

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: INTERSECTING ARCS

2l

2b

|FT| + |TG| = 2l

F G

M

QP

T

d1d2

(x, y)

= C

(a, 0)(a, 0)

+ = C

+ = 1 Ellipse if C 2a

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: STRING APPROACH

Source: wiki

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

2a

2b

QP

m

n

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Trammel of Archimedes

QP

m

n

y

x𝑦 1𝑥1

𝑙1❑

fl1

𝑥1𝑥

= 11+ 𝑓

;𝑦1𝑦

= 𝑓 ;𝑥2+𝑦2=𝑙1❑2

( 𝑥11+ 𝑓 )

2

+( 𝑦1𝑓 )2

=𝑙1❑2

R

Precise Ellipse!

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Parallelogram Method

1234567

0

8

1 2 3 4 5 6 7 8 1234567

TA: An exact or approximate Ellipse?TA: Can you identify the Conjugate Diameters?

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: Conjugate Diameter Method

A

B

C

D

Q

AB, CD: Conjugate DiametersO: Center of Ellipse

O

OQ: Perpendicular to AB

TA: Exact Ellipse?

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: FOUR CENTER METHOD

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF ELLIPSE: FOUR CENTER METHOD

C1

C2 C3

C4

TA: An exact or approximate Ellipse?

Clue: The longest diagonal first

TA: Only withRhombus?

ANUPAM SAXENATA101 LECTURE II

CONSTRUCTION OF A PARABOLA

12

34 5

6 776

54

32

1

Keep Thinking and Analyzing

Until next time...