Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk By Dr. Samah Mohamed Mabrouk .

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Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk www.smmabrouk.faculty.zu. edu.eg

Transcript of Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk By Dr. Samah Mohamed Mabrouk .

Page 1: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Lecture 5: The Auxiliary projection

By Dr. Samah Mohamed Mabrouk

www.smmabrouk.faculty.zu.edu.eg

Page 2: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

The Auxiliary projection is defined by

The auxiliary projection planes are

perpendicular planes on 1 or 2 and can be moved on

parallel or perpendicular to a geometric objects in

order to transform the positions of these geometric

objects into more simple positions, from which the

complex problems can be solved easily.

Page 3: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

1 -Auxiliary projection in a plane 3 1.

A2

A1

x12

x35

A3

zA

z A

A5

x13

Page 4: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

2 -Auxiliary projection in a plane 4 2.

A2

A1

x12

x24

yA

yA

A4

x46A6

Page 5: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Problem (1):The true length of a straight line in space .

A2

A1

x12

B2

B1

x13

A3B3 T.L.

Page 6: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

x12

h2

h1 = T.Lf1

f 2 = T.L

h // 1 f // 2

Page 7: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Problem (1):The true length of a straight line in space .

A2

A1

x12

x24

T.L.

B2

B1

A4

B4

Page 8: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

B1

T.L. of m

A1

zAB

B2

T.L. of m

A2

yAB

B3

T.L. of m

A3

xAB

Triangles of solution

Page 9: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

A2

A1

x12

B2

B1

x13

A3B3 T.L.

Problem (2):Convert a straight line into a point .

x35

A5=B5

Page 10: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

The auxiliary projection of a plane.Problem (3): Convert a plane into a line

x12

v

h

x 13

3

A2

A1

A3

1

Page 11: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Problem (4):The dihedral angle between two planes.

A2

B2

C2

B1

C1

D1

A1

D2

خط ( المشترك الخط تحويل هى األساسية الفكرةنقطة) الى التقاطع

Page 12: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Problem (4):The dihedral angle between two planes.

A2

B2

C2

B1

C1

D1

A1

D2

x12

x 35

A3

D3

C3

B3x13

A5 =B5

D5C5

T.L.

Page 13: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Example (2):Given a point R and a line m{A,B}, find d(R, AB) .

x12

R2

R1

B2

A2

B1

A1

R3

B3

A3 x13

x35

A5 =B5

R5

d(R

, AB

)

T.L.

Page 14: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

Example (4):Given the two projections of Parallelogram and the vertical projection of the point M. find the horizontal projection of point M if d(M,ABCD)=3cm also, find the true shape of the Parallelograms ABCD.

x12

M2

D1

B2

A2

B1A1

x 13

D2

C2

C1

K2

K1

B3

A3C3D3

3 cm

zM

z M

L:M3

L:M3

M3

M2

Page 15: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .

x12

D1

B1A1

x 13

D2

C2

C1

K1

B3

A3C3D3

M1

x35

D5

C5

B5

A5

T.S.

Page 16: Lecture 5: The Auxiliary projection By Dr. Samah Mohamed Mabrouk  By Dr. Samah Mohamed Mabrouk .