# 3 graphical methods: 3. graphical methods: stereographic

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3 Graphical methods: 3. Graphical methods: Stereographic projection Stereographic projection

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TRANSCRIPT 3 Graphical methods: 3. Graphical methods: Stereographic projectionStereographic projection 1) Types of projection1) Types of projection

•Projection: Mapping 3D images into 2D ones on a planar surface

•Parallel projection: parallel rays are projected to a planar surface from•Parallel projection: parallel rays are projected to a planar surface from an object. It is useful to convey measurements of distances or angles.distances or angles.

ex.) Orthographic projection (Fig.3.1, Fig 3.4), oblique projection (Fig.3.2)

• Perspective projection: nonparallel rays connecting one or more fociand a object are projected to a surfaceand a object are projected to a surface. It is useful to convey perspective views of objects.

ex.) Equal area projection, ) q p j ,equal angle (stereographic) projection 1) Equal-Area projection (등면적투영법)

Fig. 3.5

Advantage: Area of a small circle is preserved.g pDisadvantage: Shape of a small circle is distorted according to

its location on the sphere. 2) Stereographic (equal-angle, 등각) projection

Lower focal point (=upper hemisphere) projection

Advantage: Shape of a small circle (angle) is preserved g p ( g ) p(conformal).

Disadvantage: Area of a small circle changes according to its location on the sphere.  2) Stereographic projection of lines and planes

• Projection of a vector

2) Stereographic projection of lines and planes

Projection of a vector

0Define a line intersecting a focal point and a point ( , , ) on a sphere, and calculate x y z t0

when the line intersects a horizontal projection plane.

' , ' , '

' 0 at the projection plane

x xt y yt z R z R t

z

0 Rtz R

0 0,

z RRx Ryx y

z R z R • Projection of a great circle1 Define a line intersecting a focal point and a point ( 0) in a horizontalx y0 01. Define a line intersecting a focal point and a point ( , ,0) in a horizontal projection plane.2. Obtain when the line meets the sphere.

x y

t

0 0

p3. Obtain the relation between and when the linx y e meets a joint plane.

2 2 2 2

Joint plane: sin sin sin cos cos 0Sphere: Li 0 0 R 0 1

x y zx y z R

t t R t

0 0 0 0

222 2 2 2 2 2

0 0 2 2 2

Line 0,0,-R , ,0 : , , 1

2Line - sphere: 1

x y x x t y y t z R t

Rx t y t R t R tx y R

0 0

2 20 0

2 2 2 20 0 0

2 2,

x y R

R x R yx yx y R x

2 2 20 0

2 2 2 2 20 0 0

,R R x y

zy R x y R

0 0 0x y x 0 0 0y x y Li J i t l

2 2 2 2 20 0 0 0

2 2 2

Line - Joint plane:

2 sin sin 2 sin cos cos 0

2 i 2 0

R x R y R R x y

R R R

2 2 20 0 0 0

2 2 2 20 0

2 tan sin 2 tan cos 0

tan sin tan cos tan si

R x R y R x y

x R y R R

2 2 2n cos R

22 2 2 2

0 0 2 tan sin tan cos 1 tan Circlecos

Rx R y R R

R

Circle: , tan sin , tan cos , radiuscosx y

RC C R R • Projection of a small circle• Projection of a small circle

Apply the same procedure for the great circle to deriving the small circle projected.

2 2 2 2

pp y p g g p j

Joint plane: sin sin sin cos cos cosSphere:

x y z Rx y z R

0 0

Sphere: Line: , , 1

Sphere li

x y z Rx x t y y t z R t

2 2 22 20 00 02 2ne:

R R x yR x R yx y z

Sphere - li

2 2 2 2 2 2 2 2 20 0 0 0 0 0

2 2 2 2 2 2 2 2

ne: , ,

Joint plane - line:

2R sin sin 2R sin cos cos cos

x y zx y R x y R x y R

x y R R x y R R x y

0 0 0 0 0 0

2 2 20 0 0 0

2R sin sin 2R sin cos cos cos

cos cos 2 sin sin cos cos 2 sin cos cos

x y R R x y R R x y

x R x y R y R

2 2 2

cos

R sin sin R sin cos sin CircleRx y

0 0 Circlecos cos cos cos cos cos

Circle:

R sin sin R sin cos sin

x y

R

R sin sin R sin cos sin, , , radiuscos cos cos cos cos cosx y

RC C • H.W.2Draw an equatorial-net by equal-angleDraw an equatorial net by equal angle projection. Set the angle between two adjacent longitudinal/latitudinal lines 10°longitudinal/latitudinal lines 10 . 3) Stereographic projection of a joint pyramid

P j ti f h lf

3) Stereographic projection of a joint pyramid

• Projection of half spaces

Fig. 3.16 • Joint pyramid: an intersect of joint half spaces shifted to the center of a projection sphere

3Fig. 3.17 • Intersection of two joints

Arc of joint p ramid (Fi 3 17)

Fig. 3.12

• Arc of joint pyramid (Fig. 3.17)

- An arc represents a joint plane.- A concave arc to the center of a joint pyramid indicates upper

space of the joint arc when lower-focal-point projection adopted. 4) Additions• Normal to a given plane

0 0, , ,x y z x yR R

0 0,

sin sin sin cosi i i

Rx Ryx yR z R z

R RR R R

2 20 0 0 0

sin sin , sin cos , cos ,1 cos 1 cos

sinDistance from an origin to , tan

n R R R

Rx y x y R

0 0 0 0g ,1 cos 2

y y • Plane normal to a given vector (line)

, , sin sin ,sin cos ,cos ,

, tan sin , tan cos , radius from the great circle equationcos

x y z

x x

n n n

RC C R R

(X Y ) (X Y Z)

cos

• (X0, Y0) (X,Y,Z)From the procedure to obtain the great circle equation

2 2 2 2

0 0

Sphere: Line: , , 1

x y z Rx x t y y t z R t

2 2 22 220 00 0

2 2 2 2 2 2 2 2 2 2 2 20 0 0 0 0 0 0 0

2 22 , , ,R R x yR x R yRt x y z

x y R x y R x y R x y R • Center of a great circle passing two points

-Vector analysis

1 1 1 1 1 2 2 2 2 2

1 2

, in a projection plane , , on a sphere, , , ,ˆ ˆˆ

X Y x y z X Y x y zx xn n n n

1 2

1 2

, ,ˆ ˆ

C t f t i l

x y zn n n nx x

Center of a great circle:

, , sin sin ,sin cos ,cos tan sin , tan cos , yxx y z

z z

RnRnn n n R Rn n

,

z z

yx

z z

RnRnn n

1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

,R y z z y R z x x z

x y y x x y y x - Graphical procedure: refer to Fig.3.19 and Fig.3.9 with Eqn.(3.7)1) Plot an opposite vector point to one of the predefined points.2) Draw a circle passing through the 3 points.

• Orthographic projection of a vector on a plane1) Draw a great circle of a plane1) Draw a great circle of a plane.2) Plot plane normal (n) and a vector (v).3) Plot the opposite vector, -v.4) Draw a circle passing through the 3 points (n, V, and-V)5) Find out the intersection points of the two great circles.

Refer to Fig. 3.20 5) Projection of sliding direction

• Lifting

5) Projection of sliding direction

Lifting - The lifting direction is identical with the direction of resultant

fforce:

• Single-face slidingrs ˆˆ

1) Draw a great circle of a block sliding plane.2) Draw a great circle passing through a normal to the sliding plane and a

direction vector of resultant forcedirection vector of resultant force.3) Find out two intersection points (vectors) of the above two planes4) Draw a great circle whose normal vector is identical with the direction ) g

vector of resultant force.5) The sliding vector is one of the two intersection vectors which is located in

the great circle abovethe great circle above. • Sliding in two planes simultaneouslySliding in two planes simultaneously1) Draw great circles of two block sliding planes.2) Find out two intersection points (vectors) of the above two circles) p ( )3) Draw a great circle whose normal vector is identical with the direction

vector of resultant force.5) Th lidi i f h i i hi h i l d i5) The sliding vector is one of the two intersection vectors which is located in

the great circle above.