3 Graphical methods:
3. Graphical methods:
Stereographic projection
Stereographic projection
1) Types of projection 1) 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 foci and a object are projected to a surface
and 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 (
등면적 투영법)
Fig. 3.5
Advantage: Area of a small circle is preserved. g p
Disadvantage: 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
Define a line intersecting a focal point and a point ( , , ) on a sphere, and calculate x y z t00
when the line intersects a horizontal projection plane.
' , ' , '
' 0 at the projection plane
x xt y yt z R z R t
z
0 R
t z R
0 , 0
z R
Rx Ry
x y
z R z R
• Projection of a great circle
1 Define a line intersecting a focal point and a point (x y0 0 0) in a horizontal 1. 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
p
3. 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 0 Sphere:
Li 0 0 R 0 1
x y z
x y z R
t t R t
0 0 0 0
2 2
2 2 2 2 2 2
0 0 2 2 2
Line 0,0,-R , , 0 : , , 1
Line - sphere: 1 2
x y x x t y y t z R t
x t y t R t R t R
x y R
0 0
2 2
0 0
2 2 2 2
0 0 0
2 2
,
x y R
R x R y
x y
x y R x
2 02 02
2 2 2 2 2
0 0 0
, R R x y
y R z x y R
0 0 0
x y x y0 x0 y0
Li J i t l
2 2 2 2 2
0 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 2
0 0 0 0
2 2 2 2
0 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
n2 cos2
R2
2 2 2 2 2
0 0 2
tan sin tan cos 1 tan Circle
cos
x R y R R R
R
Circle: , tan sin , tan cos , radius
x y cos
C C R 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 cos Sphere:
x y z R
x y z R
0 0
Sphere:
Line: , , 1
Sphere li
x y z R
x x t y y t z R t
2 2 2
2 2
0 0
0 0
2 2
ne: R x R y R R x y
x y z
Sphere - li
2 2 2 2 2 2 2 2 2
0 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 z
x 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 2
0 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
Circle
x y R
0 0 Circle
cos cos cos cos cos cos
Circle:
R sin sin R sin cos sin
x y
R
,
R sin sin , R sin cos , radius sincos cos cos cos cos cos
x y
C C R
• H.W.2
Draw an equatorial-net by equal-angle Draw 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
3 Fig. 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
4) Additions
g p
x y z, ,
x y0, 0
R R
0 , 0
sin sin sin cos
i i i
Rx Ry
x y
R z R z
R R
R R R
0 0
02 02sin sin , sin cos , cos ,
1 cos 1 cos
Distance from an origin to , sin tan
n R R R
x y x y R 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 equation cos
x y z
x x
n n n
C C R R R
(X Y ) (X Y Z)
cos
• (X 0 , Y 0 ) (X,Y,Z)
From the procedure to obtain the great circle equation
2 2 2 2
0 0
Sphere:
Line: , , 1
x y z R
x x t y y t z R t
2 2 2
2 2
2 0 0 0 0
2 2 2 2 2 2 2 2 2 2 2 2
0 0 0 0 0 0 0 0
2 2
2R , R x , R y , R R x y
t 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 ˆ ˆ
X Y x y z X Y x y z
x x n n n n
1 21 2
, ,
ˆ ˆ
C t f t i l
x y z
n n n n
x x
Center of a great circle:
, , sin sin , sin cos , cos tan sin , tan cos x , y
x y z
z z
Rn Rn
n n n R R
n n
,
z z
x y
z z
Rn Rn
n 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 plane
1) Draw a great circle of a plane 1) 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 sliding
r s ˆ ˆ
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 force
direction vector of resultant force.
3) Find out two intersection points (vectors) of the above two planes
4) 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 above
the great circle above.
• Sliding in two planes simultaneously Sliding in two planes simultaneously
1) 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 i
5) The sliding vector is one of the two intersection vectors which is located in the great circle above.
