2019年03月23日

Circle of sphere and its applications

By Ching-Her, Day(戴清河), www.chday169.url.tw

 (1).Introduction

Many rock slope engineering and structural geology problems such as dip direction/dip (or stride/dip), graphical presentation of geological data, interpretation of drill cores are concerning with space planar equation Ax+By+Cz=D. Further more, the intersection of planes, the rotation of geological structural plane about a inclined axis (secondary tilt problems), are related with the planar equation as well. If we use the simple mathematical tools (vector operations, linear system equations of 3 unknowns, etc.), you will find those problem will be simple and easy.

截面圓(Spherical circle)是等角度,等面積投影製圖(含投影網製作)之理論基礎及空間曲線(torus knots等)管狀圖示電腦製圖之立論基礎.如能澈底瞭解截面圓之求解及製圖原理,對研讀構造地質學及岩坡工程學助益良多,真所謂小兵立大功,一招半式闖天下,粉簡單.

(2).Intersection of a sphere and a plane

If the equation of the sphere S with center at point C(x,y,z) and radius ris defined as:

(x-x)+( (y-y)+(z-z)=r………………….(1)

The sphere S is cut off by a the plane Pl-E, then the section plane is a circle ,the circle is call spherical circle. If the plane passing through spherical center, then the circle section is called great circle, otherwise, it is called small circle.

Figure 1 Relation of Plane and sphere

Let a plane with a normal N=[A,B,C], and A,B,C are not all zero ,then the plane Pl_E can be defined as

Ax+By+Cz=D……………….(2a)

Figure 1 illustrates the relation of the sphere S and plane Pl_E

Eq.(2) can be modified as:

 [A,B,C][x,y,z]=D…………………………………….(2b)

If we denote normal vector N=[A,B,C] , R=[x,y,z] then eq.(2b) can be rewritten as:

NR=D……………………………………………….(2c)

Here the magnitude of vector N is equal to

=………………………….(3)

Suppose a line L parallel to the normal and pass through Cand point C(x,y,z), then it can be defined as:

===t , i.e

x=x+At

y=y+Bt

z=z+Ct ……………….(4)

Substituting eq. (4) into eq.(2a) and solving t , getting

x=x+A

y=y+B

z=z+C……………….(5)

The distance L between points C and C is

L==

Hence the length Lof Point C(x,y,z) and an arbitrarily point P(x,y,z) on the intersected curve of the sphere and plane E is equal to

L=

The value of Lis a constant, and point P is an arbitrarily point lying on both plane E and sphere S, so that the trace of rotation of point P along line L is a circle with a radius requivalent to L, i.e

r= L=……………….(6)

From eq. (2a) we can get

A(x-x)+B(y-y)+C(z-z)=D-(Ax+By+Cz)……………….(7a)

Or

[A,B,C][ (x-x),(y-y),(z-z)]= D-(Ax+By+Cz)…………….(7b)

Here [ (x-x),(y-y),(z-z)] is a position vector Rwith initial point at sphere center and terminal point on section circle circumstance , so that it’s magnitude is equal the sphere radius. Then eq.(7b) can be rewritten as:

cos()=D-(Ax+By+Cz)…………….(7b)

Here  is the include angle between normal vector N and position vector R, so that

cos()=…………….(8)

The radius of spherical circle rcan be defined also as:

r=rsin()…………….(9)

Notes here, the diagram of spherical circle is a true circle only under the projection along in the direction parallel to the planar normal vector N, otherwise it always shown as ellipse or a point.

(3).Parametrical form of spherical circle

The parametrical form of spherical circle can be derived as following:

Denote the three orthogonal basic axe are denoted as u , v and w. Here u , v and w can not be zero vectors, w is an unit vector parallel to planar normal N, and

a=

b=

c= ………………………………….(10)

Since uv=vw=wu=0 and uv=w , vw=u ,wu=v, so if a+b0 we can set u=[b,-a,0] ,otherwise we can set u=[0,c,-b] or u=[0,-b,c],etc.

If we set u=[b,-a,0] here then

v=  ,i.e

u=[b,-a,0]/

v=[-ac,-bc, a+b]/

w=[a,b,c] ………………………………….(11)

The parametrical form of spherical circle can be derived as:

x=x+r(ucos(t)+vsin(t))

y=y+r(ucos(t)+vsin(t))

x=z+r(ucos(t)+vsin(t)) ……………………….(12)

If we want to plot the spherical circle on u-v plane, supposing the coordinates of a point on spherical is denoted as (x,y,z) then

u=[u,u,u]  [x,y,z]

= [u,u,u][ x+r(ucos(t)+vsin(t)), y+r(ucos(t)+vsin(t)),

z+r(ucos(t)+vsin(t)) ……………….(13a)

v=[v,v,v]  [x,y,z]

=[u,u,u][ x+r(ucos(t)+vsin(t)), y+r(ucos(t)+vsin(t)),

z+r(ucos(t)+vsin(t)) ……………….(13b)

Furthermore if we use the spherical coordinates (, q, p) system, then sphere S can be described as:

x=x+rcos(p)cos(q)

y=y+rsin(p)sin(q)

z=z+rcos(q) ……………………………………….(14)

Where p measures from x(N)axis in clockwise, and q is measured from y-z Plane downward with clockwise following right-hand rule.

So the downward vector N=[-cos(p)sin(q), -sin(p)sin(q),cos(q)]

Then the components of u and v are:

u=[u,u,u][ x+rcos(p)cos(q), y+rsin(p)cos(q). z+rsin(q)]

v=[v,v,v][ x+rcos(p)cos(q), y+rsin(p)cos(q). z+rsin(q)]

……………………………………………….(15)

A Circle plane section passing through C( x,y,z) ,i.e

Ax+By+C=D=0 is called great circle, if D≠0 is called small circle.

For example a plane x+2y+3z=10, and the sphere

(x-3)+ (y-4)+(z-5)=8,find the parametrical circle of the intersection of the plane and sphere?

t=-1.14296

x=x+At=3-1.14286×1=1.8571

y=y+Bt=1.7143

z=z+Ct =1.5714

N=[1,2,3]

w=n=[0.26726,0.53452,0.80178]

u=[0.89442,-0.44721,0]

v=[-0.35856,-0.71713,0.59761]

r= L==6.7612

When tp=0,Then

x=x+r(ucos(tp)+vsin(tp))=1.8571+6.7612(0.89442×1+0)=7.9046

y=y+r(ucos(tp)+vsin(tp))=1.7143+6.7612(-0.44721×1+0)=-1.3094

z=z+r(ucos(tp)+usin(tp)) =1.5714+6.7612(0×1+0)=1.5714

The points on u-v-w coordinates are

u=[7.9046,-1.3094,1.5714]  [0.89442,-0.44721,0]=7.6557

v=[7.9046,-1.3094,1.5714] [-0.35856,-0.71713,0.59761]=-0.9562

w=[7.9046,-1.3094,1.5714] [0.26726,0.53452,0.80178]=2.6726

Figure 3 Relation between plane and sphere

(Proj. on x-y plane)

Figure 4 Relation between plane and sphere

(Proj. on u-v plane)

 If the plane change to x+2y+3z=26,i.e

(x-3)+2(y-4)+3(z-5)=0,Then the spherical circle is a great circle, the diagram of Figure 3 and 4 will change to as figure 5 and 6 shown.

Figure 5 Relation between plane and sphere

(Proj. on x-y plane)

Figure 6 Relation between plane and sphere

(Proj. on u-v plane)

(4)Equal angle projection and equal area projection

    It is more convenient to use the sphere radius equal to 1 unit in equal angle projection and equal area projection. In this case equ.(1) will change to as eq.(16)

x+y+z=1………………………………….(16)

and

ax+by+cz=d………………………………….(10)

where a=,b=,c=

 The equal angle or stereographic or Wulff projection, is illustrated in figure 7. Point P is the equal angle projection of the spherical surface point A. Point P is the pierced point of a line (connecting the zenith point T and point A) and the horizontal plane (passing through the spherical center O).

Figure 7 Equal angle projection

 While the equal area projection (also known Lambwet or Schimidt projection) is shown in figure 8. Point P is the projection of a spherical surface point A by swing A in a arc centered at the contacting point of the sphere and a horizontal surface circle with a radius of upon which it stands.

Figure 8 Equal area projection

 The derived equations of equal angle and equal area projection are discussed completely in author’s publication (Chinese edition) or www.chday169.url.tw.

x=(equal angle)…………………………(17a)

y=(equal angle)…………………………(17b)

x=(equal area)………………………(18a)

y=(equal area)………………………(18b)

The equal angle and equal area projections can be obtained by projecting each spherical circle point of a plane using eq.(17) and eq.(18). Figure 11 and figure 12 are the equal angle and equal area projection plane x+2y+3z=2, respectively.

Figure 11(a) Equal angle projection of a plane

Figure 11(b) Equal angle projection of a plane

Figure 12(a) Equal area projection of a plane

Figure 12(b) Equal area projection of a plane

  In representation of geological data, we use the equal angle or equal area projection net. Those net can be easily to plot by computer.

The dot product of two unit vectors (e and d) can be denoted as:

ed=cos(ψ) ………………………………(19a)

From the definition of equal angle projection, we can obtain the spherical

 equation of a plane as eq.(18b) shown:

(x-)+(y-)=()……..(19b)

If we set e=n=[a,b,c]=[cosαcosβ, sinαcosβ,sinβ],hereα,β are dip direction and dip of a plane. Thus ifβ=90°,then eq.(18b) can be reduced to

(x-)+(y-)=tan…………………………(19c)

IF α=0 ( small circle with a normal point to north-south),eq.(18c) can be rewritten as

(x-sec)+y= tan…………………………(19d)

Using eq.(17) or eq.(19d) we can design a computer program to draw the equal angle projection net.

Similarly, we can get the equal area projection equations as:

(ex+ey)+e(1-x-y)= cos………(20a)

Ifφ=90°, then eq.(19a) can be reformed as eq.(19b)

(xcos()+ysin()=cos………(20b)

,and if set α=0 and φ=90° ,then we get

(2-x-y)x= cos…………………………(20c)

Using eq.(18) or eq.(20c) we can design a computer program to draw the equal area projection net as well.

Notes here, both equal angle and equal area projection are using plane α=90° andα=270° and β=0,2,4,6…88,90° to draw the great circle, while using α=0° andβ=0,2,4,6…..88,90° to draw the small circle.

(5)Tube plots

  Parametrical curves in 3-spaces can be difficult to visualize and draw without the help of a graphic utility. Sometime we are difficult to distinguish the points whether it in front of other or not. But using the tube plots we can overcome those problems easily. The principle of tube plots is quite simple and easy by using the parametrical form of spherical circle.

Along the path of 3D curve we can divide the path to n sections. Suppose at stage t, a point on the path denoted as P(x,y,z). At stage t point P moving to Q(x,y,z). And then, we set N=

[ (x- x),(y- y),(z-z)]. So we can get a spherical circle with a radius. Projecting those points (say 36 points) of stage tand ton spherical circle one by one to a projection plane. Drawing the spherical circle and connecting sequential point of the two spherical circle curves (3 or 4 points polygon). Picking up two adjacent projected points on stage tand tof spherical circle to form a polygon and drawing it. Following the same procedures, we can draw completely the tube plot of the 3D space curves. Knowing the principles and procedures we can design a computer program or Microsoft excel sheets to draw the parametrical curves in 3-spaces. Figures 9 and 10 are the tube plots of space helix curve and torus knots, respectively.

 x=4cos(u)/5, y=4sin(u)/5,z=u/5, (-9.42≦u≦9.42) and

 x=cos(2u)(3+cos(5u)),y= sin(2u)(3+cos(5u)),z=sin(5u),(-2pi<-u<=2pi)

Figure 9

Figure 10

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