Electromagnetics 03, Python Coding

Chapter title: Coordinate Systems

1. Example 1

1) 문제

Given the coordinate (x, y, z) in the rectangular coordinate of a point P,

a) find the coordinate (r, θ, φ) in the spherical coordinate of the point P, express θ and φ in degree unit

and

b) express base vectors at P in terms of base vectors .

2) 수식

a)

To convert θ and φ expressed in radian, multiply them with 180/π.

b)

3) 코딩

# EM-03-Python-Ex1

# Given a point P(x,y,z),

# 1) express P in the spherical coordinate

# 2) express base vectors <r>,<theta>,<phi> in terms of base vectors <x>,<y>,<z>

from math import *

pi=3.141593

while True:

x,y,z=map(float, input('x,y,z=').split())

r=sqrt(x**2+y**2+z**2)

theta=acos(z/r)

rho=sqrt(x**2+y**2)

phi=acos(x/rho)

if y < 0:

phi=pi+phi

print('(r,theta,phi)=(',r,',',theta*180/pi,',',phi*180/pi,')')

rx=sin(theta)*cos(phi)

ry=sin(theta)*sin(phi)

rz=cos(theta)

thetax=cos(theta)*cos(phi)

thetay=cos(theta)*sin(phi)

thetaz=-sin(theta)

phix=-sin(phi)

phiy=cos(phi)

phiz=0

print('<r> =(',rx,',',ry,',',rz,')')

print('<theta>=(',thetax,',',thetay,',',thetaz,')')

print('<phi> =(',phix,',',phiy,',',phiz,')')

4) 코드 수행결과

x,y,z=

1 1 1

(r,theta,phi)=( 1.7320508075688772 , 54.73560428178051 , 44.99999503803985 )

<r> =( 0.5773502691896257 , 0.5773502691896258 , 0.5773502691896258 )

<theta>=( 0.408248290463863 , 0.4082482904638631 , -0.816496580927726 )

<phi> =( -0.7071067811865476 , 0.7071067811865475 , 0 )

x,y,z=

1 1 -1

(r,theta,phi)=( 1.7320508075688772 , 125.26437587037888 , 44.99999503803985 )

<r> =( 0.5773502691896255 , 0.5773502691896256 , -0.5773502691896261 )

<theta>=( -0.4082482904638632 , -0.40824829046386324 , -0.8164965809277258 )

<phi> =( -0.7071067811865476 , 0.7071067811865475 , 0 )

x,y,z=

1 -1 1

(r,theta,phi)=( 1.7320508075688772 , 54.73560428178051 , 224.99999503803986 )

<r> =( -0.5773500691895651 , -0.5773504691896172 , 0.5773502691896258 )

<theta>=( -0.4082481490424639 , -0.4082484318852132 , -0.816496580927726 )

<phi> =( 0.7071070261355112 , -0.707106536237499 , 0 )

x,y,z=

-1 1 1

(r,theta,phi)=( 1.7320508075688772 , 54.73560428178051 , 134.9999851141195 )

<r> =( -0.5773502691896257 , 0.5773502691896258 , 0.5773502691896258 )

<theta>=( -0.408248290463863 , 0.4082482904638631 , -0.816496580927726 )

<phi> =( -0.7071067811865476 , -0.7071067811865475 , 0 )

x,y,z=

-1 -1 1

(r,theta,phi)=( 1.7320508075688772 , 54.73560428178051 , 314.9999851141195 )

<r> =( 0.5773504691896169 , -0.5773500691895654 , 0.5773502691896258 )

<theta>=( 0.40824843188521304 , -0.40824814904246415 , -0.816496580927726 )

<phi> =( 0.7071065362374993 , 0.7071070261355109 , 0 )

x,y,z=

-1 -1 -1

(r,theta,phi)=( 1.7320508075688772 , 125.26437587037888 , 314.9999851141195 )

<r> =( 0.5773504691896167 , -0.5773500691895652 , -0.5773502691896261 )

<theta>=( -0.40824843188521315 , 0.4082481490424643 , -0.8164965809277258 )

<phi> =( 0.7071065362374993 , 0.7071070261355109 , 0 )

x,y,z=** Process Stopped **

Press Enter to exit terminal

2. Example 2

1) 문제

반경 a인 구면 상의 두점 P = (a, θ₁, φ₁) and Q = (a, θ₂, φ₂)이 주어질 경우 구면 상에서 P에서 Q 로 이동할 때 최소경로 길이 L를 구하라. Theta와 phi 각은 degree 단위로 입력한다.

(참고) L을 great-circle distance, orthodromic distance, spherical distance라고 부른다.

https://en.wikipedia.org/wiki/Great-circle_distance

2) 수식

L 공식:

L = aα

R₁ : P₁의 위치벡터

R₂ : P₂의 위치벡터

α : R₁, R₂ 사이의 각도

R₁, R₂를 직각좌표계로 표현

R₁ = (x₁, y₁, z₁) ; R₂ = (x₂, y₂, z₂)

R₁, R₂의 단위벡터:

R₁, R₂ 사이의 각도 구하기:

3) 코딩

# EM-03-Python-Ex2

# Given two points on a sphere, find the great-circle distance between them.

import from math *

pi=3.141593

a=float(input('Sphere radius: a='))

while True:

t1,p1=map(float, input('Point P: theta(deg), phi(deg)=').split())

t2,p2=map(float, input('Point Q: theta(deg), phi(deg)=').split())

t1=t1*pi/180 ; p1=p1*pi/180 ; t2=t2*pi/180 ; p2=p2*pi/180 # degree to radian

x1=sin(t1)*cos(p1) ; y1=sin(t1)*sin(p1) ; z1=cos(t1)

x2=sin(t2)*cos(p2) ; y2=sin(t2)*sin(p2) ; z2=cos(t2)

r1_dot_r2=x1*x2+y1*y2+z1*z2

alpha=acos(r1_dot_r2)

print('Great-circle distance from P to Q =',a*alpha)

4) 코드 수행결과

Sphere radius: a=

1

Point P: theta, phi=

0 0

Point Q: theta, phi=

90 0

Great-circle distance from P to Q = 1.5707965

Point P: theta, phi=

0 0

Point Q: theta, phi=

90 90

Great-circle distance from P to Q = 1.5707965

Point P: theta, phi=

45 45

Point Q: theta, phi=

90 90

Great-circle distance from P to Q = 1.0471976926179716

Point P: theta, phi=** Process Stopped **

Press Enter to exit terminal

3. Example 3

1) 문제

반경 a인 원통면 상의 두점 A = (a, φ₁, z₁) and B = (a, φ₂, z₂)이 주어질 경우 원통면 상에서 A에서 B로 이동할 때 최소경로 길이 S를 구하라. φ₁과 φ₂는 도(degree) 단위로 입력된다.

2) 수식

φ₁과 φ₂를 radian으로 변환

φ₁ (rad) = φ₁ (deg) × π / 180

점 A와 점 B의 φ 각도 차이:

두 점 사이의 곡선 최소거리:

3) 코딩

# EM-03-Python-Ex3

# Given two points on a circular cylinder, find the surface distance between them.

from math import *

pi=3.141593

a=float(input('Cylinder radius: a='))

while True:

p1,z1=map(float, input('Point A: phi(deg), z=').split())

p2,z2=map(float, input('Point B: phi(deg), z=').split())

p1=p1*pi/180 ; p2=p2*pi/180

dp=abs(p1-p2)

if dp > pi:

dp=2*pi-dp

s=sqrt((z1-z2)**2+(a*dp)**2)

print('Surface distance between A and B = ',s)

4) 코드 수행결과

Cylinder radius: a=

1

Point A: phi(deg), z=

45 1

Point B: phi(deg), z=

90 2

Surface distance between A and B = 1.2715543288051292

Point A: phi(deg), z=** Process Stopped **

Press Enter to exit terminal

4. Example 4

1) 문제

평면상의 xy 좌표 기준으로 점 P = (x, y)가 주어졌다. xy 축을 반시계방향으로 θ 만큼 회전시킨 새로운 좌표축을 x'y' 좌표축이라 한다. 점 P의 좌표를 x'y' 좌표로 표시하라. θ는 도(degree) 단위로 입력된다.

2) 수식

3) 코드

# EM-03-Python-Ex4

# A point P(x, y) is given. X'Y' axis is the XY axis rotated CCW by theta. Express the coordinate of P in

# the new X'Y' coordinate system.

from math import *

pi=3.141593

while True:

x,y=map(float, input('Point P: x, y=').split())

theta=float(input('XY axis CCW rotation angle: theta(deg)='))

theta=theta*pi/180

xp=x*cos(theta)+y*sin(theta); yp=-x*sin(theta)+y*cos(theta)

print('Point P: xp, yp =',xp,yp)

4) 코드 수행결과

Point P: x, y=

1 2

XY axis CCW rotation angle: theta(deg)=

45

Point P: xp, yp = 2.121320404796886 0.7071065974747902

Point P: x, y=

1 0

XY axis CCW rotation angle: theta(deg)=

45

Point P: xp, yp = 0.7071067199492933 -0.7071068424237964

Point P: x, y=** Process Stopped **

Press Enter to exit terminal