Coordinate systems in 3-D
Although there are 11- separable coordinate systems, the most
commonly used ones are Cartesian, cylindrical, and spherical
coordinates. The relationship between Cartesian and Cylindrical
coordinates is very similar to the relationship between Cartesian
and polar coordinates in 2-D.
(i.e.)
Polar to Cartesian
 rˆ   cos  sin   xˆ 
ˆ   sin  cos   yˆ 
  
 
Cylindrical to Cartesian
 rˆ   cos  sin   0  xˆ 
ˆ   sin   cos  0  yˆ 
  
 
 zˆ   0
0
1  zˆ 
6-1
Spherical Coordinates
Suppose you have an object whose
position is defined in spherical
coordinates by the vector:

R  R, ,    RRˆ
Defined as the vector through
the line OP
One way to analyze this geometry is by
breaking the vector into its polar and
vertical components as:
Rˆ  rˆ cos   zˆ sin 
ˆ  ˆ
ˆ  rˆ sin   zˆ cos 
6-2
Knowing that…
rˆ  xˆ cos   yˆ sin 
ˆ   xˆ sin   yˆ cos 
Leads to a relationship between Spherical and Cartesian coordinates:
Rˆ  cos xˆ cos   yˆ sin   zˆ sin 
ˆ   xˆ sin   yˆ cos 
ˆ  xˆ cos   yˆ sin sin   zˆ cos 
And transformation matrices
 Rˆ   cos  cos  sin   cos  sin    xˆ 
 xˆ 
 ˆ 
  yˆ   T  yˆ 







sin

cos

0
S C  
  
 
ˆ   cos sin    sin  sin   cos   zˆ 
 zˆ 
 
  
 Rˆ 
 xˆ  cos  cos   sin    cos sin    Rˆ 
 yˆ    sin   cos  cos   sin  sin   ˆ   T 1 ˆ 
S C
 
  
 
ˆ 
 zˆ   sin  
 ˆ 
0
cos 
 
 
6-3
Transformation matrices between cylindrical and spherical
coordinates are given by:
 Rˆ   cos  0 sin    rˆ 
 ˆ 
 ˆ  T


0
1
0
S Cy
  
 
ˆ   sin   0 cos   zˆ 
 
  

 rˆ  cos  0  sin    Rˆ 
ˆ   0
 ˆ   T
1
0
S Cy
  
 
 zˆ   sin   0 cos   ˆ 
 
 Rˆ 
1  ˆ 
 
ˆ 
 


 rˆ 
ˆ
 
 zˆ 

Transformation matrices between cylindrical and Cartesian
coordinates are given by:
 rˆ   cos 
ˆ   sin  
  
 zˆ   0
sin   0  xˆ 
cos  0  yˆ   TC Cy
0
1  zˆ 

 xˆ  cos   sin   1  rˆ 
 yˆ    sin   cos  0 ˆ  T
C Cy
  
 
 zˆ   1
0
1  zˆ 

 xˆ 
 yˆ 
 
 zˆ 

 rˆ 
1  ˆ 
 
 zˆ 

6-4
Velocity and Acceleration in Spherical Coordinates
Given that the position of a particle can be described by the vector in

Spherical coordinates as:
ˆ
R  R, ,    RR
The task is to derive expressions for velocity and acceleration
(i.e.)
and
 d    ˆ

v  R  R  RR  RRˆ
dt
 d    ˆ  ˆ  ˆ

a  R  R  RR  RR  RR  RRˆ
dt
First, we must ask ourselves, what is
ˆ ?
R
6-5
For simplicity, let’s first convert the unit vector from Spherical to
Cartesian Coordinates:
 Rˆ   cos  cos  sin   cos  sin    xˆ 
 ˆ 
  yˆ 







sin

cos

0
  
 
ˆ   cos sin    sin  sin   cos   zˆ 
 
  
d
d
d

Rˆ  xˆ cos  cos   yˆ sin   cos   zˆ sin  
dt
dt
dt

Rˆ  xˆ  sin   cos   cos sin    yˆ cos  cos   sin  sin    zˆ cos 

 
 xˆ sin  
 xˆ cos sin  

Rˆ   cos  yˆ cos      yˆ sin  sin     cos ˆ  ˆ




0
zˆ cos 

Thus, the velocity vector and its magnitude in spherical coordinates
can be written as:
  ˆ
ˆ  ˆ
v  R  RR  RR  RR  R cos ˆ  Rˆ

2
2

2



v  R  R  R  R cos 
  

6-6
For the sake of completeness, let’s also write down the velocity
in cylindrical coordinates and Cartesian coordinates. For
example, previously we derived the velocity in 2-D polar
coordinates to be 

ˆ
r  rrˆ  r

In cylindrical coordinates, this becomes: r  rrˆ  rˆ  zzˆ
The magnitude of the velocity vector in cylindrical coordinates
 
is therefore:
2
2
2
 
v  r  r  r
 z
In Cartesian coordinates, the velocity vector is written as:

r  xxˆ  yyˆ  zzˆ
The magnitude of the velocity vector in Cartesian coordinates is
therefore:
 
2
2
2
v  r  x  y  z
6-7
We can perform a similar analysis for acceleration:


  d  d  ˆ
av  R
RR  R cos ˆ  Rˆ 
dt
dt
  ˆ  ˆ  


aR
R  RR  R cos ˆ  R sin  ˆ  R cos ˆ  R cos ˆ  R ˆ  Rˆ  Rˆ

Rˆ   cos  ˆ  ˆ
d
 xˆ cos sin    yˆ sin  sin    zˆ cos 
dt
 xˆ sin  sin    cos  cos   yˆ  cos sin    sin   cos   zˆ sin  

ˆ 

 
 xˆ cos  cos 
 xˆ sin  
  yˆ sin   cos    sin   yˆ cos    Rˆ   sin  ˆ
 zˆ sin   


0

ˆ 

d
 xˆ sin    yˆ cos 
dt
 xˆ cos 
  xˆ cos   yˆ sin      yˆ sin     Rˆ  cos   ˆ sin  
 zˆ0 
6-8
For good accounting, it is best to write the time derivatives of the
unit vectors in matrix form as:
 Rˆ  
0
 cos 
   Rˆ 
  
 ˆ
ˆ


0
 sin    
     cos 
ˆ    
 sin  
 ˆ 


0

 
 
Doing so, will make it easier to derive an expression for acceleration
in spherical coordinates, as shown below:
  ˆ  ˆ  


aR
R  RR  R cos ˆ  R sin  ˆ  R cos ˆ  R cos ˆ  R ˆ  Rˆ  Rˆ


  R 2 cos2    R 2


Rˆ R

 
a  ˆ 2 R  cos   R cos   2 R sin   
ˆ 2 R   R  R 2 cos sin  









Given this form, the magnitude of the acceleration vector can be written as:

a 
R  R
2
cos2    R 2
  2R  cos   Rcos   2R sin   2R   R  R
2
2
2

cos sin  
6-9
2
Again for the sake of completeness, let’s also write down the
acceleration vectors in cylindrical coordinates and Cartesian
coordinates. For example, previously we derived the
acceleration in 2-D polar coordinates to be:

 

r  r  r 2 rˆ  2r  r ˆ

2

In cylindrical coordinates, this becomes: r  r  r rˆ  2r  r ˆ  zzˆ

 

The magnitude of the velocity vector in cylindrical coordinates
is therefore:
2
 
2 2




a  r  r  r
 2r  r  z2

 

In Cartesian coordinates, the velocity vector is written as:
r  xxˆ  yyˆ  zzˆ
The magnitude of the velocity vector in Cartesian coordinates is
therefore:
 
2
2
2
a  r  x  y  z
6 - 10
Prob. in Meriam & Kraige
s
b



b  120 mm
s  100 mm
1
  sin 4t   rad
6
     rad
s
What is the velocity and acceleration of Mr. Smiley face at t=0.5 s?
6 - 11