- Source: List of common coordinate transformations
This is a list of some of the most commonly used coordinate transformations.
2-dimensional
Let
(
x
,
y
)
{\displaystyle (x,y)}
be the standard Cartesian coordinates, and
(
r
,
θ
)
{\displaystyle (r,\theta )}
the standard polar coordinates.
= To Cartesian coordinates
=From polar coordinates
x
=
r
cos
θ
y
=
r
sin
θ
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
[
cos
θ
−
r
sin
θ
sin
θ
−
r
cos
θ
]
Jacobian
=
det
∂
(
x
,
y
)
∂
(
r
,
θ
)
=
r
{\displaystyle {\begin{aligned}x&=r\cos \theta \\y&=r\sin \theta \\[5pt]{\frac {\partial (x,y)}{\partial (r,\theta )}}&={\begin{bmatrix}\cos \theta &-r\sin \theta \\\sin \theta &{\phantom {-}}r\cos \theta \end{bmatrix}}\\[5pt]{\text{Jacobian}}=\det {\frac {\partial (x,y)}{\partial (r,\theta )}}&=r\end{aligned}}}
From log-polar coordinates
x
=
e
ρ
cos
θ
,
y
=
e
ρ
sin
θ
.
{\displaystyle {\begin{aligned}x&=e^{\rho }\cos \theta ,\\y&=e^{\rho }\sin \theta .\end{aligned}}}
By using complex numbers
(
x
,
y
)
=
x
+
i
y
′
{\displaystyle (x,y)=x+iy'}
, the transformation can be written as
x
+
i
y
=
e
ρ
+
i
θ
{\displaystyle x+iy=e^{\rho +i\theta }}
That is, it is given by the complex exponential function.
From bipolar coordinates
x
=
a
sinh
τ
cosh
τ
−
cos
σ
y
=
a
sin
σ
cosh
τ
−
cos
σ
{\displaystyle {\begin{aligned}x&=a{\frac {\sinh \tau }{\cosh \tau -\cos \sigma }}\\y&=a{\frac {\sin \sigma }{\cosh \tau -\cos \sigma }}\end{aligned}}}
From 2-center bipolar coordinates
x
=
1
4
c
(
r
1
2
−
r
2
2
)
y
=
±
1
4
c
16
c
2
r
1
2
−
(
r
1
2
−
r
2
2
+
4
c
2
)
2
{\displaystyle {\begin{aligned}x&={\frac {1}{4c}}\left(r_{1}^{2}-r_{2}^{2}\right)\\[1ex]y&=\pm {\frac {1}{4c}}{\sqrt {16c^{2}r_{1}^{2}-\left(r_{1}^{2}-r_{2}^{2}+4c^{2}\right)^{2}}}\end{aligned}}}
From Cesàro equation
x
=
∫
cos
[
∫
κ
(
s
)
d
s
]
d
s
y
=
∫
sin
[
∫
κ
(
s
)
d
s
]
d
s
{\displaystyle {\begin{aligned}x&=\int \cos \left[\int \kappa (s)\,ds\right]ds\\y&=\int \sin \left[\int \kappa (s)\,ds\right]ds\end{aligned}}}
= To polar coordinates
=From Cartesian coordinates
r
=
x
2
+
y
2
θ
′
=
arctan
|
y
x
|
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=\arctan \left|{\frac {y}{x}}\right|\end{aligned}}}
Note: solving for
θ
′
{\displaystyle \theta '}
returns the resultant angle in the first quadrant (
0
<
θ
<
π
2
{\textstyle 0<\theta <{\frac {\pi }{2}}}
). To find
θ
,
{\displaystyle \theta ,}
one must refer to the original Cartesian coordinate, determine the quadrant in which
θ
{\displaystyle \theta }
lies (for example, (3,−3) [Cartesian] lies in QIV), then use the following to solve for
θ
:
{\displaystyle \theta :}
θ
=
{
θ
′
for
θ
′
in QI:
0
<
θ
′
<
π
2
π
−
θ
′
for
θ
′
in QII:
π
2
<
θ
′
<
π
π
+
θ
′
for
θ
′
in QIII:
π
<
θ
′
<
3
π
2
2
π
−
θ
′
for
θ
′
in QIV:
3
π
2
<
θ
′
<
2
π
{\displaystyle \theta ={\begin{cases}\theta '&{\text{for }}\theta '{\text{ in QI: }}&0<\theta '<{\frac {\pi }{2}}\\[1.2ex]\pi -\theta '&{\text{for }}\theta '{\text{ in QII: }}&{\frac {\pi }{2}}<\theta '<\pi \\[1.2ex]\pi +\theta '&{\text{for }}\theta '{\text{ in QIII: }}&\pi <\theta '<{\frac {3\pi }{2}}\\[1.2ex]2\pi -\theta '&{\text{for }}\theta '{\text{ in QIV: }}&{\frac {3\pi }{2}}<\theta '<2\pi \end{cases}}}
The value for
θ
{\displaystyle \theta }
must be solved for in this manner because for all values of
θ
{\displaystyle \theta }
,
tan
θ
{\displaystyle \tan \theta }
is only defined for
−
π
2
<
θ
<
+
π
2
{\textstyle -{\frac {\pi }{2}}<\theta <+{\frac {\pi }{2}}}
, and is periodic (with period
π
{\displaystyle \pi }
). This means that the inverse function will only give values in the domain of the function, but restricted to a single period. Hence, the range of the inverse function is only half a full circle.
Note that one can also use
r
=
x
2
+
y
2
θ
′
=
2
arctan
y
x
+
r
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta '&=2\arctan {\frac {y}{x+r}}\end{aligned}}}
From 2-center bipolar coordinates
r
=
r
1
2
+
r
2
2
−
2
c
2
2
θ
=
arctan
[
8
c
2
(
r
1
2
+
r
2
2
−
2
c
2
)
r
1
2
−
r
2
2
−
1
]
{\displaystyle {\begin{aligned}r&={\sqrt {\frac {r_{1}^{2}+r_{2}^{2}-2c^{2}}{2}}}\\\theta &=\arctan \left[{\sqrt {{\frac {8c^{2}(r_{1}^{2}+r_{2}^{2}-2c^{2})}{r_{1}^{2}-r_{2}^{2}}}-1}}\right]\end{aligned}}}
Where 2c is the distance between the poles.
= To log-polar coordinates from Cartesian coordinates
=ρ
=
log
x
2
+
y
2
,
θ
=
arctan
y
x
.
{\displaystyle {\begin{aligned}\rho &=\log {\sqrt {x^{2}+y^{2}}},\\\theta &=\arctan {\frac {y}{x}}.\end{aligned}}}
= Arc-length and curvature
=In Cartesian coordinates
κ
=
x
′
y
″
−
y
′
x
″
(
x
′
2
+
y
′
2
)
3
2
s
=
∫
a
t
x
′
2
+
y
′
2
d
t
{\displaystyle {\begin{aligned}\kappa &={\frac {x'y''-y'x''}{\left({x'}^{2}+{y'}^{2}\right)^{\frac {3}{2}}}}\\s&=\int _{a}^{t}{\sqrt {{x'}^{2}+{y'}^{2}}}\,dt\end{aligned}}}
In polar coordinates
κ
=
r
2
+
2
r
′
2
−
r
r
″
(
r
2
+
r
′
2
)
3
2
s
=
∫
a
φ
r
2
+
r
′
2
d
φ
{\displaystyle {\begin{aligned}\kappa &={\frac {r^{2}+2{r'}^{2}-rr''}{(r^{2}+{r'}^{2})^{\frac {3}{2}}}}\\s&=\int _{a}^{\varphi }{\sqrt {r^{2}+{r'}^{2}}}\,d\varphi \end{aligned}}}
3-dimensional
Let (x, y, z) be the standard Cartesian coordinates, and (ρ, θ, φ) the spherical coordinates, with θ the angle measured away from the +Z axis (as [1], see conventions in spherical coordinates). As φ has a range of 360° the same considerations as in polar (2 dimensional) coordinates apply whenever an arctangent of it is taken. θ has a range of 180°, running from 0° to 180°, and does not pose any problem when calculated from an arccosine, but beware for an arctangent.
If, in the alternative definition, θ is chosen to run from −90° to +90°, in opposite direction of the earlier definition, it can be found uniquely from an arcsine, but beware of an arccotangent. In this case in all formulas below all arguments in θ should have sine and cosine exchanged, and as derivative also a plus and minus exchanged.
All divisions by zero result in special cases of being directions along one of the main axes and are in practice most easily solved by observation.
= To Cartesian coordinates
=From spherical coordinates
x
=
ρ
sin
θ
cos
φ
y
=
ρ
sin
θ
sin
φ
z
=
ρ
cos
θ
∂
(
x
,
y
,
z
)
∂
(
ρ
,
θ
,
φ
)
=
(
sin
θ
cos
φ
ρ
cos
θ
cos
φ
−
ρ
sin
θ
sin
φ
sin
θ
sin
φ
ρ
cos
θ
sin
φ
ρ
sin
θ
cos
φ
cos
θ
−
ρ
sin
θ
0
)
{\displaystyle {\begin{aligned}x&=\rho \,\sin \theta \cos \varphi \\y&=\rho \,\sin \theta \sin \varphi \\z&=\rho \,\cos \theta \\{\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}&={\begin{pmatrix}\sin \theta \cos \varphi &\rho \cos \theta \cos \varphi &-\rho \sin \theta \sin \varphi \\\sin \theta \sin \varphi &\rho \cos \theta \sin \varphi &\rho \sin \theta \cos \varphi \\\cos \theta &-\rho \sin \theta &0\end{pmatrix}}\end{aligned}}}
So for the volume element:
d
x
d
y
d
z
=
det
∂
(
x
,
y
,
z
)
∂
(
ρ
,
θ
,
φ
)
d
ρ
d
θ
d
φ
=
ρ
2
sin
θ
d
ρ
d
θ
d
φ
{\displaystyle dx\,dy\,dz=\det {\frac {\partial (x,y,z)}{\partial (\rho ,\theta ,\varphi )}}\,d\rho \,d\theta \,d\varphi =\rho ^{2}\sin \theta \,d\rho \,d\theta \,d\varphi }
From cylindrical coordinates
x
=
r
cos
θ
y
=
r
sin
θ
z
=
z
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
z
)
=
(
cos
θ
−
r
sin
θ
0
sin
θ
r
cos
θ
0
0
0
1
)
{\displaystyle {\begin{aligned}x&=r\,\cos \theta \\y&=r\,\sin \theta \\z&=z\,\\{\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}&={\begin{pmatrix}\cos \theta &-r\sin \theta &0\\\sin \theta &r\cos \theta &0\\0&0&1\end{pmatrix}}\end{aligned}}}
So for the volume element:
d
V
=
d
x
d
y
d
z
=
det
∂
(
x
,
y
,
z
)
∂
(
r
,
θ
,
z
)
d
r
d
θ
d
z
=
r
d
r
d
θ
d
z
{\displaystyle dV=dx\,dy\,dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}\,dr\,d\theta \,dz=r\,dr\,d\theta \,dz}
= To spherical coordinates
=From Cartesian coordinates
ρ
=
x
2
+
y
2
+
z
2
θ
=
arctan
(
x
2
+
y
2
z
)
=
arccos
(
z
x
2
+
y
2
+
z
2
)
φ
=
arctan
(
y
x
)
=
arccos
(
x
x
2
+
y
2
)
=
arcsin
(
y
x
2
+
y
2
)
∂
(
ρ
,
θ
,
φ
)
∂
(
x
,
y
,
z
)
=
(
x
ρ
y
ρ
z
ρ
x
z
ρ
2
x
2
+
y
2
y
z
ρ
2
x
2
+
y
2
−
x
2
+
y
2
ρ
2
−
y
x
2
+
y
2
x
x
2
+
y
2
0
)
{\displaystyle {\begin{aligned}\rho &={\sqrt {x^{2}+y^{2}+z^{2}}}\\\theta &=\arctan \left({\frac {\sqrt {x^{2}+y^{2}}}{z}}\right)=\arccos \left({\frac {z}{\sqrt {x^{2}+y^{2}+z^{2}}}}\right)\\\varphi &=\arctan \left({\frac {y}{x}}\right)=\arccos \left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)=\arcsin \left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)\\{\frac {\partial \left(\rho ,\theta ,\varphi \right)}{\partial \left(x,y,z\right)}}&={\begin{pmatrix}{\frac {x}{\rho }}&{\frac {y}{\rho }}&{\frac {z}{\rho }}\\{\frac {xz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&{\frac {yz}{\rho ^{2}{\sqrt {x^{2}+y^{2}}}}}&-{\frac {\sqrt {x^{2}+y^{2}}}{\rho ^{2}}}\\{\frac {-y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}&0\\\end{pmatrix}}\end{aligned}}}
See also the article on atan2 for how to elegantly handle some edge cases.
So for the element:
d
ρ
d
θ
d
φ
=
det
∂
(
ρ
,
θ
,
φ
)
∂
(
x
,
y
,
z
)
d
x
d
y
d
z
=
1
x
2
+
y
2
x
2
+
y
2
+
z
2
d
x
d
y
d
z
{\displaystyle d\rho \,d\theta \,d\varphi =\det {\frac {\partial (\rho ,\theta ,\varphi )}{\partial (x,y,z)}}\,dx\,dy\,dz={\frac {1}{{\sqrt {x^{2}+y^{2}}}{\sqrt {x^{2}+y^{2}+z^{2}}}}}\,dx\,dy\,dz}
From cylindrical coordinates
ρ
=
r
2
+
h
2
θ
=
arctan
r
h
φ
=
φ
∂
(
ρ
,
θ
,
φ
)
∂
(
r
,
h
,
φ
)
=
(
r
r
2
+
h
2
h
r
2
+
h
2
0
h
r
2
+
h
2
−
r
r
2
+
h
2
0
0
0
1
)
det
∂
(
ρ
,
θ
,
φ
)
∂
(
r
,
h
,
φ
)
=
1
r
2
+
h
2
{\displaystyle {\begin{aligned}\rho &={\sqrt {r^{2}+h^{2}}}\\\theta &=\arctan {\frac {r}{h}}\\\varphi &=\varphi \\{\frac {\partial (\rho ,\theta ,\varphi )}{\partial (r,h,\varphi )}}&={\begin{pmatrix}{\frac {r}{\sqrt {r^{2}+h^{2}}}}&{\frac {h}{\sqrt {r^{2}+h^{2}}}}&0\\{\frac {h}{r^{2}+h^{2}}}&{\frac {-r}{r^{2}+h^{2}}}&0\\0&0&1\\\end{pmatrix}}\\\det {\frac {\partial (\rho ,\theta ,\varphi )}{\partial (r,h,\varphi )}}&={\frac {1}{\sqrt {r^{2}+h^{2}}}}\end{aligned}}}
= To cylindrical coordinates
=From Cartesian coordinates
r
=
x
2
+
y
2
θ
=
arctan
(
y
x
)
z
=
z
{\displaystyle {\begin{aligned}r&={\sqrt {x^{2}+y^{2}}}\\\theta &=\arctan {\left({\frac {y}{x}}\right)}\\z&=z\quad \end{aligned}}}
∂
(
r
,
θ
,
h
)
∂
(
x
,
y
,
z
)
=
(
x
x
2
+
y
2
y
x
2
+
y
2
0
−
y
x
2
+
y
2
x
x
2
+
y
2
0
0
0
1
)
{\displaystyle {\frac {\partial (r,\theta ,h)}{\partial (x,y,z)}}={\begin{pmatrix}{\frac {x}{\sqrt {x^{2}+y^{2}}}}&{\frac {y}{\sqrt {x^{2}+y^{2}}}}&0\\{\frac {-y}{x^{2}+y^{2}}}&{\frac {x}{x^{2}+y^{2}}}&0\\0&0&1\end{pmatrix}}}
From spherical coordinates
r
=
ρ
sin
φ
h
=
ρ
cos
φ
θ
=
θ
∂
(
r
,
h
,
θ
)
∂
(
ρ
,
φ
,
θ
)
=
(
sin
φ
ρ
cos
φ
0
cos
φ
−
ρ
sin
φ
0
0
0
1
)
det
∂
(
r
,
h
,
θ
)
∂
(
ρ
,
φ
,
θ
)
=
−
ρ
{\displaystyle {\begin{aligned}r&=\rho \sin \varphi \\h&=\rho \cos \varphi \\\theta &=\theta \\{\frac {\partial (r,h,\theta )}{\partial (\rho ,\varphi ,\theta )}}&={\begin{pmatrix}\sin \varphi &\rho \cos \varphi &0\\\cos \varphi &-\rho \sin \varphi &0\\0&0&1\\\end{pmatrix}}\\\det {\frac {\partial (r,h,\theta )}{\partial (\rho ,\varphi ,\theta )}}&=-\rho \end{aligned}}}
= Arc-length, curvature and torsion from Cartesian coordinates
=s
=
∫
0
t
x
′
2
+
y
′
2
+
z
′
2
d
t
κ
=
(
z
″
y
′
−
y
″
z
′
)
2
+
(
x
″
z
′
−
z
″
x
′
)
2
+
(
y
″
x
′
−
x
″
y
′
)
2
(
x
′
2
+
y
′
2
+
z
′
2
)
3
2
τ
=
x
‴
(
y
′
z
″
−
y
″
z
′
)
+
y
‴
(
x
″
z
′
−
x
′
z
″
)
+
z
‴
(
x
′
y
″
−
x
″
y
′
)
(
x
′
y
″
−
x
″
y
′
)
2
+
(
x
″
z
′
−
x
′
z
″
)
2
+
(
y
′
z
″
−
y
″
z
′
)
2
{\displaystyle {\begin{aligned}s&=\int _{0}^{t}{\sqrt {{x'}^{2}+{y'}^{2}+{z'}^{2}}}\,dt\\[3pt]\kappa &={\frac {\sqrt {\left(z''y'-y''z'\right)^{2}+\left(x''z'-z''x'\right)^{2}+\left(y''x'-x''y'\right)^{2}}}{\left({x'}^{2}+{y'}^{2}+{z'}^{2}\right)^{\frac {3}{2}}}}\\[3pt]\tau &={\frac {x'''\left(y'z''-y''z'\right)+y'''\left(x''z'-x'z''\right)+z'''\left(x'y''-x''y'\right)}{{\left(x'y''-x''y'\right)}^{2}+{\left(x''z'-x'z''\right)}^{2}+{\left(y'z''-y''z'\right)}^{2}}}\end{aligned}}}
See also
Geographic coordinate conversion
Transformation matrix
References
Arfken, George (2013). Mathematical Methods for Physicists. Academic Press. ISBN 978-0123846549.
Kata Kunci Pencarian:
- List of common coordinate transformations
- Geographic coordinate conversion
- Polar coordinate system
- Cartesian coordinate system
- Transformation (function)
- Astronomical coordinate systems
- Cylindrical coordinate system
- Spatial reference system
- Affine transformation
- Spherical coordinate system