Maths - Cylindrical Polar coordinates

 

Cylindrical coordinates allow points to be specified using two linear distances and one angle. These three coordinates are shown on the above diagram as:

• r = radius (distance from axis of cylinder)
• h = height (distance along axis of cylinder)
• a = or sometimes written as θ = angle around axis.

Cylinder aligned with the z axis

Cylindrical coordinates to Cartesian coordinates

 

Cx Cy Cz

=

r sin(θ) r cos(θ) h

Cartesian coordinates to Cylindrical coordinates

 

θ r h

=

atan2(Cx,Cy) √(Cx2 + Cy2) Cz

Cylinder aligned with the arbitrary axis

Cylindrical coordinates to Cartesian coordinates

We want to rotate the above so that the h axis is aligned with the arbitrary axis (Ax, Ay, Az) in other words we want to lookat the point (Ax, Ay, Az) see lookat

Cx Cy Cz

=

√(Az2 + Ay2) 0 Ax 0 √(Az2 + Ax2) Ay -Ax -Ay Az

r sin(θ) r cos(θ) h

Multiplying gives:

Cx Cy Cz

=

√(Az2 + Ay2) * r sin(θ) + h*Ax √(Az2 + Ax2) * r cos(θ) + h*Ay -Ax * r sin(θ)-Ay * r cos(θ) + h*Az

Cartesian coordinates to Cylindrical coordinates

In order to do this we need to invert the above matrix, since this is orthogonal we can invert by transposing as follows:

r sin(θ) r cos(θ) h

=

√(Az2 + Ay2) 0 -Ax 0 √(Az2 + Ax2) -Ay Ax Ay Az

Cx Cy Cz

Multiplying gives,

r sin(θ) r cos(θ) h

=

Cx * √(Az2 + Ay2) - Ax * Cz Cy * √(Az2 + Ax2) -Ay * Cz Cx * Ax + Cy * Ay +Cz * Az

So in terms of θ, r , h

θ r h

=

atan2(Cx * √(Az2 + Ay2) - Ax * Cz,Cy * √(Az2 + Ax2) -Ay * Cz) √((Cx * √(Az2 + Ay2) - Ax * Cz)2 + (Cy * √(Az2 + Ax2) -Ay * Cz)2) Cz

Using Tensors

This is more advanced stuff and only really needed if you need to do physics or advanced geometry in curvilinear coordinates.

On the curvilinear coordinates page we saw that the expression of coordinates as a linear equation:

e_i Aⁱ = e₁ A¹ + e₂ A² + e₃ A³

can be modified for curvilinear coordinates where either the basis or the components depend on the location:

e(x,y,z)_i Aⁱ = e(x,y,z)₁ A¹ + e(x,y,z)₂ A² + e(x,y,z)₃ A³…

or

e_i A(x,y,z)ⁱ = e₁ A(x,y,z)¹ + e₂ A(x,y,z)² + e₃ A(x,y,z)³…

where:

• e(x,y,z) is a basis which is a function of position.
• A(x,y,z) are terms which are functions of position.

We have already looked at the situation where the terms are expressed as a function of a global linear coordinate system. Now lets look at the situation where the linear basis is a function of position. There are two ways to do this:

• they can be built along the coordinate axes (axis-colinear) then the basis vectors transform like covariant vectors.
• they can be built to be perpendicular (normal) to the coordinate surfaces then the basis vectors transform like contravariant vectors.

Covariant Axis

e1 A1 + e2 A2 + e3 A3 =

∂x

e1

∂θ

A1 +

∂y

e2

∂r

A2 +

∂z

e3

∂h

A3

∂θ

∂x

∂r

∂y

∂h

∂z

where:

∂θ	= ∂atan2(x,y)	=y	=y
∂x	∂x	1 +(x/y)2	x2+ y 2
∂r	=√((x)2) + (y)2)	= 1
∂y	∂y
∂h	= 1
∂z

where:

∂	= cosθ	∂	-r-1sinθ	∂
∂x		∂r		∂θ
∂	= sinθ	∂	+r-1cosθ	∂
∂y		∂r		∂θ
∂	=	∂
∂z		∂h

Jacobean

J =

∂x'i ∂xj

=

x √(x2 + y2) y √(x2 + y2) 0 -y (x2 + y2) x (x2 + y2) 0 0 0 1

=

cosθ sinθ 0 -r-1sinθ r-1cosθ 0 0 0 1

Inverse Jacobean

J-1 =

∂xj ∂x'i

=

x √(x2 + y2) -y 0 -y (x2 + y2) x 0 0 0 1

=

cosθ sinθ 0 r sinθ r cosθ 0 0 0 1

Contravariant Axis

 

e1 A1 + e2 A2 + e3 A3 =

∂θ

e1

∂x1

A1 +

∂r

e2

∂x2

A2 +

∂h

e3

∂x3

A3

∂x1

∂θ

∂x2

∂r

∂x3

∂h

 

Coordinate Systems

Standards

Vector Dot Product