Converting from Cylindrical to Cartesian

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The discussion focuses on converting cylindrical coordinates to Cartesian coordinates, specifically addressing the confusion surrounding the notation used for unit vectors in cylindrical systems. The participants clarify that in cylindrical coordinates, the unit vectors are represented as ar for radial direction and aθ for tangential direction, similar to Cartesian unit vectors. There is an emphasis on the need for specific points to determine the values of angles like θ, which are essential for conversion. The conversation also touches on the differences between cylindrical and spherical coordinates, noting that the notation can be nonstandard and potentially confusing. Overall, the thread aims to clarify the conversion process and the significance of understanding the coordinate systems involved.
salman213
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1. This is not a question from the book but i think if i can get the answer to this it will clear the idea i am confused about

can i covert a cylindrical vector such as

P(1 ar, 1a\theta)

into cartesian

after using the matrix

i got

Px = cos \theta - sin \theta
Py = sin \theta + cos \thetaNow these values of phi are dependent on a point right? So i cannot find numbers until the question asks for the conversion at a specific point...

for example
at point (1,2)

then cos \theta = x/sqrt(x^2+y^2)

AM I RIGHT?OR DOES \theta = 1 from

the vector P(1 ap, 1 a\theta)
 
Last edited:
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salman213 said:
1. This is not a question from the book but i think if i can get the answer to this it will clear the idea i am confused about

can i covert a cylindrical vector such as

P(1 ap, 1aphi)
What is this? I don't understand ap and aphi. In a cylindrical coordinate system you have three coordinates: r, \theta, and z. To convert into Cartesian coordinates, you have:
x = r * cos(\theta)
y = r * sin(\theta)
z = z
salman213 said:
into cartesian

after using the matrix

i got

Px = cos phi - sin phi
Py = sin phi + cos phi


Now these values of phi are dependent on a point right? So i cannot find numbers until the question gives asks for the conversion at a specific point...

for example
at point (1,2,3)

then cosphi = x/sqrt(x^2+y^2)

AM I RIGHT?


OR DOES phi = 1 from

the vector P(1 ap, 1aphi)
 
yes that is to covert specific coordinates but to covert vectors you need to use the following for cylindrical to cartesian

Px = cos \theta - sin \theta
Py = sin \theta + cos \theta

and obviously

Pz = z

since that does not changebut what i was asking is in 2d to make it even simpler because I am very confusedSo I have a vector in cylindrical coordinate system

P ( 1 ar, 1 a\theta)
 
I'm still not understanding your notation: P ( 1 ar, 1 a\theta).
What do ar and a\theta mean?
 
they are unit vectors in the R direction and in the Theta direction

they are just representing vectors in Cylindrical System
for example

In cartesian system a vector can be represented like

1 ax + 2 ay or 1 i + 2 j etc. where ax = i and ay = j are unit vectorssimilarly my notation i guess is confusing but its just the unit vectors in the respective coordinate system

Cylindrical (r theta and z)
 
salman213 said:
they are unit vectors in the R direction and in the Theta direction
Maybe I'm missing something, but this doesn't make any sense to me. Considering 2D cartesian and polar coordinates for the moment, the point A(1, 1) in Cartesian coordinates can also be used to define a vector, OA, which can be represented as 1i + 1j. The same point in polar coordinates is (\sqrt{2}, \pi/4). I don't see how you can break up this vector into radius and angle components. I understand the idea of a unit vector in the direction of r, but I completely don't understand the concept of the direction of an angle, and particularly a unit vector with that direction. How do you define the direction of an angle? The only possibilities I can think of would be the direction of the starting ray (the x axis) or the ending ray (which is the direction along which r lies).

Are you talking about radial and tangential components of the vector?
salman213 said:
they are just representing vectors in Cylindrical System
for example

In cartesian system a vector can be represented like

1 ax + 2 ay or 1 i + 2 j etc. where ax = i and ay = j are unit vectors
Why complicate things in what seems like a needless way with "ax" and "ay"? What is the purpose of "a" in these expressions?
salman213 said:
similarly my notation i guess is confusing but its just the unit vectors in the respective coordinate system

Cylindrical (r theta and z)
 
I don't know I'm just learning this that is what I am confused about :(..

here is what it says in my textbook

http://img261.imageshack.us/img261/7998/53164335jl9.jpg

maybe you can make some sense out of the VECTOR in the cylindrical coordinate system and explain briefly to me :(
 
Last edited by a moderator:
OK, now that makes sense. Glad you included the figure to help me out.
a\phi as shown in the figure is a unit vector that is tangential, while a\rho is a unit vector in the radial direction.

The notation that is used is nonstandard from what I remember. The cylindrical coordinates for a point are (r, \theta, z) where \theta is the same angle as that shown as \phi in the figure.

The spherical coordinates for a point are (\rho, \theta, \phi), where \rho measures the distance from the origin to the point, \theta is the same as used in polar or cylindrical coordinates, and \phi is the angle measured from the positive z-axis down to the line segment from the origin to the point. The figure mixes up \theta and \phi

To answer your original question,
Now these values of phi are dependent on a point right? So i cannot find numbers until the question asks for the conversion at a specific point...
Yes, but let's call it theta for the sake of cylindrical coordinates. The value of theta and therefore a\theta depend on the point, but theta and a\theta are different things.
for example
at point (1,2)

then cos \theta = x/sqrt(x^2+y^2)

AM I RIGHT?
Yes
OR DOES \theta = 1 from

the vector P(1 ap, 1 a\theta)
I don't believe so. |a\theta| = 1, but \theta doesn't have to be 1.
 
Ok thank you :),
 

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