- #1
teslajet
- 2
- 0
So I'm trying to find out what the procedure is to convert a cartesian unit vector to a cylindrical unit vector. Any thoughts?
How to find the unit vector in cylindrical coordinates
- Thread starter teslajet
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To convert a Cartesian unit vector to a cylindrical unit vector, the procedure involves expressing the Cartesian coordinates in terms of cylindrical coordinates using the relationships between the two systems. The cylindrical unit vector e_r can be derived from the Cartesian unit vectors e_x and e_y, incorporating trigonometric functions. The transformation requires calculating the angle φ, which can be found using φ = tan^-1(y1/x1). There is confusion regarding the angle used in a specific problem, as the manual suggests using 70 degrees instead of the calculated -89 degrees. Understanding the correct angle is crucial for accurate vector transformation in cylindrical coordinates.
- #2
Jerbearrrrrr
- 124
- 0
The cylindrical unit vector is er.
x/|x|
Where x is where ever we are. (not cartesian x)
x=r(e_r)+z(e_z)
and |x|^2 = x.x
I think that works. Sorry about not underlining vectors.
e_r can be expressed in terms of e_x and e_y and some trig things.
x/|x|
Where x is where ever we are. (not cartesian x)
x=r(e_r)+z(e_z)
and |x|^2 = x.x
I think that works. Sorry about not underlining vectors.
e_r can be expressed in terms of e_x and e_y and some trig things.
- #3
teslajet
- 2
- 0
Here is my understanding,
Given a unit vector A = x1,y1,z1 in cartesian, to transform into cylindrical just use the transform
A . \rho
A . \phi
Z(cartesian)=Z(cylindrical)
my question is, since x . \rho = cos\phi, is the \phi that I am supposed to use the tan^-1(y1/X1)?
If this is the case then I don't understand what the solutions manual did with the following problem
I understand part a.) but in part b.) they use 70 degrees as \phi when according to part a.), \phi should be -89 degrees. Am I missing something?
Given a unit vector A = x1,y1,z1 in cartesian, to transform into cylindrical just use the transform
A . \rho
A . \phi
Z(cartesian)=Z(cylindrical)
my question is, since x . \rho = cos\phi, is the \phi that I am supposed to use the tan^-1(y1/X1)?
If this is the case then I don't understand what the solutions manual did with the following problem
I understand part a.) but in part b.) they use 70 degrees as \phi when according to part a.), \phi should be -89 degrees. Am I missing something?
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