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
- Context: Undergrad
- Thread starter teslajet
- Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 9K views
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.
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 . [tex]\rho[/tex]
A . [tex]\phi[/tex]
Z(cartesian)=Z(cylindrical)
my question is, since x . [tex]\rho[/tex] = cos[tex]\phi[/tex], is the [tex]\phi[/tex] 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 [tex]\phi[/tex] when according to part a.), [tex]\phi[/tex] 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 . [tex]\rho[/tex]
A . [tex]\phi[/tex]
Z(cartesian)=Z(cylindrical)
my question is, since x . [tex]\rho[/tex] = cos[tex]\phi[/tex], is the [tex]\phi[/tex] 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 [tex]\phi[/tex] when according to part a.), [tex]\phi[/tex] should be -89 degrees. Am I missing something?