# Converting from spherical to cylindrical coordinates

I have the coordinates of a hurricane at a particular point defined on the surface of a sphere i.e. longitude and latitude. Now I want to transform these coordinates into a axisymmetric representation cylindrical coordinate i.e. radial and azimuth angle.

Is there a way to do the mathematical transformation without transforming from spherical to cartesian and from cartesian to cylindrical ?

BvU
Homework Helper
Hi,
Not clear to me what you mean with the radius of your coordinate system. Are you aware that latitude is unambiguous, whereas distance to the earth axis of rotation is not ? (I mean that ##R_{\rm earth} \cos( {\text {latitude}})## is symmetrical around 0)

BvU - Thanks for your response. Very good question. So what I do know is that the center of the vortex(hurricane) is the center of the cylindrical coordinate system. So the radius to me is the distance from the center of the vortex to that grid point. I am talking of a coordinate system attached to a hurricane having it's origin at the center of the hurricane. Usually one identifies the center of the vortex to be that point with the maximum relative vorticity.

BvU
Homework Helper
Means you need at least two positions on the surface to determine length and direction of the difference vector....
and the origin of your local coordinate system is supposed to be fixed ?

I want to transform these coordinates

Again really penetrating questions( I am really learning quickly). No because the hurricane moves with time the origin of my local coordinate system is not fixed.

BvU
Homework Helper
Center isn't fixed, can be accomodated. But you still need two points as input to your trasformation function as far as I can discern: vortex and grid point.

We had a comparable (slightly more general) thread recently - maybe you can find something useful there:
Is there a way of subtracting two vectors in spherical coordinate system without first having to convert them to Cartesian or other forms?

Since I have already searched and found the difference between Two Vectors in Spherical Coordinates as,

$$|\vec{x}-\vec{x'}|=(\rho^{2}+\rho'^{2}-2\rho\rho'[\cos(\theta-\theta')+\sin\theta\sin\theta'(cos(\phi-\phi')-1))])^{\frac{1}{2}}$$

which I believe the radius of displaced vector. I still didn't get any way to find the theta (angle from positive z axis)and psi(angle from positive x axis).

For more information please refer this link. math.stackexchange.com/a/1365667/613522 This helped me with the cylindrical coordinate case. I Just want a similar solution in Spherical Coordinate system.

Center isn't fixed, can be accomodated. But you still need two points as input to your trasformation function as far as I can discern: vortex and grid point.

We had a comparable (slightly more general) thread recently - maybe you can find something useful there:
When you write "vortex and grid point" you mean the center of the vortex and grid point is that correct ? So that defines my radius vector in the local coordinate system. From which I can get the magnitude of the radius vector which is nothing but the local radius. Finally the longitude transforms as the azimuth angle unchanged in the local cylindrical coordinate system. Am I right ?

BvU
Homework Helper
When you write "vortex and grid point" you mean the center of the vortex and grid point is that correct ?
I intend to adopt your notation, but needed clarification, since in post #1 there was no mention of a grid point. "that grid point " surfaced only in #3
So the radius to me is the distance from the center of the vortex to that grid point. I am talking of a coordinate system attached to a hurricane having it's origin at the center of the hurricane. Usually one identifies the center of the vortex to be that point with the maximum relative vorticity.
And I think you are very wrong with
Finally the longitude transforms as the azimuth angle unchanged in the local cylindrical coordinate system. Am I right ?
Could you post a sketch of what you mean ?

Yes apologies what I meant by grid point is the surface of a sphere.

And yes you are right I was wrong when I wrote that the longitude transforms unchanged. I am looking for the appropriate transformation equation from a global spherical coordinate to a local cylindrical coordinate system centered on the center of the vortex.

one way to get through this problem and I would like some feedback on the approach I am taking right now - is to convert from a global spherical to a local spherical coordinate system and then convert the local spherical to local cylindrical.

BvU
Homework Helper
You don't really have a global spherical difference vector to convert -- or at least I don't think that is what you do. You have ##r_1, \theta_1, \phi_1## for the vortex and ##r_2, \theta_2, \phi_2## for the second point. With a possible simplification of staying on the surface: ##r_1 = r_2##. And you want a function ##\Phi(r_1, \theta_1, \phi_1, r_2, \theta_2, \phi_2)## to convert to local polar coordinates ##\rho, \phi_3## . Right so far ?

You don't really have a global spherical difference vector to convert -- or at least I don't think that is what you do. You have ##r_1, \theta_1, \phi_1## for the vortex and ##r_2, \theta_2, \phi_2## for the second point. With a possible simplification of staying on the surface: ##r_1 = r_2##. And you want a function ##\Phi(r_1, \theta_1, \phi_1, r_2, \theta_2, \phi_2)## to convert to local polar coordinates ##\rho, \phi_3## . Right so far ?
BvU - One hundred per cent correct.

So the motivation for doing this and it is a research level question - is given all the coordinates of a vortex(or a hurricane) - is the hurricane axisymmetric or not ? As a first step in answering that is to convert to a cylindrical coordinate system where it maybe possible to answer that question using some objective methodology.

BvU