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

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- Thread starter meteo student
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- #1

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

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BvU

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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)

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BvU

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and the origin of your local coordinate system is supposed to be fixed ?

and would (0,0) be a satisfactory answer to your original question

I want to transform these coordinates

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- #6

BvU

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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.

Please help.

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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 ?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:

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BvU

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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 #3When you write "vortex and grid point" you mean the center of the vortex and grid point is that correct ?

And I think you are very wrong withSo 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.

Could you post a sketch of what you mean ?Finally the longitude transforms as the azimuth angle unchanged in the local cylindrical coordinate system. Am I right ?

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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.

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BvU - One hundred per cent correct.

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- #14

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- rotate the global coordinate system until the vortex is at ##\phi_1' = 0## . What happens to ##\phi_2, \theta_2## ? (Note: physics spherical)
- tilt until vortex is at ##\theta_1' = 0##. What happens to ##\ \phi_2', \theta_2'## ? (Note: not so easy!)
- convert second point ##\phi_2'', \theta_2''\ \ ## to ##\ \rho, \phi''##

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