Cartesian to spherical polar coordinates

AI Thread Summary
The discussion focuses on converting Cartesian coordinates to spherical coordinates and the challenges faced in obtaining the correct theta value after transformations. The user initially struggles with the range of theta, which is limited to 0 to pi after using the arccos function, while their original THETA values range from 0 to 360 degrees. The solution involves considering the quadrants of the x and y values to determine the correct angle for phi, using the tangent function for calculations. The conversation highlights the importance of understanding the relationships between the angles and their respective ranges in spherical coordinates. Ultimately, the user receives helpful guidance on how to adjust their approach to achieve the desired results.
birdhen
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Hi there,
I am getting confused about how to work this out.
I know that to convert cartesian coordinates to spherical coordinates you can use:
theta=arccos(z)
phi=arcsin(y/sin(theta))

my problem is that I have a list of coordinates, let's call them THETA and PHI. I change them into X,Y,Z and then rotate them by 2 Euler angles.
THETA is in the range(0->2pi)
PHI is in the range (-pi/2->pi/2).

the problem is once I have completed the transforms I want the new value theta, As it is found using arccos the value returned is only in the range 0->pi, the values come back between 0 and 180, where as the THETA values are between 0 and 360, and therefore I want my transformed values to be in the range 0 to 360. I think I need to use quadrants but I have been searching the internet and can't find the info I need.

Any help would be greatly appreciated.
 
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Hi birdhen! :smile:

(have a theta: θ and a phi: φ and a pi: π and a degree: º :wink:)
birdhen said:
theta=arccos(z)
phi=arcsin(y/sin(theta))

my problem is that I have a list of coordinates, let's call them THETA and PHI. I change them into X,Y,Z and then rotate them by 2 Euler angles.
THETA is in the range(0->2pi)
PHI is in the range (-pi/2->pi/2).

I normally do it the other way round …

θ from -π/2 to π/2, and φ from 0 to 2π …

then you have x = rsinθcosφ, y = rsinθsinφ, so you can use x as well as y to work out what φ is. :smile:
 
ah, thank you,
so y/x=tanφ,
and the value of φ will depend on whether x and y are negative or positve.

Wonderful, that was the hint I needed,

Thank you!
 
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