Spherical Vector To Cartesian Vector

• Aug 20th 2010, 04:01 AM
tsa256
Spherical Vector To Cartesian Vector
Hello everyone,

I have a quick and straightforward problem, that I need some help with.

Suppose I have a spherical vector defined by a radius and two angles, denoted by <rho,theta,psi>. Where, rho is the radius, theta is an angle, and so is psi.

Which that being said, how would I convert that spherical vector into a Cartesian vector.

(Note: I know how to convert from Spherical to Cartesian;
x=rho*sin(psi) *cos(theta)
Y=rho*sin(psi) *sin(theta)
Z=rho*cos(theta)

Would these conversion also be appropriate if the input is a spherical vector, and the required output is a Cartesian vector.

Thanks for the help.
Taylor S. Amarel
Learning is living
• Aug 20th 2010, 04:08 AM
sa-ri-ga-ma
theta is an angle, and so is psi.

with respect to what? Which one is the reference line?
• Aug 20th 2010, 04:28 AM
tsa256
Theta= The angle of rotation around Y on the XZ plane
Psi= The angle of rotation around Z on the YX plane

So the reference direction would be X? Is that right, if I knew the formula I am sure I could swap variables out into their proper location for my coordinate system
• Aug 20th 2010, 04:48 AM
Ackbeet
Your equations to convert from spherical to cartesian are incorrect. The x and y must have sin(theta) in them, where theta is the polar angle (measured down from the positive z axis), and z must have cos(theta) in it. What changes in x and y is the trig function of the azimuthal angle (measured around the z axis).
• Aug 20th 2010, 05:05 AM
tsa256
So if I understand you correctly, the conversion would be as follows.
x=rho*sin(theta)*cos(psi)
y=rho*sin(theta)*sin(psi)
z=rho*cos(theta)
I am correct?
• Aug 20th 2010, 05:26 AM
Ackbeet
That is correct. Does this solve your problem?
• Aug 20th 2010, 07:00 AM
tsa256
Thank You!
Quote:

Originally Posted by Ackbeet
That is correct. Does this solve your problem?

Yes, indeed it does. Kudos for the help, and have a good day.

Thanks,
Taylor S. Amarel
Learning is Living
• Aug 20th 2010, 07:06 AM
Ackbeet
You're welcome. Have a good one!