Determine the Transformation from Cylindrical to Rectangular coordinates

[LagrangeEuler]

In physics is usually defined that in cylindrical coordinates $\varphi \in [0,2 \pi)$. In relation with Deckart coordinates it is usually written that
\varphi=\text{arctg}(\frac{y}{x}).
Problem is of course because arctg takes values from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. What is the best way to solve this?

 

[jedishrfu]

To get those $\phi$ values you'd have to have $x=0$ so avoid that condition ie consider it a singular point.

 

[LagrangeEuler]

Why do I need to have $x=0$ to get those values of $\varphi$? I do not understand? I cannot have those values of $\varphi$ with this definition. That is the problem.

 

[mathman]

arctan is periodic with period $\pi$. To get the correct quadrant, use the signs of x and y. y positive is first or second quadrant and x positive is first or fourth quadrant.

 

[pasmith]

You can't use arctan(y/x) to uniquely determine \phi; both (x,y) and (-x,-y) have the same value of y/x but have \phi values which differ by \pi. You must therefore also consider the signs of x and y:
<br /> \phi(x,y) = \begin{cases}<br /> \arctan\left (\frac yx\right) &amp; x &gt; 0, y \geq 0 \\<br /> \arctan\left (\frac yx\right) + \pi &amp; x &lt; 0 \\<br /> \arctan\left (\frac yx\right) + 2\pi &amp; x &gt; 0, y &lt; 0 \\<br /> \frac{\pi}{2} &amp; x = 0, y &gt; 0 \\<br /> \frac{3\pi}{2} &amp; x = 0 , y &lt; 0 \\<br /> \mbox{undefined} &amp; x = 0, y = 0<br /> \end{cases}<br />

 

[Svein]

I had the same problem when designing an inexpensive rotary encoder. Since the algorithm had to be implemented in hardware, I used the following algorithm set:\lvert x \rvert &gt; \lvert y\rvert ? \varphi_{0}=\arctan\frac{\lvert y\rvert}{\lvert x\rvert} :\varphi_{0}=\frac{\pi}{2}-\arctan\frac{\lvert x\rvert}{\lvert y\rvert} (sorry for the "C" terminology, but if-then-else looks silly in Tex). From there on it was just a matter of determining the quadrant based on the signs of x and y. And in our case x and y could not be 0 at the same time (but in another case I had to deal with such singularities - and decided not to update anything when passing through a singularity).