You can use Property Nodes for the Picture (right-click on Segmentkreis in the Block Diagram, choose Create Property Node). The Position Property will give you the pixel coordinates of the upper-left corner of the Picture, and the Draw Area Size will give you the width and height of the Picture. Now if you know the coordinates of the Mouse, you can express this in polar coordinates relative to the center of the Picture. If the Radius shows that the Mouse is "within the wheel", you can use the angular argument to determine in which segment you are. Note that your code has Segments and Start Angle as parameters, so your "Where's the Mouse?" sub-VI will need to know these values, as well.
Bob "It's only Math" Schor
P.S. -- of interest is that you created a Picture with a Bounding Rectangle of 0, 0 and 500, 500. As expected, the Origin property for the Picture is 0, 0. Not so expected is that the Draw Area Size is 498, 498, as though the outer-most pixels are not part of the Draw Area. A minor detail, but it may result in an "off-by-one" error when you try to calculate the center point.
Moral of the Story -- write some code and test cases, do some experimentation to be sure you understand how things like Properties work.
