Question

I need to know what the visible height of a display object will be after I change it’s rotationX value.

I have an application that allows users to lay out a floor in 3D space. I want the size of the floor to automatically stretch after a 3D rotation so that it always covers a certain area.

Anyone know a formula for working this out?

EDIT: I guess what I am really trying to do is convert degrees to pixels.

On a 2D plane say 100 x 100 pixels, a -10 degree change on rotationX means that the plane has a gap at the top where it is no longer visible. I want to know how many pixels this gap will be so that I can stretch the plane.

In Flex, the value for the display objects height property remains the same both before and after applying the rotation, which may in fact be a bug.

EDIT 2: There must be a general math formula to work this out rather than something Flash/Flex specific. When viewing an object in 3D space, if the object rotates backwards (top of object somersaults away from the viewer), what would the new visible height be based on degrees of rotation? This could be in pixels, metres, cubits or whatever.

Answer 1

I don’t have a test case, but off the top of my head I’d guess something like:

var d:DisplayObject;
var rotationRadians:Number = d.rotationX * Math.PI / 180;
var visibleHeight:Number = d.height * Math.cos(rotationRadians);

This doesn’t take any other transformations into account, though.