Re: CATransform3D perspective question

FROM : John Harper
DATE : Sat Dec 01 18:10:44 2007

Presumably you did what the samples do and set the perspective matrix 
as the sublayerTransform property of the superlayer of the layer 
you're rotating?

Both sublayerTransform and transform properties are applied relative 
to the anchor point (typically center, though appkit sets the anchor 
of layers it creates to the bottom left corner iirc) of the layer the 
property is set on. So if you have two layers, one with a perspective 
matrix, and its child rotated around the Y axis by 90°, you will only 
see the child layer exactly edge on when its center is aligned with 
the center of its superlayer. Does that explain what you're seeing?

   John

Other than that, your expectations seem correct

On Nov 7, 2007, at 11:27 AM, Nicko van Someren wrote:

> I'm having some trouble with using Core Animation's 3D transforms to 
> get a perspective effect.  Part of the problem is that the exact 
> nature of the transformations is not documented as far as I can 
> tell, and it seems not to behave the way that I've seen these 
> transforms used before.
>
> The CATransform3D type is a 4 by 4 matrix.  Typically these matrices 
> are multiplied by the vector (x,y,z,1) to return a vector (x', y', 
> z', s) and the display co-ordinates are (x'/s, y'/s), with z'/s 
> being used for z-buffer values if drawing is done with z-buffering 
> (which I don't think CA supports).
>
> My initial tests supported my view that this was the way CA was 
> going to use the matrix.  Rotations about the z axis put sin(t) and 
> cos(t) values into the m11, m12, m21 and m22 slots; translations 
> effect the m4{1,2,3} slots; scales work as expected.  Most 
> importantly, the only example I could find for applying a 
> perspective transformation was three lines of code in the CA "Layer 
> Geometry and Transforms" docs, listing 2, which uses the standard 
> technique of putting -(1/distance) into the m34 slot (this exact 
> same code also appears in the CoverFlow example).  Adjusting the 
> zPosition value for a layer zooms the layer in and out.  So far so 
> good.
>
> The problem is simply this; applying a 90 degree rotation about the 
> y axis does NOT turn the image edge-on.  I've hooked up a rotary 
> NSSlider to an action that makes a transform thus:
>     flip = CATransform3DMakeRotation(rotation, 0, 1, 0);
> I'm printing this transform and then setting the transform on a 
> layer containing an image.  Rotating the slider rotates the image 
> and if the slider is set to 90 degrees then the transform is the 
> expected: 1.0 in m13, m22, m31 and m44, with zeros everywhere else. 
> This _should_ produce a transform where the output 'x' co-ordinates 
> are invariant of the 'x' input (and in fact should be solely 
> dependent on the 'z' position).  Unfortunately, when I do this I get 
> an image which looks like it's turned by about 75 degrees, 
> definitely not edge-on, and very much with the output x co-ordinates 
> dependent on the input x value.  I have to turn the slider to about 
> 115 degrees to get the image edge-on.
>
> So, my question is what exactly is the process by which the layer co-
> ordinates are converted to the display co-ordinates?  The 
> documentation on this seems to be missing and while it looks like it 
> should be fairly standard it does not function as expected.  Any 
> help would be much appreciated.
>
>     Nicko
>
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