Convert CSS3 rotateY to PHP Imagick::distortImage

Question

I have an image that gets rotateY(-54deg) on the front-end and I need to rotate it the same way on PHP using Imagick::distortImage

$image->distortImage(Imagick::DISTORTION_PERSPECTIVE, $controlPoints, true);

Is there an easy way to convert the -54deg to the $controlPoints that distortImage() needs?

@NicolòCozzani Because I don't want to simple rotate the image, I need to change the perspective. — Ju Nogueira, Feb 17 '17 at 14:55

score 0 · Accepted Answer · answered Feb 17 '17 at 15:33

I dont think there is any (easy) way to do this.

The Imagick::DISTORTION_PERSPECTIVE is used to distort the images using perspective projection which requires 8 precalculated coefficients.

From the imagemagick docs

You can get these coefficients by looking at the -verbose output of a 'Perspective' distortion, or by calculating them yourself. If the last two perspective scaling coefficients are zero, the remaining 6 represents a transposed 'Affine Matrix'.

I dont think -54deg has got enough data inherently in it to decipher these 8 coefficients. You will have to do this the hard way, sorry :)

That's what I thought. Thanks. – Ju Nogueira Feb 17 '17 at 17:23 — Ju Nogueira, Feb 17 '17 at 17:23

Convert CSS3 rotateY to PHP Imagick::distortImage

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