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cssperspectivetranslate3d

What is the formula for proportions of elements using perspective and translateZ?


I would like to do the following: Given a container with perspective and an inner element with a translateZ value I'd like to "pull it up" with translateY in order to visually touch the top of the surrounding container: http://jsfiddle.net/8R4ym/129/

Is there some kind of formula that, given the perspective value of a container, the width and height of an element and it's Z-translation I can get to that kind of "top" calculation? I have been playing around with it but I can't seem to find some rules for it, as it seems that those are all variables.

Thanks for any help.


Solution

  • Yes!

    There's actually quite a simple formula for finding the offset - the 3d Projection article on Wikipedia has a diagram and the formula.

    The formula is bx = ax * bz / az where:

    • ax is the original distance from the transform origin point
    • az is the perspective + the negative translateZ
    • bz is the perspective

    and this will give you:

    • bx - the new distance from the transform origin point

    So, you need to know:

    • bz : the perspective (eg: 1000px)
    • ax : the offset from the transform origin point, eg: if the origin point is 50% then this needs to be the element's top relative to the center of the parent element (parent.height/2 + element.top) -- let's say -500px
    • z : the element's translateZ (eg: -600px)
    • az is then bz + z * -1, in this case: 1000 + (-600 * -1) = 1600

    so the formula is: -500 * 1000 / 1600 = -312.5

    The element is offset vertically -312.5px from the origin, whereas originally it was offset -500px, the difference between the two number is what you'll need to add to the old top value to get the equivalent new value.

    This formula also works for the Y axis.

    I put together a quick example here: http://jsfiddle.net/trolleymusic/xYRgx/