Reputation: 1483
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.
Upvotes: 3
Views: 1300
Reputation: 2277
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 pointaz
is the perspective + the negative translateZ
bz
is the perspectiveand this will give you:
bx
- the new distance from the transform origin pointSo, 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/
Upvotes: 8
Reputation: 1878
There might be (don't know offhand) but have you tried changing -webkit-transform-origin:;
first to see if you can simply apply the transformation along the top so the element appears where you want it without translating it?
Upvotes: 0